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4c13d067238106648d920d3957d2437e | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_13 | Let $A$ be an acute angle such that $\tan A = 2 \cos A.$ Find the number of positive integers $n$ less than or equal to $1000$ such that $\sec^n A + \tan^n A$ is a positive integer whose units digit is $9.$ | \[\tan A = 2 \cos A \implies \sin A = 2 \cos^2 A \implies \sin^2 A + \cos^2 A = 4 \cos^4 A + \cos^2 A = 1\] \[\implies \cos^2 A = \frac {\sqrt {17} - 1}{8}.\] \[c_n = \sec^n A + \tan^n A = \frac {1}{\cos^n A} + 2^n \cos^n A = (4\cos^2 A +1)^{\frac {n}{2}}+(4 \cos^2 A)^{\frac {n}{2}} =\] \[= \left(\frac {\sqrt {17} + 1}{2}\right)^{\frac {n}{2}}+ \left(\frac {\sqrt {17} - 1}{2}\right)^{\frac {n}{2}}.\]
It is clear, that $c_n$ is not integer if $n \ne 4k, k > 0.$
Denote $x = \frac {\sqrt {17} + 1}{2}, y = \frac {\sqrt {17} - 1}{2} \implies$ \[x \cdot y = 4, x + y = \sqrt{17}, x - y = 1 \implies x^2 + y^2 = (x - y)^2 + 2xy = 9 = c_4.\]
\[c_8 = x^4 + y^4 = (x^2 + y^2)^2 - 2x^2 y^2 = 9^2 - 2 \cdot 16 = 49.\] \[c_{4k+4} = x^{4k+4} + y^{4k+4} = (x^{4k} + y^{4k})(x^2+y^2)- (x^2 y^2)(x^{4k-2}+y^{4k-2}) = 9 c_{4k}- 16 c_{4k – 4} \implies\] \[c_{12} = 9 c_8 - 16 c_4 = 9 \cdot 49 - 16 \cdot 9 = 9 \cdot 33 = 297.\] \[c_{16} = 9 c_{12} - 16 c_8 = 9 \cdot 297 - 16 \cdot 49 = 1889.\] \[c_{12m + 4} \pmod{10} = 9 \cdot c_{12m} \pmod{10} - 16 \pmod{10} \cdot c_{12m - 4} \pmod{10} =\] \[= (9 \cdot 7 - 6 \cdot 9) \pmod{10} = (3 - 4) \pmod{10} = 9.\] \[c_{12m + 8}\pmod{10} = 9 \cdot c_{12m+4} \pmod{10} - 16 \pmod{10} \cdot c_{12m } \pmod{10} =\] \[= (9 \cdot 9 - 6 \cdot 7) \pmod{10} = (1 - 2)\pmod{10} = 9.\] \[c_{12m + 12} \pmod{10} = 9 \cdot c_{12m + 8} \pmod{10} - 16 \pmod{10} \cdot c_{12m + 4} \pmod{10} =\] \[= (9 \cdot 9 - 6 \cdot 9) \pmod{10} = (1 - 4) \pmod{10} = 7 \implies\]
The condition is satisfied iff $n = 12 k + 4$ or $n = 12k + 8.$
If $n \le N$ then the number of possible n is $\left\lfloor \frac{N}{4} \right\rfloor - \left\lfloor \frac{N}{12} \right\rfloor.$
For $N = 1000$ we get $\left\lfloor \frac{1000}{4} \right\rfloor - \left\lfloor \frac{1000}{12} \right\rfloor = 250 - 83 = \boxed{167}.$ | null | 167 |
621b577295ae4a51d4252f28466fae37 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_14 | A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangle $ABDC$ is perpendicular to $\mathcal{P},$ vertex $B$ is $2$ meters above $\mathcal{P},$ vertex $C$ is $8$ meters above $\mathcal{P},$ and vertex $D$ is $10$ meters above $\mathcal{P}.$ The cube contains water whose surface is parallel to $\mathcal{P}$ at a height of $7$ meters above $\mathcal{P}.$ The volume of water is $\frac{m}{n}$ cubic meters, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | [asy] defaultpen(linewidth(0.6)); pair A = (0, 0), B = (5.656,2), C = (-2.828, 8), D = B+C, P = 0.875*C, Q = B+0.625*C, H = (-2.828, 0), G = (5.656, 0); pair P1 = (-5, 0), P2 = (10, 0); draw(A--B--D--C--A); filldraw(A--B--Q--P--cycle,rgb(0.35,0.7,0.9)); draw(C--H, dotted); draw(B--G, dotted); draw(P1--P2); dot("$A$",A,S); dot("$B$",B,E); dot("$C$",C,N); dot("$D$",D,N); dot("$H$",H,S); dot("$G$",G,S); dot("$P$",P,SE); dot("$Q$",Q,E); label("$\mathcal P$",(11, 0)); [/asy]
Let's first view the cube from a direction perpendicular to $ABDC$ , as illustrated above. Let $x$ be the cube's side length. Since $\triangle CHA \sim \triangle AGB$ , we have \[\frac{CA}{CH} = \frac{AB}{AG}.\] We know $AB = x$ $AG = \sqrt{x^2-2^2}$ $AC = \sqrt{2}x$ $CH = 8$ . Plug them into the above equation, we get \[\frac{\sqrt{2}x}{8} = \frac{x}{\sqrt{x^2-2^2}}.\] Solving this we get the cube's side length $x = 6$ , and $AC = 6\sqrt{2}.$
Let $PQ$ be the water's surface, both $P$ and $Q$ are $7$ meters from $\mathcal P$ . Notice that $C$ is $8$ meters from $\mathcal P$ , this means \[CP = \frac{1}{8}CA = \frac{3\sqrt{2}}{4}.\] Similarly, \[DQ = \frac{3}{8}CA = \frac{9\sqrt{2}}{4}.\]
[asy] defaultpen(linewidth(0.6)); pair A = (0, 0), C = (0, 2*6), X = (6, 6), Y = (-6, 6), P = (0, 1.75*6), I = (-0.25*6, 1.75*6), J = (0.25*6, 1.75*6); pair P1 = (-8, 0), P2 = (8, 0); draw(A--X--C--Y--A); filldraw(A--X--J--I--Y--cycle,rgb(0.35,0.7,0.9)); draw(P1--P2, dotted); dot("$A$",A,S); dot("$C$",C,N); dot("$P$",P,S); label("$\mathcal P$",(10, 0)); [/asy]
Now, we realize that the 3D space inside the cube without water is a frustum, with $P$ on its smaller base and $Q$ on its larger base. To find its volume, all we need is to find the areas of both bases and the height, which is $x = 6$ . To find the smaller base, let's move our viewpoint onto the plane $ABDC$ and view the cube from a direction parallel to $ABDC$ , as shown above. The area of the smaller base is simply \[S_1 = CP^2 = \Bigl(\frac{3\sqrt{2}}{4}\Bigr)^2 = \frac{9}{8}.\] Similarly, the area of the larger base is \[S_2 = DQ^2 = \Bigl(\frac{9\sqrt{2}}{4}\Bigr)^2 = \frac{81}{8}.\]
Finally, applying the formula for a frustum's volume,
\[V = \frac{1}{3} \cdot x \cdot (S_1 + \sqrt{S_1S_2} + S_2) = \frac{1}{3} \cdot 6 \cdot \Bigl(\frac{9}{8} + \sqrt{\frac{9}{8}\cdot\frac{81}{8}} + \frac{81}{8}\Bigl) = \frac{117}{4}.\]
The water's volume is thus \[6^3 - \frac{117}{4} = \frac{747}{4},\] giving $\boxed{751}$ | null | 751 |
621b577295ae4a51d4252f28466fae37 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_14 | A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangle $ABDC$ is perpendicular to $\mathcal{P},$ vertex $B$ is $2$ meters above $\mathcal{P},$ vertex $C$ is $8$ meters above $\mathcal{P},$ and vertex $D$ is $10$ meters above $\mathcal{P}.$ The cube contains water whose surface is parallel to $\mathcal{P}$ at a height of $7$ meters above $\mathcal{P}.$ The volume of water is $\frac{m}{n}$ cubic meters, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Denote $h(X)$ the distance from point $X$ to $\mathcal{P}, h(A) = 0, h(B) = 2,$ $h(C) = 8, h(D) = 10, h(G) = h(I) = h(H) = 7, AB = a, AC = a \sqrt{2}.$
Let slope $AB$ to $\mathcal{P}$ be $\alpha.$ Notation is shown in the diagram. \[\tan \alpha = \frac {\sin \alpha}{\cos \alpha} = \frac {h(B)}{AB}\cdot \frac {AC}{h(C)} = \frac {\sqrt{2}}{4} \implies a = 6.\] Let $S = GI \cap CD \implies h(S) = h(G) = 7.$ \[h(C) – h(G) = 8 - 7 = 1, h(D)- h(I) = 10 - 7 = 3.\] \[h(E) = h(F) = \frac {h(D) +h(B)}{2} = 6 \implies\] \[\frac {DI}{DE} = \frac {h(D) - h(I)}{h(D)-h(E)} = \frac {3}{4} \implies DI = DH = \frac {9}{2}.\]
Similarly $CG = \frac {3}{2} \implies SD = 9.$
Let the volume without water be $V,$ volume of the pyramid $SCGJ$ be $U.$
It is clear that $U + V = 27U = \frac {SD}{6} \cdot DI^2 = \frac {243}{8} \implies$ $V = \frac {243 \cdot 26}{8 \cdot 27 } = \frac {117}{4} = 6^3 - \frac {747}{4}$ from which $\boxed{751}.$ | null | 751 |
621b577295ae4a51d4252f28466fae37 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_14 | A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangle $ABDC$ is perpendicular to $\mathcal{P},$ vertex $B$ is $2$ meters above $\mathcal{P},$ vertex $C$ is $8$ meters above $\mathcal{P},$ and vertex $D$ is $10$ meters above $\mathcal{P}.$ The cube contains water whose surface is parallel to $\mathcal{P}$ at a height of $7$ meters above $\mathcal{P}.$ The volume of water is $\frac{m}{n}$ cubic meters, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | We introduce a Cartesian coordinate system to the diagram.
We put the origin at $A$ . We let the $z$ -components of $B$ $C$ $D$ be positive.
We set the $x$ -axis in a direction such that $B$ is on the $x-O-z$ plane.
The coordinates of $A$ $B$ $C$ are $A = \left( 0, 0, 0 \right)$ $B = \left( x_B, 0 , 2 \right)$ $C = \left( x_C, y_C, 8 \right)$
Because $AB \perp AC$ $\overrightarrow{AB} \cdot \overrightarrow{AC} = 0$ .
Thus, \[ x_B x_C + 16 = 0 . \hspace{1cm} (1) \]
Because $AC$ is a diagonal of a face, $AC^2 = 2 AB^2$ .
Thus, \[ x_C^2 + y_C^2 + 8^2 = 2 \left( x_B^2 + 2^2 \right) . \hspace{1cm} (2) \]
Because plane $ABCD$ is perpendicular to plan $P$ $\hat z \cdot \left( \overrightarrow{AB} \times \overrightarrow{AC} \right) = 0$ .
Thus, \[ \begin{vmatrix} 0 & 0 & 1 \\ x_B & 0 & 2 \\ x_C & y_C & 8 \end{vmatrix} = 0 . \hspace{1cm} (3) \]
Jointly solving (1), (2), (3), we get one solution $x_B = 4 \sqrt{2}$ $x_C = - 2 \sqrt{2}$ $y_C = 0$ .
Thus, the side length of the cube is $|AB| = \sqrt{x_B^2 + 2^2} = 6$
Denote by $P$ and $Q$ two vertices such that $AP$ and $AQ$ are two edges, and satisfy the right-hand rule that $\widehat{AB} \times \widehat{AP} = \widehat{AQ}$ .
Now, we compute the coordinates of $P$ and $Q$
Because $|AB| = 6$ , we have $\overrightarrow{AP} \times \overrightarrow{AQ} = 6 \overrightarrow{AB}$ $\overrightarrow{AQ} \times \overrightarrow{AB} = 6 \overrightarrow{AP}$ $\overrightarrow{AB} \times \overrightarrow{AP} = 6 \overrightarrow{AQ}$
Hence, \begin{align*} \begin{bmatrix} \hat i & \hat j & \hat k \\ x_P & y_P & z_P \\ x_Q & y_Q & z_Q \end{bmatrix} & = 6 \left( 4 \sqrt{2} \hat i + 2 \hat k \right) , \\ \begin{vmatrix} \hat i & \hat j & \hat k \\ x_Q & y_Q & z_Q \\ 4 \sqrt{2} & 0 & 2 \end{vmatrix} & = 6 \left( x_P \hat i + y_P \hat j + z_P \hat k \right) , \\ \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 \sqrt{2} & 0 & 2 \\ x_P & y_P & z_P \end{vmatrix} & = 6 \left( x_Q \hat i + y_Q \hat j + z_Q \hat k \right) . \end{align*}
By solving these equations, we get $y_P^2 + y_Q^2 = 36 .$
In addition, we have $\overrightarrow{AC} = \overrightarrow{AP} + \overrightarrow{AQ}$ .
Thus, $P = \left( - \sqrt{2} , 3 \sqrt{2} , 4 \right)$ $Q = \left( - \sqrt{2} , - 3 \sqrt{2} , 4 \right)$
Therefore, the volume of the water is \begin{align*} V = & 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ z_B u + z_P v + z_Q w \leq 7 \right\} dw dv du \\ & = 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u + 4 v + 4 w \leq 7 \right\} dw dv du \\ & = 6^3 - 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u + 4 v + 4 w > 7 \right\} dw dv du . \end{align*}
Define $u' = 1 - u$ $v' = 1 - v$ $w' = 1 - w$ .
Thus, \begin{align*} V & = 6^3 - 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u' + 4 v' + 4 w' < 3 \right\} dw dv du \\ & = 6^3 - 6^3 \int_{u'=0}^1 \left( \int_{v'=0}^1 \int_{w'=0}^1 \mathbf 1 \left\{ v' + w' < \frac{3}{4} - \frac{u'}{2} \right\} dw' dv' \right) du' \\ & = 6^3 - 6^3 \int_{u'=0}^1 \frac{1}{2} \left( \frac{3}{4} - \frac{u'}{2} \right)^2 du' . \end{align*}
Define $u'' = \frac{3}{4} - \frac{u'}{2}$ .
Thus, \begin{align*} V & = 6^3 - 6^3 \int_{u'' = 1/4}^{3/4} \left( u'' \right)^2 du'' \\ & = 6^3 - 6^3 \frac{1}{3} \left( \left(\frac{3}{4}\right)^3 - \left(\frac{1}{4}\right)^3 \right) \\ & = 216 - \frac{117}{4} \\ & = \frac{747}{4} . \end{align*}
Therefore, the answer is $747 + 4 = \boxed{751}$ | null | 751 |
06d4e053d0c5703e55ae53f626c7fc73 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_15 | For each positive integer $n$ let $a_n$ be the least positive integer multiple of $23$ such that $a_n \equiv 1 \pmod{2^n}.$ Find the number of positive integers $n$ less than or equal to $1000$ that satisfy $a_n = a_{n+1}.$ | Denote $a_n = 23 b_n$ .
Thus, for each $n$ , we need to find smallest positive integer $k_n$ , such that \[ 23 b_n = 2^n k_n + 1 . \]
Thus, we need to find smallest $k_n$ , such that \[ 2^n k_n \equiv - 1 \pmod{23} . \]
Now, we find the smallest $m$ , such that $2^m \equiv 1 \pmod{23}$ .
By Fermat's Theorem, we must have $m | \phi \left( 23 \right)$ . That is, $m | 22$ .
We find $m = 11$
Therefore, for each $n$ , we need to find smallest $k_n$ , such that \[ 2^{{\rm Rem} \left( n , 11 \right)} k_n \equiv - 1 \pmod{23} . \]
We have the following results:
If \({\rm Rem} \left( n , 11 \right) = 0\), then \(k_n = 22\) and \(b_n = 1\).
If \({\rm Rem} \left( n , 11 \right) = 1\), then \(k_n = 11\) and \(b_n = 1\).
If \({\rm Rem} \left( n , 11 \right) = 2\), then \(k_n = 17\) and \(b_n = 3\).
If \({\rm Rem} \left( n , 11 \right) = 3\), then \(k_n = 20\) and \(b_n = 7\).
If \({\rm Rem} \left( n , 11 \right) = 4\), then \(k_n = 10\) and \(b_n = 7\).
If \({\rm Rem} \left( n , 11 \right) = 5\), then \(k_n = 5\) and \(b_n = 7\).
If \({\rm Rem} \left( n , 11 \right) = 6\), then \(k_n = 14\) and \(b_n = 39\).
If \({\rm Rem} \left( n , 11 \right) = 7\), then \(k_n = 7\) and \(b_n = 39\).
If \({\rm Rem} \left( n , 11 \right) = 8\), then \(k_n = 15\) and \(b_n = 167\).
If \({\rm Rem} \left( n , 11 \right) = 9\), then \(k_n = 19\) and \(b_n = 423\).
If \({\rm Rem} \left( n , 11 \right) = 10\), then \(k_n = 21\) and \(b_n = 935\).
Therefore, in each cycle, $n \in \left\{ 11 l , 11l + 1 , \cdots , 11l + 10 \right\}$ , we have $n = 11l$ $11l + 3$ $11l + 4$ $11l + 6$ , such that $b_n = b_{n+1}$ . That is, $a_n = a_{n+1}$ .
At the boundary of two consecutive cycles, $b_{11L + 10} \neq b_{11 \left(l + 1 \right)}$
We have $1000 = 90 \cdot 11 + 10$ .
Therefore, the number of feasible $n$ is $91 \cdot 4 - 1 = \boxed{363}$ | null | 363 |
d4f177b4475b273a237ded2b85f38eba | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | Let $R(x)=P(x)+Q(x).$ Since the $x^2$ -terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.
Note that \begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \end{alignat*} so the slope of $R(x)$ is $\frac{106-108}{20-16}=-\frac12.$
It follows that the equation of $R(x)$ is \[R(x)=-\frac12x+c\] for some constant $c,$ and we wish to find $R(0)=c.$
We substitute $x=20$ into this equation to get $106=-\frac12\cdot20+c,$ from which $c=\boxed{116}.$ | null | 116 |
d4f177b4475b273a237ded2b85f38eba | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | Let \begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\ Q(x) &= &\hspace{1mm}-2x^2 + cx + d, \end{alignat*} for some constants $a,b,c$ and $d.$
We are given that \begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\ Q(16) &= &\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\ P(20) &= &800 + 20a + b &= 53, &&(3) \\ Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4) \end{alignat*} and we wish to find \[P(0)+Q(0)=b+d.\] We need to cancel $a$ and $c.$ Since $\operatorname{lcm}(16,20)=80,$ we subtract $4\cdot[(3)+(4)]$ from $5\cdot[(1)+(2)]$ to get \[b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.\] ~MRENTHUSIASM | null | 116 |
d4f177b4475b273a237ded2b85f38eba | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | Like Solution 2, we can begin by setting $P$ and $Q$ to the quadratic above, giving us \begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\ Q(16) &= &\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\ P(20) &= &800 + 20a + b &= 53, &&(3) \\ Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4) \end{alignat*} We can first add $(1)$ and $(2)$ to obtain $16(a-c) + (b+d) = 108.$
Next, we can add $(3)$ and $(4)$ to obtain $20(a-c) + (b+d) = 106.$ By subtracting these two equations, we find that $4(a-c) = -2,$ so substituting this into equation $[(1) + (2)],$ we know that $4 \cdot (-2) + (b+d) = 108,$ so therefore $b+d = \boxed{116}.$ | null | 116 |
d4f177b4475b273a237ded2b85f38eba | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | Let \begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\ Q(x) &= &\hspace{1mm}-2x^2 + cx + d, \end{alignat*} By substituting $(16, 54)$ and $(20, 53)$ into these equations, we can get: \begin{align*} 2(16)^2 + 16a + b &= 54, \\ 2(20)^2 + 20a + b &= 53. \end{align*} Hence, $a = -72.25$ and $b = 698.$
Similarly, \begin{align*} -2(16)^2 + 16c + d &= 54, \\ -2(20)^2 + 20c + d &= 53. \end{align*} Hence, $c = 71.75$ and $d = -582.$
Notice that $b = P(0)$ and $d = Q(0).$ Therefore \[P(0) + Q(0) = 698 + (-582) = \boxed{116}.\] ~Littlemouse | null | 116 |
d4f177b4475b273a237ded2b85f38eba | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | Add the equations of the polynomials $y=2x^2+ax+b$ and $y=-2x^2+cx+d$ to get $2y=(a+c)x+(b+d)$ . This equation must also pass through the two points $(16,54)$ and $(20,53)$
Let $m=a+c$ and $n=b+d$ . We then have two equations: \begin{align*} 108&=16m+n, \\ 106&=20m+n. \end{align*} We are trying to solve for $n=P(0)$ . Using elimination: \begin{align*} 540&=80m+5n, \\ 424&=80m+4n. \end{align*} Subtracting both equations, we find that $n=\boxed{116}$ | null | 116 |
9aaf3e4f91ff30c798a198724532e92f | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_2 | Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. | We are given that \[100a + 10b + c = 81b + 9c + a,\] which rearranges to \[99a = 71b + 8c.\] Taking both sides modulo $71,$ we have \begin{align*} 28a &\equiv 8c \pmod{71} \\ 7a &\equiv 2c \pmod{71}. \end{align*} The only solution occurs at $(a,c)=(2,7),$ from which $b=2.$
Therefore, the requested three-digit positive integer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.$ | null | 227 |
9aaf3e4f91ff30c798a198724532e92f | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_2 | Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. | As shown in Solution 1, we get $99a = 71b+8c$
Note that $99$ and $71$ are large numbers comparatively to $8$ , so we hypothesize that $a$ and $b$ are equal and $8c$ fills the gap between them. The difference between $99$ and $71$ is $28$ , which is a multiple of $4$ . So, if we multiply this by $2$ , it will be a multiple of $8$ and thus the gap can be filled. Therefore, the only solution is $(a,b,c)=(2,2,7)$ , and the answer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}$ | null | 227 |
9aaf3e4f91ff30c798a198724532e92f | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_2 | Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. | As shown in Solution 1, we get $99a = 71b+8c.$
We list a few multiples of $99$ out: \[99,198,297,396.\] Of course, $99$ can't be made of just $8$ 's. If we use one $71$ , we get a remainder of $28$ , which can't be made of $8$ 's either. So $99$ doesn't work. $198$ can't be made up of just $8$ 's. If we use one $71$ , we get a remainder of $127$ , which can't be made of $8$ 's. If we use two $71$ 's, we get a remainder of $56$ , which can be made of $8$ 's.
Therefore we get $99\cdot2=71\cdot2+8\cdot7$ so $a=2,b=2,$ and $c=7$ . Plugging this back into the original problem shows that this answer is indeed correct. Therefore, $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.$ | null | 227 |
9aaf3e4f91ff30c798a198724532e92f | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_2 | Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. | As shown in Solution 1, we get $99a = 71b+8c$
We can see that $99$ is $28$ larger than $71$ , and we have an $8c$ . We can clearly see that $56$ is a multiple of $8$ , and any larger than $56$ would result in $c$ being larger than $9$ . Therefore, our only solution is $a = 2, b = 2, c = 7$ . Our answer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}$ | null | 227 |
9aaf3e4f91ff30c798a198724532e92f | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_2 | Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. | As shown in Solution 1, we have that $99a = 71b + 8c$
Note that by the divisibility rule for $9$ , we have $a+b+c \equiv a \pmod{9}$ . Since $b$ and $c$ are base- $9$ digits, we can say that $b+c = 0$ or $b+c=9$ . The former possibility can be easily eliminated, and thus $b+c=9$ . Next, we write the equation from Solution 1 as $99a = 63b + 8(b+c)$ , and dividing this by $9$ gives $11a = 7b+8$ . Taking both sides modulo $7$ , we have $4a \equiv 1 \pmod{7}$ . Multiplying both sides by $2$ gives $a\equiv 2 \pmod{7}$ , which implies $a=2$ . From here, we can find that $b=2$ and $c=7$ , giving an answer of $\boxed{227}$ | null | 227 |
6a4bc209462e330802e61c25a358e95f | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | We have the following diagram: [asy] /* Made by MRENTHUSIASM , modified by Cytronical */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); X = intersectionpoint(B--5*(Q-B)+B,C--D); Y = (0,6*sqrt(731)); Z = intersectionpoint(A--4*(P-A)+A,B--4*(Q-B)+B); W = intersectionpoint(A--5*(P-A)+A,C--D); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*(-1,0),linewidth(4)); dot("$Q$",Q,1.5*E,linewidth(4)); dot("$X$",X,1.5*dir(-105),linewidth(4)); dot("$Y$",Y,1.5*N,linewidth(4)); dot("$Z$",Z,4.5*dir(75),linewidth(4)); dot("$W$",W,1.5*dir(-75),linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--W^^Q--X^^Y--Z,dashed); [/asy]
Let $X$ and $W$ be the points where $AP$ and $BQ$ extend to meet $CD$ , and $YZ$ be the height of $\triangle AZB$ . As proven in Solution 2, triangles $APD$ and $DPW$ are congruent right triangles. Therefore, $AD = DW = 333$ . We can apply this logic to triangles $BCQ$ and $XCQ$ as well, giving us $BC = CX = 333$ . Since $CD = 650$ $XW = DW + CX - CD = 16$
Additionally, we can see that $\triangle XZW$ is similar to $\triangle PQZ$ and $\triangle AZB$ . We know that $\frac{XW}{AB} = \frac{16}{500}$ . So, we can say that the height of the triangle $AZB$ is $500u$ while the height of the triangle $XZW$ is $16u$ . After that, we can figure out the distance from $Y$ to $PQ: \frac{500+16}{2} = 258u$ and the height of triangle $PZQ: 500-258 = 242u$
Finally, since the ratio between the height of $PZQ$ to the height of $AZB$ is $242:500$ and $AB$ is $500$ $PQ = \boxed{242}.$ | null | 242 |
6a4bc209462e330802e61c25a358e95f | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$ . The diagram looks like this: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); P1 = intersectionpoint(A--D,P--(-300)*(Q-P)+P); Q1 = intersectionpoint(B--C,Q--300*(Q-P)+Q); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); dot("$Q$",Q,1.5*NW,linewidth(4)); dot("$P'$",P1,1.5*W,linewidth(4)); dot("$Q'$",Q1,1.5*E,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--P1^^Q--Q1,dashed); [/asy] Because the trapezoid is isosceles, by symmetry $PQ$ is parallel to $AB$ and $CD$ . Therefore, $\angle PAB \cong \angle APP'$ by interior angles and $\angle PAB \cong \angle PAD$ by the problem statement. Thus, $\triangle P'AP$ is isosceles with $P'P = P'A$ . By symmetry, $P'DP$ is also isosceles, and thus $P'A = \frac{AD}{2}$ . Similarly, the same thing is happening on the right side of the trapezoid, and thus $P'Q'$ is the midline of the trapezoid. Then, $PQ = P'Q' - (P'P + Q'Q)$
Since $P'P = P'A = \frac{AD}{2}, Q'Q = Q'B = \frac{BC}{2}$ and $AD = BC = 333$ , we have $P'P + Q'Q = \frac{333}{2} + \frac{333}{2} = 333$ . The length of the midline of a trapezoid is the average of their bases, so $P'Q' = \frac{500+650}{2} = 575$ . Finally, $PQ = 575 - 333 = \boxed{242}$ | null | 242 |
6a4bc209462e330802e61c25a358e95f | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | We have the following diagram: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); X = intersectionpoint(B--5*(Q-B)+B,C--D); Y = intersectionpoint(C--5*(Q-C)+C,A--B); Z = intersectionpoint(D--5*(P-D)+D,A--B); W = intersectionpoint(A--5*(P-A)+A,C--D); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*(-1,0),linewidth(4)); dot("$Q$",Q,1.5*E,linewidth(4)); dot("$X$",X,1.5*dir(-105),linewidth(4)); dot("$Y$",Y,1.5*N,linewidth(4)); dot("$Z$",Z,1.5*N,linewidth(4)); dot("$W$",W,1.5*dir(-75),linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--Z^^P--W^^Q--X^^Q--Y,dashed); [/asy] Extend lines $AP$ and $BQ$ to meet line $DC$ at points $W$ and $X$ , respectively, and extend lines $DP$ and $CQ$ to meet $AB$ at points $Z$ and $Y$ , respectively.
Claim: quadrilaterals $AZWD$ and $BYXC$ are rhombuses.
Proof: Since $\angle DAB + \angle ADC = 180^{\circ}$ $\angle ADP + \angle PAD = 90^{\circ}$ . Therefore, triangles $APD$ $APZ$ $DPW$ and $PZW$ are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, $\triangle PZW$ is congruent to the other three. Therefore, $AD = DW = WZ = AZ$ , so $AZWD$ is a rhombus. By symmetry, $BYXC$ is also a rhombus.
Extend line $PQ$ to meet $\overline{AD}$ and $\overline{BC}$ at $R$ and $S$ , respectively. Because of rhombus properties, $RP = QS = \frac{333}{2}$ . Also, by rhombus properties, $R$ and $S$ are the midpoints of segments $AD$ and $BC$ , respectively; therefore, by trapezoid properties, $RS = \frac{AB + CD}{2} = 575$ . Finally, $PQ = RS - RP - QS = \boxed{242}$ | null | 242 |
6a4bc209462e330802e61c25a358e95f | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); dot("$Q$",Q,1.5*NW,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); X = (-121,6*sqrt(731)); Y = (121,6*sqrt(731)); Z = (-121,-6*sqrt(731)); W = (121,-6*sqrt(731)); draw(X--Z^^Y--W,dashed); draw(rightanglemark(A,X,Z,500),red); draw(rightanglemark(B,Y,W,500),red); draw(rightanglemark(C,W,Y,500),red); draw(rightanglemark(D,Z,X,500),red); dot("$X$",X,1.5*N,linewidth(4)); dot("$Y$",Y,1.5*N,linewidth(4)); dot("$Z$",Z,1.5*S,linewidth(4)); dot("$W$",W,1.5*S,linewidth(4)); [/asy]
Let $X$ and $Y$ be the feet of the altitudes from $P$ and $Q$ , respectively, to $AB$ , and let $Z$ and $W$ be the feet of the altitudes from $P$ and $Q$ , respectively, to $CD$ . Side $AB$ is parallel to side $CD$ , so $XYWZ$ is a rectangle with width $PQ$ . Furthermore, because $CD - AB = 650-500 = 150$ and trapezoid $ABCD$ is isosceles, $WC - YB = ZD - XA = 75$
Also because $ABCD$ is isosceles, $\angle ABC + \angle BCD$ is half the total sum of angles in $ABCD$ , or $180^{\circ}$ . Since $BQ$ and $CQ$ bisect $\angle ABC$ and $\angle BCD$ , respectively, we have $\angle QBC + \angle QCB = 90^{\circ}$ , so $\angle BQC = 90^{\circ}$
Letting $BQ = 333k$ , applying Pythagoras to $\triangle BQC$ yields $QC = 333\sqrt{1-k^2}$ . We then proceed using similar triangles: $\angle BYQ = \angle BQC = 90^{\circ}$ and $\angle YBQ = \angle QBC$ , so by AA similarity $YB = 333k^2$ . Likewise, $\angle CWQ = \angle BQC = 90^{\circ}$ and $\angle WCQ = \angle QCB$ , so by AA similarity $WC = 333(1 - k^2)$ . Thus $WC + YB = 333$
Adding our two equations for $WC$ and $YB$ gives $2WC = 75 + 333 = 408$ . Therefore, the answer is $PQ = ZW = CD - 2WC = 650 - 408 = \boxed{242}$ | null | 242 |
6a4bc209462e330802e61c25a358e95f | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | This will be my first solution on AoPS. My apologies in advance for any errors.
Angle bisectors can be thought of as the locus of all points equidistant from the lines whose angle they bisect. It can thus be seen that $P$ is equidistant from $AB, AD,$ and $CD$ and $Q$ is equidistant from $AB, BC,$ and $CD.$ If we let the feet of the altitudes from $P$ to $AB, AD,$ and $CD$ be called $E, F,$ and $G$ respectively, we can say that $PE = PF = PG.$ Analogously, we let the feet of the altitudes from $Q$ to $AB, BC,$ and $CD$ be $H, I,$ and $J$ respectively. Thus, $QH = QI = QJ.$ Because $ABCD$ is an isosceles trapezoid, we can say that all of the altitudes are equal to each other.
By SA as well as SS congruence for right triangles, we find that triangles $AEP, AFP, BHQ,$ and $BIQ$ are congruent. Similarly, $DFP, DGP, CJQ,$ and $CIQ$ by the same reasoning. Additionally, $EH = GJ = PQ$ since $EHQP$ and $GJQP$ are congruent rectangles.
If we then let $x = AE = AF = BH = BI,$ let $y = CI = CJ = DG = DF,$ and let $z = EH = GJ = PQ,$ we can create the following system of equations with the given side length information: \begin{align*} 2x + z &= 500, \\ 2y + z &= 650, \\ x + y &= 333. \end{align*} Adding the first two equations, subtracting by twice the second, and dividing by $2$ yields $z = PQ = \boxed{242}.$ | null | 242 |
6a4bc209462e330802e61c25a358e95f | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$ . The diagram looks like this: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); P1 = intersectionpoint(A--D,P--(-300)*(Q-P)+P); Q1 = intersectionpoint(B--C,Q--300*(Q-P)+Q); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); dot("$Q$",Q,1.5*NW,linewidth(4)); dot("$P'$",P1,1.5*W,linewidth(4)); dot("$Q'$",Q1,1.5*E,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--P1^^Q--Q1,dashed); [/asy] Since $\angle A + \angle D=\angle B + \angle C = 180^{\circ}$ , it follows that $\angle P'AP+\angle P'DP = \angle Q'BQ + \angle Q'CQ = 90^{\circ}$ . Thus, $\angle APD = \angle BQC = 90^{\circ}$ , implying that $\triangle APD$ and $\triangle BQC$ are right triangles. Since $P'P$ and $Q'Q$ are medians, $P'P+Q'Q=\frac{333\times2}{2}=333$ . Since $P'Q'=\frac{500+650}{2}=575$ , we have $PQ+P'P+Q'Q=575$ , or $PQ=575-333=\boxed{242}$ | null | 242 |
6a4bc209462e330802e61c25a358e95f | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | Let $PQ = x$ . Note that since $AP$ bisects $\angle{A}$ and $DP$ bisects $\angle{D}$ , we have \[\angle{APD} = 180^{\circ}-\tfrac12 \angle{A}-\tfrac12 \angle{D}=90^{\circ}.\] Let $\angle{ADP}=\theta$ . We have that $\angle{ADC} = 2\theta.$ Now, drop an altitude from $A$ to $CD$ at $E$ . Notice that $DE=\tfrac{650-500}{2}=75$ . By the definition of cosine, we have \[\cos{2\theta}=1-2\cos^2{\theta}=\tfrac{75}{333}=\tfrac{25}{111} \implies \cos{\theta}=\tfrac{2\sqrt{1887}}{111}.\] Notice, however, that we can also apply this to $\triangle{APD}$ ; we have \[\cos{\theta}=\tfrac{DP}{333} \implies DP=6\sqrt{1887}.\] By the Pythagorean Theorem, we get \[AP=\sqrt{333^2-(6\sqrt{1887})^2}=3\sqrt{4773}.\] Then, drop an altitude from $P$ to $AB$ at $F$ ; if $AF=y$ , then $PQ=x=500-2y$ . Because $AP$ is an angle bisector, we see that $\angle{BAP}=\angle{DAP}=90^{\circ}-\theta$ . Again, by the definition of cosine, we have \[\cos{(90^{\circ}-\theta)}=\sin{\theta}=\tfrac{\sqrt{4773}}{111}=\tfrac{y}{3\sqrt{4773}} \implies y=129.\] Finally, $PQ=500-2y=\boxed{242}$ | null | 242 |
6a4bc209462e330802e61c25a358e95f | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 | In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$ | [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); dot("$Q$",Q,1.5*NW,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); X = (-121,6*sqrt(731)); Y = (121,6*sqrt(731)); Z = (-121,-6*sqrt(731)); W = (121,-6*sqrt(731)); draw(X--Z^^Y--W,dashed); draw(rightanglemark(A,X,Z,500),red); draw(rightanglemark(B,Y,W,500),red); draw(rightanglemark(C,W,Y,500),red); draw(rightanglemark(D,Z,X,500),red); dot("$X$",X,1.5*N,linewidth(4)); dot("$Y$",Y,1.5*N,linewidth(4)); dot("$Z$",Z,1.5*S,linewidth(4)); dot("$W$",W,1.5*S,linewidth(4)); [/asy]
As in solution 4, $\angle APD = 90^{\circ}$ . Set $k = AX$ and $x = DP$
We know that $DZ = AX + \frac{DC-AB}{2}$ , so $DZ = k + \frac{650-500}{2} = k + 75$
$\triangle DPZ \sim \triangle APD$ by AA, so we have $\frac{PD}{AD} = \frac{ZD}{PD}$ , resulting in
\[\frac{x}{333} = \frac{k+75}{x} \text{ (1)}\]
$\triangle APX \sim \triangle ADP$ by AA, so we have $\frac{AP}{AD} = \frac{AX}{AP}$ , resulting in
\[\frac{\sqrt{333^2-x^2}}{333} = \frac{k}{\sqrt{333^2-k^2}} \text{ (2)}\]
From $\text{(1)}$ , we have $x^2 = 333k + 333(75) = 333k + 24975$ . From $\text{(2)}$ , we have $333^2 - x^2 = 333k$ , or $x^2 = 333^2 - 333k$ . Thus, $333k + 24975 = 333^2 - 333k$ . Solving for $k$ yields $k = 129$
By symmetry, $YB = AX = 129$ . Thus, $PQ = XY = AB - 2AX = 500 - 2(129) = \boxed{242}$ | null | 242 |
92403c57dfe49de626a66c92cfa0dae2 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_4 | Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$ | We rewrite $w$ and $z$ in polar form: \begin{align*} w &= e^{i\cdot\frac{\pi}{6}}, \\ z &= e^{i\cdot\frac{2\pi}{3}}. \end{align*} The equation $i \cdot w^r = z^s$ becomes \begin{align*} e^{i\cdot\frac{\pi}{2}} \cdot \left(e^{i\cdot\frac{\pi}{6}}\right)^r &= \left(e^{i\cdot\frac{2\pi}{3}}\right)^s \\ e^{i\left(\frac{\pi}{2}+\frac{\pi}{6}r\right)} &= e^{i\left(\frac{2\pi}{3}s\right)} \\ \frac{\pi}{2}+\frac{\pi}{6}r &= \frac{2\pi}{3}s+2\pi k \\ 3+r &= 4s+12k \\ 3+r &= 4(s+3k). \end{align*} for some integer $k.$
Since $4\leq 3+r\leq 103$ and $4\mid 3+r,$ we conclude that \begin{align*} 3+r &\in \{4,8,12,\ldots,100\}, \\ s+3k &\in \{1,2,3,\ldots,25\}. \end{align*} Note that the values for $s+3k$ and the values for $r$ have one-to-one correspondence.
We apply casework to the values for $s+3k:$
Together, the answer is $264+306+264=\boxed{834}.$ | null | 834 |
92403c57dfe49de626a66c92cfa0dae2 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_4 | Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$ | First we recognize that $w = \operatorname{cis}(30^{\circ})$ and $z = \operatorname{cis}(120^{\circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$ , respectively. By De Moivre's theorem, $\operatorname{cis}(\theta)^n = \operatorname{cis}(n\theta)$ . When you multiply by $i$ , we can think of that as rotating the complex number $90^{\circ}$ counterclockwise in the complex plane. Therefore, by the equation we know that $30r + 90$ and $120s$ land on the same angle.
This means that \[30r + 90 \equiv 120s \pmod{360},\] which we can simplify to \[r+3 \equiv 4s \pmod{12}.\] Notice that this means that $r$ cycles by $12$ for every value of $s$ . This is because once $r$ hits $12$ , we get an angle of $360^{\circ}$ and the angle laps onto itself again. By a similar reasoning, $s$ laps itself every $3$ times, which is much easier to count. By listing the possible values out, we get the pairs $(r,s)$ \[\begin{array}{cccccccc} (1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & \ldots & (97,1) \\ (1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & \ldots & (97,4) \\ (1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & \ldots & (97,7) \\ [-1ex] \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ (1,100) & (5,98) & (9,99) & (13,100) & (17,98) & (21,99) & \ldots & (97,100) \end{array}\] We have $25$ columns in total: $34$ values for the first column, $33$ for the second, $33$ for the third, and then $34$ for the fourth, $33$ for the fifth, $33$ for the sixth, etc. Therefore, this cycle repeats every $3$ columns and our total sum is $(34+33+33) \cdot 8 + 34 = 100 \cdot 8 + 34 = \boxed{834}$ | null | 834 |
868d5da193995d533e05fa0868e64337 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_5 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D$ | Define $m$ as the number of minutes they swim for.
Let their meeting point be $A$ . Melanie is swimming against the current, so she must aim upstream from point $A$ , to compensate for this; in particular, since she is swimming for $m$ minutes, the current will push her $14m$ meters downstream in that time, so she must aim for a point $B$ that is $14m$ meters upstream from point $A$ . Similarly, Sherry is swimming downstream for $m$ minutes, so she must also aim at point $B$ to compensate for the flow of the current.
If Melanie and Sherry were to both aim at point $B$ in a currentless river with the same dimensions, they would still both meet at that point simultaneously. Since there is no current in this scenario, the distances that Melanie and Sherry travel, respectively, are $80m$ and $60m$ meters. We can draw out this new scenario, with the dimensions that we have: [asy] unitsize(0.02cm); draw((0,0)--(0,264)--(550,264)--(550,0)--cycle); pair B = (198,264); dot(B^^(0,0)^^(550,0),linewidth(5)); draw((0,0)--B,dashed); draw((550,0)--B,dashed); label("$60m$", (0,0)--B, E); label("$80m$", (550,0)--B, W); label("$264$", (0,0)--(0,264), W); label("$\frac{D}{2} - 14m$", (0,264)--B, N); label("$\frac{D}{2} + 14m$", B--(550,264), N); label("$D$", (0,0)--(550,0), S); label("$B$", B, N); label("Downstream", (350,325), E); label("Upstream", (200,325), W); draw((225,325)--(325,325), Arrows); [/asy] (While it is indeed true that the triangle above with side lengths $60m$ $80m$ and $D$ is a right triangle, we do not know this yet, so we cannot assume this based on the diagram.)
By the Pythagorean Theorem, we have \begin{align*} 264^{2} + \left( \frac{D}{2} - 14m \right) ^{2} &= 3600m^{2} \\ 264^{2} + \left( \frac{D}{2} + 14m \right) ^{2} &= 6400m^{2}. \end{align*}
Subtracting the first equation from the second gives us $28Dm = 2800m^{2}$ , so $D = 100m$ . Substituting this into our first equation, we have that \begin{align*}264^{2} + 36^{2} m^{2} &= 60^{2}m^{2} \\ 264^{2} &= 96 \cdot 24 \cdot m^{2} \\ 11^{2} &= 4 \cdot m^{2} \\ m &= \frac{11}{2}. \end{align*}
So $D = 100m = \boxed{550}$ | null | 550 |
868d5da193995d533e05fa0868e64337 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_5 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D$ | Claim
Median $AM$ and altitude $AH$ are drawn in triangle $ABC$ $AB = c, AC = b < c, BC = a$ are known. Let's denote $MH = x$
Prove that \begin{align*}2ax = c^{2} - b^{2}\end{align*}
Proof \[BH + CH = a,\] \begin{align*} BH^{2} - CH^{2} = c^{2} - b^{2}\implies BH - CH &= \frac{c^{2} - b^{2}} {a},\end{align*} \[BH = \frac{c^{2} - b^{2}}{2a} + \frac{a}{2},\] \begin{align*}MH = BH - BM &= \frac{c^{2} - b^{2}} {2a}.\end{align*}
Solution
In the coordinate system associated with water, the movement is described by the scheme in the form of a triangle, the side on which Melanie floats is $80t$ , where t is the time of Melanie's movement, the side along which Sherry floats is $60t$
The meeting point floated away at a distance of $14t$ from the midpoint between the starting points of Melanie and Sherry.
In the notation of the Claim \begin{align*} c = 80t, b = 60t, x = 14t \implies a = \frac{(80t)^2-(60t)^2}{2 \cdot 14t}=\frac{20^2}{4}\cdot \frac{16-9}{7}t = 100t.\end{align*} Hence, \begin{align*} AH = \sqrt{BC^2-BH^2}= \sqrt{(80t)^2-(50t+14t)^2}=16t \cdot \sqrt{5^2-4^2}= 48t = 264 \implies t = 5.5.\end{align*} \[D = a = 100t = \boxed{550}\] vladimir.shelomovskii@gmail.com, vvsss | null | 550 |
868d5da193995d533e05fa0868e64337 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_5 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D$ | We have the following diagram: [asy] /* Made by MRENTHUSIASM */ size(350); pair A, B, C; A = (0,264); B = (-275,0); C = (275,0); draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2)); dot("Finish",A,1.75*N,linewidth(5)); dot("Sherry",B,1.75*S,linewidth(5)); dot("Melanie",C,1.75*S,linewidth(5)); Label L1 = Label("$D$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$264$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); Label L3 = Label("Current $(14)$", position=EndPoint, filltype=Fill(3,0,white)); Label L4 = Label("$y$", align=(-1,0), position=Relative(0.4)); Label L5 = Label("$x$", align=(0,1), position=Relative(0.4)); Label L6 = Label("$y$", align=(1,0), position=Relative(0.4)); Label L7 = Label("$x$", align=(0,1), position=Relative(0.4)); draw(B-(0,75)--C-(0,75), L=L1, arrow=Arrows(),bar=Bars(15)); draw((-350,0)--(-350,264), L=L2, arrow=Arrows(),bar=Bars(15)); draw((-300,-120)--(300,-120), L=L3, arrow=EndArrow()); draw(B--B+(0,48), L=L4, arrow=EndArrow()); draw(B+(0,48)--B+(50,48), L=L5, arrow=EndArrow()); draw(C--C+(0,48), L=L6, arrow=EndArrow()); draw(C+(0,48)--C+(-50,48), L=L7, arrow=EndArrow()); [/asy] Since Melanie and Sherry swim for the same distance and the same amount of time, they swim at the same net speed.
Let $x$ and $y$ be some positive numbers. We have the following table: \[\begin{array}{c||c|c|c} & \textbf{Net Velocity Vector (m/min)} & \textbf{Natural Velocity Vector (m/min)} & \textbf{Natural Speed (m/min)} \\ \hline \hline &&& \\ [-2.25ex] \textbf{Melanie} & \langle -x,y\rangle & \langle -x-14,y\rangle & 80 \\ \hline &&& \\ [-2.25ex] \textbf{Sherry} & \langle x,y\rangle & \langle x-14,y\rangle & 60 \end{array}\] Recall that $|\text{velocity}|=\text{speed},$ so \begin{align*} (-x-14)^2 + y^2 &= 80^2, &&(1) \\ (x-14)^2 + y^2 &= 60^2. &&(2) \end{align*} We subtract $(2)$ from $(1)$ to get $56x=2800,$ from which $x=50.$ Substituting this into either equation, we have $y=48.$
It follows that Melanie and Sherry both swim for $264\div y=5.5$ minutes. Therefore, the answer is \[D=2x\cdot5.5=\boxed{550}.\] ~MRENTHUSIASM | null | 550 |
868d5da193995d533e05fa0868e64337 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_5 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D$ | We can break down movement into two components: the $x$ -component and the $y$ -component. Suppose that Melanie travels a distance of $a$ in the $x$ -direction and a distance of $c$ in the $y$ -direction in one minute when there is no current. Similarly, suppose that Sherry travels a distance of $a$ in the $x$ -direction but a distance of $b$ in the $y$ -direction in one minute when there is no current. The current only affects the $x$ -components because it goes in the $x$ -direction.
[asy] /* Ruthlessly plagiarized from MRENTHUSIASM by Curious_crow */ size(350); pair A, B, C; A = (0,264); B = (-275,0); C = (275,0); draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2)); dot("Finish",A,1.75*N,linewidth(5)); dot("Sherry",B,1.75*S,linewidth(5)); dot("Melanie",C,1.75*S,linewidth(5)); Label L1 = Label("$D$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$264$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); Label L3 = Label("Current $(14)$", position=EndPoint, filltype=Fill(3,0,white)); Label L4 = Label("$a$", align=(-1,0), position=Relative(0.4)); Label L5 = Label("$b+14$", align=(0,1), position=Relative(0.4)); Label L6 = Label("$a$", align=(1,0), position=Relative(0.4)); Label L7 = Label("$c-14$", align=(0,1), position=Relative(0.4)); draw(B-(0,75)--C-(0,75), L=L1, arrow=Arrows(),bar=Bars(15)); draw((-350,0)--(-350,264), L=L2, arrow=Arrows(),bar=Bars(15)); draw((-300,-120)--(300,-120), L=L3, arrow=EndArrow()); draw(B--B+(0,48), L=L4, arrow=EndArrow()); draw(B+(0,48)--B+(50,48), L=L5, arrow=EndArrow()); draw(C--C+(0,48), L=L6, arrow=EndArrow()); draw(C+(0,48)--C+(-50,48), L=L7, arrow=EndArrow()); [/asy]
Now, note that $a^2 + b^2 = 60^2$ because Sherry travels 60 meters in a minute. Thus, $a^2 + c^2 = 80^2$ because Melanie travels 80 meters in a minute. Also, the distance they travel with the current must be the same in one minute because they reach the point equidistant from them at the same time. That means $b + 14 = c - 14$ or $b = c - 28$ . So now we can plug that into the two equations to get: \begin{align*} a^2 + c^2 &= 80^2, \\ a^2 + (c-28)^2 &= 60^2. \end{align*} We can solve the system of equations to get $a = 48$ and $c = 64$ . From this, we can figure out that it must've taken them $5.5$ minutes to get to the other side, because $264/48 = 5.5$ . This means that there are $5.5$ lengths of $48$ in each person's travel. Also, $D$ must be equal to $11(b+14) = 11(c-14)$ because there are $(5.5)2 = 11$ lengths of $b-14$ between them, $5.5$ on each person's side. Since $c = 64$ , we have $c-14 = 50$ , so the answer is \[D=11\cdot50=\boxed{550}.\] ~Curious_crow | null | 550 |
466d7782518942fdb4ba15d1ab0dad94 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_6 | Find the number of ordered pairs of integers $(a, b)$ such that the sequence \[3, 4, 5, a, b, 30, 40, 50\] is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression. | Since $3,4,5,a$ and $3,4,5,b$ cannot be an arithmetic progression, $a$ or $b$ can never be $6$ . Since $b, 30, 40, 50$ and $a, 30, 40, 50$ cannot be an arithmetic progression, $a$ and $b$ can never be $20$ . Since $a < b$ , there are ${24 - 2 \choose 2} = 231$ ways to choose $a$ and $b$ with these two restrictions in mind.
However, there are still specific invalid cases counted in these $231$ pairs $(a,b)$ . Since \[3,5,a,b\] cannot form an arithmetic progression, $\underline{(a,b) \neq (7,9)}$ \[a,b,30,50\] cannot be an arithmetic progression, so $(a,b) \neq (-10,10)$ ; however, since this pair was not counted in our $231$ , we do not need to subtract it off. \[3,a,b,30\] cannot form an arithmetic progression, so $\underline{(a,b) \neq (12,21)}$ \[4, a, b, 40\] cannot form an arithmetic progression, so $\underline{(a,b) \neq (16,28)}$ \[5, a,b, 50\] cannot form an arithmetic progression, $(a,b) \neq 20, 35$ ; however, since this pair was not counted in our $231$ (since we disallowed $a$ or $b$ to be $20$ ), we do not to subtract it off.
Also, the sequences $(3,a,b,40)$ $(3,a,b,50)$ $(4,a,b,30)$ $(4,a,b,50)$ $(5,a,b,30)$ and $(5,a,b,40)$ will never be arithmetic, since that would require $a$ and $b$ to be non-integers.
So, we need to subtract off $3$ progressions from the $231$ we counted, to get our final answer of $\boxed{228}$ | null | 228 |
466d7782518942fdb4ba15d1ab0dad94 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_6 | Find the number of ordered pairs of integers $(a, b)$ such that the sequence \[3, 4, 5, a, b, 30, 40, 50\] is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression. | We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.
We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract $\textbf{1}$ from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50.
(Note if we take only 1 at a time, there will have to be 3 of $a, b$ , which is impossible.)
If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}.
Finally, we count the sequences that are something like (one of 3,4,5,), $a, b$ , (one of 30, 40, 50). If this is to be the case, then let $a$ be the starting value in the sequence. The sequence will be $a, a+d, a+2d, a+3d$ ; We see that if we subtract the largest term by the smallest term we have $3d$ , so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are $3,a,b,30; 4,a,b,40; 5,a,b,50$ . Of these, only the last is invalid because it gives $b = 35$ , larger than our bounds $6<a<b<30$ . Therefore, we subtract $\textbf{2}$ from this case.
Our final answer is $231 - 1 - 2 = \boxed{228}$ | null | 228 |
466d7782518942fdb4ba15d1ab0dad94 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_6 | Find the number of ordered pairs of integers $(a, b)$ such that the sequence \[3, 4, 5, a, b, 30, 40, 50\] is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression. | Denote $S = \left\{ (a, b) : 6 \leq a < b \leq 29 \right\}$
Denote by $A$ a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes $a$ but not $b$
Denote by $B$ a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes $b$ but not $a$
Hence, $C$ is a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes both $a$ and $b$
Hence, this problem asks us to compute \[ | S | - \left( | A | + | B | + | C | \right) . \]
First, we compute $| S |$
We have $| S | = \binom{29 - 6 + 1}{2} = \binom{24}{2} = 276$
Second, we compute $| A |$
$\textbf{Case 1}$ $a = 6$
We have $b = 8 , \cdots , 19, 21, 22, \cdots, 29$ .
Thus, the number of solutions is 22.
$\textbf{Case 2}$ $a = 20$
We have $b = 21, 22, \cdots , 29$ .
Thus, the number of solutions is 9.
Thus, $| A | = 22 + 9 = 31$
Third, we compute $| B |$
In $B$ , we have $b = 6, 20$ . However, because $6 \leq a < b$ , we have $b \geq 7$ .
Thus, $b = 20$
This implies $a = 7, 8, 9, 11, 12, \cdots , 19$ . Note that $(a, b)=(10, 20)$ belongs in $C$
Thus, $| B | = 12$
Fourth, we compute $| C |$
$\textbf{Case 1}$ : In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the same side of $a$ and $b$
Hence, $(a, b) = (6 , 7), (7, 9) , (10, 20)$ .
Therefore, the number solutions in this case is 3.
$\textbf{Case 2}$ : In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the opposite sides of $a$ and $b$
$\textbf{Case 2.1}$ : The arithmetic sequence is $3, a, b, 30$
Hence, $(a, b) = (12, 21)$
$\textbf{Case 2.2}$ : The arithmetic sequence is $4, a, b, 40$
Hence, $(a, b) = (16, 28)$
$\textbf{Case 2.3}$ : The arithmetic sequence is $5, a, b, 50$
Hence, $(a, b) = (20, 35)$ . However, the sequence $... 20, 35, 30, 40, 50$ is not strictly increasing.
Putting two cases together, $| C | = 65.$
Therefore, \[| S | - \left( | A | + | B | + | C | \right) = 276 - \left( 31 + 12 + 5 \right) = \boxed{228}.\] | null | 228 |
466d7782518942fdb4ba15d1ab0dad94 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_6 | Find the number of ordered pairs of integers $(a, b)$ such that the sequence \[3, 4, 5, a, b, 30, 40, 50\] is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression. | divide cases into $7\leq a<20; 21\leq a\leq28$ .(Notice that $a$ can't be equal to $6,20$ , that's why I divide them into two parts.
There are three cases that arithmetic sequence forms: $3,12,21,30;4,16,28,40;3,5,7,9$ .(NOTICE that $5,20,35,50$ IS NOT A VALID SEQUENCE!)
So when $7\leq a<20$ , there are $10+11+12+...+22-3-13=192$ possible ways( 3 means the arithmetic sequence and 13 means there are 13 "a" s and b cannot be 20)
When $21\leq a \leq 28$ , there are $1+2+\cdots+8=36$ ways.
In all, there are $192+36=\boxed{228}$ possible sequences. | null | 228 |
4b065e3314d32f64f4369d3dc064d8a1 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_7 | Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{1}{7\cdot8\cdot9}.$
If we minimize the numerator, then $a \cdot b \cdot c - d \cdot e \cdot f = 1.$ Note that $a \cdot b \cdot c \cdot d \cdot e \cdot f = (a \cdot b \cdot c) \cdot (a \cdot b \cdot c - 1) \geq 6! = 720,$ so $a \cdot b \cdot c \geq 28.$ It follows that $a \cdot b \cdot c$ and $d \cdot e \cdot f$ are consecutive composites with prime factors no other than $2,3,5,$ and $7.$ The smallest values for $a \cdot b \cdot c$ and $d \cdot e \cdot f$ are $36$ and $35,$ respectively. So, we have $\{a,b,c\} = \{2,3,6\}, \{d,e,f\} = \{1,5,7\},$ and $\{g,h,i\} = \{4,8,9\},$ from which $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} = \frac{1}{288}.$
If we do not minimize the numerator, then $a \cdot b \cdot c - d \cdot e \cdot f > 1.$ Note that $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{2}{7\cdot8\cdot9} > \frac{1}{288}.$
Together, we conclude that the minimum possible positive value of $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}$ is $\frac{1}{288}.$ Therefore, the answer is $1+288=\boxed{289}.$ | null | 289 |
4b065e3314d32f64f4369d3dc064d8a1 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_7 | Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Obviously, to find the correct answer, we need to get the largest denominator with the smallest numerator.
To bash efficiently, we can start out with $7\cdot8\cdot9$ as our denominator. This, however, leaves us with the numbers $1, 2, 3, 4, 5,$ and $6$ left. The smallest we can make out of this is $1\cdot5\cdot6 - 2\cdot3\cdot4 = 30 - 24 = 6$ . When simplified, it gives us $\frac{1}{84}$ , which gives a small answer of $85$ . Obviously there are larger answers than this.
After the first bash, we learn to bash even more efficiently, we can consider both the numerator and the denominator when guessing. We know the numerator has to be extremely small while still having a large denominator. When bashing, we soon find out the couple $(a,b,c)=(2,3,6)$ and $(d,e,f)=(1,5,7)$
This gives us a numerator of $36-35=1$ , which is by far the smallest yet. With the remaining numbers $4, 8,$ and $9$ , we get $\frac{36-35}{4\cdot8\cdot9}=\frac{1}{288}$
Finally, we add up our numerator and denominator: The answer is $1+288=\boxed{289}$ | null | 289 |
4b065e3314d32f64f4369d3dc064d8a1 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_7 | Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | To minimize the numerator, we must have $abc - def = 1$ . Thus, one of these products must be odd and the other must be even. The odd product must consist of only odd numbers. The smallest such value $(d, e, f) = (1, 3, 5)$ cannot result in a difference of $1$ , and the next smallest product, $(d, e, f) = (1, 3, 7)$ cannot either, but $(d, e, f) = (1, 5, 7)$ can if $(a, b, c) = (2, 3, 6)$ . Thus, the denominator must be $(g, h, i) = (4, 8, 9)$ , and the smallest fraction possible is $\dfrac{36 - 35}{288} = \dfrac{1}{288}$ , making the answer $1 + 288 = \boxed{289}$ | null | 289 |
0684181973c778a4c29c79a6dcc14ea8 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_8 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC,$ and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC.$ The side length of the smaller equilateral triangle can be written as $\sqrt{a} - \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$ | We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\omega_A$ is the incircle of $\triangle AB'C'$ [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C,dashed); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*(-1,0),linewidth(4)); dot("$C$",C,1.5*(1,0),linewidth(4)); dot("$B'$",B1,1.5*dir(B1),linewidth(4)); dot("$C'$",C1,1.5*dir(C1),linewidth(4)); dot("$O$",W,1.5*dir(90),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); [/asy] Since the diameter of the circle is the height of this triangle, the height of this triangle is $36$ . We can use inradius or equilateral triangle properties to get the inradius of this triangle is $12$ (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is $12$
Let $O=\omega$ be the center of the largest circle. We will set up a coordinate system with $O$ as the origin. The center of $\omega_A$ will be at $(0,-6)$ because it is directly beneath $O$ and is the length of the larger radius minus the smaller radius, or $18-12 = 6$ . By rotating this point $120^{\circ}$ around $O$ , we get the center of $\omega_B$ . This means that the magnitude of vector $\overrightarrow{O\omega_B}$ is $6$ and is at a $30$ degree angle from the horizontal. Therefore, the coordinates of this point are $(3\sqrt{3},3)$ and by symmetry the coordinates of the center of $\omega_C$ is $(-3\sqrt{3},3)$
The upper left and right circles intersect at two points, the lower of which is $X$ . The equations of these two circles are: \begin{align*} (x+3\sqrt3)^2 + (y-3)^2 &= 12^2, \\ (x-3\sqrt3)^2 + (y-3)^2 &= 12^2. \end{align*} We solve this system by subtracting to get $x = 0$ . Plugging back in to the first equation, we have $(3\sqrt{3})^2 + (y-3)^2 = 144 \implies (y-3)^2 = 117 \implies y-3 = \pm \sqrt{117} \implies y = 3 \pm \sqrt{117}$ . Since we know $X$ is the lower solution, we take the negative value to get $X = (0,3-\sqrt{117})$
We can solve the problem two ways from here. We can find $Y$ by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find $OX$ as they lie on the same vertical, $\angle XOY$ is $120$ degrees so we can make use of $30-60-90$ triangles, and $OX = OY$ because $O$ is the center of triangle $XYZ$ . We can draw the diagram as such: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C^^W--X^^W--Y^^W--midpoint(X--Y),dashed); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*(-1,0),linewidth(4)); dot("$C$",C,1.5*(1,0),linewidth(4)); dot("$B'$",B1,1.5*dir(B1),linewidth(4)); dot("$C'$",C1,1.5*dir(C1),linewidth(4)); dot("$O$",W,1.5*dir(90),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); [/asy] Note that $OX = OY = \sqrt{117} - 3$ . It follows that \begin{align*} XY &= 2 \cdot \frac{OX\cdot\sqrt{3}}{2} \\ &= OX \cdot \sqrt{3} \\ &= (\sqrt{117}-3) \cdot \sqrt{3} \\ &= \sqrt{351}-\sqrt{27}. \end{align*} Finally, the answer is $351+27 = \boxed{378}$ | null | 378 |
0684181973c778a4c29c79a6dcc14ea8 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_8 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC,$ and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC.$ The side length of the smaller equilateral triangle can be written as $\sqrt{a} - \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$ | [asy] /* Made by MRENTHUSIASM */ /* Modified by isabelchen */ size(250); pair A, B, C, W, WA, WB, WC, X, Y, Z, D, E; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); D = intersectionpoint(Circle(WA,12),A--C); E = intersectionpoints(Circle(WB,12),Circle(WC,12))[0]; filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$\omega$",W,1.5*dir(270),linewidth(4)); dot("$\omega_A$",WA,1.5*dir(-WA),linewidth(4)); dot("$\omega_B$",WB,1.5*dir(-WB),linewidth(4)); dot("$\omega_C$",WC,1.5*dir(-WC),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); draw(WC--WB^^WC--X^^WC--E^^WA--D^^A--X); [/asy] For equilateral triangle with side length $l$ , height $h$ , and circumradius $r$ , there are relationships: $h = \frac{\sqrt{3}}{2} l$ $r = \frac{2}{3} h = \frac{\sqrt{3}}{3} l$ , and $l = \sqrt{3}r$
There is a lot of symmetry in the figure. The radius of the big circle $\odot \omega$ is $R = 18$ , let the radius of the small circles $\odot \omega_A$ $\odot \omega_B$ $\odot \omega_C$ be $r$
We are going to solve this problem in $3$ steps:
$\textbf{Step 1:}$
We have $\triangle A \omega_A D$ is a $30-60-90$ triangle, and $A \omega_A = 2 \cdot \omega_A D$ $A \omega_A = 2R-r$ $\odot \omega$ and $\odot \omega_A$ are tangent), and $\omega_A D = r$ . So, we get $2R-r = 2r$ and $r = \frac{2}{3} \cdot R = 12$
Since $\odot \omega$ and $\odot \omega_A$ are tangent, we get $\omega \omega_A = R - r = \frac{1}{3} \cdot R = 6$
Note that $\triangle \omega_A \omega_B \omega_C$ is an equilateral triangle, and $\omega$ is its center, so $\omega_B \omega_C = \sqrt{3} \cdot \omega \omega_A = 6 \sqrt{3}$
$\textbf{Step 2:}$
Note that $\triangle \omega_C E X$ is an isosceles triangle, so \[EX = 2 \sqrt{(\omega_C E)^2 - \left(\frac{\omega_B \omega_C}{2}\right)^2} = 2 \sqrt{r^2 - \left(\frac{\omega_B \omega_C}{2}\right)^2} = 2 \sqrt{12^2 - (3 \sqrt{3})^2} = 2 \sqrt{117}.\]
$\textbf{Step 3:}$
In $\odot \omega_C$ , Power of a Point gives $\omega X \cdot \omega E = r^2 - (\omega_C \omega)^2$ and $\omega E = EX - \omega X = 2\sqrt{117} - \omega X$
It follows that $\omega X \cdot (2\sqrt{117} - \omega X) = 12^2 - 6^2$ . We solve this quadratic equation: $\omega X = \sqrt{117} - 3$
Since $\omega X$ is the circumradius of equilateral $\triangle XYZ$ , we have $XY = \sqrt{3} \cdot \omega X = \sqrt{3} \cdot (\sqrt{117} - 3) = \sqrt{351}-\sqrt{27}$
Therefore, the answer is $351+27 = \boxed{378}$ | null | 378 |
0684181973c778a4c29c79a6dcc14ea8 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_8 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC,$ and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC.$ The side length of the smaller equilateral triangle can be written as $\sqrt{a} - \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$ | Let $O$ be the center, $R = 18$ be the radius, and $CC'$ be the diameter of $\omega.$ Let $r$ be the radius, $E,D,F$ are the centers of $\omega_A, \omega_B,\omega_C.$ Let $KGH$ be the desired triangle with side $x.$ We find $r$ using \[CC' = 2R = C'K + KC = r + \frac{r}{\sin 30^\circ} = 3r.\] \[r = \frac{2R}{3} = 12.\] \[OE = R – r = 6.\] Triangles $\triangle DEF$ and $\triangle KGH$ – are equilateral triangles with a common center $O,$ therefore in the triangle $OEH$ $OE = 6, \angle EOH = 120^\circ, OH = \frac{x}{\sqrt3}.$
We apply the Law of Cosines to $\triangle OEH$ and get \[OE^2 + OH^2 + OE \cdot OH = EH^2.\] \[6^2 + \frac{x^2}{3} + \frac{6x}{\sqrt3} = 12^2.\] \[x^2 + 6x \sqrt{3} = 324\] \[x= \sqrt{351} - \sqrt{27} \implies 351 + 27 = \boxed{378}\] | null | 378 |
0684181973c778a4c29c79a6dcc14ea8 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_8 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC,$ and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC.$ The side length of the smaller equilateral triangle can be written as $\sqrt{a} - \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$ | Let $O$ be the center of $\omega$ $X$ be the intersection of $\omega_B,\omega_C$ further from $A$ , and $O_A$ be the center of $\omega_A$ . Define $Y, Z, O_B, O_C$ similarly. It is well-known that the $A$ -mixtilinear inradius $R_A$ is $\tfrac{r}{\cos^2\left(\frac{\angle A}{2}\right)} = \tfrac{9}{\cos^2\left(30^{\circ}\right)} = 12$ , so in particular this means that $OO_B = 18 - R_B = 6 = OO_C$ . Since $\angle O_BOO_C = \angle BOC = 120^\circ$ , it follows by Law of Cosines on $\triangle OO_BO_C$ that $O_BO_C = 6\sqrt{3}$ . Then the Pythagorean theorem gives that the altitude of $O_BO_CX$ is $\sqrt{117}$ , so $OY = OX = \text{dist}(X, YZ) - \text{dist}(O, YZ) = \sqrt{117} - 3$ and $YZ = \tfrac{O_BO_C\cdot OY}{OO_B} = \tfrac{6\sqrt{3}(\sqrt{117} - 3)}{6}=\sqrt{351} - \sqrt{27}$ so the answer is $351 + 27 = \boxed{378}$ | null | 378 |
d0f9e6c9b52692ce4b19a17bd8cd420d | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_9 | Ellina has twelve blocks, two each of red ( $\textbf{R}$ ), blue ( $\textbf{B}$ ), yellow ( $\textbf{Y}$ ), green ( $\textbf{G}$ ), orange ( $\textbf{O}$ ), and purple ( $\textbf{P}$ ). Call an arrangement of blocks $\textit{even}$ if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement \[\textbf{R B B Y G G Y R O P P O}\] is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Consider this position chart: \[\textbf{1 2 3 4 5 6 7 8 9 10 11 12}\] Since there has to be an even number of spaces between each pair of the same color, spots $1$ $3$ $5$ $7$ $9$ , and $11$ contain some permutation of all $6$ colored balls. Likewise, so do the even spots, so the number of even configurations is $6! \cdot 6!$ (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of $\frac{12!}{(2!)^6}$ possible arrangements, so the probability is: \[\frac{6!\cdot6!}{\frac{12!}{(2!)^6}} = \frac{6!\cdot2^6}{7\cdot8\cdot9\cdot10\cdot11\cdot12} = \frac{2^4}{7\cdot11\cdot3} = \frac{16}{231},\] which is in simplest form. So, $m + n = 16 + 231 = \boxed{247}$ | null | 247 |
d0f9e6c9b52692ce4b19a17bd8cd420d | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_9 | Ellina has twelve blocks, two each of red ( $\textbf{R}$ ), blue ( $\textbf{B}$ ), yellow ( $\textbf{Y}$ ), green ( $\textbf{G}$ ), orange ( $\textbf{O}$ ), and purple ( $\textbf{P}$ ). Call an arrangement of blocks $\textit{even}$ if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement \[\textbf{R B B Y G G Y R O P P O}\] is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | We can simply use constructive counting. First, let us place the red blocks; choose the first slot in $12$ ways, and the second in $6$ ways, because the number is cut in half due to the condition in the problem. This gives $12 \cdot 6$ ways to place the red blocks. Similarly, there are $10 \cdot 5$ ways to place the blue blocks, and so on, until there are $2 \cdot 1$ ways to place the purple blocks. Thus, the probability is \[\frac{12 \cdot 6 \cdot 10 \cdot 5 \cdot 8 \cdot 4 \cdot 6 \cdot 3 \cdot 4 \cdot 2 \cdot 2 \cdot 1}{12!}=\frac{16}{231},\] and the desired answer is $16+231=\boxed{247}$ | null | 247 |
12bdca34c9a92ed7cf640dc3390e6e85 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_10 | Three spheres with radii $11$ $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ | This solution refers to the Diagram section.
We let $\ell$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$ . We take a cross-section that contains $A$ and $B$ , which contains these two spheres but not the third, as shown below: [asy] size(400); pair A, B, OA, OB; B = (0,0); A = (-23.6643191,0); OB = (0,8); OA = (-23.6643191,4); draw(circle(OB,13)); draw(circle(OA,11)); draw((-48,0)--(24,0)); label("$\ell$",(-42,0),S); label("$A$",A,S); label("$B$",B,S); label("$O_A$",OA,N); label("$O_B$",OB,N); draw(A--OA); draw(B--OB); draw(OA--OB); draw(OA--(0,4)); draw(OA--(-33.9112699,0)); draw(OB--(10.2469508,0)); label("$24$",midpoint(OA--OB),N); label("$\sqrt{560}$",midpoint(A--B),S); label("$11$",midpoint(OA--(-33.9112699,0)),NW); label("$13$",midpoint(OB--(10.2469508,0)),NE); label("$r$",midpoint(midpoint(A--B)--A),S); label("$r$",midpoint(midpoint(A--B)--B),S); label("$r$",midpoint(A--(-33.9112699,0)),S); label("$r$",midpoint(B--(10.2469508,0)),S); label("$x$",midpoint(midpoint(B--OB)--OB),W); label("$D$",midpoint(B--OB),E); [/asy] Because the plane cuts out congruent circles, they have the same radius and from the given information, $AB = \sqrt{560}$ . Since $ABO_BO_A$ is a trapezoid, we can drop an altitude from $O_A$ to $BO_B$ to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is $\sqrt{560}$ and let the distance from $O_B$ to $D$ be $x$ . Then we have $x^2 = 576-560 \implies x = 4$
We have $AO_A = BD$ because of the rectangle, so $\sqrt{11^2-r^2} = \sqrt{13^2-r^2}-4$ .
Squaring, we have $121-r^2 = 169-r^2 + 16 - 8 \cdot \sqrt{169-r^2}$ .
Subtracting, we get $8 \cdot \sqrt{169-r^2} = 64 \implies \sqrt{169-r^2} = 8 \implies 169-r^2 = 64 \implies r^2 = 105$ .
We also notice that since we had $\sqrt{169-r^2} = 8$ means that $BO_B = 8$ and since we know that $x = 4$ $AO_A = 4$
We take a cross-section that contains $A$ and $C$ , which contains these two spheres but not the third, as shown below: [asy] size(400); pair A, C, OA, OC, M; C = (0,0); A = (-27.4954541697,0); OC = (0,16); OA = (-27.4954541697,4); M = midpoint(A--C); draw(circle(OC,19)); draw(circle(OA,11)); draw((-48,0)--(24,0)); label("$\ell$",(-42,0),S); label("$A$",A,S); label("$C$",C,S); label("$O_A$",OA,N); label("$O_C$",OC,N); draw(A--OA); draw(C--OC); draw(OA--OC); draw(OA--(0,4)); draw(OA--(-37.8877590151,0)); draw(OC--(10.2469508,0)); label("$30$",midpoint(OA--OC),NW); label("$11$",midpoint(OA--(-37.8877590151,0)),NW); label("$19$",midpoint(OC--(10.2469508,0)),NE); label("$r$",midpoint(midpoint(M--A)--A),S); label("$r$",midpoint(midpoint(M--C)--C),S); label("$r$",midpoint(A--(-37.8877590151,0)),S); label("$r$",midpoint(C--(10.2469508,0)),S); label("$E$",(0,4),E); [/asy] We have $CO_C = \sqrt{19^2-r^2} = \sqrt{361 - 105} = \sqrt{256} = 16$ . Since $AO_A = 4$ , we have $EO_C = 16-4 = 12$ . Using Pythagorean theorem, $O_AE = \sqrt{30^2 - 12^2} = \sqrt{900-144} = \sqrt{756}$ . Therefore, $O_AE^2 = AC^2 = \boxed{756}$ | null | 756 |
12bdca34c9a92ed7cf640dc3390e6e85 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_10 | Three spheres with radii $11$ $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ | Let the distance between the center of the sphere to the center of those circular intersections as $a,b,c$ separately.
According to the problem, we have $a^2-11^2=b^2-13^2=c^2-19^2; (11+13)^2-(b-a)^2=560.$ After solving we have $b-a=4,$ plug this back to $11^2-a^2=13^2-b^2,$ we have $a=4, b=8,$ and $c=16.$
The desired value is $(11+19)^2-(16-4)^2=\boxed{756}.$ | null | 756 |
12bdca34c9a92ed7cf640dc3390e6e85 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_10 | Three spheres with radii $11$ $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ | Denote by $r$ the radius of three congruent circles formed by the cutting plane.
Denote by $O_A$ $O_B$ $O_C$ the centers of three spheres that intersect the plane to get circles centered at $A$ $B$ $C$ , respectively.
Because three spheres are mutually tangent, $O_A O_B = 11 + 13 = 24$ $O_A O_C = 11 + 19 = 30$
We have $O_A A^2 = 11^2 - r^2$ $O_B B^2 = 13^2 - r^2$ $O_C C^2 = 19^2 - r^2$
Because $O_A A$ and $O_B B$ are perpendicular to the plane, $O_A AB O_B$ is a right trapezoid, with $\angle O_A A B = \angle O_B BA = 90^\circ$
Hence, \begin{align*} O_B B - O_A A & = \sqrt{O_A O_B^2 - AB^2} \\ & = 4 . \hspace{1cm} (1) \end{align*}
Recall that \begin{align*} O_B B^2 - O_A A^2 & = \left( 13^2 - r^2 \right) - \left( 11^2 - r^2 \right) \\ & = 48 . \hspace{1cm} (2) \end{align*}
Hence, taking $\frac{(2)}{(1)}$ , we get \[ O_B B + O_A A = 12 . \hspace{1cm} (3) \]
Solving (1) and (3), we get $O_B B = 8$ and $O_A A = 4$
Thus, $r^2 = 11^2 - O_A A^2 = 105$
Thus, $O_C C = \sqrt{19^2 - r^2} = 16$
Because $O_A A$ and $O_C C$ are perpendicular to the plane, $O_A AC O_C$ is a right trapezoid, with $\angle O_A A C = \angle O_C CA = 90^\circ$
Therefore, \begin{align*} AC^2 & = O_A O_C^2 - \left( O_C C - O_A A \right)^2 \\ & = \boxed{756} | null | 756 |
5e8e6ab8df063ef422b177389473e3aa | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_12 | For any finite set $X$ , let $| X |$ denote the number of elements in $X$ . Define \[S_n = \sum | A \cap B | ,\] where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\left\{ 1 , 2 , 3, \cdots , n \right\}$ with $|A| = |B|$ .
For example, $S_2 = 4$ because the sum is taken over the pairs of subsets \[(A, B) \in \left\{ (\emptyset, \emptyset) , ( \{1\} , \{1\} ), ( \{1\} , \{2\} ) , ( \{2\} , \{1\} ) , ( \{2\} , \{2\} ) , ( \{1 , 2\} , \{1 , 2\} ) \right\} ,\] giving $S_2 = 0 + 1 + 0 + 0 + 1 + 2 = 4$ .
Let $\frac{S_{2022}}{S_{2021}} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p + q$ is divided by
1000. | Let's try out for small values of $n$ to get a feel for the problem. When $n=1, S_n$ is obviously $1$ . The problem states that for $n=2, S_n$ is $4$ . Let's try it out for $n=3$
Let's perform casework on the number of elements in $A, B$
$\textbf{Case 1:} |A| = |B| = 1$
In this case, the only possible equivalencies will be if they are the exact same element, which happens $3$ times.
$\textbf{Case 2:} |A| = |B| = 2$
In this case, if they share both elements, which happens $3$ times, we will get $2$ for each time, and if they share only one element, which also happens $6$ times, we will get $1$ for each time, for a total of $12$ for this case.
$\textbf{Case 3:} |A| = |B| = 3$
In this case, the only possible scenario is that they both are the set $\{1,2,3\}$ , and we have $3$ for this case.
In total, $S_3 = 18$
Now notice, the number of intersections by each element $1 \ldots 3$ , or in general, $1 \ldots n$ is equal for each element because of symmetry - each element when $n=3$ adds $6$ to the answer. Notice that $6 = \binom{4}{2}$ - let's prove that $S_n = n \cdot \binom{2n-2}{n-1}$ (note that you can assume this and answer the problem if you're running short on time in the real test).
Let's analyze the element $k$ - to find a general solution, we must count the number of these subsets that $k$ appears in. For $k$ to be in both $A$ and $B$ , we need both sets to contain $k$ and another subset of $1$ through $n$ not including $k$ . ( $A = \{k\} \cup A'| A' \subset \{1,2,\ldots,n\} \land A' \not \subset \{k\}$ and $B = \{k\} \cup B'| B' \subset \{1,2,\ldots,n\} \land B' \not \subset \{k\}$
For any $0\leq l \leq n-1$ that is the size of both $A'$ and $B'$ , the number of ways to choose the subsets $A'$ and $B'$ is $\binom{n-1}{l}$ for both subsets, so the total number of ways to choose the subsets are $\binom{n-1}{l}^2$ .
Now we sum this over all possible $l$ 's to find the total number of ways to form sets $A$ and $B$ that contain $k$ . This is equal to $\sum_{l=0}^{n-1} \binom{n-1}{l}^2$ . This is a simplification of Vandermonde's identity, which states that $\sum_{k=0}^{r} \binom{m}{k} \cdot \binom{n}{r-k} = \binom{m+n}{r}$ . Here, $m$ $n$ and $r$ are all $n-1$ , so this sum is equal to $\binom{2n-2}{n-1}$ . Finally, since we are iterating over all $k$ 's for $n$ values of $k$ , we have $S_n = n \cdot \binom{2n-2}{n-1}$ , proving our claim.
We now plug in $S_n$ to the expression we want to find. This turns out to be $\frac{2022 \cdot \binom{4042}{2021}}{2021 \cdot \binom{4040}{2020}}$ . Expanding produces $\frac{2022 \cdot 4042!\cdot 2020! \cdot 2020!}{2021 \cdot 4040! \cdot 2021! \cdot 2021!}$
After cancellation, we have \[\frac{2022 \cdot 4042 \cdot 4041}{2021 \cdot 2021 \cdot 2021} \implies \frac{4044\cdot 4041}{2021 \cdot 2021}\]
$4044$ and $4041$ don't have any common factors with $2021$ , so we're done with the simplification. We want to find $4044 \cdot 4041 + 2021^2 \pmod{1000} \equiv 44 \cdot 41 + 21^2 \pmod{1000} \equiv 1804+441 \pmod{1000} \equiv 2245 \pmod{1000} \equiv \boxed{245}$ | null | 245 |
5e8e6ab8df063ef422b177389473e3aa | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_12 | For any finite set $X$ , let $| X |$ denote the number of elements in $X$ . Define \[S_n = \sum | A \cap B | ,\] where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\left\{ 1 , 2 , 3, \cdots , n \right\}$ with $|A| = |B|$ .
For example, $S_2 = 4$ because the sum is taken over the pairs of subsets \[(A, B) \in \left\{ (\emptyset, \emptyset) , ( \{1\} , \{1\} ), ( \{1\} , \{2\} ) , ( \{2\} , \{1\} ) , ( \{2\} , \{2\} ) , ( \{1 , 2\} , \{1 , 2\} ) \right\} ,\] giving $S_2 = 0 + 1 + 0 + 0 + 1 + 2 = 4$ .
Let $\frac{S_{2022}}{S_{2021}} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p + q$ is divided by
1000. | We take cases based on the number of values in each of the subsets in the pair. Suppose we have $k$ elements in each of the subsets in a pair (for a total of n elements in the set). The expected number of elements in any random pair will be $n \cdot \frac{k}{n} \cdot \frac{k}{n}$ by linearity of expectation because for each of the $n$ elements, there is a $\frac{k}{n}$ probability that the element will be chosen. To find the sum over all such values, we multiply this quantity by $\binom{n}{k}^2$ . Summing, we get \[\sum_{k=1}^{n} \frac{k^2}{n} \binom{n}{k}^2\] Notice that we can rewrite this as \[\sum_{k=1}^{n} \frac{1}{n} \left(\frac{k \cdot n!}{(k)!(n - k)!}\right)^2 = \sum_{k=1}^{n} \frac{1}{n} n^2 \left(\frac{(n-1)!}{(k - 1)!(n - k)!}\right)^2 = n \sum_{k=1}^{n} \binom{n - 1}{k - 1}^2 = n \sum_{k=1}^{n} \binom{n - 1}{k - 1}\binom{n - 1}{n - k}\] We can simplify this using Vandermonde's identity to get $n \binom{2n - 2}{n - 1}$ . Evaluating this for $2022$ and $2021$ gives \[\frac{2022\binom{4042}{2021}}{2021\binom{4040}{2020}} = \frac{2022 \cdot 4042 \cdot 4041}{2021^3} = \frac{2022 \cdot 2 \cdot 4041}{2021^2}\] Evaluating the numerators and denominators mod $1000$ gives $804 + 441 = 1\boxed{245}$ | null | 245 |
5e8e6ab8df063ef422b177389473e3aa | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_12 | For any finite set $X$ , let $| X |$ denote the number of elements in $X$ . Define \[S_n = \sum | A \cap B | ,\] where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\left\{ 1 , 2 , 3, \cdots , n \right\}$ with $|A| = |B|$ .
For example, $S_2 = 4$ because the sum is taken over the pairs of subsets \[(A, B) \in \left\{ (\emptyset, \emptyset) , ( \{1\} , \{1\} ), ( \{1\} , \{2\} ) , ( \{2\} , \{1\} ) , ( \{2\} , \{2\} ) , ( \{1 , 2\} , \{1 , 2\} ) \right\} ,\] giving $S_2 = 0 + 1 + 0 + 0 + 1 + 2 = 4$ .
Let $\frac{S_{2022}}{S_{2021}} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p + q$ is divided by
1000. | For each element $i$ , denote $x_i = \left( x_{i, A}, x_{i, B} \right) \in \left\{ 0 , 1 \right\}^2$ , where $x_{i, A} = \Bbb I \left\{ i \in A \right\}$ (resp. $x_{i, B} = \Bbb I \left\{ i \in B \right\}$ ).
Denote $\Omega = \left\{ (x_1, \cdots , x_n): \sum_{i = 1}^n x_{i, A} = \sum_{i = 1}^n x_{i, B} \right\}$
Denote $\Omega_{-j} = \left\{ (x_1, \cdots , x_{j-1} , x_{j+1} , \cdots , x_n): \sum_{i \neq j} x_{i, A} = \sum_{i \neq j} x_{i, B} \right\}$
Hence, \begin{align*} S_n & = \sum_{(x_1, \cdots , x_n) \in \Omega} \sum_{i = 1}^n \Bbb I \left\{ x_{i, A} = x_{i, B} = 1 \right\} \\ & = \sum_{i = 1}^n \sum_{(x_1, \cdots , x_n) \in \Omega} \Bbb I \left\{ x_{i, A} = x_{i, B} = 1 \right\} \\ & = \sum_{i = 1}^n \sum_{(x_1, \cdots , x_{i-1} , x_{i+1} , \cdots , x_n) \in \Omega_{-i}} 1 \\ & = \sum_{i = 1}^n \sum_{j=0}^{n-1} \left( \binom{n-1}{j} \right)^2 \\ & = n \sum_{j=0}^{n-1} \left( \binom{n-1}{j} \right)^2 \\ & = n \sum_{j=0}^{n-1} \binom{n-1}{j} \binom{n-1}{n-1-j} \\ & = n \binom{2n-2}{n-1} . \end{align*}
Therefore, \begin{align*} \frac{S_{2022}}{S_{2021}} & = \frac{2022 \binom{4042}{2021}}{2021 \binom{4040}{2020}} \\ & = \frac{4044 \cdot 4041}{2021^2} . \end{align*}
This is in the lowest term.
Therefore, modulo 1000, \begin{align*} p + q & \equiv 4044 \cdot 4041 + 2021^2 \\ & \equiv 44 \cdot 41 + 21^2 \\ & \equiv \boxed{245} | null | 245 |
72175051ea49eab70ce72da98718e976 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_14 | Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the $\textit{splitting line}$ of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive integers. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC},$ respectively, and suppose that the splitting lines of $\triangle ABC$ through $M$ and $N$ intersect at $30^\circ.$ Find the perimeter of $\triangle ABC.$ | We now need to solve $a^2+ab+b^2 = 3^2\cdot 73^2$ . A quick $(\bmod 9)$ check gives that $3\mid a$ and $3\mid b$ . Thus, it's equivalent to solve $x^2+xy+y^2 = 73^2$
Let $\omega$ be one root of $\omega^2+\omega+1=0$ . Then, recall that $\mathbb Z[\omega]$ is the ring of integers of $\mathbb Q[\sqrt{-3}]$ and is a unique factorization domain. Notice that $N(x-y\omega) = (x-y\omega)(x-y\omega^2) = x^2+xy+y^2$ . Therefore, it suffices to find an element of $\mathbb Z[\omega]$ with the norm $73^2$
To do so, we factor $73$ in $\mathbb Z[\omega]$ . Since it's $1\pmod 3$ , it must split. A quick inspection gives $73 = (8-\omega)(8-\omega^2)$ . Thus, $N(8-\omega) = 73$ , so \begin{align*} 73^2 &= N((8-\omega)^2) \\ &= N(64 - 16\omega + \omega^2) \\ &= N(64 - 16\omega + (-1-\omega)) \\ &= N(63 - 17\omega), \end{align*} giving the solution $x=63$ and $y=17$ , yielding $a=189$ and $b=51$ , so the sum is $\boxed{459}$ . Since $8-\omega$ and $8-\omega^2$ are primes in $\mathbb Z[\omega]$ , the solution must divide $73^2$ . One can then easily check that this is the unique solution. | null | 459 |
72175051ea49eab70ce72da98718e976 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_14 | Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the $\textit{splitting line}$ of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive integers. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC},$ respectively, and suppose that the splitting lines of $\triangle ABC$ through $M$ and $N$ intersect at $30^\circ.$ Find the perimeter of $\triangle ABC.$ | Denote $BC = a$ $CA = b$ $AB = c$
Let the splitting line of $\triangle ABC$ through $M$ (resp. $N$ ) crosses $\triangle ABC$ at another point $X$ (resp. $Y$ ).
WLOG, we assume $c \leq b$
$\textbf{Case 1}$ $a \leq c \leq b$
We extend segment $AB$ to $D$ , such that $BD = a$ .
We extend segment $AC$ to $E$ , such that $CE = a$
In this case, $X$ is the midpoint of $AE$ , and $Y$ is the midpoint of $AD$
Because $M$ and $X$ are the midpoints of $AB$ and $AE$ , respectively, $MX \parallel BE$ .
Because $N$ and $Y$ are the midpoints of $AC$ and $AD$ , respectively, $NY \parallel CD$
Because $CB = CE$ $\angle CBE =\angle CEB = \frac{\angle ACB}{2}$ .
Because $BC = BD$ $\angle BCD = \angle BDC = \frac{\angle ABC}{2}$
Let $BE$ and $CD$ intersect at $O$ .
Because $MX \parallel BE$ and $NY \parallel CD$ , the angle formed between lines $MX$ and $NY$ is congruent to $\angle BOD$ . Hence, $\angle BOD = 30^\circ$ or $150^\circ$
We have \begin{align*} \angle BOD & = \angle CBE + \angle BCD \\ & = \frac{\angle ACB}{2} + \frac{\angle ABC}{2} \\ & = 90^\circ - \frac{\angle A}{2} . \end{align*}
Hence, we must have $\angle BOD = 30^\circ$ , not $150^\circ$ .
Hence, $\angle A = 120^\circ$
This implies $a > b$ and $a >c$ . This contradicts the condition specified for this case.
Therefore, this case is infeasible.
$\textbf{Case 2}$ $c \leq a \leq b$
We extend segment $CB$ to $D$ , such that $BD = c$ .
We extend segment $AC$ to $E$ , such that $CE = a$
In this case, $X$ is the midpoint of $AE$ , and $Y$ is the midpoint of $CD$
Because $M$ and $X$ are the midpoints of $AB$ and $AE$ , respectively, $MX \parallel BE$ .
Because $N$ and $Y$ are the midpoints of $AC$ and $CD$ , respectively, $NY \parallel AD$
Because $CB = CE$ $\angle CBE =\angle CEB = \frac{\angle ACB}{2}$ .
Because $BA = BD$ $\angle BAD = \angle BDA = \frac{\angle ABC}{2}$
Let $O$ be a point of $AC$ , such that $BO \parallel AD$ .
Hence, $\angle OBC = \angle BDA = \frac{B}{2}$
Because $MX \parallel BE$ and $NY \parallel AD$ and $AD \parallel BO$ , the angle formed between lines $MX$ and $NY$ is congruent to $\angle OBE$ . Hence, $\angle OBE = 30^\circ$ or $150^\circ$
We have \begin{align*} \angle OBE & = \angle OBC + \angle CBE \\ & = \frac{\angle ABC}{2} + \frac{\angle ACB}{2} \\ & = 90^\circ - \frac{\angle A}{2} . \end{align*}
Hence, we must have $\angle OBE = 30^\circ$ , not $150^\circ$ .
Hence, $\angle A = 120^\circ$
This implies $a > b$ and $a >c$ . This contradicts the condition specified for this case.
Therefore, this case is infeasible.
$\textbf{Case 3}$ $c \leq b \leq a$
We extend segment $CB$ to $D$ , such that $BD = c$ .
We extend segment $BC$ to $E$ , such that $CE = b$
In this case, $X$ is the midpoint of $BE$ , and $Y$ is the midpoint of $CD$
Because $M$ and $X$ are the midpoints of $AB$ and $BE$ , respectively, $MX \parallel AE$ .
Because $N$ and $Y$ are the midpoints of $AC$ and $CD$ , respectively, $NY \parallel AD$
Because $CA = CE$ $\angle CAE =\angle CEB = \frac{\angle ACB}{2}$ .
Because $BA = BD$ $\angle BAD = \angle BDA = \frac{\angle ABC}{2}$
Because $MX \parallel AE$ and $NY \parallel AD$ , the angle formed between lines $MX$ and $NY$ is congruent to $\angle DAE$ . Hence, $\angle DAE = 30^\circ$ or $150^\circ$
We have \begin{align*} \angle DAE & = \angle BAD + \angle CAE + \angle BAC \\ & = \frac{\angle ABC}{2} + \frac{\angle ACB}{2} + \angle BAC \\ & = 90^\circ + \frac{\angle BAC}{2} . \end{align*}
Hence, we must have $\angle OBE = 150^\circ$ , not $30^\circ$ .
Hence, $\angle BAC = 120^\circ$
In $\triangle ABC$ , by applying the law of cosines, we have \begin{align*} a^2 & = b^2 + c^2 - 2bc \cos \angle BAC\\ & = b^2 + c^2 - 2bc \cos 120^\circ \\ & = b^2 + c^2 + bc . \end{align*}
Because $a = 219$ , we have \[ b^2 + c^2 + bc = 219^2 . \]
Now, we find integer solution(s) of this equation with $c \leq b$
Multiplying this equation by 4, we get \[ \left( 2 c + b \right)^2 + 3 b^2 = 438^2 . \hspace{1cm} (1) \]
Denote $d = 2 c + b$ . Because $c \leq b$ $b < d \leq 3 b$
Because $438^2 - 3 b^2 \equiv 0 \pmod{3}$ $d^2 \equiv 0 \pmod{3}$ .
Thus, $d \equiv 0 \pmod{3}$ .
This implies $d^2 \equiv 0 \pmod{9}$
We also have $438^2 \equiv 0 \pmod{9}$ .
Hence, $3 b^2 \equiv 0 \pmod{9}$ .
This implies $b \equiv 0 \pmod{3}$
Denote $b = 3 p$ and $d = 3 q$ . Hence, $p < q \leq 3 p$ .
Hence, Equation (1) can be written as \[ q^2 + 3 p^2 = 146^2 . \hspace{1cm} (2) \]
Now, we solve this equation.
First, we find an upper bound of $q$
We have $q^2 + 3 p^2 \geq q^2 + 3 \left( \frac{q}{3} \right)^2 = \frac{4 q^2}{3}$ .
Hence, $\frac{4 q^2}{3} \leq 146^2$ .
Hence, $q \leq 73 \sqrt{3} < 73 \cdot 1.8 = 131.4$ .
Because $q$ is an integer, we must have $q \leq 131$
Second, we find a lower bound of $q$
We have $q^2 + 3 p^2 < q^2 + 3 q^2 = 4 q^2$ .
Hence, $4 q^2 > 146^2$ .
Hence, $q > 73$ .
Because $q$ is an integer, we must have $q \geq 74$
Now, we find the integer solutions of $p$ and $q$ that satisfy Equation (2) with $74 \leq q \leq 131$
First, modulo 9, \begin{align*} q^2 & \equiv 146^2 - 3 p^2 \\ & \equiv 4 - 3 \cdot ( 0 \mbox{ or } 1 ) \\ & \equiv 4 \mbox{ or } 1 . \end{align*}
Hence $q \equiv \pm 1, \pm 2 \pmod{9}$
Second, modulo 5, \begin{align*} q^2 & \equiv 146^2 - 3 p^2 \\ & \equiv 1 + 2 p^2 \\ & \equiv 1 + 2 \cdot ( 0 \mbox{ or } 1 \mbox{ or } -1 ) \\ & \equiv 1 \mbox{ or } 3 \mbox{ or } - 1 . \end{align*}
Because $q^2 \equiv 0 \mbox{ or } 1 \mbox{ or } - 1$ , we must have $q^2 \equiv 1 \mbox{ or } - 1$ .
Hence, $5 \nmid q$
Third, modulo 7, \begin{align*} q^2 & \equiv 146^2 - 3 p^2 \\ & \equiv 1 - 3 \cdot ( 0 \mbox{ or } 1 \mbox{ or } 5 \mbox{ or } 2 ) \\ & \equiv 1 \mbox{ or } 2 \mbox{ or } 3 \mbox{ or } 5 . \end{align*}
Because $q^2 \equiv 0 \mbox{ or } 1 \mbox{ or } 2 \mbox{ or } 4 \pmod{ 7 }$ , we must have $q^2 \equiv 1 \mbox{ or } 2 \pmod{7}$ .
Hence, $q \equiv 1, 3, 4, 6 \pmod{7}$
Given all conditions above, the possible $q$ are 74, 83, 88, 92, 97, 101, 106, 109, 116, 118, 127.
By testing all these numbers, we find that the only solution is $q = 97$ .
This implies $p = 63$
Hence, $b = 3p = 189$ and $d = 3q = 291$ .
Hence, $c = \frac{d - b}{2} = 51$
Therefore, the perimeter of $\triangle ABC$ is $b + c + a = 189 + 51 + 219 = \boxed{459}$ | null | 459 |
72175051ea49eab70ce72da98718e976 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_14 | Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the $\textit{splitting line}$ of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive integers. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC},$ respectively, and suppose that the splitting lines of $\triangle ABC$ through $M$ and $N$ intersect at $30^\circ.$ Find the perimeter of $\triangle ABC.$ | We wish to solve the Diophantine equation $a^2+ab+b^2=3^2 \cdot 73^2$ . It can be shown that $3|a$ and $3|b$ , so we make the substitution $a=3x$ and $b=3y$ to obtain $x^2+xy+y^2=73^2$ as our new equation to solve for.
Notice that $r^2+r+1=(r-\omega)(r-{\omega}^2)$ , where $\omega=e^{i\frac{2\pi}{3}}$ . Thus, \[x^2+xy+y^2 = y^2((x/y)^2+(x/y)+1) = y^2 (\frac{x}{y}-\omega)(\frac{x}{y}-{\omega}^2) = (x-y\omega)(x-y{\omega}^2).\]
Note that $8^2+1^2+8 \cdot 1=73$ . Thus, $(8-\omega)(8-{\omega}^2)=73$ . Squaring both sides yields \begin{align} (8-\omega)^2(8-{\omega}^2)^2&=73^2\\ (63-17\omega)(63-17{\omega}^2)&=73^2. \end{align} Thus, by $(2)$ $(63, 17)$ is a solution to $x^2+xy+y^2=73^2$ . This implies that $a=189$ and $b=51$ , so our final answer is $189+51+219=\boxed{459}$ | null | 459 |
2308c0c14a9aed11cd98a1ee1fe22ad2 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_15 | Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | First, let define a triangle with side lengths $\sqrt{2x}$ $\sqrt{2z}$ , and $l$ , with altitude from $l$ 's equal to $\sqrt{xz}$ $l = \sqrt{2x - xz} + \sqrt{2z - xz}$ , the left side of one equation in the problem.
Let $\theta$ be angle opposite the side with length $\sqrt{2x}$ . Then the altitude has length $\sqrt{2z} \cdot \sin(\theta) = \sqrt{xz}$ and thus $\sin(\theta) = \sqrt{\frac{x}{2}}$ , so $x=2\sin^2(\theta)$ and the side length $\sqrt{2x}$ is equal to $2\sin(\theta)$
We can symmetrically apply this to the two other equations/triangles.
By law of sines, we have $\frac{2\sin(\theta)}{\sin(\theta)} = 2R$ , with $R=1$ as the circumradius, same for all 3 triangles.
The circumcircle's central angle to a side is $2 \arcsin(l/2)$ , so the 3 triangles' $l=1, \sqrt{2}, \sqrt{3}$ , have angles $120^{\circ}, 90^{\circ}, 60^{\circ}$ , respectively.
This means that by half angle arcs, we see that we have in some order, $x=2\sin^2(\alpha)$ $y=2\sin^2(\beta)$ , and $z=2\sin^2(\gamma)$ (not necessarily this order, but here it does not matter due to symmetry), satisfying that $\alpha+\beta=180^{\circ}-\frac{120^{\circ}}{2}$ $\beta+\gamma=180^{\circ}-\frac{90^{\circ}}{2}$ , and $\gamma+\alpha=180^{\circ}-\frac{60^{\circ}}{2}$ . Solving, we get $\alpha=\frac{135^{\circ}}{2}$ $\beta=\frac{105^{\circ}}{2}$ , and $\gamma=\frac{165^{\circ}}{2}$
We notice that \[[(1-x)(1-y)(1-z)]^2=[\sin(2\alpha)\sin(2\beta)\sin(2\gamma)]^2=[\sin(135^{\circ})\sin(105^{\circ})\sin(165^{\circ})]^2\] \[=\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{6}-\sqrt{2}}{4} \cdot \frac{\sqrt{6}+\sqrt{2}}{4}\right)^2 = \left(\frac{\sqrt{2}}{8}\right)^2=\frac{1}{32} \to \boxed{033}. \blacksquare\] | null | 033 |
2308c0c14a9aed11cd98a1ee1fe22ad2 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_15 | Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | (This eventually whittles down to the same concept as Solution 1)
Note that in each equation in this system, it is possible to factor $\sqrt{x}$ $\sqrt{y}$ , or $\sqrt{z}$ from each term (on the left sides), since each of $x$ $y$ , and $z$ are positive real numbers. After factoring out accordingly from each terms one of $\sqrt{x}$ $\sqrt{y}$ , or $\sqrt{z}$ , the system should look like this: \begin{align*} \sqrt{x}\cdot\sqrt{2-y} + \sqrt{y}\cdot\sqrt{2-x} &= 1 \\ \sqrt{y}\cdot\sqrt{2-z} + \sqrt{z}\cdot\sqrt{2-y} &= \sqrt2 \\ \sqrt{z}\cdot\sqrt{2-x} + \sqrt{x}\cdot\sqrt{2-z} &= \sqrt3. \end{align*}
This should give off tons of trigonometry vibes. To make the connection clear, $x = 2\cos^2 \alpha$ $y = 2\cos^2 \beta$ , and $z = 2\cos^2 \theta$ is a helpful substitution:
\begin{align*} \sqrt{2\cos^2 \alpha}\cdot\sqrt{2-2\cos^2 \beta} + \sqrt{2\cos^2 \beta}\cdot\sqrt{2-2\cos^2 \alpha} &= 1 \\ \sqrt{2\cos^2 \beta}\cdot\sqrt{2-2\cos^2 \theta} + \sqrt{2\cos^2 \theta}\cdot\sqrt{2-2\cos^2 \beta} &= \sqrt2 \\ \sqrt{2\cos^2 \theta}\cdot\sqrt{2-2\cos^2 \alpha} + \sqrt{2\cos^2 \alpha}\cdot\sqrt{2-2\cos^2 \theta} &= \sqrt3. \end{align*}
From each equation $\sqrt{2}^2$ can be factored out, and when every equation is divided by 2, we get:
\begin{align*} \sqrt{\cos^2 \alpha}\cdot\sqrt{1-\cos^2 \beta} + \sqrt{\cos^2 \beta}\cdot\sqrt{1-\cos^2 \alpha} &= \frac{1}{2} \\ \sqrt{\cos^2 \beta}\cdot\sqrt{1-\cos^2 \theta} + \sqrt{\cos^2 \theta}\cdot\sqrt{1-\cos^2 \beta} &= \frac{\sqrt2}{2} \\ \sqrt{\cos^2 \theta}\cdot\sqrt{1-\cos^2 \alpha} + \sqrt{\cos^2 \alpha}\cdot\sqrt{1-\cos^2 \theta} &= \frac{\sqrt3}{2}. \end{align*}
which simplifies to (using the Pythagorean identity $\sin^2 \phi + \cos^2 \phi = 1 \; \forall \; \phi \in \mathbb{C}$ ):
\begin{align*} \cos \alpha\cdot\sin \beta + \cos \beta\cdot\sin \alpha &= \frac{1}{2} \\ \cos \beta\cdot\sin \theta + \cos \theta\cdot\sin \beta &= \frac{\sqrt2}{2} \\ \cos \theta\cdot\sin \alpha + \cos \alpha\cdot\sin \theta &= \frac{\sqrt3}{2}. \end{align*}
which further simplifies to (using sine addition formula $\sin(a + b) = \sin a \cos b + \cos a \sin b$ ):
\begin{align*} \sin(\alpha + \beta) &= \frac{1}{2} \\ \sin(\beta + \theta) &= \frac{\sqrt2}{2} \\ \sin(\alpha + \theta) &= \frac{\sqrt3}{2}. \end{align*}
Taking the inverse sine ( $0\leq\theta\frac{\pi}{2}$ ) of each equation yields a simple system:
\begin{align*} \alpha + \beta &= \frac{\pi}{6} \\ \beta + \theta &= \frac{\pi}{4} \\ \alpha + \theta &= \frac{\pi}{3} \end{align*}
giving solutions:
\begin{align*} \alpha &= \frac{\pi}{8} \\ \beta &= \frac{\pi}{24} \\ \theta &= \frac{5\pi}{24} \end{align*}
Since these unknowns are directly related to our original unknowns, there are consequent solutions for those:
\begin{align*} x &= 2\cos^2\left(\frac{\pi}{8}\right) \\ y &= 2\cos^2\left(\frac{\pi}{24}\right) \\ z &= 2\cos^2\left(\frac{5\pi}{24}\right) \end{align*}
When plugging into the expression $\left[ (1-x)(1-y)(1-z) \right]^2$ , noting that $-\cos 2\phi = 1 - 2\cos^2 \phi\; \forall \; \phi \in \mathbb{C}$ helps to simplify this expression into:
\begin{align*} \left[ (-1)^3\left(\cos \left(2\cdot\frac{\pi}{8}\right)\cos \left(2\cdot\frac{\pi}{24}\right)\cos \left(2\cdot\frac{5\pi}{24}\right)\right)\right]^2 \\ = \left[ (-1)\left(\cos \left(\frac{\pi}{4}\right)\cos \left(\frac{\pi}{12}\right)\cos \left(\frac{5\pi}{12}\right)\right)\right]^2 \end{align*}
Now, all the cosines in here are fairly standard:
\begin{align*} \cos \frac{\pi}{4} &= \frac{\sqrt{2}}{2} \\ \cos \frac{\pi}{12} &=\frac{\sqrt{6} + \sqrt{2}}{4} & (= \cos{\frac{\frac{\pi}{6}}{2}} ) \\ \cos \frac{5\pi}{12} &= \frac{\sqrt{6} - \sqrt{2}}{4} & (= \cos\left({\frac{\pi}{6} + \frac{\pi}{4}} \right) ) \end{align*}
With some final calculations:
\begin{align*} &(-1)^2\left(\frac{\sqrt{2}}{2}\right)^2\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 \\ =& \left(\frac{1}{2}\right) \left(\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)\right)^2 \\ =&\frac{1}{2} \frac{4^2}{16^2} = \frac{1}{32} \end{align*}
This is our answer in simplest form $\frac{m}{n}$ , so $m + n = 1 + 32 = \boxed{033}$ | null | 033 |
2308c0c14a9aed11cd98a1ee1fe22ad2 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_15 | Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Let $1-x=a;1-y=b;1-z=c$ , rewrite those equations
$\sqrt{(1-a)(1+b)}+\sqrt{(1+a)(1-b)}=1$
$\sqrt{(1-b)(1+c)}+\sqrt{(1+b)(1-c)}=\sqrt{2}$
$\sqrt{(1-a)(1+c)}+\sqrt{(1-c)(1+a)}=\sqrt{3}$
and solve for $m/n = (abc)^2 = a^2b^2c^2$
Square both sides and simplify, to get three equations:
$2ab-1=2\sqrt{(1-a^2)(1-b^2)}$
$2bc~ ~ ~ ~ ~ ~=2\sqrt{(1-b^2)(1-c^2)}$
$2ac+1=2\sqrt{(1-c^2)(1-a^2)}$
Square both sides again, and simplify to get three equations:
$a^2+b^2-ab=\frac{3}{4}$
$b^2+c^2~ ~ ~ ~ ~ ~=1$
$a^2+c^2+ac=\frac{3}{4}$
Subtract first and third equation, getting $(b+c)(b-c)=a(b+c)$ $a=b-c$
Put it in first equation, getting $b^2-2bc+c^2+b^2-b(b-c)=b^2+c^2-bc=\frac{3}{4}$ $bc=\frac{1}{4}$
Since $a^2=b^2+c^2-2bc=\frac{1}{2}$ $m/n = a^2b^2c^2 = a^2(bc)^2 = \frac{1}{2}\left(\frac{1}{4}\right)^2=\frac{1}{32}$ and so the final answer is $\boxed{033}$ | null | 033 |
2308c0c14a9aed11cd98a1ee1fe22ad2 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_15 | Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Denote $u = 1 - x$ $v = 1 - y$ $w = 1 - z$ .
Hence, the system of equations given in the problem can be written as \begin{align*} \sqrt{(1-u)(1+v)} + \sqrt{(1+u)(1-v)} & = 1 \hspace{1cm} (1) \\ \sqrt{(1-v)(1+w)} + \sqrt{(1+v)(1-w)} & = \sqrt{2} \hspace{1cm} (2) \\ \sqrt{(1-w)(1+u)} + \sqrt{(1+w)(1-u)} & = \sqrt{3} . \hspace{1cm} (3) \end{align*}
Each equation above takes the following form: \[ \sqrt{(1-a)(1+b)} + \sqrt{(1+a)(1-b)} = k . \]
Now, we simplify this equation by removing radicals.
Denote $p = \sqrt{(1-a)(1+b)}$ and $q = \sqrt{(1+a)(1-b)}$
Hence, the equation above implies \[ \left\{ \begin{array}{l} p + q = k \\ p^2 = (1-a)(1+b) \\ q^2 = (1+a)(1-b) \end{array} \right.. \]
Hence, $q^2 - p^2 = (1+a)(1-b) - (1-a)(1+b) = 2 (a-b)$ .
Hence, $q - p = \frac{q^2 - p^2}{p+q} = \frac{2}{k} (a-b)$
Because $p + q = k$ and $q - p = \frac{2}{k} (a-b)$ , we get $q = \frac{a-b}{k} + \frac{k}{2}$ .
Plugging this into the equation $q^2 = (1+a)(1-b)$ and simplifying it, we get \[ a^2 + \left( k^2 - 2 \right) ab + b^2 = k^2 - \frac{k^4}{4} . \]
Therefore, the system of equations above can be simplified as \begin{align*} u^2 - uv + v^2 & = \frac{3}{4} \\ v^2 + w^2 & = 1 \\ w^2 + wu + u^2 & = \frac{3}{4} . \end{align*}
Denote $w' = - w$ .
The system of equations above can be equivalently written as \begin{align*} u^2 - uv + v^2 & = \frac{3}{4} \hspace{1cm} (1') \\ v^2 + w'^2 & = 1 \hspace{1cm} (2') \\ w'^2 - w'u + u^2 & = \frac{3}{4} \hspace{1cm} (3') . \end{align*}
Taking $(1') - (3')$ , we get \[ (v - w') (v + w' - u) = 0 . \]
Thus, we have either $v - w' = 0$ or $v + w' - u = 0$
$\textbf{Case 1}$ $v - w' = 0$
Equation (2') implies $v = w' = \pm \frac{1}{\sqrt{2}}$
Plugging $v$ and $w'$ into Equation (2), we get contradiction. Therefore, this case is infeasible.
$\textbf{Case 2}$ $v + w' - u = 0$
Plugging this condition into (1') to substitute $u$ , we get \[ v^2 + v w' + w'^2 = \frac{3}{4} \hspace{1cm} (4) . \]
Taking $(4) - (2')$ , we get \[ v w' = - \frac{1}{4} . \hspace{1cm} (5) . \]
Taking (4) + (5), we get \[ \left( v + w' \right)^2 = \frac{1}{2} . \]
Hence, $u^2 = \left( v + w' \right)^2 = \frac{1}{2}$
Therefore, \begin{align*} \left[ (1-x)(1-y)(1-z) \right]^2 & = u^2 (vw)^2 \\ & = u^2 (vw')^2 \\ & = \frac{1}{2} \left( - \frac{1}{4} \right)^2 \\ & = \frac{1}{32} . \end{align*}
Therefore, the answer is $1 + 32 = \boxed{033}$ | null | 033 |
2308c0c14a9aed11cd98a1ee1fe22ad2 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_15 | Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | In given equations, $0 \leq x,y,z \leq 2,$ so we define some points: \[\bar {O} = (0, 0), \bar {A} = (1, 0), \bar{M} = \left(\frac {1}{\sqrt{2}},\frac {1}{\sqrt{2}}\right),\] \[\bar {X} = \left(\sqrt {\frac {x}{2}}, \sqrt{1 – \frac{x}{2}}\right), \bar {Y'} = \left(\sqrt {\frac {y}{2}}, \sqrt{1 – \frac{y}{2}}\right),\] \[\bar {Y} = \left(\sqrt {1 – \frac{y}{2}},\sqrt{\frac {y}{2}}\right), \bar {Z} = \left(\sqrt {1 – \frac{z}{2}},\sqrt{\frac {z}{2}}\right).\] Notice, that \[\mid \vec {AO} \mid = \mid \vec {MO} \mid = \mid \vec {XO} \mid =\mid \vec {YO} \mid = \mid \vec {Y'O} \mid =\mid \vec {ZO} \mid = 1\] and each points lies in the first quadrant.
We use given equations and get some scalar products: \[(\vec {XO} \cdot \vec {YO}) = \frac {1}{2} = \cos \angle XOY \implies \angle XOY = 60 ^\circ,\] \[(\vec {XO} \cdot \vec {ZO}) = \frac {\sqrt{3}}{2} = \cos \angle XOZ \implies \angle XOZ = 30^\circ,\] \[(\vec {Y'O} \cdot \vec {ZO}) = \frac {1}{\sqrt{2}} = \cos \angle Y'OZ \implies \angle Y'OZ = 45^\circ.\] So $\angle YOZ = \angle XOY – \angle XOZ = 60 ^\circ – 30 ^\circ = 30 ^\circ, \angle Y'OY = \angle Y'OZ + \angle YOZ = 45^\circ + 30 ^\circ = 75^\circ.$
Points $Y$ and $Y'$ are symmetric with respect to $OM.$
Case 1 \[\angle YOA = \frac{90^\circ – 75^\circ}{2} = 7.5^\circ, \angle ZOA = 30^\circ + 7.5^\circ = 37.5^\circ, \angle XOA = 60^\circ + 7.5^\circ = 67.5^\circ .\] \[1 – x = \left(\sqrt{1 – \frac{x}{2}} \right)^2– \left(\sqrt{\frac {x}{2}}\right)^2 = \sin^2 \angle XOA – \cos^2 \angle XOA = –\cos 2 \angle XOA = –\cos 135^\circ,\] \[1 – y = \cos 15^\circ, 1 – z = \cos 75^\circ \implies \left[ (1–x)(1–y)(1–z) \right]^2 = \left[ \sin 45^\circ \cdot \cos 15^\circ \cdot \sin 15^\circ \right]^2 =\] \[=\left[ \frac {\sin 45^\circ \cdot \sin 30^\circ}{2} \right]^2 = \frac {1}{32} \implies \boxed{033}.\] Case 2 | null | 033 |
1a8903115fe86637b463038816695b7b | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_1 | Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. | Let $x$ be the number of people at the party before the bus arrives. We know that $x\equiv 0\pmod {12}$ , as $\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x + 50 \equiv 0 \pmod{25}$ , as $\frac{11}{25}$ of the people at the party are adults after the bus arrives. $x + 50 \equiv 0 \pmod{25}$ can be reduced to $x \equiv 0 \pmod{25}$ , and since we are looking for the minimum amount of people, $x$ is $300$ . That means there are $350$ people at the party after the bus arrives, and thus there are $350 \cdot \frac{11}{25} = \boxed{154}$ adults at the party. | null | 154 |
1a8903115fe86637b463038816695b7b | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_1 | Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. | Since at the beginning, adults make up $\frac{5}{12}$ of the concert, the amount of people must be a multiple of 12.
Call the amount of people in the beginning $x$ .Then $x$ must be divisible by 12, in other words: $x$ must be a multiple of 12.
Since after 50 more people arrived, adults make up $\frac{11}{25}$ of the concert, $x+50$ is a multiple of 25. This means $x+50$ must be a multiple of 5.
Notice that if a number is divisible by 5, it must end with a 0 or 5. Since 5 is impossible (obviously, since multiples of 12 end in 2, 4, 6, 8, 0,...), $x$ must end in 0.
Notice that the multiples of 12 that end in 0 are: 60, 120, 180, etc.. By trying out, you can clearly see that $x=300$ is the minimum number of people at the concert.
So therefore, after 50 more people arrive, there are $300+50=350$ people at the concert, and the number of adults is $350*\frac{11}{25}=154$ . Therefore the answer is $\boxed{154}$ | null | 154 |
1a8903115fe86637b463038816695b7b | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_1 | Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. | Let $a$ be the number of adults before the bus arrived and $x$ be the total number of people at the concert. So, $\frac{a}{x}=\frac{5}{12}$ . Solving for $x$ in terms of $a$ $x = \frac{12}{5}a$ . After the bus arrives, let's say there are an additional $y$ adults out of the 50 more people who enter the concert. From that, we get $\frac{a+y}{x+50}=\frac{11}{25}$ . Replacing $x$ with the value of $a$ , the second equation becomes $\frac{a+y}{\frac{12}{5}a+50}=\frac{11}{25}$
By cross-multiplying and simplifying, we get that $25(y-22)=\frac{7a}{5}$
Observe that we must make sure $y-22$ is positive and divisible by $7$ to have an integer value of $a$ . The smallest possible value of $y$ that satisfies this conditions is $29$ . Plugging this into the equation, $a = 125$ . The question asks for the minimum number of adults that are there after the bus arrives, which is $a+y$ . Thus, the answer is simply $125+29=\boxed{154}$ | null | 154 |
8a0bfccb7f6e8fc0759e5a34a40c5bfa | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_2 | Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability $\frac23$ . When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability $\frac34$ . Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Let $A$ be Azar, $C$ be Carl, $J$ be Jon, and $S$ be Sergey. The $4$ circles represent the $4$ players, and the arrow is from the winner to the loser with the winning probability as the label.
2022AIMEIIP2.png
This problem can be solved by using $2$ cases.
$\textbf{Case 1:}$ $C$ 's opponent for the semifinal is $A$
The probability $C$ 's opponent is $A$ is $\frac13$ . Therefore the probability $C$ wins the semifinal in this case is $\frac13 \cdot \frac13$ . The other semifinal game is played between $J$ and $S$ , it doesn't matter who wins because $C$ has the same probability of winning either one. The probability of $C$ winning in the final is $\frac34$ , so the probability of $C$ winning the tournament in case 1 is $\frac13 \cdot \frac13 \cdot \frac34$
$\textbf{Case 2:}$ $C$ 's opponent for the semifinal is $J$ or $S$
It doesn't matter if $C$ 's opponent is $J$ or $S$ because $C$ has the same probability of winning either one. The probability $C$ 's opponent is $J$ or $S$ is $\frac23$ . Therefore the probability $C$ wins the semifinal in this case is $\frac23 \cdot \frac34$ . The other semifinal game is played between $A$ and $J$ or $S$ . In this case it matters who wins in the other semifinal game because the probability of $C$ winning $A$ and $J$ or $S$ is different.
$\textbf{Case 2.1:}$ $C$ 's opponent for the final is $A$
For this to happen, $A$ must have won $J$ or $S$ in the semifinal, the probability is $\frac34$ . Therefore, the probability that $C$ won $A$ in the final is $\frac34 \cdot \frac13$
$\textbf{Case 2.2:}$ $C$ 's opponent for the final is $J$ or $S$
For this to happen, $J$ or $S$ must have won $A$ in the semifinal, the probability is $\frac14$ . Therefore, the probability that $C$ won $J$ or $S$ in the final is $\frac14 \cdot \frac34$
In Case 2 the probability of $C$ winning the tournament is $\frac23 \cdot \frac34 \cdot (\frac34 \cdot \frac13 + \frac14 \cdot \frac34)$
Adding case 1 and case 2 together we get $\frac13 \cdot \frac13 \cdot \frac34 + \frac23 \cdot \frac34 \cdot (\frac34 \cdot \frac13 + \frac14 \cdot \frac34) = \frac{29}{96},$ so the answer is $29 + 96 = \boxed{125}$ | null | 125 |
df23ee5c7f1cf21a0a412b91db11da3e | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_3 | A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius $\frac{6}{\sqrt{2}} = 3\sqrt{2}$ ). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: $l$ . Because of the symmetrical property of the pyramid,
we can imagine that the line of the apex and the (sphere's) center will intersect the square at the (base's) center.
Since the volume is $54 = \frac{1}{3} \cdot S \cdot h = \frac{1}{3} \cdot 6^2 \cdot h$ , where $h=\frac{9}{2}$ is the height of this pyramid, we have: $l^2=\left(\frac{9}{2}-l\right)^2+\left(3\sqrt{2}\right)^2$ according to the Pythagorean theorem.
Solve this equation will give us $l = \frac{17}{4},$ therefore $m+n=\boxed{021}.$ | null | 021 |
df23ee5c7f1cf21a0a412b91db11da3e | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_3 | A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | To start, we find the height of the pyramid. By the volume of a pyramid formula, we have \[\frac13 \cdot 6^2 \cdot h=54 \implies h=\frac92.\] Next, let us find the length of the non-base sides of the pyramid. By the Pythagorean Theorem, noting that the distance from one vertex of the base to the center of the base is $\frac12 \cdot 6\sqrt2=3\sqrt2$ , we have \[x=\sqrt{\left(\frac92\right)^2+(3\sqrt2)^2}=\sqrt{\frac{153}4}=\frac{3\sqrt{17}}2.\] Taking the cross section of the pyramid and transforming the problem into $2$ -d, it suffices to find the radius of the circumcircle of a triangle of side lengths $\frac{3\sqrt{17}}2$ $\frac{3\sqrt{17}}2$ $6\sqrt2$ . This turns out to be easy by the formula $R=\frac{abc}{4A}$ , and through computing this value (the work has been left out) we find that $R=\frac{17}4$ , so our answer is $\boxed{021}$ | null | 021 |
df23ee5c7f1cf21a0a412b91db11da3e | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_3 | A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | By the volume of a pyramid formula, we have that the height of the pyramid is $\frac{9}{2}$ . Since the base is a square with side length 6, the simplest way to place it in the coordinate plane is to put the center of the square at the origin and let the base be on the $xy$ plane. Then, the vertices of the base would be $(3,3,0), (3,-3,0), (-3,3,0), (-3,-3,0)$ in some order. Also, let the vertex be $(0,0,\frac{9}{2})$ . Recall that the formula for a sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ where the center is $(a,b,c)$ and the radius is $r$ . Symmetry gives that $a=b=0$ . Plug in $(3,3,0)$ and $(0,0,\frac{9}{2})$ and you get the system of equation
$18+c^2=r^2$
$(\frac{9}{2}-c)^2=r^2$
Solving gives $c=1/4$ and $r=17/4$ , so our answer is $17+4=\boxed{021}$ .~ Ddk001 | null | 021 |
f6928c8c6da0dedaf27031e04e9a23ba | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Define $a$ to be $\log_{20x} (22x) = \log_{2x} (202x)$ , what we are looking for. Then, by the definition of the logarithm, \[\begin{cases} (20x)^{a} &= 22x \\ (2x)^{a} &= 202x. \end{cases}\] Dividing the first equation by the second equation gives us $10^a = \frac{11}{101}$ , so by the definition of logs, $a = \log_{10} \frac{11}{101}$ . This is what the problem asked for, so the fraction $\frac{11}{101}$ gives us $m+n = \boxed{112}$ | null | 112 |
f6928c8c6da0dedaf27031e04e9a23ba | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We could assume a variable $v$ which equals to both $\log_{20x} (22x)$ and $\log_{2x} (202x)$
So that $(20x)^v=22x \textcircled{1}$ and $(2x)^v=202x \textcircled{2}$
Express $\textcircled{1}$ as: $(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot \left(10^v\right)=22x \textcircled{3}$
Substitute $\textcircled{{2}}$ to $\textcircled{3}$ $202x \cdot (10^v)=22x$
Thus, $v=\log_{10} \left(\frac{22x}{202x}\right)= \log_{10} \left(\frac{11}{101}\right)$ , where $m=11$ and $n=101$
Therefore, $m+n = \boxed{112}$ | null | 112 |
f6928c8c6da0dedaf27031e04e9a23ba | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We have \begin{align*} \log_{20x} (22x) & = \frac{\log_k 22x}{\log_k 20x} \\ & = \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} . \end{align*}
We have \begin{align*} \log_{2x} (202x) & = \frac{\log_k 202x}{\log_k 2x} \\ & = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . \end{align*}
Because $\log_{20x} (22x)=\log_{2x} (202x)$ , we get \[ \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . \]
We denote this common value as $\lambda$
By solving the equality $\frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} = \lambda$ , we get $\log_k x = \frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1}$
By solving the equality $\frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} = \lambda$ , we get $\log_k x = \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1}$
By equating these two equations, we get \[ \frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1} = \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1} . \]
Therefore, \begin{align*} \log_{20x} (22x) & = \lambda \\ & = \frac{\log_k 22 - \log_k 202}{\log_k 20 - \log_k 2} \\ & = \frac{\log_k \frac{11}{101}}{\log_k 10} \\ & = \log_{10} \frac{11}{101} . \end{align*}
Therefore, the answer is $11 + 101 = \boxed{112}$ | null | 112 |
f6928c8c6da0dedaf27031e04e9a23ba | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $a$ be the exponent such that $(20x)^a = 22x$ and $(2x)^a = 202x$ . Dividing, we get \begin{align*} \dfrac{(20x)^a}{(2x)^a} &= \dfrac{22x}{202x}. \\ \left(\dfrac{20x}{2x}\right)^a &= \dfrac{22x}{202x}. \\ 10^a &= \dfrac{11}{101}. \\ \end{align*} Thus, we see that $\log_{10} \left(\dfrac{11}{101}\right) = a = \log_{20x} 22x$ , so the answer is $11 + 101 = \boxed{112}$ | null | 112 |
f6928c8c6da0dedaf27031e04e9a23ba | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | By the change of base rule, we have $\frac{\log 22x}{\log 20x}=\frac{\log 202x}{\log 2x}$ , or $\frac{\log 22 +\log x}{\log 20 +\log x}=\frac{\log 202 +\log x}{\log 2 +\log x}=k$ . We also know that if $a/b=c/d$ , then this also equals $\frac{a-c}{b-d}$ . We use this identity and find that $k=\frac{\log 202 -\log 22}{\log 2 -\log 20}=-\log\frac{202}{22}=\log\frac{11}{101}$ . The requested sum is $11+101=\boxed{112}.$ | null | 112 |
f6928c8c6da0dedaf27031e04e9a23ba | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | By change of base formula, \[\frac{\log_{2x} 22x}{\log_{2x} 20x} = \frac{{\log_{2x} 11} + 1}{{\log_{2x} 10} + 1} = {\log_{2x} 101} + 1\] \[\log_{2x} 11 + 1 = (\log_{2x} 10)(\log_{2x} 101) + \log{2x} 1010 + 1\] \[\frac{\log_{2x} \frac{11}{1010}}{\log_{2x} 10} = \log_{2x} 101\] \[\log_{10} {\frac{11}{1010}} = \log_{2x} 101\] \[\log_{10} {\frac{11}{1010}} + 1 = \log_{2x} 101 + 1 = \log_{2x} 202x = \log_{20x} {22x}\] Thus, \[\log_{20x} 22x = \log_{10} \left( \frac{11}{1010} \times 10 \right) = \log_{10} \frac{11}{101}\] The requested answer is $11 + 101 = \boxed{112}$ | null | 112 |
16584ab68e29a347e520c153998152da | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_5 | Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points. | Let $a$ $b$ , and $c$ be the vertex of a triangle that satisfies this problem, where $a > b > c$ \[a - b = p_1\] \[b - c = p_2\] \[a - c = p_3\]
$p_3 = a - c = a - b + b - c = p_1 + p_2$ . Because $p_3$ is the sum of two primes, $p_1$ and $p_2$ $p_1$ or $p_2$ must be $2$ . Let $p_1 = 2$ , then $p_3 = p_2 + 2$ . There are only $8$ primes less than $20$ $2, 3, 5, 7, 11, 13, 17, 19$ . Only $3, 5, 11, 17$ plus $2$ equals another prime. $p_2 \in \{ 3, 5, 11, 17 \}$
Once $a$ is determined, $a = b+2$ and $b = c + p_2$ . There are $18$ values of $a$ where $a+2 \le 20$ , and $4$ values of $p_2$ . Therefore the answer is $18 \cdot 4 = \boxed{072}$ | null | 072 |
16584ab68e29a347e520c153998152da | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_5 | Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points. | As above, we must deduce that the sum of two primes must be equal to the third prime. Then, we can finish the solution using casework.
If the primes are $2,3,5$ , then the smallest number can range between $1$ and $15$ .
If the primes are $2,5,7$ , then the smallest number can range between $1$ and $13$ .
If the primes are $2,11,13$ , then the smallest number can range between $1$ and $7$ .
If the primes are $2,17,19$ , then the smallest number can only be $1$
Adding all cases gets $15+13+7+1=36$ . However, due to the commutative property, we must multiply this by 2. For example, in the $2,17,19$ case the numbers can be $1,3,20$ or $1,18,20$ . Therefore the answer is $36\cdot2=\boxed{072}$ | null | 072 |
b6f804b3c99881c188eca0da2b1ce300 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_6 | Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | To find the greatest value of $x_{76} - x_{16}$ $x_{76}$ must be as large as possible, and $x_{16}$ must be as small as possible. If $x_{76}$ is as large as possible, $x_{76} = x_{77} = x_{78} = \dots = x_{100} > 0$ . If $x_{16}$ is as small as possible, $x_{16} = x_{15} = x_{14} = \dots = x_{1} < 0$ . The other numbers between $x_{16}$ and $x_{76}$ equal to $0$ . Let $a = x_{76}$ $b = x_{16}$ . Substituting $a$ and $b$ into $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ we get: \[25a - 16b = 1\] \[25a + 16b = 0\] $a = \frac{1}{50}$ $b = -\frac{1}{32}$
$x_{76} - x_{16} = a - b = \frac{1}{50} + \frac{1}{32} = \frac{41}{800}$ $m+n = \boxed{841}$ | null | 841 |
b6f804b3c99881c188eca0da2b1ce300 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_6 | Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Define $s_N$ to be the sum of all the negatives, and $s_P$ to be the sum of all the positives.
Since the sum of the absolute values of all the numbers is $1$ $|s_N|+|s_P|=1$
Since the sum of all the numbers is $0$ $s_N=-s_P\implies |s_N|=|s_P|$
Therefore, $|s_N|=|s_P|=\frac 12$ , so $s_N=-\frac 12$ and $s_P=\frac 12$ since $s_N$ is negative and $s_P$ is positive.
To maximize $x_{76}-x_{16}$ , we need to make $x_{16}$ as small of a negative as possible, and $x_{76}$ as large of a positive as possible.
Note that $x_{76}+x_{77}+\cdots+x_{100}=\frac 12$ is greater than or equal to $25x_{76}$ because the numbers are in increasing order.
Similarly, $x_{1}+x_{2}+\cdots+x_{16}=-\frac 12$ is less than or equal to $16x_{16}$
So we now know that $\frac 1{50}$ is the best we can do for $x_{76}$ , and $-\frac 1{32}$ is the least we can do for $x_{16}$
Finally, the maximum value of $x_{76}-x_{16}=\frac 1{50}+\frac 1{32}=\frac{41}{800}$ , so the answer is $\boxed{841}$ | null | 841 |
c0c733b0e322980c50b9bf5c8dfd99a0 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_7 | A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles. | [asy] //Created by isabelchen size(12cm, 12cm); draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); dot((72/5, 96/5)); label("$A$",(72/5, 96/5),NE); dot((168/5, 24/5)); label("$B$",(168/5, 24/5),NE); dot((24,0)); label("$C$",(24,0),NW); dot((40, 0)); label("$D$",(40, 0),NE); dot((24, 12)); label("$E$",(24, 12),NE); dot((24, -12)); label("$F$",(24, -12),SE); dot((54/5, 72/5)); label("$G$",(54/5, 72/5),NW); dot((0, 0)); label("$O_1$",(0, 0),S); dot((30, 0)); label("$O_2$",(30, 0),S); [/asy]
$r_1 = O_1A = 24$ $r_2 = O_2B = 6$ $AG = BO_2 = r_2 = 6$ $O_1G = r_1 - r_2 = 24 - 6 = 18$ $O_1O_2 = r_1 + r_2 = 30$
$\triangle O_2BD \sim \triangle O_1GO_2$ $\frac{O_2D}{O_1O_2} = \frac{BO_2}{GO_1}$ $\frac{O_2D}{30} = \frac{6}{18}$ $O_2D = 10$
$CD = O_2D + r_2 = 10 + 6 = 16$
$EF = 2EC = EA + EB = AB = GO_2 = \sqrt{(O_1O_2)^2-(O_1G)^2} = \sqrt{30^2-18^2} = 24$
$DEF = \frac12 \cdot EF \cdot CD = \frac12 \cdot 24 \cdot 16 = \boxed{192}$ | null | 192 |
c0c733b0e322980c50b9bf5c8dfd99a0 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_7 | A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles. | Let the center of the circle with radius $6$ be labeled $A$ and the center of the circle with radius $24$ be labeled $B$ . Drop perpendiculars on the same side of line $AB$ from $A$ and $B$ to each of the tangents at points $C$ and $D$ , respectively. Then, let line $AB$ intersect the two diagonal tangents at point $P$ . Since $\triangle{APC} \sim \triangle{BPD}$ , we have \[\frac{AP}{AP+30}=\frac14 \implies AP=10.\] Next, throw everything on a coordinate plane with $A=(0, 0)$ and $B = (30, 0)$ . Then, $P = (-10, 0)$ , and if $C = (x, y)$ , we have \[(x+10)^2+y^2=64,\] \[x^2+y^2=36.\] Combining these and solving, we get $(x, y)=\left(-\frac{18}5, \frac{24}5\right)$ . Notice now that $P$ $C$ , and the intersections of the lines $x=6$ (the vertical tangent) with the tangent containing these points are collinear, and thus every slope between a pair of points will have the same slope, which in this case is $\frac{-\frac{18}5+10}{\frac{24}5}=\frac34$ . Thus, the other two vertices of the desired triangle are $(6, 12)$ and $(6, -12)$ . By the Shoelace Formula, the area of a triangle with coordinates $(-10, 0)$ $(6, 12)$ , and $(6, -12)$ is \[\frac12|-120-0-72-72+0-120|=\boxed{192}.\] | null | 192 |
c0c733b0e322980c50b9bf5c8dfd99a0 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_7 | A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles. | [asy] //Created by isabelchen and edited by afly size(12cm, 12cm); draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); draw((30,0)--(30,15/2)); dot((72/5, 96/5)); label("$A$",(72/5, 96/5),NE); dot((168/5, 24/5)); label("$B$",(168/5, 24/5),NE); dot((24,0)); label("$C$",(24,0),NW); dot((40, 0)); label("$D$",(40, 0),NE); dot((24, 12)); label("$E$",(24, 12),NE); dot((24, -12)); label("$F$",(24, -12),SE); dot((54/5, 72/5)); label("$G$",(54/5, 72/5),NW); dot((0, 0)); label("$O_1$",(0, 0),S); dot((30, 0)); label("$O_2$",(30, 0),S); dot((30,15/2)); label("$H$",(30,15/2),N); label("$x$",(30,0)--(40,0),N); [/asy]
First, we want to find $O_2D$ . We know that $\angle O_1AD = \angle O_2BD = 90^{\circ}$ , so by AA similarity, $\triangle O_1AD \sim \triangle O_2BD$ . We want to find the length of $x$ , and using the similar triangles, we write an equation: $\frac{30 + x}{4} = x$ . Solving, we get $x=10$ . Therefore, $CD = 10 + 6 = 16$ . Next, we find that using AA similarity, $\triangle O_2BD \sim \triangle HO_2D \sim \triangle ECD$ and they are 3-4-5 triangles. We can quickly compute $EF = 2EC = 2 \cdot \left( \frac{3}{4} \cdot 16 \right) = 2 \cdot 12 = 24$ . Therefore, the area is $\frac{1}{2} \cdot 16 \cdot 24 = \boxed{192}$ | null | 192 |
8d365e0d567b2e799e99925d7cc7c7d5 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_8 | Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$ | We need to find all numbers between $1$ and $600$ inclusive that are multiples of $4$ $5$ , and/or $6$ which are also multiples of $4$ $5$ , and/or $6$ when $1$ is added to them.
We begin by noting that the LCM of $4$ $5$ , and $6$ is $60$ . We can therefore simplify the problem by finding all such numbers described above between $1$ and $60$ and multiplying the quantity of such numbers by $10$ $600$ $60$ $10$ ).
After making a simple list of the numbers between $1$ and $60$ and going through it, we see that the numbers meeting this condition are $4$ $5$ $15$ $24$ $35$ $44$ $54$ , and $55$ . This gives us $8$ numbers. $8$ $10$ $\boxed{080}$ | null | 080 |
8d365e0d567b2e799e99925d7cc7c7d5 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_8 | Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$ | This is Solution 1 with a slick element included. Solution 1 uses the concept that $60k+l$ is a solution for $n$ if $60k+l$ is a multiple of $3$ $4$ , and/or $5$ and $60k+l+1$ is a multiple of $3$ $4$ , and/or $5$ for positive integer values of $l$ and essentially any integer value of $k$ . But keeping the same conditions in mind for $k$ and $l$ , we can also say that if $60k+l$ is a solution, then $60k-l-1$ is a solution! Therefore, one doesn't have to go as far as determining the number of values between $1$ and $60$ and then multiplying by $10$ . One only has to determine the number of values between $1$ and $30$ and then multiply by $20$ . The values of $n$ that work between $1$ and $30$ are $4$ $5$ $15$ , and $24$ . This gives us $4$ numbers. $4$ $20$ $\boxed{080}$ | null | 080 |
8d365e0d567b2e799e99925d7cc7c7d5 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_8 | Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$ | 1. For $n$ to be uniquely determined, $n$ AND $n + 1$ both need to be a multiple of $4, 5,$ or $6.$ Since either $n$ or $n + 1$ is odd, we know that either $n$ or $n + 1$ has to be a multiple of $5.$ We can state the following cases:
1. $n$ is a multiple of $4$ and $n+1$ is a multiple of $5$
2. $n$ is a multiple of $6$ and $n+1$ is a multiple of $5$
3. $n$ is a multiple of $5$ and $n+1$ is a multiple of $4$
4. $n$ is a multiple of $5$ and $n+1$ is a multiple of $6$
Solving for each case, we see that there are $30$ possibilities for cases 1 and 3 each, and $20$ possibilities for cases 2 and 4 each. However, we over-counted the cases where
1. $n$ is a multiple of $24$ and $n+1$ is a multiple of $5$
2. $n$ is a multiple of $5$ and $n+1$ is a multiple of $24$
Each case has $10$ possibilities.
Adding all the cases and correcting for over-counting, we get $30 + 20 + 30 + 20 - 10 - 10 = \boxed{080}.$ | null | 080 |
8d365e0d567b2e799e99925d7cc7c7d5 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_8 | Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$ | Here is a detailed solution for Solution 2.
Over-counted cases:
$30 + 20 + 30 + 20 - 10 - 10 = \boxed{080}$ | null | 080 |
8d365e0d567b2e799e99925d7cc7c7d5 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_8 | Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$ | The problem is the same as asking how many unique sets of values of $\lfloor\frac{n}{4}\rfloor$ $\lfloor\frac{n}{5}\rfloor$ , and $\lfloor\frac{n}{6}\rfloor$ can be produced by one and only one value of $n$ for positive integers $n$ less than or equal to 600.
Seeing that we are dealing with the unique values of the floor function, we ought to examine when it is about to change values, for instance, when $n$ is close to a multiple of 4 in $\lfloor\frac{n}{4}\rfloor$
For a particular value of $n$ , let $a$ $b$ , and $c$ be the original values of $\lfloor\frac{n}{4}\rfloor$ $\lfloor\frac{n}{5}\rfloor$ , and $\lfloor\frac{n}{6}\rfloor$ , respectively.
Notice when $n$ $\equiv0\mod4$ and $n$ $\equiv4\mod5$ , the value of $\lfloor\frac{n-1}{4}\rfloor$ will be 1 less than the original $a$ . The value of $\lfloor\frac{n+1}{5}\rfloor$ will be 1 greater than the original value of $b$
More importantly, this means that no other value less than or greater than $n$ will be able to produce the set of original values of $a$ $b$ , and $c$ , since they make either $a$ or $b$ differ by at least 1.
Generalizing, we find that $n$ must satisfy:
Where $j$ and $k$ are pairs of distinct values of 4, 5, and 6.
Plugging in the values of $j$ and $k$ , finding the solutions to the 6 systems of linear congruences, and correcting for the repeated values, we find that there are $\boxed{080}$ solutions of $n$ | null | 080 |
8d365e0d567b2e799e99925d7cc7c7d5 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_8 | Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$ | By Chinese Remainder Theorem , the general solution of systems of $2$ linear congruences is:
$lcm[4, 5, 6] = 60$ , we solve the number of values for $n \le 60$ , then multiply by $10$ to get the number of values for $n \le 600$ . We are going to solve the following $6$ systems of linear congruences:
$n \in \{ 4, 5, 15, 24, 35, 44, 54, 55 \}$ , there are $8$ values for $n \le 60$ . For $n \le 600$ , the answer is $8 \cdot 10 = \boxed{080}$ | null | 080 |
796add53a34d4adcecac5b998347c674 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_9 | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ [asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy] | We can use recursion to solve this problem:
1. Fix 7 points on $\ell_A$ , then put one point $B_1$ on $\ell_B$ . Now, introduce a function $f(x)$ that indicates the number of regions created, where x is the number of points on $\ell_B$ . For example, $f(1) = 6$ because there are 6 regions.
2. Now, put the second point $B_2$ on $\ell_B$ . Join $A_1~A_7$ and $B_2$ will create $7$ new regions (and we are not going to count them again), and split the existing regions. Let's focus on the spliting process: line segment formed between $B_2$ and $A_1$ intersect lines $\overline{B_1A_2}$ $\overline{B_1A_3}$ , ..., $\overline{B_1A_7}$ at $6$ points $\Longrightarrow$ creating $6$ regions (we already count one region at first), then $5$ points $\Longrightarrow$ creating $5$ regions (we already count one region at first), 4 points, etc. So, we have: \[f(2) = f(1) + 7 + (6+5+...+1) = 34.\]
3. If you still need one step to understand this: $A_1~A_7$ and $B_3$ will still create $7$ new regions. Intersecting \[\overline{A_2B_1}, \overline{A_2B_2};\] \[\overline{A_3B_1}, \overline{A_3B_2};\] \[...\] \[\overline{A_7B_1}, \overline{A_7B_2}\] at $12$ points, creating $12$ regions, etc. Thus, we have: \[f(3) = f(2)+7+(12+10+8+...+2)=34+7+6\cdot 7=83.\]
Yes, you might already notice that: \[f(n+1) = f(n)+7+(6+5+...+1)\cdot n = f(n) + 7 + 21n.\]
5. Finally, we have $f(4) = 153$ , and $f(5)=244$ . Therefore, the answer is $\boxed{244}$ | null | 244 |
796add53a34d4adcecac5b998347c674 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_9 | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ [asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy] | We want to derive a general function $f(m,n)$ that indicates the number of bounded regions. Observing symmetry, we know this is a symmetric function about $m$ and $n$ . Now let's focus on $f(m+1, n)-f(m, n)$ , which is the difference caused by adding one point to the existing $m$ points of line $\ell_A$ . This new point, call it #m, when connected to point #1 on $\ell_B$ , crosses $m*(n-1)$ lines, thus making additional $m*(n-1)+1$ bounded regions; when connected to point #2 on $\ell_B$ , it crosses $m*(n-2)$ lines, thus making additional $m*(n-2)+1$ bounded regions; etc. By simple algebra/recursion methods, we see
$f(m+1, n)-f(m, n)=m*\frac{n(n-1)}{2} +n$
Notice $f(1,n)=n-1$ . Not very difficult to figure out:
$f(m, n)=\frac{m(m-1)n(n-1)}{4} +mn-1$
The fact that $f(3,2)=8$ makes us more confident about the formula. Now plug in $m=5, n=7$ , we get the final answer of $\boxed{244}$ | null | 244 |
796add53a34d4adcecac5b998347c674 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_9 | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ [asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy] | Let some number of segments be constructed. We construct a new segment. We start from the straight line $l_B.$ WLOG from point $B_3.$ Segment will cross several existing segments (points $A,B,C,...$ ) and enter one of the points of the line $l_A (A_1).$
Each of these points adds exactly 1 new bounded region (yellow bounded regions).
The exception is the only first segment $(A_1 B_1),$ which does not create any bounded region.
Thus, the number of bounded regions is $1$ less than the number of points of intersection of the segments plus the number of points of arrival of the segments to $l_A.$
Each point of intersection of two segments is determined uniquely by the choice of pairs of points on each line.
The number of such pairs is $\dbinom{n}{2} \cdot \dbinom{m}{2}.$
Exactly one segment comes to each of the $n$ points of the line $l_A$ from each of the $m$ points of the line $l_B.$ The total number of arrivals is equal to $mn.$ Hence, the total number of bounded regions is $N = \dbinom{n}{2} \cdot \dbinom{m}{2} + mn – 1.$
We plug in $m=5, n=7$ , we get the final answer of $\boxed{244}$ | null | 244 |
796add53a34d4adcecac5b998347c674 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_9 | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ [asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy] | When a new point is added to a line, the number of newly bounded regions it creates with each line segment will be one more than the number of intersection points the line makes with other lines.
Case 1: If a new point $P$ is added to the right on a line when both lines have an equal amount of points.
WLOG, let the point be on line $\ell_A$ . We consider the complement, where new lines don't intersect other line segments. Simply observing, we see that the only line segments that don't intersect with the new lines are lines attached to some point that a new line does not pass through. If we look at a series of points on line $\ell_B$ from left to right and a line connects $P$ to an arbitrary point, then the lines formed with that point and with remaining points on the left of that point never intersect with the line with $P$ . Let there be $s$ points on lines $\ell_A$ and $\ell_B$ before $P$ was added. For each of the $s$ points on $\ell_B$ , we subtract the total number of lines formed, which is $s^2$ , not counting $P$ . Considering all possible points on $\ell_B$ , we get $(s^2-s)+(s^2-2s)\cdots(s^2-s^2)$ total intersections. However, for each of the lines, there is one more bounded region than number of intersections, so we add $s$ . Simplifying, we get $s^3-s\sum_{i=1}^{s}{i}+s\Longrightarrow s(s^2-\sum_{i=1}^{s}{i}+1)$ . Note that this is only a recursion formula to find the number of new regions added for a new point $P$ added to $\ell_A$
Case 2: If a new point $P$ is added to the right of a line that has one less point than the other line.
Continuing on case one, let this point $P$ be on line $\ell_B$ . With similar reasoning, we see that the idea remains the same, except $s+1$ lines are formed with $P$ instead of just $s$ lines. Once again, each line from $P$ to a point on line $\ell_A$ creates $s$ non-intersecting lines for that point and each point to its left. Subtracting from $s(s+1)$ lines and considering all possible lines created by $P$ , we get $(s(s+1)-s)+(s(s+1)-2s)\cdots(s(s+1)-s(s+1)$ intersections. However, the number of newly bounded regions is the number of intersections plus the number of points on line $\ell_A$ . Simplying, we get $s(s+1)^2-s\sum_{i=1}^{s+1}{i}+(s+1)$ newly bounded regions.
For the base case $s=2$ for both lines, there are $4$ bounded regions. Next, we plug in $s=2,3,4$ for both formulas and plug $s=5$ for the first formula to find the number of regions when $m=6$ and $n=5$ . Notice that adding a final point on $\ell_A$ is a variation of our Case 1. The only difference is for each of the $s$ lines formed by $P$ , there are $s+1$ points that can form a non-intersecting line. Therefore, we are subtracting a factor of $s+1$ lines instead of $s$ lines from a total of $s(s+1)$ lines. However, the number of lines formed by $P$ remains the same so we still add $s$ at the end when considering intersection points. Thus, the recursive equation becomes $(s(s+1)-(s+1))+(s(s+1)-2(s+1))\cdots(s(s+1)-s(s+1))+s\Longrightarrow s^2(s+1)-(s+1)\sum_{i=1}^{s}{i}+s$ . Plugging $s=5$ into this formula and adding the values we obtained from the other formulas, the final answer is $4+4+9+12+22+28+45+55+65=\boxed{244}$ | null | 244 |
796add53a34d4adcecac5b998347c674 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_9 | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ [asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy] | We know the by Euler's Formula for planar graphs that $F-E+V=2$ , where $F$ is the number of bounded faces, plus the outer region, $E$ is the number of edges, and $V$ is the number of vertices. Temporarily disregarding the intersections between the lines, we can easily calculate that:
$V_{i}=7+5=12$
$E_{i}=6+4+7\cdot5=45$
However, the resulting graph is not planar, as the edges clearly intersect. To account for this, we must turn all intersection points into vertices, and update our values accordingly.
Observe that each intersection point can be mapped to two points on either line, and analogously, two points on either line can be mapped to one intersection point, uniquely. Thus, to count intersection points, we simply calculate:
${7 \choose 2}{5 \choose 2} = 210$
And thus,
$V=V_{i}+210=222$
We must also account for the edges. Observe that each intersection point turns the two edges that make it into four, that is, each intersection point adds $2$ to the number of edges. Therefore,
$E=E_{i}+2\cdot210=465$
Plugging these into Euler's Formula we get: $F-E+V=2$
$F-465+222=2$
$F=245$
Disregarding the outer region, we conclude that our answer is $F-1=245-1=\boxed{244}$ | null | 244 |
bbf67cfcdad23a60be0d2a50265a81d5 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 | Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$ | [asy] defaultpen(fontsize(12)+0.6); size(300); pair A,B,C,D,M,H; real xb=71, xd=121; A=origin; D=(7,0); B=2*dir(xb); C=3*dir(xd)+D; M=(B+C)/2; H=foot(M,A,D); path c=CR(D,3); pair A1=bisectorpoint(D,A,B), D1=bisectorpoint(C,D,A), Bp=IP(CR(A,2),A--H), Cp=IP(CR(D,3),D--H); draw(B--A--D--C--B); draw(A--M--D^^M--H^^Bp--M--Cp, gray+0.4); draw(rightanglemark(A,H,M,5)); dot("$A$",A,SW); dot("$D$",D,SE); dot("$B$",B,NW); dot("$C$",C,NE); dot("$M$",M,up); dot("$H$",H,down); dot("$B'$",Bp,down); dot("$C'$",Cp,down); [/asy]
According to the problem, we have $AB=AB'=2$ $DC=DC'=3$ $MB=MB'$ $MC=MC'$ , and $B'C'=7-2-3=2$
Because $M$ is the midpoint of $BC$ , we have $BM=MC$ , so: \[MB=MB'=MC'=MC.\]
Then, we can see that $\bigtriangleup{MB'C'}$ is an isosceles triangle with $MB'=MC'$
Therefore, we could start our angle chasing: $\angle{MB'C'}=\angle{MC'B'}=180^\circ-\angle{MC'D}=180^\circ-\angle{MCD}$
This is when we found that points $M$ $C$ $D$ , and $B'$ are on a circle. Thus, $\angle{BMB'}=\angle{CDC'} \Rightarrow \angle{B'MA}=\angle{C'DM}$ . This is the time we found that $\bigtriangleup{AB'M} \sim \bigtriangleup{MC'D}$
Thus, $\frac{AB'}{B'M}=\frac{MC'}{C'D} \Longrightarrow (B'M)^2=AB' \cdot C'D = 6$
Point $H$ is the midpoint of $B'C'$ , and $MH \perp AD$ $B'H=HC'=1 \Longrightarrow MH=\sqrt{B'M^2-B'H^2}=\sqrt{6-1}=\sqrt{5}$
The area of this quadrilateral is the sum of areas of triangles: \[S_{\bigtriangleup{ABM}}+S_{\bigtriangleup{AB'M}}+S_{\bigtriangleup{CDM}}+S_{\bigtriangleup{CD'M}}+S_{\bigtriangleup{B'C'M}}\] \[=S_{\bigtriangleup{AB'M}}\cdot 2 + S_{\bigtriangleup{B'C'M}} + S_{\bigtriangleup{C'DM}}\cdot 2\] \[=2 \cdot \frac{1}{2} \cdot AB' \cdot MH + \frac{1}{2} \cdot B'C' \cdot MH + 2 \cdot \frac{1}{2} \cdot C'D \cdot MH\] \[=2\sqrt{5}+\sqrt{5}+3\sqrt{5}=6\sqrt{5}\]
Finally, the square of the area is $(6\sqrt{5})^2=\boxed{180}$ | null | 180 |
bbf67cfcdad23a60be0d2a50265a81d5 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 | Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$ | Denote by $M$ the midpoint of segment $BC$ .
Let points $P$ and $Q$ be on segment $AD$ , such that $AP = AB$ and $DQ = DC$
Denote $\angle DAM = \alpha$ $\angle BAD = \beta$ $\angle BMA = \theta$ $\angle CMD = \phi$
Denote $BM = x$ . Because $M$ is the midpoint of $BC$ $CM = x$
Because $AM$ is the angle bisector of $\angle BAD$ and $AB = AP$ $\triangle BAM \cong \triangle PAM$ .
Hence, $MP = MB$ and $\angle AMP = \theta$ .
Hence, $\angle MPD = \angle MAP + \angle PMA = \alpha + \theta$
Because $DM$ is the angle bisector of $\angle CDA$ and $DC = DQ$ $\triangle CDM \cong \triangle QDM$ .
Hence, $MQ = MC$ and $\angle DMQ = \phi$ .
Hence, $\angle MQA = \angle MDQ + \angle QMD = \beta + \phi$
Because $M$ is the midpoint of segment $BC$ $MB = MC$ .
Because $MP = MB$ and $MQ = MC$ $MP = MQ$
Thus, $\angle MPD = \angle MQA$
Thus, \[ \alpha + \theta = \beta + \phi . \hspace{1cm} (1) \]
In $\triangle AMD$ $\angle AMD = 180^\circ - \angle MAD - \angle MDA = 180^\circ - \alpha - \beta$ .
In addition, $\angle AMD = 180^\circ - \angle BMA - \angle CMD = 180^\circ - \theta - \phi$ .
Thus, \[ \alpha + \beta = \theta + \phi . \hspace{1cm} (2) \]
Taking $(1) + (2)$ , we get $\alpha = \phi$ .
Taking $(1) - (2)$ , we get $\beta = \theta$
Therefore, $\triangle ADM \sim \triangle AMB \sim \triangle MDC$
Hence, $\frac{AD}{AM} = \frac{AM}{AB}$ and $\frac{AD}{DM} = \frac{DM}{CD}$ .
Thus, $AM = \sqrt{AD \cdot AD} = \sqrt{14}$ and $DM = \sqrt{AD \cdot CD} = \sqrt{21}$
In $\triangle ADM$ , by applying the law of cosines, $\cos \angle AMD = \frac{AM^2 + DM^2 - AD^2}{2 AM \cdot DM} = - \frac{1}{\sqrt{6}}$ .
Hence, $\sin \angle AMD = \sqrt{1 - \cos^2 \angle AMD} = \frac{\sqrt{5}}{\sqrt{6}}$ .
Hence, ${\rm Area} \ \triangle ADM = \frac{1}{2} AM \cdot DM \dot \sin \angle AMD = \frac{7 \sqrt{5}}{2}$
Therefore, \begin{align*} {\rm Area} \ ABCD & = {\rm Area} \ \triangle AMD + {\rm Area} \ \triangle ABM + {\rm Area} \ \triangle MCD \\ & = {\rm Area} \ \triangle AMD \left( 1 + \left( \frac{AM}{AD} \right)^2 + \left( \frac{MD}{AD} \right)^2 \right) \\ & = 6 \sqrt{5} . \end{align*}
Therefore, the square of ${\rm Area} \ ABCD$ is $\left( 6 \sqrt{5} \right)^2 = \boxed{180}$ | null | 180 |
bbf67cfcdad23a60be0d2a50265a81d5 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 | Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$ | Claim
In the triangle $ABC, AB = 2AC, M$ is the midpoint of $AB. D$ is the point of intersection of the circumcircle and the bisector of angle $A.$ Then $DM = BD.$
Proof
Let $A = 2\alpha.$ Then $\angle DBC = \angle DCB = \alpha.$
Let $E$ be the intersection point of the perpendicular dropped from $D$ to $AB$ with the circle.
Then the sum of arcs $\overset{\Large\frown} {BE} + \overset{\Large\frown}{AC} + \overset{\Large\frown}{CD} = 180^\circ.$ \[\overset{\Large\frown} {BE} = 180^\circ – 2\alpha – \overset{\Large\frown}{AC}.\]
Let $E'$ be the point of intersection of the line $CM$ with the circle. $CM$ is perpendicular to $AD, \angle AMC = 90^\circ – \alpha,$ the sum of arcs $\overset{\Large\frown}{A}C + \overset{\Large\frown}{BE'} = 180^\circ – 2\alpha \implies E'$ coincides with $E.$
The inscribed angles $\angle DEM = \angle DEB, M$ is symmetric to $B$ with respect to $DE, DM = DB.$
Solution
Let $AB' = AB, DC' = DC, B'$ and $C'$ on $AD.$
Then $AB' = 2, DC' = 3, B'C' = 2 = AB'.$
Quadrilateral $ABMC'$ is cyclic.
Let $\angle A = 2\alpha.$ Then $\angle MBC' = \angle MC'B = \alpha.$
Circle $BB'C'C$ centered at $M, BC$ is its diameter, $\angle BC'C = 90^\circ.$ $\angle DMC' = \angle MC'B,$ since they both complete $\angle MC'C$ to $90^\circ.$
$\angle MB'A = \angle MC'D,$ since they are the exterior angles of an isosceles $\triangle MB'C'.$ $\triangle AMB' \sim \triangle MDC'$ by two angles. $\frac {AB'}{MC'} = \frac {MB'}{DC'}, MC' =\sqrt{AB' \cdot C'D} = \sqrt{6}.$
The height dropped from $M$ to $AD$ is $\sqrt{MB'^2 - (\frac{B'C'}{2})^2} =\sqrt{6 - 1} = \sqrt{5}.$
The areas of triangles $\triangle AMB'$ and $\triangle MC'B'$ are equal to $\sqrt{5},$ area of $\triangle MC'D$ is $\frac{3}{2} \sqrt{5}.$
\[\triangle AMB' = \triangle AMB, \triangle MC'D = \triangle MCD \implies\] The area of $ABCD$ is $(1 + 2 + 3) \sqrt{5} = 6\sqrt{5} \implies 6^2 \cdot 5 = \boxed{180}.$ | null | 180 |
bbf67cfcdad23a60be0d2a50265a81d5 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 | Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$ | Extend $AB$ and $CD$ so they intersect at a point $X$ . Then note that $M$ is the incenter of $\triangle{XAD}$ , implying that $M$ is on the angle bisector of $X$ . Now because $XM$ is both an angle bisector and a median of $\triangle{XBC}$ $\triangle{XBC}$ is isosceles. Then we can start angle chasing:
Let $\angle{BAM}=a, \angle{CDM}=b,$ and $\angle{XBC}=c$ . Then $\angle{AMD}=\pi-(a+b), \angle{ABM}=\pi-c, \angle{DCM}=\pi-c$ , implying that $\angle{BMA}+\angle{CMD}=a+b$ , implying that $2c-(a+b)=(a+b)$ , or that $c=a+b$ . Substituting this into the rest of the diagram, we find that $\triangle{BMA} \sim \triangle{CDM} \sim \triangle{MDA}$
Then $\frac{AB}{BM}=\frac{MC}{CD}$ , or $BM=CM=\sqrt{6}$ . Moreover, $\frac{AB}{AM}=\frac{AM}{AD}$ , or $AM=\sqrt{14}$ . Similarly, $\frac{CD}{MD}=\frac{MD}{AD}$ , or $DM=\sqrt{21}$ . Then using Law of Cosines on $\triangle{AMD}$ , to get that $cos\angle{AMD}=-\frac{\sqrt{6}}{6}$ , or $sin\angle{AMD}=\frac{\sqrt{30}}{6}$
We finish by using the formula $K=\frac{1}{2}absinC$ , as follows:
$[ABCD]=[ABM]+[CDM]+[ADM]=\frac{\frac{\sqrt{30}}{6}(2\sqrt{6}+3\sqrt{6}+7\sqrt{6})}{2}=6\sqrt{5}$
$(6\sqrt{5})^2=\boxed{180}$ | null | 180 |
bbf67cfcdad23a60be0d2a50265a81d5 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 | Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$ | As shown in paragraph one of solution 4, extending $AB$ and $CD$ to $X$ , we realize that $\triangle{XBC}$ is isosceles, thus $XM \perp BC$ . Let $XB = XC = x$ . And, midpoint $M$ is the incenter of $\triangle{XAD}$ . Construct perpendiculars $ME, MF, MG$ to sides $AD, AX, DX$ respectively (constructing the radii of the incircle). Let $EM = FM = GM = r$ . The semiperimeter $s = \frac{2x + 2 + 3 + 7}{2} = x+6$ . Since $FX$ is the tangent off the incircle, $FX = s - AD = x-1$ . So, $BF = BX - FX = 1$ . Because $\triangle{BFM} \sim \triangle{MFX}$
\[\frac{BF}{FM} = \frac{FM}{FX} \implies {FM}^2 = BF \cdot FX = x - 1 \implies r^2 = x - 1\]
By Heron's formula and the inradius area formula,
\[(x+6)r = \sqrt{(x+6)\cdot 4 \cdot 3 \cdot (x-1)} \implies (x+6)r^2 = 12(x-1) \implies (x+6)(x-1) = 12(x-1) \implies x=6\]
Then, $r^2 = x - 1 = 5 \implies r = \sqrt{5}$ . Finally,
\[[ABCD] = [ABM] + [CDM] + [AMD] = \frac{AB \cdot FM}{2} + \frac{CD \cdot GM}{2} + \frac{AD \cdot EM}{2} = \frac{2r}{2} + \frac{3r}{2} + \frac{7r}{2} = 6r = 6\sqrt{5}\]
Thus, our answer is $(6\sqrt{5})^{2} = \boxed{180}$ | null | 180 |
bbf67cfcdad23a60be0d2a50265a81d5 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 | Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$ | Let the midpoint of $BC$ be $M$ . Angle-chase and observe that $\Delta AMD~\Delta ABM~\Delta MCD$ . Let $BM=CM=a$ and $AM=x$ and $DM=y$ . As a result of this similarity, we write
\[\dfrac2a=\dfrac a3,\]
which gives $a=\sqrt 6$ . Similarly, we write
\[\dfrac2x=\dfrac x7\]
and
\[\dfrac3y=\dfrac y7\]
to get $x=\sqrt{14}$ and $y=\sqrt{21}$
We now have all required side lengths; we can find the area of $\Delta AMD$ with Heron's formula. Doing so yields $\dfrac72\sqrt5$ . We could also bash out the areas of the other two triangles since we know all their side lengths (this is what I did :sob:), but a more intelligent method is to recall the triangles' similarity. The ratio of similarlity between $\Delta AMD$ and $\Delta ABM$ is $\dfrac{\sqrt{14}}7=\sqrt{\dfrac27}$ , and between $\Delta AMD$ and $\Delta MCD$ is $\dfrac{\sqrt{21}}7=\sqrt{\dfrac37}$ . Thus, the area ratios are $\dfrac27$ and $\dfrac37$ , respectively, so adding together we have $\dfrac27+\dfrac37=\dfrac57$ . Multiplying this by our $\dfrac72\sqrt5$ , we have $\dfrac52\sqrt5$ as their total area. Adding this to our original area, we have $\dfrac52\sqrt5+\dfrac72\sqrt5=\sqrt5\left(\dfrac52+\dfrac72\right)=\sqrt5\left(\dfrac{12}2\right)=6\sqrt5$
The square of this is $\boxed{180}$ | null | 180 |
bbf67cfcdad23a60be0d2a50265a81d5 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 | Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$ | As in solution 4, let $X=AB\cap CD$ , so $M$ is the incenter of $ADX$ and $XB=XC$ . Let $XB=XC=x$ . Then the normalized barycentric coordinates of $B$ $C$ , and $M$ with respect to $ADX$ are $\left[\frac{x}{x+2}:0:\frac{2}{x+2}\right]$ $\left[0:\frac{x}{x+3}:\frac{3}{x+3}\right]$ , and $\left[\frac{x+3}{2x+12}:\frac{x+2}{2x+12}:\frac{7}{2x+12}\right]$ . So we have $\frac{1}{2}\frac{x}{x+2}=\frac{x+3}{2x+12}$ giving $x=6$ . The sidelengths of $ADX$ are thus $AD=7$ $AX=8$ , and $DX=9$ giving $[ADX]=12\sqrt 5$ . Also, we have $[BCX]=\frac{6}{8}\cdot\frac{6}{9}[ADX]=6\sqrt 5$ so that $[ABCD]=[ADX]-[BCX]=6\sqrt 5$ . The area squared is thus $\boxed{180}$ .
~~ golue3120 | null | 180 |
70ddf93a3ee80c0a16bf9cd746352971 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_12 | Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that \[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\] Find the least possible value of $a+b.$ | Denote $P = \left( x , y \right)$
Because $\frac{x^2}{a^2}+\frac{y^2}{a^2-16} = 1$ $P$ is on an ellipse whose center is $\left( 0 , 0 \right)$ and foci are $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$
Hence, the sum of distance from $P$ to $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$ is equal to twice the major axis of this ellipse, $2a$
Because $\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2} = 1$ $P$ is on an ellipse whose center is $\left( 20 , 11 \right)$ and foci are $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$
Hence, the sum of distance from $P$ to $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$ is equal to twice the major axis of this ellipse, $2b$
Therefore, $2a + 2b$ is the sum of the distance from $P$ to four foci of these two ellipses.
To make this minimized, $P$ is the intersection point of the line that passes through $\left( - 4 , 0 \right)$ and $\left( 20 , 10 \right)$ , and the line that passes through $\left( 4 , 0 \right)$ and $\left( 20 , 12 \right)$
The distance between $\left( - 4 , 0 \right)$ and $\left( 20 , 10 \right)$ is $\sqrt{\left( 20 + 4 \right)^2 + \left( 10 - 0 \right)^2} = 26$
The distance between $\left( 4 , 0 \right)$ and $\left( 20 , 12 \right)$ is $\sqrt{\left( 20 - 4 \right)^2 + \left( 12 - 0 \right)^2} = 20$
Hence, $2 a + 2 b = 26 + 20 = 46$
Therefore, $a + b = \boxed{023}.$ | null | 023 |
36567689b9645b4a28ab0daf56eac1f1 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_13 | There is a polynomial $P(x)$ with integer coefficients such that \[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\] holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$ | Because $0 < x < 1$ , we have \begin{align*} P \left( x \right) & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \binom{6}{a} x^{2310a} \left( - 1 \right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\ & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} x^{2310 a + 105 b + 70 c + 42 d + 30 e} . \end{align*}
Denote by $c_{2022}$ the coefficient of $P \left( x \right)$ .
Thus, \begin{align*} c_{2022} & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} \Bbb I \left\{ 2310 a + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-0} \Bbb I \left\{ 2310 \cdot 0 + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \Bbb I \left\{ 105 b + 70 c + 42 d + 30 e = 2022 \right\} . \end{align*}
Now, we need to find the number of nonnegative integer tuples $\left( b , c , d , e \right)$ that satisfy \[ 105 b + 70 c + 42 d + 30 e = 2022 . \hspace{1cm} (1) \]
Modulo 2 on Equation (1), we have $b \equiv 0 \pmod{2}$ .
Hence, we can write $b = 2 b'$ . Plugging this into (1), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c , d , e \right)$ that satisfy \[ 105 b' + 35 c + 21 d + 15 e = 1011 . \hspace{1cm} (2) \]
Modulo 3 on Equation (2), we have $2 c \equiv 0 \pmod{3}$ .
Hence, we can write $c = 3 c'$ . Plugging this into (2), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d , e \right)$ that satisfy \[ 35 b' + 35 c' + 7 d + 5 e = 337 . \hspace{1cm} (3) \]
Modulo 5 on Equation (3), we have $2 d \equiv 2 \pmod{5}$ .
Hence, we can write $d = 5 d' + 1$ . Plugging this into (3), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d' , e \right)$ that satisfy \[ 7 b' + 7 c' + 7 d' + e = 66 . \hspace{1cm} (4) \]
Modulo 7 on Equation (4), we have $e \equiv 3 \pmod{7}$ .
Hence, we can write $e = 7 e' + 3$ . Plugging this into (4), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d' , e' \right)$ that satisfy \[ b' + c' + d' + e' = 9 . \hspace{1cm} (5) \]
The number of nonnegative integer solutions to Equation (5) is $\binom{9 + 4 - 1}{4 - 1} = \binom{12}{3} = \boxed{220}$ | null | 220 |
36567689b9645b4a28ab0daf56eac1f1 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_13 | There is a polynomial $P(x)$ with integer coefficients such that \[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\] holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$ | Note that $2022 = 210\cdot 9 +132$ . Since the only way to express $132$ in terms of $105$ $70$ $42$ , or $30$ is $132 = 30+30+30+42$ , we are essentially just counting the number of ways to express $210*9$ in terms of these numbers. Since $210 = 2*105=3*70=5*42=7*30$ , it can only be expressed as a sum in terms of only one of the numbers ( $105$ $70$ $42$ , or $30$ ). Thus, the answer is (by sticks and stones) \[\binom{12}{3} = \boxed{220}\] | null | 220 |
7c1488f718916da7d4e81a0aab855795 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_14 | For positive integers $a$ $b$ , and $c$ with $a < b < c$ , consider collections of postage stamps in denominations $a$ $b$ , and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$ | This should be $\lfloor \frac{1000}{c} \rfloor$ . The current function breaks when $c \mid 1000$ and $b \mid c$ . Take $c = 200$ and $b = 20$ . Then, we have $\lfloor \frac{999}{200} \rfloor = 4$ stamps of value 200, $\lfloor \frac{199}{20} \rfloor = 9$ stamps of value b, and 19 stamps of value 1. The maximum such a collection can give is $200 \cdot 4 + 20 \cdot 9 +19 \cdot 1 = 999$ , just shy of the needed 1000. As for the rest of solution, proceed similarly, except use $1000$ instead of $999$
Also, some explanation: $b-1$ one cent stamps cover all residues module $b$ . Having $\lfloor \frac{c-1}{b} \rfloor$ stamps of value b covers all residue classes modulo $c$ . Finally, we just need $\lfloor \frac{1000}{c} \rfloor$ to cover everything up to 1000.
In addition, note that this function sometimes may not always minimize the number of stamps required. This is due to the fact that the stamps of value $b$ and of value $1$ have the capacity to cover values greater than or equal to $c$ (which occurs when $c-1$ has a remainder less than $b-1$ when divided by $b$ ). Thus, in certain cases, not all $\lfloor \frac{1000}{c} \rfloor$ stamps of value c may be necessary, because the stamps of value $b$ and 1 can replace one $c$
CrazyVideoGamez
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Therefore using $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$ $\lfloor \frac{c-1}{b} \rfloor$ stamps of value $b$ , and $b-1$ stamps of value $1$ , all values up to $1000$ can be represented in sub-collections, while minimizing the number of stamps.
So, $f(a, b, c) = \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1$
$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$ . We can get the answer by solving this equation.
$c > \lfloor \frac{c-1}{b} \rfloor + b-1$
$\frac{999}{c} + c > \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$
$c^2 - 97c + 999 > 0$ $c > 85.3$ or $c < 11.7$
$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 > \frac{999}{c}$
$97 > \frac{999}{c}$ $c>10.3$
The $3$ least values of $c$ are $11$ $88$ $89$ $11 + 88+ 89 = \boxed{188}$ | null | 188 |
335ad41c1cde963664b13052bbcaf023 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_15 | Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ $O_1O_2 = 15$ $CD = 16$ , and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. [asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("$\omega_1$",8*dir(110),SW); label("$\omega_2$",5*dir(70)+(15,0),SE); label("$O_1$",O1,W); label("$O_2$",O2,E); label("$B$",B,N+1/2*E); label("$A$",A,N+1/2*W); label("$C$",C,S+1/4*W); label("$D$",D,S+1/4*E); label("$15$",midpoint(O1--O2),N); label("$16$",midpoint(C--D),N); label("$2$",midpoint(A--B),S); label("$\Omega$",o.C+(o.r-1)*dir(270)); [/asy] | First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$ . Let points $A'$ and $B'$ be the reflections of $A$ and $B$ , respectively, about the perpendicular bisector of $\overline{O_1O_2}$ . Then quadrilaterals $ABO_1O_2$ and $B'A'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $A'B'O_1CDO_2$ have the same area. Furthermore, triangles $DO_2A'$ and $B'O_1C$ are congruent, so $A'D = B'C$ and quadrilateral $A'B'CD$ is an isosceles trapezoid. [asy] import olympiad; size(180); defaultpen(linewidth(0.7)); pair Ap = dir(105), Bp = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175); draw(unitcircle^^Ap--Bp--O1--C--D--O2--cycle); label("$A'$",Ap,dir(origin--Ap)); label("$B'$",Bp,dir(origin--Bp)); label("$O_1$",O1,dir(origin--O1)); label("$C$",C,dir(origin--C)); label("$D$",D,dir(origin--D)); label("$O_2$",O2,dir(origin--O2)); draw(O2--O1,linetype("4 4")); draw(Ap--D^^Bp--C,linetype("2 2")); [/asy] Next, remark that $B'O_1 = DO_2$ , so quadrilateral $B'O_1DO_2$ is also an isosceles trapezoid; in turn, $B'D = O_1O_2 = 15$ , and similarly $A'C = 15$ . Thus, Ptolmey's theorem on $A'B'CD$ yields $A'D\cdot B'C + 2\cdot 16 = 15^2$ , whence $A'D = B'C = \sqrt{193}$ . Let $\alpha = \angle A'B'D$ . The Law of Cosines on triangle $A'B'D$ yields \[\cos\alpha = \frac{15^2 + 2^2 - (\sqrt{193})^2}{2\cdot 2\cdot 15} = \frac{36}{60} = \frac 35,\] and hence $\sin\alpha = \tfrac 45$ . Thus the distance between bases $A’B’$ and $CD$ is $12$ (in fact, $\triangle A'B'D$ is a $9-12-15$ triangle with a $7-12-\sqrt{193}$ triangle removed), which implies the area of $A'B'CD$ is $\tfrac12\cdot 12\cdot(2+16) = 108$
Now let $O_1C = O_2A' = r_1$ and $O_2D = O_1B' = r_2$ ; the tangency of circles $\omega_1$ and $\omega_2$ implies $r_1 + r_2 = 15$ . Furthermore, angles $A'O_2D$ and $A'B'D$ are opposite angles in cyclic quadrilateral $B'A'O_2D$ , which implies the measure of angle $A'O_2D$ is $180^\circ - \alpha$ . Therefore, the Law of Cosines applied to triangle $\triangle A'O_2D$ yields \begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \tfrac45r_1r_2\\ &= (r_1+r_2)^2 - \tfrac45 r_1r_2 = 225 - \tfrac45r_1r_2. \end{align*}
Thus $r_1r_2 = 40$ , and so the area of triangle $A'O_2D$ is $\tfrac12r_1r_2\sin\alpha = 16$
Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\cdot 16 = \boxed{140}$ | null | 140 |
335ad41c1cde963664b13052bbcaf023 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_15 | Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ $O_1O_2 = 15$ $CD = 16$ , and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. [asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("$\omega_1$",8*dir(110),SW); label("$\omega_2$",5*dir(70)+(15,0),SE); label("$O_1$",O1,W); label("$O_2$",O2,E); label("$B$",B,N+1/2*E); label("$A$",A,N+1/2*W); label("$C$",C,S+1/4*W); label("$D$",D,S+1/4*E); label("$15$",midpoint(O1--O2),N); label("$16$",midpoint(C--D),N); label("$2$",midpoint(A--B),S); label("$\Omega$",o.C+(o.r-1)*dir(270)); [/asy] | Denote by $O$ the center of $\Omega$ .
Denote by $r$ the radius of $\Omega$
We have $O_1$ $O_2$ $A$ $B$ $C$ $D$ are all on circle $\Omega$
Denote $\angle O_1 O O_2 = 2 \theta$ .
Denote $\angle O_1 O B = \alpha$ .
Denote $\angle O_2 O A = \beta$
Because $B$ and $C$ are on circles $\omega_1$ and $\Omega$ $BC$ is a perpendicular bisector of $O_1 O$ . Hence, $\angle O_1 O C = \alpha$
Because $A$ and $D$ are on circles $\omega_2$ and $\Omega$ $AD$ is a perpendicular bisector of $O_2 O$ . Hence, $\angle O_2 O D = \beta$
In $\triangle O O_1 O_2$ \[ O_1 O_2 = 2 r \sin \theta . \]
Hence, \[ 2 r \sin \theta = 15 . \]
In $\triangle O AB$ \begin{align*} AB & = 2 r \sin \frac{2 \theta - \alpha - \beta}{2} \\ & = 2 r \sin \theta \cos \frac{\alpha + \beta}{2} - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\ & = 15 \cos \frac{\alpha + \beta}{2} - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} . \end{align*}
Hence, \[ 15 \cos \frac{\alpha + \beta}{2} - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} = 2 . \hspace{1cm} (1) \]
In $\triangle O CD$ \begin{align*} CD & = 2 r \sin \frac{360^\circ - 2 \theta - \alpha - \beta}{2} \\ & = 2 r \sin \left( \theta + \frac{\alpha + \beta}{2} \right) \\ & = 2 r \sin \theta \cos \frac{\alpha + \beta}{2} + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\ & = 15 \cos \frac{\alpha + \beta}{2} + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} . \end{align*}
Hence, \[ 15 \cos \frac{\alpha + \beta}{2} + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} = 16 . \hspace{1cm} (2) \]
Taking $\frac{(1) + (2)}{30}$ , we get $\cos \frac{\alpha + \beta}{2} = \frac{3}{5}$ .
Thus, $\sin \frac{\alpha + \beta}{2} = \frac{4}{5}$
Taking these into (1), we get $2 r \cos \theta = \frac{35}{4}$ .
Hence, \begin{align*} 2 r & = \sqrt{ \left( 2 r \sin \theta \right)^2 + \left( 2 r \cos \theta \right)^2} \\ & = \frac{5}{4} \sqrt{193} . \end{align*}
Hence, $\cos \theta = \frac{7}{\sqrt{193}}$
In $\triangle O O_1 B$ \[ O_1 B = 2 r \sin \frac{\alpha}{2} . \]
In $\triangle O O_2 A$ , by applying the law of sines, we get \[ O_2 A = 2 r \sin \frac{\beta}{2} . \]
Because circles $\omega_1$ and $\omega_2$ are externally tangent, $B$ is on circle $\omega_1$ $A$ is on circle $\omega_2$ \begin{align*} O_1 O_2 & = O_1 B + O_2 A \\ & = 2 r \sin \frac{\alpha}{2} + 2 r \sin \frac{\beta}{2} \\ & = 2 r \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right) . \end{align*}
Thus, $\sin \frac{\alpha}{2} + \sin \frac{\beta}{2} = \frac{12}{\sqrt{193}}$
Now, we compute $\sin \alpha$ and $\sin \beta$
Recall $\cos \frac{\alpha + \beta}{2} = \frac{3}{5}$ and $\sin \frac{\alpha + \beta}{2} = \frac{4}{5}$ .
Thus, $e^{i \frac{\alpha}{2}} e^{i \frac{\beta}{2}} = e^{i \frac{\alpha + \beta}{2}} = \frac{3}{5} + i \frac{4}{5}$
We also have \begin{align*} \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} & = \frac{1}{2i} \left( e^{i \frac{\alpha}{2}} - e^{-i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} - e^{-i \frac{\beta}{2}} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} } \right) \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right) . \end{align*}
Thus, \begin{align*} \sin \alpha + \sin \beta & = \frac{1}{2i} \left( e^{i \alpha} - e^{-i \alpha} + e^{i \beta} - e^{-i \beta} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) \left( e^{i \alpha} + e^{i \beta} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) \left( \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right)^2 - 2 e^{i \frac{\alpha + \beta}{2}} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) \left( \left( \frac{2 i \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)}{1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} }} \right)^2 - 2 e^{i \frac{\alpha + \beta}{2}} \right) \\ & = - \frac{1}{i} \left( e^{i \frac{\alpha + \beta}{2}} - e^{-i \frac{\alpha + \beta}{2}} \right) \left( \frac{2 \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)^2} {e^{i \frac{\alpha + \beta}{2}} + e^{-i \frac{\alpha + \beta}{2}} - 2 } + 1 \right) \\ & = \frac{167 \cdot 8}{193 \cdot 5 } . \end{align*}
Therefore, \begin{align*} {\rm Area} \ ABO_1CDO_2 & = {\rm Area} \ \triangle O_3 AB + {\rm Area} \ \triangle O_3 BO_1 + {\rm Area} \ \triangle O_3 O_1 C \\ & \quad + {\rm Area} \ \triangle O_3 C D + {\rm Area} \ \triangle O_3 D O_2 + {\rm Area} \ \triangle O_3 O_2 A \\ & = \frac{1}{2} r^2 \left( \sin \left( 2 \theta - \alpha - \beta \right) + \sin \alpha + \sin \alpha + \sin \left( 360^\circ - 2 \theta - \alpha - \beta \right) + \sin \beta + \sin \beta \right) \\ & = \frac{1}{2} r^2 \left( \sin \left( 2 \theta - \alpha - \beta \right) - \sin \left( 2 \theta + \alpha + \beta \right) + 2 \sin \alpha + 2 \sin \beta \right) \\ & = r^2 \left( - \cos 2 \theta \sin \left( \alpha + \beta \right) + \sin \alpha + \sin \beta \right) \\ & = r^2 \left( \left( 1 - 2 \cos^2 \theta \right) 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2} + \sin \alpha + \sin \beta \right) \\ & = \boxed{140} | null | 140 |