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a068f8d52bea29a125a622180ddb4e63 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_6 | Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$ . Find the degree measure of $\angle BAG$
[asy] pair A,B,C,D,E,F,G,H,I,O; O=(0,0); C=dir(90); B=dir(70); A=dir(50); D=dir(110); E=dir(130); draw(arc(O,1,50,130)); real x=2*sin(20*pi/180); F=x*dir(228)+C; G=x*dir(256)+C; H=x*dir(284)+C; I=x*dir(312)+C; draw(arc(C,x,200,340)); label("$A$",A,dir(0)); label("$B$",B,dir(75)); label("$C$",C,dir(90)); label("$D$",D,dir(105)); label("$E$",E,dir(180)); label("$F$",F,dir(225)); label("$G$",G,dir(260)); label("$H$",H,dir(280)); label("$I$",I,dir(315)); [/asy] | Let $\angle GAH = \varphi \implies \overset{\Large\frown} {GH} = 2\varphi \implies$ \[\overset{\Large\frown} {EF} = \overset{\Large\frown} {FG} = \overset{\Large\frown} {HI} = \overset{\Large\frown} {IA} = 2\varphi \implies\] \[\angle AGH = 2\varphi, \angle ACE = 10 \varphi.\]
\[BD||GH \implies \angle AJB = \angle AGH = 2 \varphi.\] \[\triangle AHG: \hspace{10mm} \angle AHG = \beta = 180^\circ – 3 \varphi.\] $\hspace{10mm} \triangle ABJ: \hspace{10mm} \angle BAG + \angle ABD = \alpha + \gamma = 180^\circ + 2 \varphi.$
Let arc $\overset{\Large\frown} {AB} = 2\psi \implies$
$\angle ACE = \frac {360^\circ – 8 \psi}{2}= 180^\circ – 4 \psi, \angle ABD = \gamma =\frac {360^\circ – 6 \psi}{2} =180^\circ – 3 \psi.$ $\gamma – \beta = 3(\varphi – \psi) = 12^\circ \implies \psi = \varphi – 4^\circ \implies 10 \varphi = 180^\circ – 4(\varphi – 4^\circ) \implies 14 \varphi = 196^\circ \implies \varphi = 14^\circ.$
Therefore $\gamma = 180^\circ – 3 \cdot (14^\circ – 4^\circ) = 150^\circ \implies \alpha = 180^\circ + 2 \cdot 14^\circ – 150^\circ = \boxed{058}.$ | null | 058 |
c8520f28d1ce610f27c7867bacbd4d16 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_7 | In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$ . Points $F$ and $G$ lie on $\overline{CE}$ , and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$ , respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$ , and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$
[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy] | Let us find the proportion of the side length of $KLMN$ and $FJGH$ . Let the side length of $KLMN=y$ and the side length of $FJGH=x$
[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy]
Now, examine $BC$ . We know $BC=BJ+JC$ , and triangles $\Delta BHJ$ and $\Delta JFC$ are similar to $\Delta EDC$ since they are $1-2-\sqrt{5}$ triangles. Thus, we can rewrite $BC$ in terms of the side length of $FJGH$ \[BJ=\frac{1}{\sqrt{5}}HJ=\frac{x}{\sqrt{5}}=\frac{x\sqrt{5}}{5}, JC=\frac{\sqrt{5}}{2}JF=\frac{x\sqrt{5}}{2}\Rightarrow BC=\frac{7x\sqrt{5}}{10}\]
Now examine $AB$ . We can express this length in terms of $x,y$ since $AB=AN+NH+HB$ . By using similar triangles as in the first part, we have \[AB=\frac{1}{\sqrt{5}}y+\frac{\sqrt{5}}{2}y+\frac{2}{\sqrt{5}}x\] \[AB=BC\Rightarrow \frac{7y\sqrt{5}}{10}+\frac{2x\sqrt{5}}{5}=\frac{7x\sqrt{5}}{10}\Rightarrow \frac{7y\sqrt{5}}{10}=\frac{3x\sqrt{5}}{10}\Rightarrow 7y=3x\]
Now, it is trivial to see that $[FJGH]=\left(\frac{x}{y}\right)^2[KLMN]=\left(\frac{7}{3}\right)^2\cdot 99=\boxed{539}.$ | null | 539 |
c8520f28d1ce610f27c7867bacbd4d16 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_7 | In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$ . Points $F$ and $G$ lie on $\overline{CE}$ , and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$ , respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$ , and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$
[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy] | [asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L); draw(P--E); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); label("$P$",P,dir(235)); [/asy]
We begin by denoting the length $ED$ $a$ , giving us $DC = 2a$ and $EC = a\sqrt5$ . Since angles $\angle DCE$ and $\angle FCJ$ are complementary, we have that $\triangle CDE \sim \triangle JFC$ (and similarly the rest of the triangles are $1-2-\sqrt5$ triangles). We let the sidelength of $FGHJ$ be $b$ , giving us:
\[JC = \sqrt5 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}\] and \[BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}\]
Since $BC = CJ + BJ$
\[2a = \frac{b\sqrt 5}{2} + \frac{b}{\sqrt5}\]
Solving for $b$ in terms of $a$ yields \[b = \frac{4a\sqrt5}{7}\]
We now use the given that $[KLMN] = 99$ , implying that $KL = LM = MN = NK = 3\sqrt{11}$ . We also draw the perpendicular from $E$ to $ML$ and label the point of intersection $P$ as in the diagram at the top
This gives that \[AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}\] and \[ME = \sqrt5 \cdot MP = \sqrt5 \cdot \frac{EP}{2} = \sqrt5 \cdot \frac{LG}{2} = \sqrt5 \cdot \frac{HG - HK - KL}{2} = \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2}\]
Since $AE$ $AM + ME$ , we get
\[2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2} = a\]
\[\Rightarrow 12\sqrt{11} + 5(\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}) = 2\sqrt5a\]
\[\Rightarrow \frac{-21}{2}\sqrt{11} + \frac{20a\sqrt5}{7} = 2\sqrt5a\]
\[\Rightarrow -21\sqrt{11} = 2\sqrt5a\frac{14 - 20}{7}\]
\[\Rightarrow \frac{49\sqrt{11}}{4} = \sqrt5a\]
\[\Rightarrow 7\sqrt{11} = \frac{4a\sqrt{5}}{7}\]
So our final answer is $(7\sqrt{11})^2 = \boxed{539}$ | null | 539 |
c8520f28d1ce610f27c7867bacbd4d16 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_7 | In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$ . Points $F$ and $G$ lie on $\overline{CE}$ , and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$ , respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$ , and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$
[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy] | This is a relatively quick solution but a fakesolve. We see that with a ruler, $KL = \frac{3}{2}$ cm and $HG = \frac{7}{2}$ cm. Thus if $KL$ corresponds with an area of $99$ , then $HG$ $FGHJ$ 's area) would correspond with $99*(\frac{7}{3})^2 = \boxed{539}$ - aops5234 | null | 539 |
97416c03179ecb6c48feb7a40f5b8816 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | You know whatever $n$ is, it has to have 3 digits, because if it had only two, the maximum of $s(n)$ is 18.
Now let $n=100a_2+10a_1+a_0$
So first we know, $a_2+a_1+a_0=20$ . Okay now we have to split into cases based on which digit gets carried. This meaning, when you add a 3 digit number to 864, we have to know when to carry the digits. Note that if you don't understand any of the steps I take, just try adding any 3-digit number to 864 regularly (using the old-fashioned "put one number over the other" method, not mental calculation), and observe what you do at each step.
(1) $\textcolor{red}{*}$ None of the digits get carried over to the next space:
So this means $a_2<2, a_1<4$ and $a_0<6$ . So
$s(864+n)=(8+a_2)+(6+a_1)+(4+a_0)=38$ So it doesn't work. Now:
(2) $a_2+8$ is the only one that carries over
So this means $a_2>1, a_1<4$ and $a_0<6$ . So
$s(864+n)=1+(8+a_2-10)+(6+a_1)+(a_0+4)=29$
(3) $\textcolor{red}{*}$ $a_0+4$ is the only one that carries over.
So
$s(864+n)=(8+a_2)+(6+a_1+1)+(4+a_0-10)=29$
(4)The first and second digit carry over (but not the third)
$s(864+n)=1+(8+a_2-10+1)+(6+a_1-10)+(4+a_0)=20$
Aha! This case works but we still have to make sure it's possible for $a_2+a_1+a_0=20$ (We assumed this is true, so we have to find a number that works.) Since only the second and first digit carry over, $a_2>0, a_1>3$ and $a_0<6$ . The smallest value we can get with this is 695. Let's see if we can find a smaller one:
(5)The first and third digit carry over (but not the second)
$s(864+n)=1+(8+a_2-10)+(7+a_1)+(4+a_0-10)=20$
The largest value for the middle digit is 2, so the other digits have to be both 9's. So the smallest possible value is 929
(6) All the digits carry over
$s(864+n)=1+(9+a_2-10)+(7+a_1-10)+(4+a_0-10)=\text{Way less than 20}$
So the answer is $\boxed{695}$ which after a quick test, does indeed work. | null | 695 |
97416c03179ecb6c48feb7a40f5b8816 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | First, it is easy to verify that $695$ works and that no other numbers beginning with the digit 6 work (i.e. $686, 677, 668, 659$ do not work).
Suppose by contradiction that there is a smaller valid $n$ , where the leading digit of the three-digit number $n$ is 5 or less. (Two-digit $n$ obviously do not work because 9 + 9 < 20.) Clearly $n > 200$ because the smallest three-digit number whose digits sum to 20 is $299$ . Also, because the second digit is at most 9, the units digit is at least 6, which means that the addition $N = n + 864$ regroups in the ones place. Then the units digit of $N$ is clearly less than 4. But as $1000 < 200 + 864 < N < 600 + 864 = 1464$ , the sum of the thousands digit and the hundredth digit is at most 5. Because the second digit is at most 9, the sum of the digits of $N$ is at most $5 + 9 + 4 < 20$ , contradiction. Hence $\boxed{695}$ is the answer. | null | 695 |
97416c03179ecb6c48feb7a40f5b8816 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | First of all, notice that the smallest $n$ with $s(n) = 20$ is $299$ . Also, if $s(n + 864) = 20$ $s(n - 136) = 19$ (because subtracting $1000$ from the number removes the $1$ in the thousands place). After checking $s(n - 136)$ for various $n$ with $s(n) = 20$ , we see that we need to have a carry when subtracting $136$ . To have this, we must either have a $2$ in the tens place or a $5$ in the units place. The minimum $n$ for the former is $929$ , and for the latter it is $695$ . We check and see that $s(695-136) = s(559) = 19$ , so our answer is $\boxed{695}$ | null | 695 |
97416c03179ecb6c48feb7a40f5b8816 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | Observation (Lemma)Â : If r is the number of regroups in the addition of n+k, $S(n+k) = S(n)+S(k)-9r$
Proof : When you add two numbers, and you do a carry, you are taking away 10 from 1 column, and adding 1 to another column, giving a net loss of 9 to the total.
Thus, we can see that we need to regroup exactly twice when we add 864. And, the lowest possible n is 299, so let's start from there.
299 gets three regroups, so we are going to need to take away from digits, and dump the excess in the hundred's place, since the hundreds are going to regroup anyways.
So, if we take away from the tens digit, we need to take away until we get 2 in the tens digit(since the ones will regroup). So, we get the number 929, which works (929+864 = 1793), but is not the smallest.
If we take away from the ones digit, we only have to take away 4, turning the unit's place to 5. 5+4 is 9, so it won't regroup. Dump the ones into the hundred's place, and we get the number $\boxed{695}$ | null | 695 |
97416c03179ecb6c48feb7a40f5b8816 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | Although this solution doesn't directly solve the problem, it greatly hastens the bashing process.
Call the three digits a, b, and c.
When you add each of 8, 6, and 4 to a, b, and c the resultant will either get smaller or larger, depending on the original number.
e.g. If c is 7, then adding 4 will reduce the 7 to a 1, whilst leaving a one for b.
If c is 3, then adding 4 will simply add four to the total, and make the 3 a 7.
Each of 8, 6, and 4 all can reduce the original number by a certain amount and can increase the original number by a certain amount.
8 can reduce by 2 for all numbers greater than 1, 6 can reduce numbers by 4, and 4 can reduce numbers by 6.
Possibilities:
Also, realize that if the number is reduced, then a one will be carried to the following decimal place on the left, consequently reducing that amount they reduce. It's like a puzzle! Within no time you should find that if you add 4 to c, subtract 4 from b, subtract 1 from a, and leave a 1 in the thousands place, the total is equated to zero. This is optimal because most of the addition is kept to the left, where the effect to real value is less. (e.g. 299 is smaller than 992)
Now you have +1, -1, -4, +4 in the decimals, and a VERY fast trial and error gives $\boxed{695}.$ | null | 695 |
97416c03179ecb6c48feb7a40f5b8816 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | First, note that to compute $s(m+n)$ (for any positive integers $m$ and $n$ ), one can simply find the sum of $s(m)$ and $s(n)$ minus 9 times the number of times one regroups when adding $m$ to $n$ . One can see why this is by noticing that if one were to "forget" to regroup, and leave, say, a 10 in the ones' place, the sum of the digits would be 9 higher than if one did regroup. Anyway, one can see that the smallest 3-digit number (on AIME, all the answers are integers from 0 to 999) whose digits sum to 20 is 299. If we add this number to 864, we have to regroup 3 times, so $s(299+864)$ will be $9=9\cdot3-(8+6+4)$ smaller than $s(299)$ . We want this difference to be 0, so we need to find a way to only regroup two times.
We now notice that regrouping the hundreds is inevitable, so we must either prevent regrouping the ones or the tens. Preventing regrouping the tens would require moving many of the tens to the hundreds' place (the ones' place is already full), which is bad when we are trying to minimize the number, but preventing regrouping the ones requires moving fewer ones to the hundreds' place.
We find that to preventing regrouping the ones, the ones' place of our number must be at most 5 (a larger number would sum to ten when added to 4). Because we want to minimize the number of ones we move to the hundreds' place, we leave exactly 5 by moving four ones to the hundreds' place: $299\to\boxed{695}$ | null | 695 |
97416c03179ecb6c48feb7a40f5b8816 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | Bashing out modulo $9$ and getting lucky we get that if the $8$ and $6$ carry over when adding $n$ and $864$ , that $100a+10b+c \equiv 1+100(a-1)+10(b-4)+c+4 \pmod{9}$ such that $n=100a+10b+c$ and after maximizing $b$ and $c$ such that $c<6$ to not make the $4$ carry over to minimize $a$ we get that $\boxed{695}$ is our answer after confirming there are no lesser solutions. | null | 695 |
d5871247f23377a1b9bf18243cbf82a3 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_9 | Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$ . Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$ . Find the number of such sequences for which $a_n=0$ for some $n$ | Let $a_1=x, a_2=y, a_3=z$ . First note that if any absolute value equals 0, then $a_n=0$ .
Also note that if at any position, $a_n=a_{n-1}$ , then $a_{n+2}=0$ .
Then, if any absolute value equals 1, then $a_n=0$ .
Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the criteria.
Assume that to be the only way the criteria is met.
To prove, let $|y-x|>1$ , and $|z-y|>1$ . Then, $a_4 \ge 2z$ $a_5 \ge 4z$ , and $a_6 \ge 4z$ .
However, since the minimum values of $a_5$ and $a_6$ are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be $z=1$ $|y-x|=2$ . Again assume that any other scenario will not meet criteria.
To prove, divide the other scenarios into two cases: $z>1$ $|y-x|>1$ , and $|z-y|>1$ ; and $z=1$ $|y-x|>2$ , and $|z-y|>1$ .
For the first one, $a_4 \ge 2z$ $a_5 \ge 4z$ $a_6 \ge 8z$ , and $a_7 \ge 16z$ , by which point we see that this function diverges.
For the second one, $a_4 \ge 3$ $a_5 \ge 6$ $a_6 \ge 18$ , and $a_7 \ge 54$ , by which point we see that this function diverges.
Therefore, the only scenarios where $a_n=0$ is when any of the following are met: $|y-x|<2$ (280 options) $|z-y|<2$ (280 options, 80 of which coincide with option 1) $z=1$ $|y-x|=2$ . (16 options, 2 of which coincide with either option 1 or option 2)
Adding the total number of such ordered triples yields $280+280-80+16-2=\boxed{494}$ | null | 494 |
d5871247f23377a1b9bf18243cbf82a3 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_9 | Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$ . Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$ . Find the number of such sequences for which $a_n=0$ for some $n$ | Note that the only way for a $0$ to be produced at $a_n$ is if either $a_{n-1} = 0$ or $a_{n-2} = a_{n-3}$ . Since the first one will eventually get to the first three assuming that there is no $a_{n-2} = a_{n-3}$ for any $n$ , that is not possible because $a_1 , a_2 , a_3 >= 1$ . Therefore, we need $a_{n - 2} = a_{n - 3}$
If $2$ consecutive numbers out of $a_1 , a_2 , a_3$ are equal, then those cases work( $a_1$ and $a_2$ or $a_2$ and $a_3$ $\textbf{NOT}$ $a_1$ and $a_3$ ). This is simply $10 \cdot 10 + 10 \cdot 10 - 10 = 190$ by PIE.
Now, note that if any of the first three numbers have difference of $1$ , we have another working case. First, we calculate how many there are given exactly one of $a_1,a_2$ or $a_2,a_3$ have difference $1$ . Given $3$ numbers such that the first $2$ have difference $1$ , exactly $4$ permutations work(assuming the numbers are $x,y,$ and $z$ such that $|x-y| = 1$ ): $x,y,z$ $y,x,z$ $z,x,y$ ; and $z,y,x$ . If the two consecutive numbers are $1$ and $2$ , then the last number has $7$ possiblities: $4,5,6,\cdots , 10$ . This is symmetric for $9$ and $10$ . If the consecutive numbers are $(2,3),\cdots , (8,9)$ , there are $6$ possibilities( $10$ minus the numbers themselves and the numbers directly above and below). Note that we are not counting any cases already counted in the first case. Therefore, this case gives you $4(7 + 6 * 7 + 7) = 224$ . Now we consider the case that there both adjacent $a$ s have increments of $1$ \[+1 , +1 \rightarrow 8\] \[-1 , -1 \rightarrow 8\] \[+1 , -1 \rightarrow 9\] \[-1 , +1 \rightarrow 9\] Therefore this gives $224 + 8 + 8 + 9 + 9 = 258$ . However, note that we have to add the case where you have $3$ consecutive numbers in arrangement such that only $2$ consecutive numbers have difference $1$ . For example, $1,3,2$ is one such triple. There are $8$ triples of consecutive numbers and $4$ ways to arrange each one(e.x: $(1,3,2);(3,1,2);(2,1,3);(2,3,1)$ ). This adds on 32 working cases, so this case gives $258 + 32 = 290$
Note that there is an ultraspecial(Yes, I know that's not a word) case where we generate a pair of $a_i$ that have difference one. This can only happen if $a_3 = 1$ and $a_1$ and $a_2$ have difference $2$ . This contributes $14$ cases( $2 * 8$ and then subtract $2$ because of the cases $3,1,1$ and $4,2,1$ ).
Therefore, our answer is $190 + 290 + 14 = \boxed{494}$ .
Solution by hyxue | null | 494 |
f57a8ef2b7ee8506bd536526a0dc46d6 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_10 | Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$ | Let $f(x)$ $ax^3+bx^2+cx+d$ .
Since $f(x)$ is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up.
By drawing a coordinate axis, and two lines representing $12$ and $-12$ , it is easy to see that $f(1)=f(5)=f(6)$ , and $f(2)=f(3)=f(7)$ ; otherwise more bends would be required in the graph. Since only the absolute value of $f(0)$ is required, there is no loss of generalization by stating that $f(1)=12$ , and $f(2)=-12$ . This provides the following system of equations. \[a + b + c + d = 12\] \[8a + 4b + 2c + d = -12\] \[27a + 9b + 3c + d = -12\] \[125a + 25b + 5c + d = 12\] \[216a + 36b + 6c + d = 12\] \[343a + 49b + 7c + d = -12\] Using any four of these functions as a system of equations yields $d = |f(0)| = \boxed{072}$ | null | 072 |
f57a8ef2b7ee8506bd536526a0dc46d6 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_10 | Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$ | By drawing the function, and similar to Solution 1, WLOG let $f(1)=f(5)=f(6)=12$ . Then, $f(2)=f(3)=f(7)$ . Set $g(x)+12=f(x)$ . Then the roots of $g(x)$ are $1,5,6$ . So, $g(x)=a(x-1)(x-5)(x-6)$ . Plug in $x=2$ to find a. We know \[-24=-12-12=f(2)-12=g(2)=a(1)(-3)(-4)=12a\] .
So, $a=-2$ . Thus, $f(x)=g(x)+12=-2(x-1)(x-5)(x-6)+12$ , and then $|f(0)|=60+12=\boxed{072}$ | null | 072 |
f57a8ef2b7ee8506bd536526a0dc46d6 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_10 | Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$ | Without loss of generality, let $f(1) = 12$ . (If $f(1) = -12$ , then take $-f(x)$ as the polynomial, which leaves $|f(0)|$ unchanged.) Because $f$ is third-degree, write \[f(x) - 12 = a(x - 1)(x - b)(x - c)\] \[f(x) + 12 = a(x - d)(x - e)(x - f)\] where $\{b, c, d, e, f \}$ clearly must be a permutation of $\{2, 3, 5, 6, 7\}$ from the given condition. Thus $b + c + d + e + f = 2 + 3 + 5 + 6 + 7 = 23.$ However, subtracting the two equations gives $-24 = a[(x - 1)(x - b)(x - c) - (x - d)(x - e)(x - f)]$ , so comparing $x^2$ coefficients gives $1 + b + c = d + e + f$ and thus both values equal to $\dfrac{24}{2} = 12$ . As a result, $\{b, c \} = \{5, 6 \}$ . As a result, $-24 = a (12)$ and so $a = -2$ . Now, we easily deduce that $f(0) = (-2) \cdot (-1) \cdot (-5) \cdot (-6) + 12 = 72,$ and so removing the without loss of generality gives $|f(0)| = \boxed{072}$ , which is our answer. | null | 072 |
f57a8ef2b7ee8506bd536526a0dc46d6 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_10 | Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$ | The following solution is similar to solution 3, but assumes nothing. Let $g(x)=(f(x))^2-144$ . Since $f$ has degree 3, $g$ has degree 6 and has roots 1,2,3,5,6, and 7. Therefore, $g(x)=k(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)$ for some $k$ . Hence $|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}$ . Note that $g(x)=(f(x)+12)(f(x)-12)$ . Since $f$ has degree 3, so do $f(x)+12$ and $f(x)-12$ ; and both have the same leading coefficient. Hence $f(x)+12=a(x-q)(x-r)(x-s)$ and $f(x)-12=a(x-t)(x-u)(x-v)$ for some $a\neq 0$ (else $f$ is not cubic) where $\{q,r,s,t,u,v\}$ is the same as the set $\{1,2,3,5,6,7\}$ . Subtracting the second equation from the first, expanding, and collecting like terms, we have that \[24=a((t+u+v-(q+r+s))x^2-a(tu+uv+tv-(qr+qs+rs))x+a(tuv-qrs)\] which must hold for all $x$ . Since $a\neq 0$ we have that (1) $t+u+v=q+r+s$ , (2) $tu+uv+tv=qr+qs+rs$ and (3) $a(tuv-qrs)=24$ . Since $q+r+s+t+u+v$ is the sum of 1,2,3,5,6, and 7, we have $q+r+s+t+u+v=24$ so that by (1) we have $q+r+s=12$ and $t+u+v=12$ . We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be $\geq 12$ with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be $\{2,3,7\}$ and $\{1,5,6\}$ . Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that $|a|=\frac{24}{|tuv-qrs|}=\frac{24}{12}=2$ . Since $a$ is the leading coefficient of $f(x)$ , the leading coefficient of $(f(x))^2$ is $a^2=|a|^2=2^2=4$ . Thus the leading coefficient of $g(x)$ is 4, i.e. $k=4$ . Then from earlier, $|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}=\sqrt{1260\cdot4+144}=\sqrt{5184}=72$ so that the answer is $\boxed{072}$ | null | 072 |
f57a8ef2b7ee8506bd536526a0dc46d6 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_10 | Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$ | Express $f(x)$ in terms of powers of $(x-4)$ \[f(x) = a(x-4)^3 + b(x-4)^2 + c(x-4) + d\] By the same argument as in the first Solution, we see that $f(x)$ is an odd function about the line $x=4$ , so its coefficients $b$ and $d$ are 0. From there it is relatively simple to solve $f(2)=f(3)=-12$ (as in the above solution, but with a smaller system of equations): \[a(1)^3 + c(1) = -12\] \[a(2)^3 + c(2) = -12\] $a=2$ and $c=-14$ \[|f(0)| = |2(-4)^3 - 14(-4)| = \boxed{072}\] | null | 072 |
f57a8ef2b7ee8506bd536526a0dc46d6 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_10 | Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$ | Because a cubic must come in a "wave form" with two points of inflection, we can see that $f(1)=f(5)=f(6)$ , and $f(2)=f(3)=f(7)$ . By symmetry, $f(4)=0$ . Now, WLOG let $f(1)=12$ , and $f(2)=f(3)=-12$ . Then, we can use finite differences to get that the third (constant) difference is $12$ , and therefore $f(0)=12+(24+(24+12))=\boxed{072}$ | null | 072 |
f57a8ef2b7ee8506bd536526a0dc46d6 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_10 | Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$ | We can rewrite our function as two different cubics, $f(x)=k(x-a)(x-b)(x-c)\pm12=k(x-d)(x-e)(x-f)\mp12$ . Note that $k$ is the same in both because they must be equal, so their leading terms must be. We must then, following Vieta's, choose our roots such that $a+b+c=d+e+f$ and verify that the other Vieta's formulas hold. Additionally, a cubic must only cross the x-axis thrice, restricting our choices for roots further. Choosing $a=1$ $b=5$ $c=6$ $d=2$ $e=3$ $f=7$ yields: \[kx^3-12kx^2+41kx-30k\pm12=kx^3-12kx^2+41kx-42k\mp12\] For the constant terms to have a difference of 24 ( $|\pm12-\mp12|$ ), we must have $k=\pm2$ , so the constant term of our polynomial is $\pm72$ , the absolute value of which is $\boxed{072}$ . -- Solution by eiis1000 | null | 072 |
84efcc6d3276d31d3ac0c320f018c84f | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_11 | Triangle $ABC$ has positive integer side lengths with $AB=AC$ . Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$ . Suppose $BI=8$ . Find the smallest possible perimeter of $\triangle ABC$ | Let $D$ be the midpoint of $\overline{BC}$ . Then by SAS Congruence, $\triangle ABD \cong \triangle ACD$ , so $\angle ADB = \angle ADC = 90^o$
Now let $BD=y$ $AB=x$ , and $\angle IBD = \dfrac{\angle ABD}{2} = \theta$
Then $\mathrm{cos}{(\theta)} = \dfrac{y}{8}$
and $\mathrm{cos}{(2\theta)} = \dfrac{y}{x} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{y^2-32}{32}$
Cross-multiplying yields $32y = x(y^2-32)$
Since $x,y>0$ $y^2-32$ must be positive, so $y > 5.5$
Additionally, since $\triangle IBD$ has hypotenuse $\overline{IB}$ of length $8$ $BD=y < 8$
Therefore, given that $BC=2y$ is an integer, the only possible values for $y$ are $6$ $6.5$ $7$ , and $7.5$
However, only one of these values, $y=6$ , yields an integral value for $AB=x$ , so we conclude that $y=6$ and $x=\dfrac{32(6)}{(6)^2-32}=48$
Thus the perimeter of $\triangle ABC$ must be $2(x+y) = \boxed{108}$ | null | 108 |
84efcc6d3276d31d3ac0c320f018c84f | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_11 | Triangle $ABC$ has positive integer side lengths with $AB=AC$ . Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$ . Suppose $BI=8$ . Find the smallest possible perimeter of $\triangle ABC$ | Angle bisectors motivate trig bash.
Define angle $IBC = x$ . Foot of perpendicular from $I$ to $BC$ is point $P$ $\overline{BC} = 2\overline{BP} = 2(8\cos(x)) = N$ , where $N$ is an integer. Thus, $\cos(x) = \frac{N}{16}$ . Via double angle, we calculate $\overline{AB}$ to be $\frac{8\cos(x)}{2\cos(x)^2 - 1} = \frac{64N}{N^2 - 128}$ . This is to be an integer. We can bound $N$ now, as $N > 11$ to avoid negative values and $N < 16$ due to triangle inequality. Testing, $N = 12$ works, giving $\overline{AB} = 48, \overline{BC} = 12$ .
Our answer is $2 * 48 + 12 = \boxed{108}$ .
- whatRthose | null | 108 |
84efcc6d3276d31d3ac0c320f018c84f | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_11 | Triangle $ABC$ has positive integer side lengths with $AB=AC$ . Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$ . Suppose $BI=8$ . Find the smallest possible perimeter of $\triangle ABC$ | Let $M$ be midpoint $BC, BM = x, AB = y, \angle IBM = \alpha.$
$BI$ is the bisector of $\angle ABM$ in $\triangle ABM.$ $BI = \frac {2 xy \cos \alpha}{x+y} = 8, \cos \alpha = \frac {x}{8} \implies \frac {x^2 y}{x+y} = 32.$ \[y = \frac {32 x} {x^2 - 32}.\] $BC = 2x$ is integer, $5.5^2 < 32 \implies x \ge 6.$ $BM < BI \implies x =\{ 6, 6.5, 7, 7.5 \}.$
If $x > 6$ then $y$ is not integer. \[x = 6 \implies y = 48 \implies 2(x+y) = \boxed{108}.\] vladimir.shelomovskii@gmail.com, vvsss | null | 108 |
b7deca4eeb02cf3f6bd724dd5a108c02 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_12 | Consider all 1000-element subsets of the set $\{1, 2, 3, ... , 2015\}$ . From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Let $M$ be the desired mean. Then because $\dbinom{2015}{1000}$ subsets have 1000 elements and $\dbinom{2015 - i}{999}$ have $i$ as their least element, \begin{align*} \binom{2015}{1000} M &= 1 \cdot \binom{2014}{999} + 2 \cdot \binom{2013}{999} + \dots + 1016 \cdot \binom{999}{999} \\ &= \binom{2014}{999} + \binom{2013}{999} + \dots + \binom{999}{999} \\ & + \binom{2013}{999} + \binom{2012}{999} + \dots + \binom{999}{999} \\ & \dots \\ & + \binom{999}{999} \\ &= \binom{2015}{1000} + \binom{2014}{1000} + \dots + \binom{1000}{1000} \\ &= \binom{2016}{1001}. \end{align*} Using the definition of binomial coefficient and the identity $n! = n \cdot (n-1)!$ , we deduce that \[M = \frac{2016}{1001} = \frac{288}{143}.\] The answer is $\boxed{431}.$ | null | 431 |
b7deca4eeb02cf3f6bd724dd5a108c02 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_12 | Consider all 1000-element subsets of the set $\{1, 2, 3, ... , 2015\}$ . From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Each 1000-element subset $\left\{ a_1, a_2,a_3,...,a_{1000}\right\}$ of $\left\{1,2,3,...,2015\right\}$ with $a_1<a_2<a_3<...<a_{1000}$ contributes $a_1$ to the sum of the least element of each subset. Now, consider the set $\left\{a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ . There are $a_1$ ways to choose a positive integer $k$ such that $k<a_1+1<a_2+1,a_3+1<...<a_{1000}+1$ $k$ can be anything from $1$ to $a_1$ inclusive). Thus, the number of ways to choose the set $\left\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ is equal to the sum. But choosing a set $\left\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ is the same as choosing a 1001-element subset from $\left\{1,2,3,...,2016\right\}$
Thus, the average is $\frac{\binom{2016}{1001}}{\binom{2015}{1000}}=\frac{2016}{1001}=\frac{288}{143}$ . Our answer is $p+q=288+143=\boxed{431}$ | null | 431 |
b7deca4eeb02cf3f6bd724dd5a108c02 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_12 | Consider all 1000-element subsets of the set $\{1, 2, 3, ... , 2015\}$ . From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Let $p$ be the size of the large set and $q$ be the size of the subset (i.e. in this problem, $p = 2015$ and $q = 1000$ ). We can easily find the answers for smaller values of $p$ and $q$
For $p = 2$ and $q = 2$ , the answer is $1$
For $p = 3$ and $q = 2$ , the answer is $\frac43$
For $p = 4$ and $q = 2$ , the answer is $\frac53$
For $p = 3$ and $q = 3$ , the answer is $1$
For $p = 4$ and $q = 3$ , the answer is $\frac54$
For $p = 5$ and $q = 3$ , the answer is $\frac32$
At this point, we can see a pattern: our desired answer is always $\frac{p+1}{q+1}$ . Plugging in $p = 2015$ and $q = 1000$ , the answer is $\frac{2016}{1001}=\frac{288}{143}$ , so $288 + 143 = \boxed{431}$ | null | 431 |
360b96ede298332a689db896687f1e16 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_13 | With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$ , where $m$ and $n$ are integers greater than 1. Find $m+n$ | Let $x = \cos 1^\circ + i \sin 1^\circ$ . Then from the identity \[\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},\] we deduce that (taking absolute values and noticing $|x| = 1$ \[|2\sin 1| = |x^2 - 1|.\] But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ - z)$ , if we let our product be $M$ then \[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ\] \[= \frac{1}{2^{90}} |x^2 - 1| |x^6 - 1| |x^{10} - 1| \dots |x^{354} - 1| |x^{358} - 1|\] because $\sin$ is positive in the first and second quadrants. Now, notice that $x^2, x^6, x^{10}, \dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z - x^2)(z - x^6)\dots (z - x^{358}) = z^{90} + 1$ , and so \[\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.\] It is easy to see that $M = 2^{89}$ and that our answer is $2 + 89 = \boxed{91}$ | null | 91 |
360b96ede298332a689db896687f1e16 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_13 | With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$ , where $m$ and $n$ are integers greater than 1. Find $m+n$ | Let $p=\sin1\sin3\sin5...\sin89$
\[p=\sqrt{\sin1\sin3\sin5...\sin177\sin179}\]
\[=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{\sin2\sin4\sin6\sin8...\sin176\sin178}}\]
\[=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{(2\sin1\cos1)\cdot(2\sin2\cos2)\cdot(2\sin3\cos3)\cdot....\cdot(2\sin89\cos89)}}\]
\[=\sqrt{\frac{1}{2^{89}}\frac{\sin90\sin91\sin92\sin93...\sin177\sin178\sin179}{\cos1\cos2\cos3\cos4...\cos89}}\]
$=\sqrt{\frac{1}{2^{89}}}$ because of the identity $\sin(90+x)=\cos(x)$
we want $\frac{1}{p^2}=2^{89}$
Thus the answer is $2+89=\boxed{091}$ | null | 091 |
360b96ede298332a689db896687f1e16 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_13 | With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$ , where $m$ and $n$ are integers greater than 1. Find $m+n$ | Similar to Solution $2$ , so we use $\sin{2\theta}=2\sin\theta\cos\theta$ and we find that: \begin{align*}\sin(4)\sin(8)\sin(12)\sin(16)\cdots\sin(84)\sin(88)&=(2\sin(2)\cos(2))(2\sin(4)\cos(4))(2\sin(6)\cos(6))(2\sin(8)\cos(8))\cdots(2\sin(42)\cos(42))(2\sin(44)\cos(44))\\ &=(2\sin(2)\sin(88))(2\sin(4))\sin(86))(2\sin(6)\sin(84))(2\sin(8)\sin(82))\cdots(2\sin(42)\sin(48))(2\sin(44)\sin(46))\\ &=2^{22}(\sin(2)\sin(88)\sin(4)\sin(86)\sin(6)\sin(84)\sin(8)\sin(82)\cdots\sin(42)\sin(48)\sin(44)\sin(46))\\ &=2^{22}(\sin(2)\sin(4)\sin(6)\sin(8)\cdots\sin(82)\sin(84)\sin(86)\sin(88))\end{align*} Now we can cancel the sines of the multiples of $4$ \[1=2^{22}(\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86))\] So $\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86)=2^{-22}$ and we can apply the double-angle formula again: \begin{align*}2^{-22}&=(\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86)\\ &=(2\sin(1)\cos(1))(2\sin(3)\cos(3))(2\sin(5)\cos(5))(2\sin(7)\cos(7))\cdots(2\sin(41)\cos(41))(2\sin(43)\cos(43))\\ &=(2\sin(1)\sin(89))(2\sin(3)\sin(87))(2\sin(5)\sin(85))(2\sin(7)\sin(87))\cdots(2\sin(41)\sin(49))(2\sin(43)\sin(47))\\ &=2^{22}(\sin(1)\sin(89)\sin(3)\sin(87)\sin(5)\sin(85)\sin(7)\sin(83)\cdots\sin(41)\sin(49)\sin(43)\sin(47))\\ &=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(41)\sin(43))(\sin(47)\sin(49)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\end{align*} Of course, $\sin(45)=2^{-\frac{1}{2}}$ is missing, so we multiply it to both sides: \[2^{-22}\sin(45)=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(41)\sin(43))(\sin(45))(\sin(47)\sin(49)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\] \[\left(2^{-22}\right)\left(2^{-\frac{1}{2}}\right)=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\] \[2^{-\frac{45}{2}}=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\] Now isolate the product of the sines: \[\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89)=2^{-\frac{89}{2}}\] And the product of the squares of the cosecants as asked for by the problem is the square of the inverse of this number: \[\csc^2(1)\csc^2(3)\csc^2(5)\csc^2(7)\cdots\csc^2(83)\csc^2(85)\csc^2(87)\csc^2(89)=\left(\frac{1}{2^{-\frac{89}{2}}}\right)^2=\left(2^{\frac{89}{2}}\right)^2=2^{89}\] The answer is therefore $m+n=(2)+(89)=\boxed{091}$ | null | 091 |
360b96ede298332a689db896687f1e16 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_13 | With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$ , where $m$ and $n$ are integers greater than 1. Find $m+n$ | Let $p=\prod_{k=1}^{45} \csc^2(2k-1)^\circ$
Then, $\sqrt{\frac{1}{p}}=\prod_{k=1}^{45} \sin(2k-1)^\circ$
Since $\sin\theta=\cos(90^{\circ}-\theta)$ , we can multiply both sides by $\frac{\sqrt{2}}{2}$ to get $\sqrt{\frac{1}{2p}}=\prod_{k=1}^{23} \sin(2k-1)^\circ\cos(2k-1)^\circ$
Using the double-angle identity $\sin2\theta=2\sin\theta\cos\theta$ , we get $\sqrt{\frac{1}{2p}}=\frac{1}{2^{23}}\prod_{k=1}^{23} \sin(4k-2)^\circ$
Note that the right-hand side is equal to $\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \sin(4k)^\circ$ , which is equal to $\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} 2\sin(2k)^\circ\cos(2k)^\circ$ , again, from using our double-angle identity.
Putting this back into our equation and simplifying gives us $\sqrt{\frac{1}{2p}}=\frac{1}{2^{45}}\prod_{k=23}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \cos(2k)^\circ$
Using the fact that $\sin\theta=\cos(90^{\circ}-\theta)$ again, our equation simplifies to $\sqrt{\frac{1}{2p}}=\frac{\sin90^\circ}{2^{45}}$ , and since $\sin90^\circ=1$ , it follows that $2p = 2^{90}$ , which implies $p=2^{89}$ . Thus, $m+n=2+89=\boxed{091}$ | null | 091 |
360b96ede298332a689db896687f1e16 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_13 | With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$ , where $m$ and $n$ are integers greater than 1. Find $m+n$ | Recall that $\sin\alpha\cdot \sin(60^{\circ}-\alpha)\cdot \sin(60^{\circ}+\alpha)=\frac{1}{4}\cdot \sin3\alpha$ Since it is in csc, we can write in sin and then take reciprocal.
We can group them by threes, $P=(\sin1^{\circ}\cdot \sin59^{\circ}\cdot \sin61^{\circ})\cdots(\sin29^{\circ}\cdot \sin31^{\circ}\cdot \sin89^{\circ})$ . Thus \begin{align*} P &=\frac{1}{4^{15}}\cdot \sin3^{\circ}\cdot \sin9^{\circ}\cdots\sin87^{\circ}\\ &=\frac{1}{4^{20}}\cdot \sin9^{\circ}\cdot \sin27^{\circ}\cdot \sin45^{\circ}\cdot \sin63^{\circ}\cdot \sin81^{\circ}\\ &=\frac{1}{4^{20}}\cdot \frac{\sqrt{2}}{2}\cdot \sin9^{\circ}\cdot \cos9^{\circ}\cdot \sin27^{\circ}\cdot \cos27^{\circ}\\ &=\frac{1}{4^{21}}\cdot \frac{\sqrt{2}}{2}\cdot \sin18^{\circ}\cdot \cos36^{\circ}=\frac{\sqrt{2}}{2^{45}} \end{align*} So we take reciprocal, $\frac 1P=2^{\frac{89}{2}}$ , the desired answer is $\frac{1}{P^2}=2^{89}$ leads to answer $\boxed{091}$ | null | 091 |
360b96ede298332a689db896687f1e16 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_13 | With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$ , where $m$ and $n$ are integers greater than 1. Find $m+n$ | We have
\[\prod_{k=1}^{45} \csc^2(2k-1)^\circ = \left(\frac{1}{\sin1^\circ \cdot \sin3^\circ \cdots \sin89^\circ}\right)^2.\]
Multiplying by $\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}$ gives
\[\left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin1^\circ \sin2^\circ \cdot \sin3^\circ \cdots \sin88^\circ \cdot \sin89^\circ}\right)^2\]
\[= \left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin1^\circ \sin2^\circ \cdot \sin3^\circ \cdots \sin 45^\circ \cdot \cos 44^\circ \cdot \cos 43^\circ \cdots \cos1^\circ}\right)^2.\]
Using $\sin\alpha \cos\alpha = \frac{1}{2}\sin{2\alpha}$ gives
\[\left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\frac{1}{2} \sin2^\circ \cdot \frac{1}{2} \sin4^\circ \cdots \frac{1}{2} \sin88^\circ \cdot \sin45^\circ}\right) ^2\]
\[= \left(\frac{1}{(\frac{1}{2})^{44} \cdot \frac{\sqrt{2}}{2}}\right)^2\]
\[= 2^{89}.\]
Thus, the answer is $2+89 = \boxed{091}.$ | null | 091 |
df7668d6d5f214ea2f5c1acbf9e1b049 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_14 | For each integer $n \ge 2$ , let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$ , where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$ . Find the number of values of $n$ with $2\le n \le 1000$ for which $A(n)$ is an integer. | Let $n\ge 2$ and define $a(n) = \left\lfloor \sqrt n \right\rfloor$ . For $2\le n \le 1000$ , we have $1\le a(n)\le 31$
For $a^2 \le x < (a+1)^2$ we have $y=ax$ . Thus $A(n+1)-A(n)=a(n+\tfrac 12) = \Delta_n$ (say), and $\Delta_n$ is an integer if $a$ is even; otherwise $\Delta_n$ is an integer plus $\tfrac 12$
If $a=1$ $n\in \{1,2,3\}$ and $\Delta_n$ is of the form $k(n)+\tfrac 12$ so $A(n)$ is an integer when $n$ is even.
If $a=2$ $n\in\{4,\ldots , 8\}$ and $\Delta_n$ is an integer for all $n$ . Since $A(3)$ is not an integer, so $A(n)$ is not an integer for any $n$
If $a=3$ $n\in\{9,\ldots , 15\}$ and $\Delta_n$ is of the form $k(n)+\tfrac 12$ . Since $A(8)$ is of the form $k+\tfrac 12$ so $A(n)$ is an integer only when $n$ is odd.
If $a=4$ $n\in\{16,\ldots , 24\}$ and $\Delta_n$ is an integer for all $n$ . Since $A(15)$ is an integer so $A(n)$ is an integer for all $n$
Now we are back to where we started; i.e., the case $a=5$ will be the same as $a=1$ and so on. Thus, \begin{align} a(n)\equiv 1\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for even } n, \\ a(n)\equiv 2\pmod 4 \qquad &\Longrightarrow \qquad A(n) \not\in \mathbb{Z} \textrm{ for any } n, \\ a(n)\equiv 3\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for odd } n, \\ a(n)\equiv 0\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for all } n. \end{align}
For each $a$ there are $2a+1$ corresponding values of $n$ : i.e., $n\in \{a^2, \ldots , (a+1)^2-1\}$
Thus, the number of values of $n$ corresponding to $(4)$ (i.e., $a(n)\equiv 0\pmod 4$ ) is given by \[\sum_{\substack{a=4k \\ a\le 31}}(2a+1) = \sum_{k=1}^7 (8k+1)=231.\]
The cases $(1)$ and $(3)$ combine to account for half the values of $n$ corresponding to odd values of $a(n)$ ; i.e., \[\frac 12 \cdot \sum_{\substack{a=2k+1 \\ a\le 31}} (2a+1) = \sum_{k=0}^{15} (2k+\tfrac 32) = 264\] However, this also includes the odd integers in $\{1001, \ldots , 1023\}$ . Subtracting $12$ to account for these, we get the number of values of $n$ corresponding to cases $(1)$ and $(3)$ to be $264-12=252$
Adding the contributions from all cases we get our answer to be $231+252= \boxed{483}$ | null | 483 |
df7668d6d5f214ea2f5c1acbf9e1b049 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_14 | For each integer $n \ge 2$ , let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$ , where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$ . Find the number of values of $n$ with $2\le n \le 1000$ for which $A(n)$ is an integer. | By considering the graph of this function, it is shown that the graph is composed of trapezoids ranging from $a^2$ to $(a+1)^2$ with the top made of diagonal line $y=ax$ . The width of each trapezoid is $3, 5, 7$ , etc. Whenever $a$ is odd, the value of $A(n)$ increases by an integer value, plus $\frac{1}{2}$ . Whenever $a$ is even, the value of $A(n)$ increases by an integer value. Since each trapezoid always has an odd width, every value of $n$ is not an integer when $a \pmod{4} \equiv 2$ , and is an integer when $a \pmod{4} \equiv 0$ . Every other value is an integer when $a$ is odd. Therefore, it is simply a matter of determining the number of values of $n$ where $a \pmod{4} \equiv 0$ $(5^2-4^2)+(9^2-8^2)+...+(29^2-28^2)$ ), and adding the number of values of $n$ where $a$ is odd ( $\frac{(2^2-1^2)+(4^2-3^2)+...+(30^2-29^2)+(1000-31^2)}{2}$ ). Adding the two values gives $231+252=\boxed{483}$ | null | 483 |
df7668d6d5f214ea2f5c1acbf9e1b049 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_14 | For each integer $n \ge 2$ , let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$ , where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$ . Find the number of values of $n$ with $2\le n \le 1000$ for which $A(n)$ is an integer. | First, draw a graph of the function. Note that it is just a bunch of line segments. Since we only need to know whether or not $A(n)$ is an integer, we can take the area of each piece from some $x$ to $x+1$ (mod 1), aka the piece from $2$ to $3$ has area $\frac{1}{2} (\mod 1)$ . There are some patterns. Every time we increase $n$ starting with $2$ , we either add $0 (\mod 1)$ or $\frac{1}{2} (\mod 1)$ . We look at $\lfloor \sqrt{x} \rfloor$ for inspiration. Every time this floor (which is really the slope) is odd, there is always an addition of $\frac{1}{2} (\mod 1)$ , and whenever that slope is even, that addition is zero.
Take a few cases. For slope $=1$ , we see that only one value satisfies. Because the last value, $n=4$ , fails, and the numbers $n$ which have a slope of an even number don't change this modulus, all these do not satisfy the criterion. The pattern then comes back to the odds, and this time $\lfloor \frac{7}{2} \rfloor + 1 = 4$ values work. Since the work/fail pattern alternates, all the $n$ s with even slope, $[17, 25]$ , satisfy the criterion. This pattern is cyclic over period 4 of slopes.
Even summation of working cases: $9+17+25+...+57 = 231$ .
Odd summation: $1+4+5+8+9+12...+29$ and plus the $20$ cases from $n=[962, 1000]$ $252$ . Answer is $\boxed{483}$ | null | 483 |
90cb7e66feea30923e3fac2c38080ed0 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_15 | A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$ , and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\overarc{AB}$ on that face measures $120^\circ$ . The block is then sliced in half along the plane that passes through point $A$ , point $B$ , and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\cdot\pi + b\sqrt{c}$ , where $a$ $b$ , and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$
[asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3); picture lpic, rpic; size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8)); size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,"$A$",(-3,3*sqrt(3),0),W); label(rpic,"$B$",(-3,-3*sqrt(3),0),W); add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy] | Label the points where the plane intersects the top face of the cylinder as $C$ and $D$ , and the center of the cylinder as $O$ , such that $C,O,$ and $A$ are collinear. Let $N$ be the center of the bottom face, and $M$ the midpoint of $\overline{AB}$ . Then $ON=4$ $MN=3$ (because of the 120 degree angle), and so $OM=5$
Project $C$ and $D$ onto the bottom face to get $X$ and $Y$ , respectively. Then the section $ABCD$ (whose area we need to find), is a stretching of the section $ABXY$ on the bottom face. The ratio of stretching is $\frac{OM}{MN}=\frac{5}{3}$ , and we do not square this value when finding the area because it is only stretching in one direction. Using 30-60-90 triangles and circular sectors, we find that the area of the section $ABXY$ is $18\sqrt{3}\ + 12 \pi$ . Thus, the area of section $ABCD$ is $20\pi + 30\sqrt{3}$ , and so our answer is $20+30+3=\boxed{053}$ | null | 053 |
90cb7e66feea30923e3fac2c38080ed0 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_15 | A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$ , and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\overarc{AB}$ on that face measures $120^\circ$ . The block is then sliced in half along the plane that passes through point $A$ , point $B$ , and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\cdot\pi + b\sqrt{c}$ , where $a$ $b$ , and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$
[asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3); picture lpic, rpic; size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8)); size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,"$A$",(-3,3*sqrt(3),0),W); label(rpic,"$B$",(-3,-3*sqrt(3),0),W); add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy] | Label the points same as in the first sentence above. Consider a view of the cylinder such that height is disregarded, i.e. a top view. From this view, note that Cylinder $O$ has become a circle with $\overarc{AB}$ $\overarc{CD}$ $120^\text{o}$ . Using 30-60-90 triangles, we get rectangle $ABCD$ to have a horizontal component of $6$ . Now, consider a side view, such that $A$ and $B$ coincide at the bottom of the diagram. From this view, consider the right triangle composed of hypotenuse $AD$ and a point along the base of the viewpoint, which will be labeled as $E$ . From the top view, $AE = 6$ . Because of the height of the cylinder, $DE$ is equal to $8$ . This makes $AD$ equal to $10$
Now, the use of simple calculus is required. Conceptualize an infinite number of lines perpendicular to $AE$ intersecting both $AE$ and $AD$ . Consider the area between point $A$ and the first vertical line. Label the point where the line intersects AE as E', and the point where the line intersects AD as D'. The area of the part of the initial unpainted face within these two positions approaches a rectangle with length AD' and width $w$ . The area of the base within these two positions approaches a rectangle with length AE' and width $w$ . The ratio of AD':AE' is 10:6, since the ratio of AD:AE is 10:6. This means that the area of the initial unpainted surface within these two positions to the area of the base within these two positions is equal to 10 $w$ :6 $w$ = 10:6. Through a similar argument, the areas between each set of vertical lines also maintains a ratio of 10:6. Therefore, the ratio of the area we wish to find to the area of the base between AB and CD (from the top perspective) is 10:6. Using 30-60-90 triangles and partial circles, the area of the base between AB and CD is calculated to be $18\sqrt{3}\ + 12 \pi$ . The area of the unpainted surface therefore becomes $20\pi + 30\sqrt{3}$ , and so our answer is $\boxed{053}$ | null | 053 |
90cb7e66feea30923e3fac2c38080ed0 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_15 | A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$ , and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\overarc{AB}$ on that face measures $120^\circ$ . The block is then sliced in half along the plane that passes through point $A$ , point $B$ , and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\cdot\pi + b\sqrt{c}$ , where $a$ $b$ , and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$
[asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3); picture lpic, rpic; size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8)); size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,"$A$",(-3,3*sqrt(3),0),W); label(rpic,"$B$",(-3,-3*sqrt(3),0),W); add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy] | This problem can be calculus-bashed (for those like me who never noticed the surface was merely a stretch of its projection). Label points as in the first paragraph of Solution 1 ( $A$ and $B$ as given, $M$ the midpoint of $AB$ $O$ the center of the cylinder, $T$ the center of the bottom face of the cylinder). Because of the 120 degrees and right triangle calculations, we find $MT$ = 3, $OT$ = 4, $OM$ = 5). We will be integrating with respect to the y-coordinate which we define as distance downwards from $O$ (in this system, the $y$ -coordinate of the bottom face would be 4).
We note that by similar triangles, we have that the length from $O$ to the point on the unpainted surface of coordinate $y$ is $\ell = \frac{5}{4} y$ , and therefore $d\ell = \frac{5}{4} dy$ . Define the segment $A'B'$ to be the intersection of the painted surface with the circular cross section of the cylinder of coordinate $y$ , with endpoints $A'$ and $B'$ and midpoint $M'$ , with $T'$ the center of this circular cross section. Then, by similar triangles, $T'M' = \frac{3}{4} y$ and thus \[A'B' = 2A' M' = 2 \sqrt{ 6^2 - \left( \frac{3}{4}y\right)^2 } = \frac{3}{2} \sqrt{ 64 - y^2 }\] . We know that $A'B'$ is perpendicular to $OM$
Now we can set up our integral: we will integrate $y$ from 0 to 4 and multiply by two because the total height is 8. \[A = 2\int_0^4 \left(\frac{3}{2}\sqrt{ 64 - y^2 }\right) \left( \frac{5}{4} dy\right)\] \[A = \frac{15}{4} \int_0^4 \sqrt{ 64 - y^2 }dy\]
Then we substitute $y = 8\sin \theta$ with $dy = 8 \cos \theta d \theta$ , changing the bounds to 0 to $\frac{\pi}{6}$ as appropriate. \[A = \frac{15}{4} \int_0^\frac{\pi}{6} \sqrt{ 64 - 64\sin^2 \theta }\cdot 8\cos\theta d\theta\] \[A = 240 \int_0^\frac{\pi}{6} \cos^2 \theta d\theta\] \[A = 240 \left[ \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right]_0^\frac{\pi}{6} = 240 \left[ \frac{\pi}{12} + \frac{\sqrt{3}}{8} \right] = 20{\pi} + 30\sqrt{3}\]
Therefore, $a + b + c = 20 + 30 + 3 = \boxed{053}$ | null | 053 |
ddae1a520ac6396d61a3cf4f1ba2ba0d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_1 | Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$ | If $N$ is $22$ percent less than one integer $k$ , then $N=\frac{78}{100}k=\frac{39}{50}k$ . In addition, $N$ is $16$ percent greater than another integer $m$ , so $N=\frac{116}{100}m=\frac{29}{25}m$ . Therefore, $k$ is divisible by $50$ and $m$ is divisible by $25$ . Setting these two equal, we have $\frac{39}{50}k=\frac{29}{25}m$ . Multiplying by $50$ on both sides, we get $39k=58m$
The smallest integers $k$ and $m$ that satisfy this are $k=1450$ and $m=975$ , so $N=1131$ . The answer is $\boxed{131}$ | null | 131 |
ddae1a520ac6396d61a3cf4f1ba2ba0d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_1 | Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$ | Continuing from Solution 1, we have $N=\frac{39}{50}k$ and $N=\frac{29}{25}m$ . It follows that $k=\frac{50}{39}N$ and $m=\frac{25}{29}N$ . Both $m$ and $k$ have to be integers, so, in order for that to be true, $N$ has to cancel the denominators of both $\frac{50}{39}$ and $\frac{25}{29}$ . In other words, $N$ is a multiple of both $29$ and $39$ . That makes $N=\operatorname{lcm}(29,39)=29\cdot39=1131$ . The answer is $\boxed{131}$ | null | 131 |
ddae1a520ac6396d61a3cf4f1ba2ba0d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_1 | Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$ | We can express $N$ as $0.78a$ and $1.22b$ , where $a$ and $b$ are some positive integers. $N=0.78a=1.22b\implies100N=78a=122b.$ Let us try to find the smallest possible value of $100N$ , first ignoring the integral constraint.
Obviously, we are just trying to find the LCM of $78$ and $116.$ They have no common factors but $2$ , so we multiply $78$ and $116$ and divide by $2$ to get $4524.$ This is obviously not divisible by $100$ , so this is not possible, as it would imply that $N=\dfrac{4524}{100},$ which is not an integer. This can be simplified to $\dfrac{1131}{25}$
We know that any possible value of $N$ will be an integral multiple of this value; the smallest such $N$ is achieved by multiplying this value by $25.$ We arrive at $1131$ , which is $\boxed{131}\mod1000.$ | null | 131 |
b3a29b85f8b20e27dbab712777bc52f9 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_2 | In a new school $40$ percent of the students are freshmen, $30$ percent are sophomores, $20$ percent are juniors, and $10$ percent are seniors. All freshmen are required to take Latin, and $80$ percent of the sophomores, $50$ percent of the juniors, and $20$ percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We see that $40\% \cdot 100\% + 30\% \cdot 80\% + 20\% \cdot 50\% + 10\% \cdot 20\% = 76\%$ of students are learning Latin. In addition, $30\% \cdot 80\% = 24\%$ of students are sophomores learning Latin. Thus, our desired probability is $\dfrac{24}{76}=\dfrac{6}{19}$ and our answer is $6+19=\boxed{025}$ | null | 025 |
b3a29b85f8b20e27dbab712777bc52f9 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_2 | In a new school $40$ percent of the students are freshmen, $30$ percent are sophomores, $20$ percent are juniors, and $10$ percent are seniors. All freshmen are required to take Latin, and $80$ percent of the sophomores, $50$ percent of the juniors, and $20$ percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Assume that there are $100$ students in the school. There are $40$ freshmen taking Latin, $24$ sophomores taking Latin, $10$ juniors taking Latin, and $2$ seniors taking Latin. We get the probability to be the number of sophomores taking Latin over the total number of students taking Latin, or $\dfrac{24}{76}$ . Simplifying, we get $\dfrac{6}{19}$ . Adding, we get $\boxed{025}$ | null | 025 |
caebe9ff3da70c84c00e1f205624005c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3 | Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ | The three-digit integers divisible by $17$ , and their digit sum: \[\begin{array}{c|c} m & s(m)\\ \hline 102 & 3 \\ 119 & 11\\ 136 & 10\\ 153 & 9\\ 170 & 8\\ 187 & 16\\ 204 & 6\\ 221 & 5\\ 238 & 13\\ 255 & 12\\ 272 & 11\\ 289 & 19\\ 306 & 9\\ 323 & 8\\ 340 & 7\\ 357 & 15\\ 374 & 14\\ 391 & 13\\ 408 & 12\\ 425 & 11\\ 442 & 10\\ 459 & 18\\ 476 & 17 \end{array}\]
Thus the answer is $\boxed{476}$ | null | 476 |
caebe9ff3da70c84c00e1f205624005c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3 | Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ | We can do the same thing as solution 1, except note the following fact: $102$ is a multiple of $17$ and its digits sum to $3$
Therefore, we can add it onto an existing multiple of $17$ that we know of to have $s(m) = 14$ , shown in the right-hand column, provided that its units digit is less than $8$ and its hundreds digit is less than $9$ . Unfortunately, $68$ does not fit the criteria, but $374$ does, meaning that, instead of continually adding multiples of $17$ , we can stop here and simply add $102$ to reach our final answer of $\boxed{476}$ | null | 476 |
caebe9ff3da70c84c00e1f205624005c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3 | Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ | The digit sum of a base $10$ integer $m$ is just $m\pmod{9}$ . In this problem, we know $17\mid m$ , or $m=17k$ for a positive integer $k$
Also, we know that $m\equiv 17\equiv -1\pmod{9}$ , or $17k\equiv -k\equiv -1\pmod{9}$
Obviously $k=1$ is a solution. This means in general, $k=9x+1$ is a solution for non-negative integer $x$
Checking the first few possible solutions, we find that $m=\boxed{476}$ is the first solution that has $s(m)=17$ , and we're done. | null | 476 |
caebe9ff3da70c84c00e1f205624005c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3 | Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ | Since the sum of the digits in the base-10 representation of $m$ is $17$ , we must have $m\equiv 17 \pmod{9}$ or $m\equiv -1\pmod{9}$ .
We also know that since $m$ is divisible by 17, $m\equiv 0 \pmod{17}$
To solve this system of linear congruences, we can use the Chinese Remainder Theorem. If we set $m\equiv (-1)(17)(8)\pmod {153}$ , we find that $m\equiv 0\pmod{17}$ and $m\equiv -1\pmod{9}$ , because $17\cdot 8\equiv 136 \equiv 1\pmod{9}$ . The trick to getting here was to find the number $x$ such that $17x\equiv 1\pmod{9}$ , so that when we take things $\pmod{9}$ , the $17$ goes away. We can do this using the Extended Euclidean Algorithm or by guess and check to find that $x\equiv 8\pmod{9}$
Finally, since $m\equiv 17\pmod{153}$ , we repeatedly add multiples of $153$ until we get a number in which its digits sum to 17, which first happens when $m=\boxed{476}$ | null | 476 |
caebe9ff3da70c84c00e1f205624005c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3 | Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ | We proceed by casework on the number of digits. Clearly the answer must have at least two digits, seeing as the maximum digit sum for a one-digit number is 9. The answer must also have less than 4 digits, because this is the AIME.
Case 1: The answer is a 2-digit number.
Represent the number as $10a + b$ , where $0 < a \leq 9$ and $0 \leq b \leq 9$ . The conditions of the problem translated into algebra are: \[17|10a+b\] \[a+b=17\] By the Euclidean Algorithm, this is equivalent to: \[17|9a\] 9 is not a factor of 17, so $17|a$ . So $a$ must be a multiple of 17, but this is impossible because of the conditions we placed on $a$ and $b$ .
(Alternatively, note that the only possible options are 89 and 98, and neither works.)
Case 2: The answer is a 3-digit number.
Represent the number as $100a+10b+c$ , where $0 < a \leq 9$ and $0 \leq b,c \leq 9$ . Translating the conditions again: \[17|100a+10b+c\] \[a+b+c=17\] \[17|99a+9b\] \[17|9(11a+b)\] \[17|11a+b\] Testing multiples of 17 yields $(4, 7, 6)$ as the minimal solution for $(a, b, c)$ and thus the answer is $\boxed{476}$ | null | 476 |
491d9e8a9b38bf61600124d38856948d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_4 | In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$ , and the altitude to these bases has length $\log 16$ . The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$ , where $p$ and $q$ are positive integers. Find $p + q$ | Call the trapezoid $ABCD$ with $AB$ as the smaller base and $CD$ as the longer. Let the point where an altitude intersects the larger base be $E$ , where $E$ is closer to $D$
Subtract the two bases and divide to find that $ED$ is $\log 8$ . The altitude can be expressed as $\frac{4}{3} \log 8$ . Therefore, the two legs are $\frac{5}{3} \log 8$ , or $\log 32$
The perimeter is thus $\log 32 + \log 32 + \log 192 + \log 3$ which is $\log 2^{16} 3^2$ . So $p + q = \boxed{018}$ | null | 018 |
491d9e8a9b38bf61600124d38856948d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_4 | In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$ , and the altitude to these bases has length $\log 16$ . The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$ , where $p$ and $q$ are positive integers. Find $p + q$ | Set the base of the log as 2. Then call the trapezoid $ABCD$ with $CD$ as the longer base. Then have the two feet of the altitudes be $E$ and $F$ , with $E$ and $F$ in position from left to right respectively. Then, $CF$ and $ED$ are $\log 192 - \log 3 = \log 64$ (from the log subtraction identity. Then $CF=EF=3$ (isosceles trapezoid and $\log 64$ being 6. Then the 2 legs of the trapezoid is $\sqrt{3^2+4^2}=5=\log 32$
And we have the answer:
$\log 192 + \log 32 + \log 32 + \log 3 = \log(192 \cdot 32 \cdot 32 \cdot 3) = \log(2^6 \cdot 3 \cdot 2^5 \cdot 2^5 \cdot 3) = \log(2^{16} \cdot 3^2) \Rightarrow 16+2 = \boxed{18}$ | null | 18 |
491d9e8a9b38bf61600124d38856948d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_4 | In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$ , and the altitude to these bases has length $\log 16$ . The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$ , where $p$ and $q$ are positive integers. Find $p + q$ | Let $ABCD$ be the trapezoid, where $\overline{AB} || \overline{CD}$ and $AB = \log 3$ and $CD = \log 192$ . Draw altitudes from $A$ and $B$ to $\overline{CD}$ with feet at $E$ and $F$ , respectively. $AB = \log 3$ , so $EF = \log 3$ . Now, we attempt to find $DE + FC$ , or what's left of $CD$ after we take out $EF$ . We make use of the two logarithmic rules:
\[\log(xy) = \log x + \log y\]
\[\log(x^a) = a\log(x)\]
\[CD = \log 192 = \log (3 \cdot 2^6) = \log 3 + \log(2^6) = \log 3 + 6\log 2\]
Thus, since $CD = DE + EF + FC = \log 3 + 6\log 2$ $CD - EF = \log 3 + 6\log 2 - \log 3 = 6\log 2 = DE + FC$
Now, why was finding $DE + FC$ important? Absolutely no reason! Just kidding, lol 🤣 Now, we essentially "glue" triangles $\triangle DAE$ and $\triangle BFC$ together to get $\triangle XC'D'$ , where $X$ is the point where $A$ and $B$ became one. Note we can do this because $\triangle DAE$ and $\triangle BFC$ are both right triangles with a common leg length (the altitude of trapezoid $ABCD$ ).
Triangle $XC'D'$ has a base of $C'D'$ , which is just equal to $DE + FC = 6\log 2$ . It is equal to $DE + FC$ because when we brought triangles $\triangle DAE$ and $\triangle BFC$ together, the length of $CD$ was not changed except for taking out $EF$
$XC' = XD'$ since $AD = BC$ because the problem tells us we have an isosceles trapezoid. Drop and altitude from $X$ to $C'D'$ The altitude has length $\log 16 = 4\log 2$ . The altitude also bisects $C'D'$ since $\triangle XC'D'$ is isosceles. Let the foot of the altitude be $M$ . Then $MD' = 3\log 2$ (Remember that C'D' was $6\log 2$ , and then it got bisected by the altitude). Thus, the hypotenuse, $XD'$ must be $5\log 2$ from the Pythagorean Theorem or by noticing that you have a 3-4-5 right triangle with a similarity ratio of $\log 2$ . Since $XD' = XC' = BC = AD$ $BC = AD = 5\log 2 = \log 2^5$
Now, we have $CD = \log (3 \cdot 2^6)$ $AB = \log 3$ , and $BC = AD = \log 2^5$ . Thus, their sum is
\[\log (3 \cdot 2^6) + \log 3 + \log 2^5 + \log 2^5 = \log (2^16 \cdot 3^2)\]
Thus, $p + q = 16 + 2 = \boxed{18}$ . ~Extremelysupercooldude | null | 18 |
836d0371a0ac8741a7bb5424f88dbb43 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_5 | Two unit squares are selected at random without replacement from an $n \times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than $\frac{1}{2015}$ | Call the given grid "Grid A". Consider Grid B, where the vertices of Grid B fall in the centers of the squares of Grid A; thus, Grid B has dimensions $(n-1) \times (n-1)$ . There is a one-to-one correspondence between the edges of Grid B and the number of adjacent pairs of unit squares in Grid A. The number of edges in Grid B is $2n(n-1)$ , and the number of ways to pick two squares out of Grid A is $\dbinom{n^2}{2}$ . So, the probability that the two chosen squares are adjacent is $\frac{2n(n-1)}{\binom{n^2}{2}} = \frac{2n(n-1)}{\frac{n^2(n^2-1)}{2}} = \frac{4}{n(n+1)}$ . We wish to find the smallest positive integer $n$ such that $\frac{4}{n(n+1)} < \frac{1}{2015}$ , and by inspection the first such $n$ is $\boxed{090}$ | null | 090 |
836d0371a0ac8741a7bb5424f88dbb43 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_5 | Two unit squares are selected at random without replacement from an $n \times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than $\frac{1}{2015}$ | Consider a $3 \times 3$ grid, where there are $4$ corner squares, $4$ edge squares, and $1$ center square. A $4 \times 4$ grid has $4$ corner squares, $8$ edge squares, and $4$ center squares. By examining simple cases, we can conclude that for a grid that is $n \times n$ , there are always $4$ corners squares, $4(n-2)$ edge squares, and $n^2-4n+4$ center squares.
Each corner square is adjacent to $2$ other squares, edge squares to $3$ other squares, and center squares to $4$ other squares. In the problem, the second square can be any square that is not the first, which means there are $n^2-1$ to choose from. With this information, we can conclude that the probability that second unit square is adjacent to the first is $\frac{2}{n^2-1}(\frac{4}{n^2}) +\frac{3}{n^2-1}(\frac{4(n-2)}{n^2}) +\frac{4}{n^2-1}(\frac{n^2-4n+4}{n^2})$
Simplifying, we get $\frac{4}{n(n+1)}$ which we can set to be less than $\frac{1}{2015}$ . By inspection, we find that the first such $n$ is $\boxed{090}$ | null | 090 |
836d0371a0ac8741a7bb5424f88dbb43 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_5 | Two unit squares are selected at random without replacement from an $n \times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than $\frac{1}{2015}$ | There are 3 cases in this problem. Number one, the center squares. Two, the edges, and three, the corners. The probability of getting one center square and an adjacent square is $\frac{(n-2)(n-2)}{n^2}$ multiplied by $\frac{4}{n^2 -1}$ . Add that to the probability of an edge and an adjacent square( $\frac{4n-8}{n^2}$ multiplied by $\frac{3}{n^2-1}$ ) and the probability of a corner and an adjacent square( $\frac{4}{n^2}$ multiplied by $\frac{2}{n^2-1}$ ) to get $\frac{4n^2-4n}{n^4-n^2}$ . Simplify to get $\frac{4}{n^2+n}$ . With some experimentation, we realize that the smallest value of n is $\boxed{090}$ | null | 090 |
80585bc6acc4fa7d6254be8f4080276c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_6 | Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve says, "You're right. Here is the value of $a$ ." He writes down a positive integer and asks, "Can you tell me the value of $c$ ?"
Jon says, "There are still two possible values of $c$ ."
Find the sum of the two possible values of $c$ | We call the three roots (some may be equal to one another) $x_1$ $x_2$ , and $x_3$ . Using Vieta's formulas, we get $x_1+x_2+x_3 = a$ $x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}$ , and $x_1 \cdot x_2 \cdot x_3 = \frac{c}{2}$
Squaring our first equation we get $x_1^2+x_2^2+x_3^2+2 \cdot x_1 \cdot x_2+2 \cdot x_1 \cdot x_3+2 \cdot x_2 \cdot x_3 = a^2$
We can then subtract twice our second equation to get $x_1^2+x_2^2+x_3^2 = a^2-2 \cdot \frac{a^2-81}{2}$
Simplifying the right side:
\begin{align*} a^2-2 \cdot \frac{a^2-81}{2} &= a^2-a^2+81\\ &= 81.\\ \end{align*}
So, we know $x_1^2+x_2^2+x_3^2 = 81$
We can then list out all the triples of positive integers whose squares sum to $81$
We get $(1, 4, 8)$ $(3, 6, 6)$ , and $(4, 4, 7)$
These triples give $a$ values of $13$ $15$ , and $15$ , respectively, and $c$ values of $64$ $216$ , and $224$ , respectively.
We know that Jon still found two possible values of $c$ when Steve told him the $a$ value, so the $a$ value must be $15$ . Thus, the two $c$ values are $216$ and $224$ , which sum to $\boxed{440}$ | null | 440 |
80585bc6acc4fa7d6254be8f4080276c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_6 | Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve says, "You're right. Here is the value of $a$ ." He writes down a positive integer and asks, "Can you tell me the value of $c$ ?"
Jon says, "There are still two possible values of $c$ ."
Find the sum of the two possible values of $c$ | First things first. Vietas gives us the following:
\begin{align} x_1+x_2+x_3 = a\\ x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}\\ x_1 \cdot x_2 \cdot x_3 = \frac{c}{2} \end{align}
From $(2)$ $a$ must have odd parity, meaning $a^2-81$ must be a multiple of $4$ , which implies that both sides of $(2)$ are even. Then, from $(1)$ , we see that an odd number of $x_1$ $x_2$ , and $x_3$ must be odd, because we have already deduced that $a$ is odd. In order for both sides of $(2)$ to be even, there must only be one odd number and two even numbers.
Now, the theoretical maximum value of the left side of $(2)$ is $3 \cdot \biggl(\frac{a}{3}\biggr)^2=\frac{a^2}{3}$ . That means that the maximum bound of $a$ is where \[\frac{a^2}{3} > \frac{a^2-81}{2},\] which simplifies to $\sqrt{243} > a$ , meaning \[16 > a.\] So now we have that $9<a$ from $(2)$ $a<16$ , and $a$ is odd from $(2)$ . This means that $a$ could equal $11$ $13$ , or $15$ . At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of $(1, 4, 8)$ $(3, 6, 6)$ , and $(4, 4, 7)$ , of which the last two return equal $a$ values. Then, $2(3 \cdot 6 \cdot 6+4 \cdot 4 \cdot 7)=\boxed{440}$ AWD. | null | 440 |
80585bc6acc4fa7d6254be8f4080276c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_6 | Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve says, "You're right. Here is the value of $a$ ." He writes down a positive integer and asks, "Can you tell me the value of $c$ ?"
Jon says, "There are still two possible values of $c$ ."
Find the sum of the two possible values of $c$ | Since each of the roots is positive, the local maximum of the function must occur at a positive value of $x$ . Taking $\frac{d}{dx}$ of the polynomial yields $6x^2-4ax+a^2-81$ , which is equal to $0$ at the local maximum. Since this is a quadratic in $a$ , we can find an expression for $a$ in terms of $x$ . The quadratic formula gives $a=\frac{4x\pm\sqrt{324-8x^2}}{2}$ , which simplifies to $a=2x\pm\sqrt{81-2x^2}$ . We know that $a$ is a positive integer, and testing small positive integer values of $x$ yields $a=15$ or $a=1$ when $x=4$ , and $a=15$ or $a=9$ when $x=6$ . Because the value of $a$ alone does not determine the polynomial, $a$ $a$ must equal $15$
Now our polynomial equals $2x^3-30x^2+144x-c$ . Because one root is less than (or equal to) the $x$ value at the local maximum (picture the graph of a cubic equation), it suffices to synthetically divide by small integer values of $x$ to see if the resulting quadratic also has positive integer roots. Dividing by $x=3$ leaves a quotient of $2x^2-24x+72=2(x-6)^2$ , and dividing by $x=4$ leaves a quotient of $2x^2-22x+56=2(x-4)(x-7)$ . Thus, $c=2\cdot 3\cdot 6\cdot 6=216$ , or $c=2\cdot 4\cdot 4\cdot 7=224$ . Our answer is $216+224=\boxed{440}$ ~bad_at_mathcounts | null | 440 |
9a043889a2081a0cc987dbf2a449bfe3 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7 | Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]
Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | If $\omega = 25$ , the area of rectangle $PQRS$ is $0$ , so
\[\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0\]
and $\alpha = 25\beta$ . If $\omega = \frac{25}{2}$ , we can reflect $APQ$ over $PQ$ $PBS$ over $PS$ , and $QCR$ over $QR$ to completely cover rectangle $PQRS$ , so the area of $PQRS$ is half the area of the triangle. Using Heron's formula, since $s = \frac{12 + 17 + 25}{2} = 27$
\[[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90\]
so
\[45 = \alpha\omega - \beta\omega^2 = \frac{625}{2} \beta - \beta\frac{625}{4} = \beta\frac{625}{4}\]
and
\[\beta = \frac{180}{625} = \frac{36}{125}\]
so the answer is $m + n = 36 + 125 = \boxed{161}$ | null | 161 |
9a043889a2081a0cc987dbf2a449bfe3 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7 | Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]
Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [asy] unitsize(20); pair A,B,C,E,F,P,Q,R,S; A=(48/5,36/5); B=(0,0); C=(25,0); E=(48/5,0); F=(48/5,18/5); P=(24/5,18/5); Q=(173/10,18/5); S=(24/5,0); R=(173/10,0); draw(A--B--C--cycle); draw(P--Q); draw(Q--R); draw(R--S); draw(S--P); draw(A--E,dashed); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$E$",E,SE); label("$F$",F,NE); label("$P$",P,NW); label("$Q$",Q,NE); label("$R$",R,SE); label("$S$",S,SW); draw(rightanglemark(B,E,A,12)); dot(E); dot(F); [/asy]
Similar triangles can also solve the problem.
First, solve for the area of the triangle. $[ABC] = 90$ . This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $AC$ and solving. (The $8$ side would be collinear with line $AB$
After finding the area, solve for the altitude to $BC$ . Let $E$ be the intersection of the altitude from $A$ and side $BC$ . Then $AE = \frac{36}{5}$ .
Solving for $BE$ using the Pythagorean Formula, we get $BE = \frac{48}{5}$ . We then know that $CE = \frac{77}{5}$
Now consider the rectangle $PQRS$ . Since $SR$ is collinear with $BC$ and parallel to $PQ$ $PQ$ is parallel to $BC$ meaning $\Delta APQ$ is similar to $\Delta ABC$
Let $F$ be the intersection between $AE$ and $PQ$ . By the similar triangles, we know that $\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}$ . Since $PF+FQ=PQ=\omega$ . We can solve for $PF$ and $FQ$ in terms of $\omega$ . We get that $PF=\frac{48}{125} \omega$ and $FQ=\frac{77}{125} \omega$
Let's work with $PF$ . We know that $PQ$ is parallel to $BC$ so $\Delta APF$ is similar to $\Delta ABE$ . We can set up the proportion:
$\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}$ . Solving for $AF$ $AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega$
We can solve for $PS$ then since we know that $PS=FE$ and $FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega$
Therefore, $[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2$
This means that $\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}$ | null | 161 |
9a043889a2081a0cc987dbf2a449bfe3 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7 | Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]
Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Using the diagram from Solution 2 above, label $AF$ to be $h$ . Through Heron's formula, the area of $\triangle ABC$ turns out to be $90$ , so using $AE$ as the height and $BC$ as the base yields $AE=\frac{36}{5}$ . Now, through the use of similarity between $\triangle APQ$ and $\triangle ABC$ , you find $\frac{w}{25}=\frac{h}{36/5}$ . Thus, $h=\frac{36w}{125}$ . To find the height of the rectangle, subtract $h$ from $\frac{36}{5}$ to get $\left(\frac{36}{5}-\frac{36w}{125}\right)$ , and multiply this by the other given side $w$ to get $\frac{36w}{5}-\frac{36w^2}{125}$ for the area of the rectangle. Finally, $36+125=\boxed{161}$ | null | 161 |
9a043889a2081a0cc987dbf2a449bfe3 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7 | Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]
Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Using the diagram as shown in Solution 2, let $AE=h$ and $AP=L$ Now, by Heron's formula, we find that the $[ABC]=90$ . Hence, $h=\frac{36}{5}$
Now, we see that $\sin{B}=\frac{PS}{12-L}\implies PS=\sin{B}(12-L)$ We easily find that $\sin{B}=\frac{3}{5}$
Hence, $PS=\frac{3}{5}(12-L)$
Now, we see that $[PQRS]=\frac{3}{5}(12-L)(w)$
Now, it is obvious that we want to find $L$ in terms of $W$
Looking at the diagram, we see that because $PQRS$ is a rectangle, $\triangle{APQ}\sim{\triangle{ABC}}$
Hence.. we can now set up similar triangles.
We have that $\frac{AP}{AB}=\frac{PQ}{BC}\implies \frac{L}{12}=\frac{W}{25}\implies 25L=12W\implies L=\frac{12W}{25}$
Plugging back in..
$[PQRS]=\frac{3w}{5}(12-(\frac{12W}{25}))\implies \frac{3w}{5}(\frac{300-12W}{25})\implies \frac{900W-36W^2}{125}$
Simplifying, we get $\frac{36W}{5}-\frac{36W^2}{125}$
Hence, $125+36=\boxed{161}$ | null | 161 |
9a043889a2081a0cc987dbf2a449bfe3 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7 | Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \alpha \omega - \beta \omega^2.\]
Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Proceed as in solution 1. When $\omega$ is equal to zero, $\alpha - \beta\omega=\alpha$ is equal to the altitude. This means that $25\beta$ is equal to $\frac{36}{5}$ , so $\beta = \frac{36}{125}$ , yielding $\boxed{161}$ | null | 161 |
7f85b9cf81b4df6e21a21c2aa8efa4ba | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_8 | Let $a$ and $b$ be positive integers satisfying $\frac{ab+1}{a+b} < \frac{3}{2}$ . The maximum possible value of $\frac{a^3b^3+1}{a^3+b^3}$ is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Let us call the quantity $\frac{a^3b^3+1}{a^3+b^3}$ as $N$ for convenience. Knowing that $a$ and $b$ are positive integers, we can legitimately rearrange the given inequality so that $a$ is by itself, which makes it easier to determine the pairs of $(a, b)$ that work. Doing so, we have \[\frac{ab+1}{a+b} < \frac{3}{2}\] \[\implies 2ab + 2 < 3a + 3b \implies 2ab - 3a < 3b - 2\] \[\implies a < \frac{3b - 2}{2b - 3}.\] Now, observe that if $b = 1$ we have that $N = \frac{a^3 + 1}{a^3 + 1} = 1$ , regardless of the value of $a$ . If $a = 1$ , we have the same result: that $N = \frac{b^3 + 1}{b^3 + 1} = 1$ , regardless of the value of $b$ . Hence, we want to find pairs of positive integers $(a, b)$ existing such that neither $a$ nor $b$ is equal to $1$ , and that the conditions given in the problem are satisfied in order to check that the maximum value for $N$ is not $1$
To avoid the possibility that $a = 1$ , we want to find values of $b$ such that $\frac{3b - 2}{2b - 3} > 2$ . If we do this, we will have that $a < \frac{3b - 2}{2b - 3} = k$ , where $k$ is greater than $2$ , and this allows us to choose values of $a$ greater than $1$ . Again, since $b$ is a positive integer, and we want $b > 1$ , we can legitimately multiply both sides of $\frac{3b - 2}{2b - 3} > 2$ by $2b - 3$ to get $3b - 2 > 4b - 6 \implies b < 4$ . For $b = 3$ , we have that $a < \frac{7}{3}$ , so the only possibility for $a$ greater than $1$ is obviously $2$ . Plugging these values into $N$ , we have that $N = \frac{8(27) + 1}{8 + 27} = \frac{217}{35} = \frac{31}{5}$ . For $b = 2$ , we have that $a < \frac{4}{1} = 4$ . Plugging $a = 3$ and $b = 2$ in for $N$ yields the same result of $\frac{31}{5}$ , but plugging $a = 2$ and $b = 2$ into $N$ yields that $N = \frac{8(8) + 1}{8 + 8} = \frac{65}{16}$ . Clearly, $\frac{31}{5}$ is the largest value we can have for $N$ , so our answer is $31 + 5 = \boxed{036}$ | null | 036 |
7f85b9cf81b4df6e21a21c2aa8efa4ba | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_8 | Let $a$ and $b$ be positive integers satisfying $\frac{ab+1}{a+b} < \frac{3}{2}$ . The maximum possible value of $\frac{a^3b^3+1}{a^3+b^3}$ is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | \[\frac{ab + 1}{a + b} < \frac{3}{2} \rightarrow 2ab + 2 < 3a + 3b \rightarrow 4ab - 6a - 6b + 4 < 0 \rightarrow (2a - 3)(2b - 3) < 5.\]
\[2a - 3, 2b - 3 \in \{x \neq 2k, k \in Z \} \rightarrow\] $(2a-3)$ and $(2b-3)$ each cannot be even or else $a$ and $b$ will not be integers \[(2a - 3)(2b - 3) = 1, 3 \rightarrow (2a - 3, 2b - 3) = (1, 1), (1, 3), (3, 1).\] \[(a, b) = (2, 2), (2, 3), (3, 2).\] \[\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{65}{16}, \frac{31}{5}.\] \[\frac{31}{5} \rightarrow \boxed{036}.\] | null | 036 |
7f85b9cf81b4df6e21a21c2aa8efa4ba | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_8 | Let $a$ and $b$ be positive integers satisfying $\frac{ab+1}{a+b} < \frac{3}{2}$ . The maximum possible value of $\frac{a^3b^3+1}{a^3+b^3}$ is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Notice that for $\frac{a^3b^3+1}{a^3+b^3}$ to be maximized, $\frac{ab+1}{a+b}$ has to be maximized. We simplify as above to $2ab + 2 < 3a + 3b$ , which is $(a-\frac{3}{2})(b-\frac{3}{2}) < \frac{5}{4}$ . To maximize, $a$ has to be as close to $b$ as possible, making $a$ close to $\frac{3+\sqrt{5}}{2}$ . Because $a$ and $b$ are positive integers, $a = 3$ , and checking back gives $b = 2$ as the maximum or the other way around, which the answer is thus $\frac{216+1}{27+8} = \frac{217}{35} = \frac{31}{5} \rightarrow \boxed{036}$ | null | 036 |
4bd17797fe494e1d481dd271ae74c031 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9 | A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$
[asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy] | Our aim is to find the volume of the part of the cube submerged in the cylinder.
In the problem, since three edges emanate from each vertex, the boundary of the cylinder touches the cube at three points. Because the space diagonal of the cube is vertical, by the symmetry of the cube, the three points form an equilateral triangle. Because the radius of the circle is $4$ , by the Law of Cosines, the side length s of the equilateral triangle is
\[s^2 = 2\cdot(4^2) - 2\cdot(4^2)\cos(120^{\circ}) = 3(4^2)\]
so $s = 4\sqrt{3}$ .* Again by the symmetry of the cube, the volume we want to find is the volume of a tetrahedron with right angles on all faces at the submerged vertex, so since the lengths of the legs of the tetrahedron are $\frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}$ (the three triangular faces touching the submerged vertex are all $45-45-90$ triangles) so
\[v = \frac{1}{3}(2\sqrt{6})\left(\frac{1}{2} \cdot (2\sqrt{6})^2\right) = \frac{1}{6} \cdot 48\sqrt{6} = 8\sqrt{6}\]
so
\[v^2 = 64 \cdot 6 = \boxed{384}.\] | null | 384 |
4bd17797fe494e1d481dd271ae74c031 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9 | A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$
[asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy] | Visualizing the corner which is submerged in the cylinder, we can see that it's like slicing off a corner, where they slice an equal part off every edge, to make a tetrahedron where there are 3 right isosceles triangles and one equilateral triangle. With this out of the way, we can now just find the area of that equilateral triangle, using the fact that the circle of radius $4$ is the circumcircle of the equilateral triangle. Using equilateral triangle properties, you can find that the height of the triangle is $6$ , and the side length is $\frac{6}{\sqrt{3}}=2\sqrt{3} \cdot 2=4\sqrt{3}$ . As the other faces are right isosceles triangles, they are $\frac{4\sqrt{3}}{\sqrt{2}}=2\sqrt{6}$ . Therefore the volume of this tetrahedron is
\[\left(\frac{2\sqrt{6}}{2}\right)^2=12 \ \cdot \ (2\sqrt{6})=24\sqrt{6} \implies \frac{24\sqrt{6}}{3}=8\sqrt{6} \implies (8\sqrt{6})^2=\boxed{384}\] | null | 384 |
4bd17797fe494e1d481dd271ae74c031 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9 | A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$
[asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy] | We can use the same method as in Solution 2 to find the side length of the equilateral triangle, which is $4\sqrt3$ . From here, its area is \[\dfrac{\bigl(4\sqrt3\bigr)^2\sqrt3}4=12\sqrt3.\] The leg of the isosceles right triangle is $\dfrac{4\sqrt3}{\sqrt2}=2\sqrt6$ , and the horizontal distance from the vertex to the base of the tetrahedron is $4$ (the radius of the cylinder), so we can find the height, as shown in the diagram. [asy] import olympiad; pair V, T, B; V = (-4, 0); B = origin; T = (0, 2*sqrt(2)); draw(V--B--T--cycle); draw(rightanglemark(V, B, T)); label("Vertex", V, W); label("Tip", T, N); label("Base", B, SE); label("$4$", V--B, S); label("$2\sqrt6$", V--T, NW); [/asy] The height from the tip to the base is $2\sqrt2$ , so the volume is $\dfrac{12\sqrt3\cdot2\sqrt2}3=8\sqrt6$ , and thus the answer is $\boxed{384}$ | null | 384 |
4bd17797fe494e1d481dd271ae74c031 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9 | A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$
[asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy] | Since the diagonal is perpendicular to the base of the cylinder, all three edges and faces can be treated symmetrically.
The cross-section of the cube with the top face of the cylinder is an equilateral triangle inscribed in a circle with radius $4$ . This means the medians of the triangle have length $\frac{3}{2} \cdot 4=6$ , because the circumcenter is also the centroid, and the centroid divides the medians into lengths of ratio $2:1$ . Using $30-60-90$ triangles, the side length of the triangle is $4\sqrt{3}$ , and its area is $\frac{(4\sqrt{3})^2\sqrt{3}}{4}=12\sqrt{3}$
Next, consider the submerged triangular sections of the faces. Each is a $45-45-90$ triangle with leg length $x$ . The area of each is then $\frac{x^2}{2}$ . By De Gua's Theorem on the submerged pyramid (which we can apply because it has a right-angled corner), $3\left( \frac{x^2}{2} \right) ^2=(12\sqrt{3})^2$ . Solving yields $x=2\sqrt{6}$
The height of the pyramid is then $\sqrt{(2\sqrt{6})^2-4^2}=2\sqrt{2}$ , by the Pythagorean Theorem (using the slant height and circumradius). The volume is then $v=\frac{1}{3}\cdot 12\sqrt{3} \cdot 2\sqrt{2}=8\sqrt{6}$ , and the requested answer is $v^2=(8\sqrt{6})^2=\boxed{384}$ . ~bad_at_mathcounts | null | 384 |
2d5bb236f8e74077503ea355f86715b5 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_10 | Call a permutation $a_1, a_2, \ldots, a_n$ of the integers $1, 2, \ldots, n$ quasi-increasing if $a_k \leq a_{k+1} + 2$ for each $1 \leq k \leq n-1$ . For example, 53421 and 14253 are quasi-increasing permutations of the integers $1, 2, 3, 4, 5$ , but 45123 is not. Find the number of quasi-increasing permutations of the integers $1, 2, \ldots, 7$ | The simple recurrence can be found.
When inserting an integer $n$ into a string with $n - 1$ integers, we notice that the integer $n$ has 3 spots where it can go: before $n - 1$ , before $n - 2$ , and at the very end.
EXAMPLE:
Putting 4 into the string 123:
4 can go before the 2: 1423,
Before the 3: 1243,
And at the very end: 1234.
Only the addition of the next number, n, will change anything.
Thus the number of permutations with n elements is three times the number of permutations with $n-1$ elements.
Start with $n=3$ since all $6$ permutations work. And go up: $18, 54, 162, 486$
Thus for $n=7$ there are $2*3^5=\boxed{486}$ permutations. | null | 486 |
f560f5b5f50adaafd8e29aef58ba9c8c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_11 | The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Call $M$ and $N$ the feet of the altitudes from $O$ to $BC$ and $AB$ , respectively. Let $OB = r$ . Notice that $\triangle{OMB} \sim \triangle{QOB}$ because both are right triangles, and $\angle{OBQ} \cong \angle{OBM}$ . By $\frac{MB}{BO}=\frac{BO}{BQ}$ $MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}$ . However, since $O$ is the circumcenter of triangle $ABC$ $OM$ is a perpendicular bisector by the definition of a circumcenter. Hence, $\frac{r^2}{4.5} = 2 \implies r = 3$ . Since we know $BN=\frac{5}{2}$ and $\triangle BOP \sim \triangle BNO$ , we have $\frac{BP}{3} = \frac{3}{\frac{5}{2}}$ . Thus, $BP = \frac{18}{5}$ $m + n=\boxed{023}$ | null | 023 |
f560f5b5f50adaafd8e29aef58ba9c8c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_11 | The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Minor arc $BC = 2A$ so $\angle{BOC}=2A$ . Since $\triangle{BOC}$ is isosceles ( $BO$ and $OC$ are radii), $\angle{CBO}=(180-2A)/2=90-A$ $\angle{CBO}=90-A$ , so $\angle{BQO}=A$ . From this we get that $\triangle{BPQ}\sim \triangle{BCA}$ . So $\dfrac{BP}{BC}=\dfrac{BQ}{BA}$ , plugging in the given values we get $\dfrac{BP}{4}=\dfrac{4.5}{5}$ , so $BP=\dfrac{18}{5}$ , and $m+n=\boxed{023}$ | null | 023 |
f560f5b5f50adaafd8e29aef58ba9c8c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_11 | The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $r=BO$ . Drawing perpendiculars, $BM=MC=2$ and $BN=NA=2.5$ . From there, \[OM=\sqrt{r^2-4}\] Thus, \[OQ=\frac{\sqrt{4r^2+9}}{2}\] Using $\triangle{BOQ}$ , we get $r=3$ . Now let's find $NP$ . After some calculations with $\triangle{BON}$ $\triangle{OPN}$ ${NP=11/10}$ . Therefore, \[BP=\frac{5}{2}+\frac{11}{10}=18/5\] $18+5=\boxed{023}$ | null | 023 |
f560f5b5f50adaafd8e29aef58ba9c8c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_11 | The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $\angle{BQO}=\alpha$ . Extend $OB$ to touch the circumcircle at a point $K$ . Then, note that $\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha$ . But since $BK$ is a diameter, $\angle{KAB}=90^\circ$ , implying $\angle{CAB}=\alpha$ . It follows that $APCQ$ is a cyclic quadrilateral.
Let $BP=x$ . By Power of a Point, \[5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.\] The answer is $18+5=\boxed{023}$ | null | 023 |
f560f5b5f50adaafd8e29aef58ba9c8c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_11 | The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | $\textit{Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.}$
Denote the circumradius of $ABC$ to be $R$ , the circumcircle of $ABC$ to be $O$ , and the shortest distance from $Q$ to circle $O$ to be $x$
Using Power of a Point on $Q$ relative to circle $O$ , we get that $x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}$ . Using Pythagorean Theorem on triangle $QOB$ to get $(x + r)^2 + r^2 = \frac{81}{4}$ . Subtracting the first equation from the second, we get that $2r^2 = 18$ and therefore $r = 3$ . Now, set $\cos{ABC} = y$ . Using law of cosines on $ABC$ to find $AC$ in terms of $y$ and plugging that into the extended law of sines, we get $\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6$ . Squaring both sides and cross multiplying, we get $36x^2 - 40x + 5 = 0$ . Now, we get $x = \frac{10 \pm \sqrt{55}}{18}$ using quadratic formula. If you drew a decent diagram, $B$ is acute and therefore $x = \frac{10 + \sqrt{55}}{18}$ (You can also try plugging in both in the end and seeing which gives a rational solution). Note that $BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.$ Using the cosine addition formula and then plugging in what we know about $QBO$ , we get that $BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}$ . Now, the hard part is to find what $\sin{B}$ is. We therefore want $\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}$ . For the numerator, by inspection $(a + b\sqrt{55})^2$ will not work for integers $a$ and $b$ . The other case is if there is $(a\sqrt{5} + b\sqrt{11})^2$ . By inspection, $5\sqrt{5} - 2\sqrt{11}$ works. Therefore, plugging all this in yields the answer, $\frac{18}{5} \rightarrow \boxed{23}$ . Solution by hyxue | null | 23 |
f560f5b5f50adaafd8e29aef58ba9c8c | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_11 | The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.7673964645097335, xmax = 9.475267639476614, ymin = -1.6884766592324019, ymax = 6.385449160754665; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((0.7129306199257198,2.4781596958650733), 3.000319171815248), linewidth(2) + wrwrwr); draw((0.7129306199257198,2.4781596958650733)--(3.178984115621537,0.7692140299269852), linewidth(2) + wrwrwr); draw((xmin, 1.4430262733614363*xmin + 1.4493820802284032)--(xmax, 1.4430262733614363*xmax + 1.4493820802284032), linewidth(2) + wrwrwr); /* line */ draw((xmin, -0.020161290322580634*xmin + 0.8333064516129032)--(xmax, -0.020161290322580634*xmax + 0.8333064516129032), linewidth(2) + wrwrwr); /* line */ draw((xmin, -8.047527437688247*xmin + 26.352175924366414)--(xmax, -8.047527437688247*xmax + 26.352175924366414), linewidth(2) + wrwrwr); /* line */ draw((xmin, -2.5113572383524088*xmin + 8.752778799300463)--(xmax, -2.5113572383524088*xmax + 8.752778799300463), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.12426176956126818*xmin + 2.389569675458691)--(xmax, 0.12426176956126818*xmax + 2.389569675458691), linewidth(2) + wrwrwr); /* line */ draw(circle((1.9173376033752174,4.895608471162773), 0.7842529827808445), linewidth(2) + wrwrwr); /* dots and labels */ dot((-1.82,0.87),dotstyle); label("$A$", (-1.7801363959463627,0.965838014692327), NE * labelscalefactor); dot((3.178984115621537,0.7692140299269852),dotstyle); label("$B$", (3.2140445236332655,0.8641046996638531), NE * labelscalefactor); dot((2.6857306099246263,4.738685150758791),dotstyle); label("$C$", (2.7238749148597092,4.831703985774336), NE * labelscalefactor); dot((0.7129306199257198,2.4781596958650733),linewidth(4pt) + dotstyle); label("$O$", (0.7539479965810783,2.556577122410283), NE * labelscalefactor); dot((-0.42105034508654754,0.8417953698606159),linewidth(4pt) + dotstyle); label("$P$", (-0.38361543510094825,0.9195955987702934), NE * labelscalefactor); dot((2.6239558409689123,5.235819298886746),linewidth(4pt) + dotstyle); label("$Q$", (2.6591355325688624,5.312625111363486), NE * labelscalefactor); dot((1.3292769824200672,5.414489427724579),linewidth(4pt) + dotstyle); label("$A'$", (1.3643478867519216,5.488346291867214), NE * labelscalefactor); dot((1.8469115849379867,4.11452402186953),linewidth(4pt) + dotstyle); label("$P'$", (1.8822629450786978,4.184310162865866), NE * labelscalefactor); dot((2.5624172335003985,5.731052930966743),linewidth(4pt) + dotstyle); label("$D$", (2.603644633462422,5.802794720137042), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy] Reflect $A$ $P$ across $OB$ to points $A'$ and $P'$ , respectively with $A'$ on the circle and $P, O, P'$ collinear. Now, $\angle A'CQ = 180^{\circ} - \angle A'CB = \angle A'AB = \angle P'PB$ by parallel lines. From here, $\angle P'PB = \angle PP'B = \angle A'P'Q$ as $P, P', Q$ collinear. From here, $A'P'QC$ is cyclic, and by power of a point we obtain $\frac{18}{5} \implies \boxed{023}$ .
~awang11's sol | null | 023 |
4cf1294ae0da292f791082ccf451f912 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | Let $a_{n}$ be the number of ways to form $n$ -letter strings made up of As and Bs such that no more than $3$ adjacent letters are identical.
Note that, at the end of each $n$ -letter string, there are $3$ possibilities for the last letter chain: it must be either $1$ $2$ , or $3$ letters long. Removing this last chain will make a new string that is $n-1$ $n-2$ , or $n-3$ letters long, respectively.
Therefore we can deduce that $a_{n}=a_{n-1}+a_{n-2}+a_{n-3}$
We can see that \[a_{1}=2\] \[a_{2}=2^{2}=4\] \[a_{3}=2^{3}=8\] so using our recursive relation we find \[a_{4}=14\] \[a_{5}=26\] \[a_{6}=48\] \[a_{7}=88\] \[a_{8}=162\] \[a_{9}=298\] \[a_{10}=\boxed{548}\] | null | 548 |
4cf1294ae0da292f791082ccf451f912 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | The solution is a simple recursion:
We have three cases for the ending of a string: three in a row, two in a row, and a single:
...AAA $(1)$ ...BAA $(2)$ ...BBA or ...ABA $(3)$
(Here, WLOG each string ends with A. This won't be the case when we actually solve for values in recursion.)
For case $(1)$ , we could only add a B to the end, making it a case $(3)$ .
For case $(2)$ , we could add an A or a B to the end, making it a case $(1)$ if you add an A, or a case $(3)$ if you add a B.
For case $(3)$ , we could add an A or a B to the end, making it a case $(2)$ or a case $(3)$
Let us create three series to represent the number of permutations for each case: $\{a\}$ $\{b\}$ , and $\{c\}$ representing case $(1)$ $(2)$ , and $(3)$ respectively.
The series have the following relationship:
$a_n=b_{n-1}$ $b_n=c_{n-1}$ $c_n=c_{n-1}+a_{n-1}+b_{n-1}$
For $n=3$ $a_3$ and $b_3$ both equal $2$ $c_3=4$ . With some simple math, we have: $a_{10}=88$ $b_{10}=162$ , and $c_{10}=298$ .
Summing the three up we have our solution: $88+162+298=\boxed{548}$ | null | 548 |
4cf1294ae0da292f791082ccf451f912 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | This is a recursion problem. Let $a_n$ be the number of valid strings of $n$ letters, where the first letter is $A$ . Similarly, let $b_n$ be the number of valid strings of $n$ letters, where the first letter is $B$
Note that $a_n=b_{n-1}+b_{n-2}+b_{n-3}$ for all $n\ge4$
Similarly, we have $b_n=a_{n-1}+a_{n-2}+a_{n-3}$ for all $n\ge4$
Here is why: every valid strings of $n$ letters $(n\ge4)$ where the first letter is $A$ must begin with one of the following:
$AAAB$ - and the number of valid ways is $b_{n-3}$
$AAB$ - and the number of valid ways is $b_{n-2}$
$AB$ - and there are $b_{n-1}$ ways.
We know that $a_1=1$ $a_2=2$ , and $a_3=4$ . Similarly, we have $b_1=1$ $b_2=2$ , and $b_3=4$ . We can quickly check our recursion to see if our recursive formula works. By the formula, $a_4=b_3+b_2+b_1=7$ , and listing out all $a_4$ , we can quickly verify our formula.
Therefore, we have the following:
$\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c}Value of n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\\hline a & 1 & 2 & 4 & 7 & 13 & 24 & 44 & 81 & 149 & 274\\b & 1 & 2 & 4 & 7 & 13 & 24 & 44 & 81 & 149 & 274\end{tabular}$
The total number of valid $10$ letter strings is equal to $a_{10}+b_{10}=274+274=\boxed{548}$ | null | 548 |
4cf1294ae0da292f791082ccf451f912 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | Playing around with strings gives this approach: We have a certain number of As, then Bs, and so on. Therefore, what if we denoted each solution with numbers like $(3,3,3,1)$ to denote AAABBBAAAB or vice versa (starting with Bs)? Every string can be represented like this!
We can have from 4 to 10 numbers in our parentheses. For each case, we will start with the largest number possible, usually a bunch of 3s, then go down systematically. Realize also that if we are left with just 2s and 1s, there is only one number of 2s and 1s that adds up to the leftover amount. Our final answer is the sum of all of these parenthetical sets [each set multiplied by its permutations, as order matters] multiplied by two [starting with either A or B, and alternating as we go along].
$4 \rightarrow (3,3,3,1) = 4, (3,3,2,2) = 6$
$5 \rightarrow (3, 3, 2, 1, 1) = 30, (3, 2, 2, 2, 1) = 20, (2,2,2,2,2)=1$
$6 \rightarrow (3,3,1,1,1,1) = 15, (3,2,2,1,1,1) = 60, (2,2,2,2,1,1) = 15$
$7 \rightarrow (3,2,1,1,1,1,1) = 42, (2,2,2,1,1,1,1) = 35$
$8 \rightarrow (3,1...1) = 8, (2,2,1...1) = 28$
$9 \rightarrow (2,1...1) = 9$
$10 \rightarrow (1,1....1) =1$
Adding them all up gives you 274; multiplying by 2 gives $\boxed{548}$ | null | 548 |
4cf1294ae0da292f791082ccf451f912 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | We are going to build the string, 1 character at a time. And, we are going to only care about the streak of letters at the end of the string.
Let $a_n$ be the number of strings of length n that satisfy the problem statement and also has a "streak" of length 1 at the end. ABABBA has a streak of length 1.
Let $b_n$ be the number of strings of length n that satisfy the problem statement and also has of length 2 at the end. ABABBAA has a streak of length 2. There are 2 "A" s at the end of the string.
Let $c_n$ be the number of string of length n that that satisfy the problem statement and also has a "streak" of length 3 at the end. ABABBAAA has a streak of length 3. There are 3 "A" s at the end of the string.
Let's establish a recursive relationship. $a_{n+1} = a_n+b_n+c_n$ , since you can simply break the streak. $b_{n+1} = a_n$ , and $c_{n+1} = b_n$ Since you can just add to the streak.
We can log everything using a table.
$\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c}Value of n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\\hline a & 2 & 2 & 4 & 8 & 14 & 26 & 48 & 88 & 162 & 298\\b & 0 & 2 & 2 & 4 & 8 & 14 & 26 & 48 & 88 & 162 \\c & 0 & 0 & 2 & 2 &4 & 8 & 14 & 26 & 48 & 88\end{tabular}$
Adding $a_{10}$ $b_{10}$ $c_{10}$ gets the total number of numbers that doesn't have more than 3 concecutive letters.
That gets a total of $298+162+88 = \boxed{548}$ | null | 548 |
4cf1294ae0da292f791082ccf451f912 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | Let $S_n$ be the number of n-letter strings satisfying the problem criteria. Then we can easily see $S_1 = 2, S_2 = 4, S_3 = 8, S_4 = 16-2 =14$ . For $n \ge 4$ consider the last three elements of the list. For example to get $S_5$ we could try to take $S_4$ and multiply by 2 since for each $S_4$ string we have two choices for the fifth character. However we have to be careful if the last three characters of $S_4$ are all the same, as in that case we would be overcounting, (i.e if we have a string $\dots aaa$ , we can only add $b$ to the end so that the problem criteria is satisfied. Additionally strings such as $\dots aaa$ or $\dots bbb$ should only be counted once as we only have one choice for the $n$ th character to add ( $b$ and $a$ respectively).
Thus to compute $S_n$ we start with $S_{n-1}$ then take out the strings ending in $\dots bbb$ or $\dots aaa$ . There are $S_{n-4}$ remaining valid strings (i.e if we pick any of the $S_{n-4}$ valid strings, examine the last character, if it is a $b$ then we append $aaa$ to it, and if if it an $a$ we append $bbb$ to it, hence the number of $S_{n-1}$ strings are in one-to-one correspondence with the number of strings in $S_{n-4}$ . Thus we have the recursion $S_n = 2(S_{n-1} - S_{n-4}) + S_{n-4}$ (where we are first taking away $S_{n-4}$ , doubling the result, then adding $S_{n-4}$ back in signifying that we only count it once, as described above).
The recursion simplifies to $S_n = 2S_{n-1} - S_{n-4}$ and we can now quickly compute the remaining values: \[S_5 = 2S_4 - S_1 = 26, S_6 = 2(26)-4 = 48, S_7 = 2(48) - 8 = 88, S_8 = 2(88) - 14 = 162, S_9 = 2(162) - 26 = 298, S_{10} = 2S_9 - 48 = \boxed{548}\] | null | 548 |
4cf1294ae0da292f791082ccf451f912 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | Let us define a "run" as a set of consecutive letters that are all the same and let a "maximal run" be a run that is not the proper subset of any other run, in other words, it cannot be expanded.
From now on, we consider all runs to be maximal.
Note that the minimum number of runs in the $10$ -letter string is $4$ (or else the Pigeonhole Principle tells us that at least $1$ run has $4$ or more letters in it, contradiction), and the maximum number of runs is clearly $10.$
Since an arbitrary run can have anywhere from $1$ to $3$ letters in it, it follows that the number of $10$ -letter strings with $n$ runs (where $4 \leq n \leq 10$ ) is \[2[x^{10}] (x+x^2+x^3)^n,\] i.e. twice the coefficient of $x^{10}$ in the expansion of $(x+x^2+x^3)^n.$ (Note that we multiplied by $2$ because there are two choices for which letter the first run has and then the rest are fixed).
Hence, we wish to find \[2\sum_{n=4}^{10} [x^{10}] (x+x^2+x^3)^n.\] First, we can rewrite this as
\begin{align*}
\sum_{n=4}^{10} [x^{10}] (x+x^2+x^3)^n &= \sum_{n=4}^{10} [x^{10}] \ x^n \cdot (1+x+x^2)^n \\
&= \sum_{n=4}^{10} [x^{10-n}] \left(\frac{1-x^3}{1-x}\right)^n \\
&= \sum_{n=4}^{10} [x^{10-n}] \frac{(1-x^3)^n}{(1-x)^n}. \\
\end{align*}
Now, we proceed case by case, utilizing the Binomial Theorem for the numerator in all of the cases:
For $n=4,$ we have
\begin{align*}
[x^6] \frac{(1-x^3)^4}{(1-x)^4} &= \binom{4}{0} [x^6] (1-x)^{-4} - \binom{4}{1} [x^3] (1-x)^{-4} + \binom{4}{2} [x^0] (1-x)^{-4} \\
&= 1\cdot\binom{9}{3} - 4\cdot\binom{6}{3} + 6\cdot\binom{3}{3} \\
&= 1\cdot84-4\cdot20+6\cdot1 \\
&= 84-80+6 \\
&= \underline{10}.
\end{align*}
For $n=5,$ we have
\begin{align*}
[x^5] \frac{(1-x^3)^5}{(1-x)^5} &= \binom{5}{0} [x^5] (1-x)^{-5} - \binom{5}{1}[x^2] (1-x)^{-5} \\
&= 1\cdot\binom{9}{4}-5\cdot\binom{6}{4} \\
&= 1\cdot126-5\cdot15 \\
&= 126-75 \\
&= \underline{51}.
\end{align*}For $n=6,$ we have
\begin{align*}
[x^4] \frac{(1-x^3)^6}{(1-x)^6} &= \binom{6}{0} [x^4](1-x)^{-6} - \binom{6}{1}[x^1](1-x)^{-6} \\
&= 1 \cdot \binom{9}{5} - 6 \cdot \binom{6}{5} \\
&= 1 \cdot 126 - 6 \cdot 6 \\
&= 126-36 \\
&= \underline{90}.
\end{align*}For $n=7,$ we have
\begin{align*}
[x^3] \frac{(1-x^3)^7}{(1-x)^7} &= \binom{7}{0}[x^3](1-x)^{-7} - \binom{7}{1}[x^0](1-x)^{-7} \\
&= 1\cdot \binom{9}{6} - 7\cdot\binom{6}{6} \\
&= 1\cdot84-7\cdot1\\
&= 84-7 \\
&= \underline{77}.
\end{align*}For $n=8,$ we have
\begin{align*}
[x^2] \frac{(1-x^3)^8}{(1-x)^8} &=\binom{8}{0} [x^2](1-x)^{-8} \\
&= 1\cdot\binom{9}{7} \\
&= 1\cdot36 \\
&= \underline{36}.
\end{align*}For $n=9,$ we have
\begin{align*}
[x^1] \frac{(1-x^3)^9}{(1-x)^9} &= \binom{9}{0} [x^1] (1-x)^{-9}\\
&= 1\cdot\binom{9}{8} \\
&= 1\cdot9 \\
&= \underline{9}.
\end{align*}For $n=10,$ we have
\begin{align*}
[x^0] \frac{(1-x^3)^{10}}{(1-x)^{10}} &= [x^0] (1-x)^{-10} \\
&= 1\cdot\binom{9}{9} \\
&= 1\cdot1 \\
&= \underline{1}.
\end{align*}Hence, the answer is
\begin{align*}
2\sum_{n=4}^{10} [x^{10-n}] \frac{(1-x^3)^n}{(1-x)^n} &= 2(10+51+90+77+36+9+1) \\
&= 2\cdot274 \\
&= \boxed{548}.
\end{align*} | null | 548 |
ec8ad5e73005e4b8ec686ef7389e2833 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_13 | Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$ | If $n = 1$ $a_n = \sin(1) > 0$ . Then if $n$ satisfies $a_n < 0$ $n \ge 2$ , and \[a_n = \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{2\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{2\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].\] Since $2\sin 1$ is positive, it does not affect the sign of $a_n$ . Let $b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)$ . Now since $\cos(0) + \cos(1) = 2\cos\left(\cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)$ and $\cos(n) + \cos(n + 1) = 2\cos\left(n + \cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)$ $b_n$ is negative if and only if $\cos\left(\cfrac{1}{2}\right) < \cos\left(n + \cfrac{1}{2}\right)$ , or when $n \in [2k\pi - 1, 2k\pi]$ . Since $\pi$ is irrational, there is always only one integer in the range, so there are values of $n$ such that $a_n < 0$ at $2\pi, 4\pi, \cdots$ . Then the hundredth such value will be when $k = 100$ and $n = \lfloor 200\pi \rfloor = \lfloor 628.318 \rfloor = \boxed{628}$ | null | 628 |
ec8ad5e73005e4b8ec686ef7389e2833 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_13 | Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$ | Notice that $a_n$ is the imaginary part of $\sum_{k=1}^n e^{ik}$ , by Euler's formula. Using the geometric series formula, we find that this sum is equal to \[\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}\] We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have \[\frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}\] We only need to look at the imaginary part, which is \[\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}\] Since $\cos 1 < 1$ $2-2 \cos 1 > 0$ , so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have $\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(n + \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( n + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),$ by sum to product. This only holds when $n$ is between $2\pi k - 1$ and $2\pi k$ for integer $k$ [continuity proof here], and since this has exactly one integer solution for every such interval, the $100$ th such $n$ is $\lfloor 200\pi \rfloor = \boxed{628}$ | null | 628 |
ec8ad5e73005e4b8ec686ef7389e2833 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_13 | Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$ | Similar to solution 2, we set a complex number $z=\cos 1+i\sin 1$ . We start from $z$ instead of $1$ because $k$ starts from $1$ : be careful.
The sum of $z+z^2+z^3+z^4+z^5\dots=\frac{z-z^{n+1}}{1-z}=\frac{z}{z-1}\left(z^n-1\right)$
We are trying to make $n$ so that the imaginary part of this expression is negative.
The argument of $z$ is $1$ . The argument of $z-1$ , however, is a little more tricky. $z-1$ is on a circle centered on $(-1,0)$ with radius $1$ . The change in angle due to $z$ is $1$ with respect to the center, but the angle that $z-1$ makes with the $y$ -axis is $half$ the change, due to Circle Theorems (this intercepted arc is the argument of $z$ ), because the $y$ - axis is tangent to the circle at the origin. So $\text{arg}(z-1)=\frac{\pi+1}{2}$ . Dividing $z$ by $z-1$ subtracts the latter argument from the former, so the angle of the quotient with the $x$ -axis is $\frac{1-\pi}{2}$
We want the argument of the whole expression $-\pi<\theta<0$ . This translates into $\frac{-\pi-1}{2}<\text{arg}\left(z^n-1\right)<\frac{\pi-1}{2}$ $z^n-1$ also consists of points on the circle centered at $(-1,0)$ , so we deal with this argument similarly: the argument of $z^n$ is twice the angle $z^n-1$ makes with the $y$ -axis. Since $z^n-1$ is always negative, $\frac{-3\pi}{2}<\text{arg}\left(z^n-1\right)<\frac{-\pi}{2}$ , and the left bound is the only one that is important. Either way, the line (the line consists of both bounds) makes a $\frac{\pi}{2}-\frac{\pi-1}{2}=\frac{-1}{2}$ angle with the $y$ -axis both ways.
So the argument of $z^n$ must be in the bound $-1<\theta<0$ by doubling, namely the last $z^n$ negative before another rotation. Since there is always one $z^n$ in this category per rotation because $\pi$ is irrational, $n_{100}\equiv z^{628}$ and the answer is $\boxed{628}$ | null | 628 |
d5e74aebbcaa9b209beea8f75d706e4d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | The expression we want to find is $2(x^3+y^3) + x^3y^3$
Factor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x^3+y^3)=945$ , respectively. Dividing the latter by the former equation yields $\frac{x^2-xy+y^2}{xy} = \frac{945}{810}$ . Adding 3 to both sides and simplifying yields $\frac{(x+y)^2}{xy} = \frac{25}{6}$ . Solving for $x+y$ and substituting this expression into the first equation yields $\frac{5\sqrt{6}}{6}(xy)^{\frac{9}{2}} = 810$ . Solving for $xy$ , we find that $xy = 3\sqrt[3]{2}$ , so $x^3y^3 = 54$ . Substituting this into the second equation and solving for $x^3+y^3$ yields $x^3+y^3=\frac{35}{2}$ . So, the expression to evaluate is equal to $2 \times \frac{35}{2} + 54 = \boxed{089}$ | null | 089 |
d5e74aebbcaa9b209beea8f75d706e4d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | Factor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x+y)(x^2-xy+y^2)=945$ , respectively. By the first equation, $x+y=\frac{810}{x^4y^4}$ . Plugging this in to the second equation and simplifying yields $(\frac{x}{y}-1+\frac{y}{x})=\frac{7}{6}$ . Now substitute $\frac{x}{y}=a$ . Solving the quadratic in $a$ , we get $a=\frac{x}{y}=\frac{2}{3}$ or $\frac{3}{2}$ As both of the original equations were symmetric in $x$ and $y$ , WLOG, let $\frac{x}{y}=\frac{2}{3}$ , so $x=\frac{2}{3}y$ . Now plugging this in to either one of the equations, we get the solutions $y=\frac{3(2^{\frac{2}{3}})}{2}$ $x=2^{\frac{2}{3}}$ . Now plugging into what we want, we get $8+54+27=\boxed{089}$ | null | 089 |
d5e74aebbcaa9b209beea8f75d706e4d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | Add three times the first equation to the second equation and factor to get $(xy)^3(x^3+3x^2y+3xy^2+y^3)=(xy)^3(x+y)^3=3375$ . Taking the cube root yields $xy(x+y)=15$ . Noting that the first equation is $(xy)^3\cdot(xy(x+y))=810$ , we find that $(xy)^3=\frac{810}{15}=54$ . Plugging this into the second equation and dividing yields $x^3+y^3 = \frac{945}{54} = \frac{35}{2}$ . Thus the sum required, as noted in Solution 1, is $54+\frac{35}{2}\cdot2 = \boxed{089}$ | null | 089 |
d5e74aebbcaa9b209beea8f75d706e4d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | As with the other solutions, factor. But this time, let $a=xy$ and $b=x+y$ . Then $a^4b=810$ . Notice that $x^3+y^3 = (x+y)(x^2-xy+y^2) = b(b^2-3a)$ , so we are looking for $2b(b^2-3a)+a^3$ . Now, if we divide the second equation by the first one, we get $7/6 = \frac{b^2-3a}{a}$ ; then $\frac{b^2}{a}=\frac{25}{6}$ . Therefore, $a = \frac{6}{25}b^2$ . Substituting $\frac{6}{25}b^2$ for $a$ in equation 1, simplifying, and then taking the cube root gives us $b^3 = \frac{5^3}{2}.$ Finding $a^3$ by cubing $a = \frac{6}{25}b^2$ on both sides and simplifying using our previous substitution, we get $a^3 = 54$ . Substituting this into the first equation and then dividing by $27$ , we get $2b(b^2 - 3a) = 35$ . Our final answer is $35+54=\boxed{089}$ | null | 089 |
d5e74aebbcaa9b209beea8f75d706e4d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | Factor the given equations as: \[x^4y^4(x+y)=810\] \[x^3y^3(x^3+y^3)=x^3y^3(x+y)(x^2-xy+y^2)=945\] We note that these expressions (as well as the desired expression) can be written exclusively in terms of $x+y$ and $xy$ . We make the substitution $s=x+y$ and $p=xy$ (for sum and product, respectively).
\[x^4y^4(x+y)=p^4s=810\] \[x^3y^3(x+y)(x^2-xy+y^2)=p^3(s)(s^2-3p)=s^3p^3-3p^4s=945\]
We see that $p^4s$ shows up in both equations, so we can eliminate it and find $sp$ , after which we can get $p^3$ from the first equation. If you rewrite the desired expression using $s$ and $p$ , it becomes clear that you don't need to actually find the values of $s$ and $p$ , but I will do so for the sake of completion.
\[s^3p^3=945+3p^4s\] \[s^3p^3=945+3(810)=3375\] \[sp=15\]
\[p^3=\frac{810}{sp}=54\] \[p=3\cdot2^{1/3}\] \[s=\frac{15}{p}=5\cdot2^{-1/3}\]
The desired expression can be written as: \[2(x^3+y^3)+(xy)^3=2(x+y)(x^2-xy+y^2)+(xy)^3\] \[2(s)(s^2-3p)+p^3=2s^3-6sp+p^3\]
Plugging in $s$ and $p$ , we get: \[2(5\cdot2^{-1/3})^3-6(15)+54=125-90+54=\boxed{089}\] | null | 089 |
d5e74aebbcaa9b209beea8f75d706e4d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | Factor the first and second equations as $(xy)^4(x+y)=810$ and $(xy)^3(x+y)(x^2-xy+y^2)=945$ , respectively. Dividing them (allowed, since neither are $0$ ), we have \[\frac{xy}{x^2-xy+y^2}=\frac67\] or \[x^2-\frac{13}{6}xy+y^2=0.\] Plugging into the quadratic formula and solving for $x$ in terms of $y,$ we have \[x=\frac{\frac{13y}{6}\pm \sqrt{\frac{169y^2}{36}-4y^2}}{2}=\frac{2y}3 , \frac{3y}2 .\] WLOG, let $x=\frac{3y}2 .$ Plugging into our first equation, we have \[\left(\frac{3}{2}y\right)^4\left(\frac52 y\right)=810 \implies y^3 = 4 .\] Plugging this result (and the one for $x$ in terms of $y$ ) into our desired expression, we have
\begin{align*} 2x^3+(xy)^3+2y^3 &= \frac{27}{4}y^3 + \left(\frac{3}{2} y^2\right)^3 +2y^3 \\ &= \frac{35}{4}y^3 +\frac{27}{8}(y^3)^2 \\ &= 35+54 \\ &= \boxed{089} ~ASAB | null | 089 |
d5e74aebbcaa9b209beea8f75d706e4d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | Take $w=x+y$ and $z=xy$ . Remark that \begin{align*} ~&2x^{3}+(xy)^{3}+2y^{3} \\ =~&2(x^{3}+y^{3})+(xy)^{3} \\ =~&2\left[(x+y)^{3}-3xy(x+y)\right]+(xy)^{3} \\ =~&2(w^{3}-3wz)+z^{3} \\ =~& 2w^{3}-6wz+z^{3}.\end{align*} The given equations imply that \[wz^{4}=810~~~\text{and}~~~(wz)^{3}-3wz^{4}=945.\] Substituting the first equation into the second, we have that $(wz)^{3}=945+3\cdot 810=3375$ , thus $wz=\sqrt[3]{3375}=15$ . Now \[z^{3}=\frac{wz^{4}}{wz}=\frac{810}{15}=54\] and \[w^{3}=\frac{3375}{54}=\frac{125}{2}.\] Thus \begin{align*}2w^{3}-6wz+z^{3}&=2\left(\frac{125}{2}\right)-6(15)+54 \\ &=125-90+54 \\ &=\boxed{89} | null | 89 |
d5e74aebbcaa9b209beea8f75d706e4d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | $x^4y^4(x+y)=810; x^3y^3(x^3+y^3)=945, \frac{x^2-xy+y^2}{xy}=\frac{7}{6}, \frac{x^2+y^2}{xy}=\frac{13}{6}$
Let $x^2+y^2=13k; xy=6k$ , then we can see $(x+y)^2-12k=13k, x+y=5\sqrt{k}$ , now, we see $x^4y^4\cdot (x+y)=1296k^4\cdot 5\sqrt{k}, k=\frac{1}{\sqrt[3]{4}}$
The rest is easy, $2(x^3+y^3)+x^3y^3=216k^3+2[(x+y)^3-3xy(x+y)]=216k^3+2\cdot 35k^{\frac{3}{2}}=\boxed{89}$ | null | 89 |
d5e74aebbcaa9b209beea8f75d706e4d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | Let's first put the left hand sides of the equations in factored forms. Doing this we obtain $(xy)^4 \cdot (x + y) = 810$ and $(xy)(x + y)(x^2 - xy + y^2) = 945$ . Now, we will subtract and add the equations to gather information on $x$ and $y$ . When we subtract the equations and clean it up via factoring, we yield $(xy)^3 \cdot (x + y) \cdot (x - y)^2 = 115$ , and when we add them, we yield $(xy)^3 \cdot (x + y) \cdot (x^2 + y^2) = 1755$ . Now with some intuition, you should divide the equations to obtain $\frac{(x^2 + y^2)}{(x - y)^2} = 13$ . Now, we clean this up to obtain the following factoring of $0 = 2 \cdot (2x - 3y) \cdot (3x - 2y)$ . This implies that $x = \frac{3y}{2}$ . We plug that into the target expression to reduce it down to one variable, and get that target expression is $2x^3 + (xy)^3 + 2y^3 = \frac{27}{4} \cdot y^3 + \frac{27}{8} \cdot y^6 + 2y^3$ . This means that if we can find a way to get $y^3$ , then the rest is trivial. We get $y^3$ by plugging in $x = \frac{3y}{2}$ into $x^3 \cdot y^6 + y^3 \cdot x^6 = 945$ . However, this time we only factor as $(xy)^3 \cdot (x^3 + y^3)$ because we particularly want a cubic degree on $y$ . Plugging in $x = \frac{3y}{2}$ we get $y^3 = 4$ . Now lets plug this into our target expression to get $\frac{27}{4} \cdot 4 + \frac{27}{8} \cdot 4^2 + 2 \cdot 4 = \boxed{89}$ | null | 89 |
d5e74aebbcaa9b209beea8f75d706e4d | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | Let $x, y$ be roots of some polynomial $P(\alpha)$ . Then we have that $P(\alpha) = (\alpha - x)(\alpha - y) \implies P(\alpha) = \alpha^2 - (x + y) \alpha + xy$ . Now from the first equation we have that $x + y = 810/(xy)^4$ . For convenience, denote $z = xy$
Define $P_k = x^k + y^k$ , where $x, y$ are roots of the above polynomial. Then we have that by Newton's formulas $P_k = (x + y)P_{k - 1} - xyP_{k - 2} = 810/z^4P_{k - 1} - zP_{k - 2}$ , where $P_0 = 2$ and $P_1 = x + y = 810/z^4$ . We desire $2P_3 + z^3$ . Now building up this recurrence;
\[P_2 = \frac{810}{z^4}P_1 - zP_0 = \frac{810^2}{z^8} - 2z.\] Then we have that \[P_3 = \frac{810}{z^4} \left(\frac{810^2}{z^8} - 3z \right).\] Plugging this value of $P_3$ into the second equation $z^3P_3 = 810$ , yields $z^9 = 164025$ . Now $z^3 = 54$ . To compute $P_3$ is now trivial, as returning this new value of $z$ into the second equation yields $P_3 = 35/2$ . Hence $z^3 + 2P_3 = 54 + 35 = \boxed{089}$ | null | 089 |
5910271583103986a1b6749ec6d74d50 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy] | Let $M$ be the intersection of $\overline{BC}$ and the common internal tangent of $\mathcal P$ and $\mathcal Q.$ We claim that $M$ is the circumcenter of right $\triangle{ABC}.$ Indeed, we have $AM = BM$ and $BM = CM$ by equal tangents to circles, and since $BM = CM, M$ is the midpoint of $\overline{BC},$ implying that $\angle{BAC} = 90.$ Now draw $\overline{PA}, \overline{PB}, \overline{PM},$ where $P$ is the center of circle $\mathcal P.$ Quadrilateral $PAMB$ is cyclic, and by Pythagorean Theorem $PM = \sqrt{5},$ so by Ptolemy on $PAMB$ we have \[AB \sqrt{5} = 2 \cdot 1 + 2 \cdot 1 = 4 \iff AB = \dfrac{4 \sqrt{5}}{5}.\] Do the same thing on cyclic quadrilateral $QAMC$ (where $Q$ is the center of circle $\mathcal Q$ and get $AC = \frac{8 \sqrt{5}}{5}.$
Let $\angle A = \angle{DAB}.$ By Law of Sines, $BD = 2R \sin A = 2 \sin A.$ Note that $\angle{D} = \angle{ABC}$ from inscribed angles, so \begin{align*} [ABD] &= \dfrac{1}{2} BD \cdot AB \cdot \sin{\angle B} \\ &= \dfrac{1}{2} \cdot \dfrac{4 \sqrt{5}}{5} \cdot 2 \sin A \sin{\left(180 - \angle A - \angle D\right)} \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \sin{\left(\angle A + \angle D\right)} \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos D + \cos A \sin D\right) \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos{\angle{ABC}} + \cos A \sin{\angle{ABC}}\right) \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\dfrac{\sqrt{5} \sin A}{5} + \dfrac{2 \sqrt{5} \cos A}{5}\right) \\ &= \dfrac{4}{5} \cdot \sin A \left(\sin A + 2 \cos A\right) \end{align*} after angle addition identity.
Similarly, $\angle{EAC} = 90 - \angle A,$ and by Law of Sines $CE = 8 \sin{\angle{EAC}} = 8 \cos A.$ Note that $\angle{E} = \angle{ACB}$ from inscribed angles, so \begin{align*} [ACE] &= \dfrac{1}{2} AC \cdot CE \sin{\angle C} \\ &= \dfrac{1}{2} \cdot \dfrac{8 \sqrt{5}}{5} \cdot 8 \cos A \sin{\left[180 - \left(90 - \angle A\right) - \angle E\right]} \\ &= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \sin{\left[\left(90 - \angle A\right) + \angle{ACB}\right]} \\ &= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \left(\dfrac{2 \sqrt{5} \cos A}{5} + \dfrac{\sqrt{5} \sin A}{5}\right) \\ &= \dfrac{32}{5} \cdot \cos A \left(\sin A + 2 \cos A\right) \end{align*} after angle addition identity.
Setting the two areas equal, we get \[\tan A = \frac{\sin A}{\cos A} = 8 \iff \sin A = \frac{8}{\sqrt{65}}, \cos A = \frac{1}{\sqrt{65}}\] after Pythagorean Identity. Now plug back in and the common area is $\frac{64}{65} \iff \boxed{129}.$ | null | 129 |
5910271583103986a1b6749ec6d74d50 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy] | [asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label("$A$",A,NE);label("$O_1$",O_1,NE);label("$O_2$",O_2,NE);label("$B$",B,SW);label("$C$",C,SW);label("$D$",D,NE);label("$E$",E,NE);label("$N$",N,W);label("$K$",(-24/15,0.2));label("$L$",(24/15,0.2));label("$n$",(-0.8,-0.12));label("$p$",((29/15,-48/15)));label("$\mathcal{P}$",(-1.6,1.1));label("$\mathcal{Q}$",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); [/asy]
Call $O_1$ and $O_2$ the centers of circles $\mathcal{P}$ and $\mathcal{Q}$ , respectively, and extend $CB$ and $O_2O_1$ to meet at point $N$ . Call $K$ and $L$ the feet of the altitudes from $B$ to $O_1N$ and $C$ to $O_2N$ , respectively. Using the fact that $\triangle{O_1BN} \sim \triangle{O_2CN}$ and setting $NO_1 = k$ , we have that $\frac{k+5}{k} = \frac{4}{1} \implies k=\frac{5}{3}$ . We can do some more length chasing using triangles similar to $O_1BN$ to get that $AK = AL = \frac{24}{15}$ $BK = \frac{12}{15}$ , and $CL = \frac{48}{15}$ . Now, consider the circles $\mathcal{P}$ and $\mathcal{Q}$ on the coordinate plane, where $A$ is the origin. If the line $\ell$ through $A$ intersects $\mathcal{P}$ at $D$ and $\mathcal{Q}$ at $E$ then $4 \cdot DA = AE$ . To verify this, notice that $\triangle{AO_1D} \sim \triangle{EO_2A}$ from the fact that both triangles are isosceles with $\angle{O_1AD} \cong \angle{O_2AE}$ , which are corresponding angles. Since $O_2A = 4\cdot O_1A$ , we can conclude that $4 \cdot DA = AE$
Hence, we need to find the slope $m$ of line $\ell$ such that the perpendicular distance $n$ from $B$ to $AD$ is four times the perpendicular distance $p$ from $C$ to $AE$ . This will mean that the product of the bases and heights of triangles $ACE$ and $DBA$ will be equal, which in turn means that their areas will be equal. Let the line $\ell$ have the equation $y = -mx \implies mx + y = 0$ , and let $m$ be a positive real number so that the negative slope of $\ell$ is preserved. Setting $A = (0,0)$ , the coordinates of $B$ are $(x_B, y_B) = \left(\frac{-24}{15}, \frac{-12}{15}\right)$ , and the coordinates of $C$ are $(x_C, y_C) = \left(\frac{24}{15}, \frac{-48}{15}\right)$ . Using the point-to-line distance formula and the condition $n = 4p$ , we have \[\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}\] \[\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.\] If $m > 2$ , then clearly $B$ and $C$ would not lie on the same side of $\ell$ . Thus since $m > 0$ , we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have \[\frac{6m}{15} + \frac{3}{15} = \frac{48}{15} - \frac{24m}{15} \implies 2m = 3 \implies m = \frac{3}{2}.\] Thus, the equation of $\ell$ is $y = -\frac{3}{2}x$
Then we can find the coordinates of $D$ by finding the point $(x,y)$ other than $A = (0,0)$ where the circle $\mathcal{P}$ intersects $\ell$ $\mathcal{P}$ can be represented with the equation $(x + 1)^2 + y^2 = 1$ , and substituting $y = -\frac{3}{2}x$ into this equation yields $x = 0, -\frac{8}{13}$ as solutions. Discarding $x = 0$ , the $y$ -coordinate of $D$ is $-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}$ . The distance from $D$ to $A$ is then $\frac{4}{\sqrt{13}}.$ The perpendicular distance from $B$ to $AD$ or the height of $\triangle{DBA}$ is $\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.$ Finally, the common area is $\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}$ , and $m + n = 64 + 65 = \boxed{129}$ | null | 129 |
5910271583103986a1b6749ec6d74d50 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy] | By homothety , we deduce that $AE = 4 AD$ . (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of $P$ and $Q$ to $l$ .) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from $B$ to $l$ is four times that from $C$ to $l$ . Let the distance from $C$ be $x$ and the distance from $B$ be $4x$
Let $P$ and $Q$ be the centers of their respective circles. Then dropping a perpendicular from $P$ to $Q$ creates a $3-4-5$ right triangle, from which $BC = 4$ and, if $\alpha = \angle{AQC}$ , that $\cos \alpha = \dfrac{3}{5}$ . Then $\angle{BPA} = 180^\circ - \alpha$ , and the Law of Cosines on triangles $APB$ and $AQC$ gives $AB = \dfrac{4}{\sqrt{5}}$ and $AC = \dfrac{8}{\sqrt{5}}.$
Now, using the Pythagorean Theorem to express the length of the projection of $BC$ onto line $l$ gives \[\sqrt{\frac{16}{5} - 16x^2} + \sqrt{\frac{64}{5} - x^2} = \sqrt{16 - 9x^2}.\] Squaring and simplifying gives \[\sqrt{\left(\frac{1}{5} - x^2\right)\left(\frac{64}{5} - x^2\right)} = x^2,\] and squaring and solving gives $x = \dfrac{8}{5\sqrt{13}}.$
By the Law of Sines on triangle $ABD$ , we have \[\frac{BD}{\sin A} = 2.\] But we know $\sin A = \dfrac{4x}{AB}$ , and so a small computation gives $BD = \dfrac{16}{\sqrt{65}}.$ The Pythagorean Theorem now gives \[AD = \sqrt{BD^2 - (4x)^2} + \sqrt{AB^2 - (4x)^2} = \frac{4}{\sqrt{13}},\] and so the common area is $\dfrac{1}{2} \cdot \frac{4}{\sqrt{13}} \cdot \frac{32}{5\sqrt{13}} = \frac{64}{65}.$ The answer is $\boxed{129}.$ | null | 129 |
5910271583103986a1b6749ec6d74d50 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy] | $DE$ goes through $A$ , the point of tangency of both circles. So $DE$ intercepts equal arcs in circle $P$ and $Q$ homothety . Hence, $AE=4AD$ . We will use such similarity later.
The diagonal distance between the centers of the circles is $4+1=5$ . The difference in heights is $4-1=3$ . So $BC=\sqrt{5^2-3^2}=4$
The triangle connecting the centers with a side parallel to $BC$ is a $3-4-5$ right triangle. Since $O_PA=1$ , the height of $A$ is $1+3/5=8/5$ . Drop an altitude from $A$ to $BC$ and call it $I$ $IB=4/5$ and $IC=4-4/5=32/5$ . Since right $\triangle AIB\sim\triangle CIB$ $ABC$ is a right triangle also; $IB:IA:IC$ form a geometric progression $\times 2$
Extend $BA$ through $A$ to a point $G$ on the other side of $\circ Q$ . By homothety $\triangle DAB\sim\triangle EAG$ . By angle chasing $\triangle DAB$ through right triangle $ABC$ , we deduce that $\angle CEG$ is a right angle. Since $ACEG$ is cyclic, $\angle GAC$ is also right. So $CG$ is a diameter of $\circ G$ . Because of this, $CG \perp BC$ , the tangent line. $\triangle BCG$ is right and $\triangle BCG\sim\triangle ABC\sim\triangle CAG$
$AC=\sqrt{(8/5)^2+(16/5)^2}=8\sqrt{5}/5$ so $AG=2AC=16\sqrt{5}/5$ and $[\triangle CAG]=64/5$
Since $[\triangle DAB]=[\triangle ACE]$ , the common area is $[ACEG]/17$ $16[\triangle DAB]=[\triangle GAE]$ because the triangles are similar with a ratio of $1:4$ . So we only need to find $[\triangle CEG]$ now.
Extend $DE$ through $E$ to intersect the tangent at $F$ . Because $4DA=AE$ , the altitude from $B$ to $AD$ is $4$ times the height from $C$ to $EA$ . So $BC=3/4BF$ and $BF=16/3$ . We look at right triangle $\triangle AIF$ $IF=68/15$ and $AI=8/5$ $\triangle AIF$ is a $17-6-5\sqrt{13}$ right triangle. Hypotenuse $AF$ intersects $CG$ at a point, we call it $H$ $CH=4/3\div 68/15\cdot 8/5=8/17$ . So $HG=8-8/17=128/17$
By Power of a Point $CH\cdot HG=AH\cdot HE$ $AH=16/5\cdot 5\sqrt{13}/17=16\sqrt{13}/17.$ So $HE=1024/289\cdot 17/(16\sqrt{13})=64/(17\sqrt{13})$ . The height from $E$ to $CG$ is $17/(5\sqrt{13})\cdot 64/(17\sqrt{13})=64/65$
Thus, $[\triangle CEG]=64/65\cdot 8\div 2=256/65$ . The area of the whole cyclic quadrilateral is $64/5+256/65=(832+256)/65=1088/65$ . Lastly, the common area is $1/17$ the area of the quadrilateral, or $64/65$ . So $64+65=\boxed{129}$ | null | 129 |