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e68d1807a0fd51eae4980b82f1f35968 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_6 | In triangle $ABC,$ $AB = 13,$ $BC = 14,$ $AC = 15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area of the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$ | First, find the area of $\Delta ABC$ either like the first solution or by using Heron’s Formula. Then, draw the medians from $G$ to each of $A, B, C, A’, B’,$ and $C’$ . Since the medians of a triangle divide the triangle into 6 triangles with equal area, we can find that each of the 6 outer triangles have equal area. (Proof: Since I’m too lazy to draw out a diagram, I’ll just have you borrow the one above. Draw medians $GA$ and $GB’$ , and let’s call the points that $GA$ intersects $C’B’$ $H$ ” and the point $GB’$ intersects $AC$ $I$ ”. From the previous property and the fact that both $\Delta ABC$ and $\Delta A’B’C’$ are congruent, $\Delta GHB’$ has the same area as $\Delta GIA$ . Because of that, both “half” triangles created also have the same area. The same logic can be applied to all other triangles).
Also, since the centroid of a triangle divides each median with the ratio $2:1$ , along with the previous fact, each outer triangle has $1/9$ the area of $\Delta ABC$ and $\Delta A’B’C’$ . Thus, the area of the region required is $\frac{4}{3}$ times the area of $\Delta ABC$ which is $\boxed{112}$ | null | 112 |
e68d1807a0fd51eae4980b82f1f35968 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_6 | In triangle $ABC,$ $AB = 13,$ $BC = 14,$ $AC = 15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area of the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$ | $[ABC]$ can be calculated as 84 using Heron's formula or other methods. Since a $180^{\circ}$ rotation is equivalent to reflection through a point, we have a homothety with scale factor $-1$ from $ABC$ to $A'B'C'$ through the centroid $G$ . Let $M$ be the midpoint of $BC$ which maps to $M'$ and note that $A'G=AG=2GM,$ implying that $GM=MA'.$ Similarly, we have $AM'=M'G=GM=MA'.$ Also let $D$ and $E$ be the intersections of $BC$ with $A'B'$ and $A'C',$ respectively. The homothety implies that we must have $DE || B'C',$ so there is in fact another homothety centered at $A'$ taking $A'DE$ to $A'B'C'$ . Since $A'M'=3A'M,$ the scale factor of this homothety is 3 and thus $[A'DE]=\frac{1}{9}[A'B'C']=\frac{1}{9}[ABC].$ We can apply similar reasoning to the other small triangles in $A'B'C'$ not contained within $ABC$ , so our final answer is $[ABC]+3\cdot\frac{1}{9}[ABC]=\boxed{112.}$ | null | 112. |
fa4459b3bd9886ca18d3641717e97567 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_7 | Find the area of rhombus $ABCD$ given that the circumradii of triangles $ABD$ and $ACD$ are $12.5$ and $25$ , respectively. | The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD $a$ and half of diagonal AC $b$ . The length of the four sides of the rhombus is $\sqrt{a^2+b^2}$
The area of any triangle can be expressed as $\frac{a\cdot b\cdot c}{4R}$ , where $a$ $b$ , and $c$ are the sides and $R$ is the circumradius. Thus, the area of $\triangle ABD$ is $ab=2a(a^2+b^2)/(4\cdot12.5)$ . Also, the area of $\triangle ABC$ is $ab=2b(a^2+b^2)/(4\cdot25)$ . Setting these two expressions equal to each other and simplifying gives $b=2a$ . Substitution yields $a=10$ and $b=20$ , so the area of the rhombus is $20\cdot40/2=\boxed{400}$ | null | 400 |
fa4459b3bd9886ca18d3641717e97567 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_7 | Find the area of rhombus $ABCD$ given that the circumradii of triangles $ABD$ and $ACD$ are $12.5$ and $25$ , respectively. | Let $\theta=\angle BDA$ . Let $AB=BC=CD=x$ . By the extended law of sines, \[\frac{x}{\sin\theta}=25\] Since $AC\perp BD$ $\angle CAD=90-\theta$ , so \[\frac{x}{\sin(90-\theta)=\cos\theta}=50\] Hence $x=25\sin\theta=50\cos\theta$ . Solving $\tan\theta=2$ $\sin\theta=\frac{2}{\sqrt{5}}, \cos\theta=\frac{1}{\sqrt{5}}$ . Thus \[x=25\frac{2}{\sqrt{5}}\implies x^2=500\] The height of the rhombus is $x\sin(2\theta)=2x\sin\theta\cos\theta$ , so we want \[2x^2\sin\theta\cos\theta=\boxed{400}\] | null | 400 |
4f0c67827bfa09407cf42707a3e44a91 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_8 | Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences. | If you multiply the corresponding terms of two arithmetic sequences, you get the terms of a quadratic function. Thus, we have a quadratic $ax^2+bx+c$ such that $f(1)=1440$ $f(2)=1716$ , and $f(3)=1848$ . Plugging in the values for x gives us a system of three equations:
$a+b+c=1440$
$4a+2b+c=1716$
$9a+3b+c=1848$
Solving gives $a=-72, b=492,$ and $c=1020$ . Thus, the answer is $-72(8)^2+492\cdot8+1020= \boxed{348}.$ | null | 348 |
4f0c67827bfa09407cf42707a3e44a91 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_8 | Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences. | Use the same rationale as in solution 1; instead of using terms $1,2,3$ , we use $-1,0,1$ and solve the $6$ th term.
$a-b+c=1440$
$c=1716$
$a+b+c=1848$
Accordingly we will solve
$a=-72, b=204, c=1716$
$36a+6b+c= \boxed{348}.$ | null | 348 |
4f0c67827bfa09407cf42707a3e44a91 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_8 | Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences. | Setting one of the sequences as $a+nr_1$ and the other as $b+nr_2$ , we can set up the following equalities
$ab = 1440$
$(a+r_1)(b+r_2)=1716$
$(a+2r_1)(b+2r_2)=1848$
We want to find $(a+7r_1)(b+7r_2)$
Foiling out the two above, we have
$ab + ar_2 + br_1 + r_1r_2 = 1716$ and $ab + 2ar_2 + 2br_1 + 4r_1r_2 = 1848$
Plugging in $ab=1440$ and bringing the constant over yields
$ar_2 + br_1 + r_1r_2 = 276$
$ar_2 + br_1 + 2r_1r_2 = 204$
Subtracting the two yields $r_1r_2 = -72$ and plugging that back in yields $ar_2 + br_1 = 348$
Now we find
$(a+7r_1)(b+7r_2) = ab + 7(ar_2 + br_1) + 49r_1r_2 = 1440 + 7(348) + 49(-72) = \boxed{348}$ | null | 348 |
f18e460d0cbfbabe15788dab748a36c2 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_9 | Consider the polynomials $P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x$ and $Q(x) = x^{4} - x^{3} - x^{2} - 1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x) = 0,$ find $P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).$ | When we use long division to divide $P(x)$ by $Q(x)$ , the remainder is $x^2-x+1$
So, since $z_1$ is a root, $P(z_1)=(z_1)^2-z_1+1$
Now this also follows for all roots of $Q(x)$ Now \[P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1\]
Now by Vieta's we know that $-z_4-z_3-z_2-z_1=-1$ ,
so by Newton's Sums we can find $z_1^2+z_2^2+z_3^2+z_4^2$
$a_ns_2+a_{n-1}s_1+2a_{n-2}=0$
$(1)(s_2)+(-1)(1)+2(-1)=0$
$s_2-1-2=0$
$s_2=3$
So finally $P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{006}.$ | null | 006 |
f18e460d0cbfbabe15788dab748a36c2 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_9 | Consider the polynomials $P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x$ and $Q(x) = x^{4} - x^{3} - x^{2} - 1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x) = 0,$ find $P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).$ | Let $S_k=z_1^k+z_2^k+z_3^k+z_4^k$ then by Vieta's Formula we have \[S_{-1}=\frac{z_1z_2z_3+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}=0\] \[S_0=4\] \[S_1=1\] \[S_2=3\] By Newton's Sums we have \[a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0\]
Applying the formula couples of times yields $P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{6}$ | null | 6 |
02f91493a9311b8a52badaff36d7dc7f | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_10 | Two positive integers differ by $60$ . The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers? | Call the two integers $b$ and $b+60$ , so we have $\sqrt{b}+\sqrt{b+60}=\sqrt{c}$ . Square both sides to get $2b+60+2\sqrt{b^2+60b}=c$ . Thus, $b^2+60b$ must be a square, so we have $b^2+60b=n^2$ , and $(b+n+30)(b-n+30)=900$ . The sum of these two factors is $2b+60$ , so they must both be even. To maximize $b$ , we want to maximixe $b+n+30$ , so we let it equal $450$ and the other factor $2$ , but solving gives $b=196$ , which is already a perfect square, so we have to keep going. In order to keep both factors even, we let the larger one equal $150$ and the other $6$ , which gives $b=48$ . This checks, so the solution is $48+108=\boxed{156}$ | null | 156 |
8d965e09ab2b2a61234d0b9f219e5e81 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_11 | Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may be expressed as $\frac {m\sqrt {n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m + n + p.$ | We use the Pythagorean Theorem on $ABC$ to determine that $AB=25.$
Let $N$ be the orthogonal projection from $C$ to $AB.$ Thus, $[CDM]=\frac{(DM)(MN)} {2}$ $MN=BN-BM$ , and $[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.$
From the third equation, we get $CN=\frac{168} {25}.$
By the Pythagorean Theorem in $\Delta BCN,$ we have
$BN=\sqrt{\left(\frac{24 \cdot 25} {25}\right)^2-\left(\frac{24 \cdot 7} {25}\right)^2}=\frac{24} {25}\sqrt{25^2-7^2}=\frac{576} {25}.$
Thus, $MN=\frac{576} {25}-\frac{25} {2}=\frac{527} {50}.$
In $\Delta ADM$ , we use the Pythagorean Theorem to get $DM=\sqrt{15^2-\left(\frac{25} {2}\right)^2}=\frac{5} {2} \sqrt{11}.$
Thus, $[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}= \frac{527\sqrt{11}} {40}.$
Hence, the answer is $527+11+40=\boxed{578}.$ | null | 578 |
8d965e09ab2b2a61234d0b9f219e5e81 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_11 | Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may be expressed as $\frac {m\sqrt {n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m + n + p.$ | Let $E$ be the intersection of lines $BC$ and $DM$ . Since triangles $\Delta CME$ and $\Delta CMD$ share a side and height, the area of $\Delta CDM$ is equal to $\frac{DM}{EM}$ times the area of $\Delta CME$ .
By AA similarity, $\Delta EMB$ is similar to $\Delta ACB$ $\frac{EM}{AC}=\frac{MB}{CB}$ . Solving yields $EM=\frac{175}{48}$ . Using the same method but for $EB$ yields $EB=\frac{625}{48}$ . As in previous solutions, by the Pythagorean Theorem $DM=\frac{5\sqrt{11}}{2}$ . So, $\frac{DM}{EM}=\frac{24\sqrt{11}}{35}$ .
Now, since we know both $CB$ and $EB$ , we can find that $CE=\frac{527}{48}$ . The height of $\Delta CME$ is the length from point $M$ to $CB$ . Since $M$ is the midpoint of $AB$ , the height is just $\frac{1}{2}\cdot7=\frac{7}{2}$ . Using this, we can find that the area of $\Delta CMD$ is $\frac{1}{2}\cdot\frac{7}{2}\cdot\frac{527}{48}\cdot\frac{24\sqrt{11}}{35}=\frac{527\sqrt{11}}{40}$ , giving our answer of $\boxed{578}$ | null | 578 |
ddacb223c30c0937ca40ebbc6a2ca1bf | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_12 | The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate. What was the smallest possible number of members of the committee? | Let $v_i$ be the number of votes candidate $i$ received, and let $s=v_1+\cdots+v_{27}$ be the total number of votes cast. Our goal is to determine the smallest possible $s$
Candidate $i$ got $\frac{v_i}s$ of the votes, hence the percentage of votes they received is $\frac{100v_i}s$ . The condition in the problem statement says that $\forall i: \frac{100v_i}s + 1 \leq v_i$ . ( $\forall$ means "for all", so this means "For all $i$ $\frac{100v_i}s + 1 \leq v_i$ is true")
Obviously, if some $v_i$ would be $0$ or $1$ , the condition would be false. Thus $\forall i: v_i\geq 2$ . We can then rewrite the above inequality as $\forall i: s\geq\frac{100v_i}{v_i-1}$
If for some $i$ we have $v_i=2$ , then from the inequality we just derived we would have $s\geq 200$
If for some $i$ we have $v_i=3$ , then $s\geq 150$
And if for some $i$ we have $v_i=4$ , then $s\geq \frac{400}3 = 133\frac13$ , and hence $s\geq 134$
Is it possible to have $s<134$ ? We just proved that to have such $s$ , all $v_i$ have to be at least $5$ . But then $s=v_1+\cdots+v_{27}\geq 27\cdot 5 = 135$ , which is a contradiction. Hence the smallest possible $s$ is at least $134$
Now consider a situation where $26$ candidates got $5$ votes each, and one candidate got $4$ votes. In this situation, the total number of votes is exactly $134$ , and for each candidate the above inequality is satisfied. Hence the minimum number of committee members is $s=\boxed{134}$ | null | 134 |
ddacb223c30c0937ca40ebbc6a2ca1bf | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_12 | The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate. What was the smallest possible number of members of the committee? | Let there be $N$ members of the committee.
Suppose candidate $n$ gets $a_n$ votes.
Then $a_n$ as a percentage out of $N$ is $100\frac{a_n}{N}$ . Setting up the inequality $a_n \geq 1 + 100\frac{a_n}{N}$ and simplifying, $a_n \geq \lceil(\frac{N}{N - 100})\rceil$ (the ceiling function is there because $a_n$ is an integer.
Note that if we set all $a_i$ equal to $\lceil(\frac{N}{100 - N})\rceil$ we have $N \geq 27\lceil(\frac{N}{100 - N})\rceil$ . Clearly $N = 134$ is the least such number that satisfies this inequality. Now we must show that we can find suitable $a_i$ . We can let 26 of them equal to $5$ and one of them equal to $4$ . Therefore, $N = \boxed{134}$ is the answer.
- whatRthose | null | 134 |
ddacb223c30c0937ca40ebbc6a2ca1bf | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_12 | The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate. What was the smallest possible number of members of the committee? | Let $n$ be the total number of people in the committee, and $a_i$ be the number of votes candidate $i$ gets where $1 \leq i \leq 27$ . The problem tells us that \[\frac{100a_i}{n} \leq a_i - 1 \implies 100a_i \leq na_i - n \implies a_i \geq \frac{n}{n-100}.\] Therefore, \[\sum^{27}_{i=1} a_i = n \geq \sum^{27}_{i=1} \frac{n}{n-100} = \frac{27n}{n-100},\] and so $n(n-127) \geq 0 \implies n \geq 127$ . Trying $n = 127$ , we get that \[a_i \geq \frac{127}{27} \approx 4.7 \implies a_i \geq 5 \implies \sum^{27}_{a_i} a_i \geq 5 \cdot 27 = 135 \geq 127,\] a contradiction. Bashing out a few more, we find that $\boxed{134}$ works perfectly fine. | null | 134 |
640607f324d1dfa6f0bb460824f86809 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_13 | A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$ | Let $P_n$ represent the probability that the bug is at its starting vertex after $n$ moves. If the bug is on its starting vertex after $n$ moves, then it must be not on its starting vertex after $n-1$ moves. At this point it has $\frac{1}{2}$ chance of reaching the starting vertex in the next move. Thus $P_n=\frac{1}{2}(1-P_{n-1})$ $P_0=1$ , so now we can build it up:
$P_1=0$ $P_2=\frac{1}{2}$ $P_3=\frac{1}{4}$ $P_4=\frac{3}{8}$ $P_5=\frac{5}{16}$ $P_6=\frac{11}{32}$ $P_7=\frac{21}{64}$ $P_8=\frac{43}{128}$ $P_9=\frac{85}{256}$ $P_{10}=\frac{171}{512}$
Thus the answer is $171+512=$ $\boxed{683}$ | null | 683 |
640607f324d1dfa6f0bb460824f86809 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_13 | A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$ | Consider there to be a clockwise and a counterclockwise direction around the triangle. Then, in order for the ant to return to the original vertex, the net number of clockwise steps must be a multiple of 3, i.e., $\#CW - \#CCW \equiv 0 \pmod{3}$ . Since $\#CW + \#CCW = 10$ , it is only possible that $(\#CW,\, \#CCW) = (5,5), (8,2), (2,8)$
In the first case, we pick $5$ out of the ant's $10$ steps to be clockwise, for a total of ${10 \choose 5}$ paths. In the second case, we choose $8$ of the steps to be clockwise, and in the third case we choose $2$ to be clockwise. Hence the total number of ways to return to the original vertex is ${10 \choose 5} + {10 \choose 8} + {10 \choose 2} = 252 + 45 + 45 = 342$ . Since the ant has two possible steps at each point, there are $2^{10}$ routes the ant can take, and the probability we seek is $\frac{342}{2^{10}} = \frac{171}{512}$ , and the answer is $\boxed{683}$ | null | 683 |
640607f324d1dfa6f0bb460824f86809 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_13 | A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$ | Label the vertices of the triangle $A,B,C$ with the ant starting at $A$ . We will make a table of the number of ways to get to $A,B,C$ in $n$ moves $n\leq10$ . The values of the table are calculated from the fact that the number of ways from a vertex say $A$ in $n$ steps equals the number of ways to get to $B$ in $n-1$ steps plus the number of ways to get to $C$ in $n-1$ steps.
\[\begin{array}{|l|ccc|} \multicolumn{4}{c}{\text{Table}}\\\hline \text{Step}&A&B&C \\\hline 1 &0 &1 &1 \\ 2 &2 &1 &1 \\ 3 &2 &3 &3\\ \vdots &\vdots&\vdots&\vdots \\ 10 &342 &341 &341 \\\hline \end{array}\] Therefore, our answer is $512+171=\boxed{683}.$ | null | 683 |
640607f324d1dfa6f0bb460824f86809 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_13 | A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$ | As a note, do NOT do this on the exam as it will eat up your time, but feel free to experiment around with this if you have a good enough understanding of linear algebra. This writeup will be extremely lengthy because I am assuming just very basic concepts of linear algebra. The concepts here extend to higher levels of mathematics, so feel free to explore more in depth so that you can end up solving almost any variation of this classic problem.
There are a possible of 3 states for this problem, so we can model the problem as a stochastic process. The resulting process has a transition matrix of:
\[\hat{T} = \begin{bmatrix} 0 & 0.5 & 0.5\\ 0.5 & 0 & 0.5\\ 0.5 & 0.5 & 0 \end{bmatrix}\]
We aim to diagonalize this transition matrix to make it easier to exponentiate by converting it into what's known as it's Jordan Canonical Form.
In order to do this, we must extract the eigenvalues and eigenvectors of the matrix. The eigenpolynomial for this matrix is obtained by calculating this matrix's determinant with $0-\lambda$ about it's main diagonal like so: \[\hat{T}_{\lambda} = \begin{bmatrix} 0-\lambda & 0.5 & 0.5\\ 0.5 & 0-\lambda & 0.5\\ 0.5 & 0.5 & 0-\lambda \end{bmatrix}\]
We have the matrix's eigenpolynomial to be $\lambda^3 - \frac{3\lambda}{4} + \frac{1}{4}$ , and extracting eigenvalues by setting the polynomial equal to $0$ , we have 2 eigenvalues: $\lambda_1 = 1$ of multiplicity 1, and $\lambda_2 = -\frac{1}{2}$ of multiplicity 2. To extract the eigenvectors, we must assess the kernel of this matrix (also known as the null space), or the linear subspace of the domain of $\hat{T}$ where everything gets mapped to the null vector.
We first do this for $\lambda_1$ . Taking $-\lambda_1$ across the diagonals to get $\hat{T}_{\lambda_1} = \begin{bmatrix} -1 & 0.5 & 0.5\\ 0.5 & -1 & 0.5\\ 0.5 & 0.5 & -1 \end{bmatrix}$ , we first reduce it to reduced row echelon form, which is $\begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & -1\\ 0 & 0 & 0 \end{bmatrix}$ . From here, we compute the kernel by setting \[\begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & -1\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = 0\] . So if we take our free variable $x_3 = 0 = t$ , then that means that in the same fashion, $x_1 - x_1 = x_2 - x_2 = 0 = t$ , so hence, the kernel of $\hat{T}_{\lambda} = \begin{bmatrix} t\\ t\\ t \end{bmatrix}$ , or more simply, $t\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ $\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ is the eigenvector corresponding to $\lambda_1$ . We do the same computations for our second unique eigenvalue, but I will save the computation to you. There are actually 2 eigenvectors for $\lambda_2$ , because the reduced row echelon form for $\hat{T}_{\lambda_2}$ has 2 free variables instead of 1, so our eigenvectors for $\lambda_2$ are $\begin{bmatrix} -1\\ 1\\ 0 \end{bmatrix}, \begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$ . We are now ready to begin finding the Jordan canonical form
In linear algebra, the JCF (which also goes by the name of Jordan Normal Form) is an upper triangular matrix representing a linear transformation over a finite-dimensional complex vector space. Any square matrix that has a Jordan Canonical Form has its field of coefficients extended into a field containing all it's eigenvalues. You can find more information about them on google, as well as exactly how to find them but for now let's get on with the problem. I will skip the computation in this step, largley because this writeup is already gargantuan for a simple AIME problem, and because there are countless resources explaining how to do so.
We aim to decompose $\hat{T}$ into the form $\hat{T} = SJS^{-1}$ , where $S$ is a matrix whose columns consist of the eigenvectors of $\hat{T}$ $J$ is the Jordan matrix, and $S^{-1}$ is, well, the inverse of $S$ . We have 1 eigenvalue of multiplicity 1, and 1 of multiplicity 2, so based on this info, we set our eigenvalues along the diagonals like so.
We have: \[J = \begin{bmatrix} -\frac{1}{2} & 0 & 0\\ 0 & -\frac{1}{2} & 0\\ 0 & 0 & 1 \end{bmatrix}, S = \begin{bmatrix} -1 & -1 & 1\\ 0 & 1 & 1\\ 1 & 0 & 1 \end{bmatrix}, S^{-1} = \frac{1}{3}\begin{bmatrix} -1 & -1 & 2\\ -1 & 2 & -1\\ 1 & 1 & 1 \end{bmatrix}\] and so: \[\hat{T} = \begin{bmatrix} 0 & 0.5 & 0.5\\ 0.5 & 0 & 0.5\\ 0.5 & 0.5 & 0 \end{bmatrix} = SJS^{-1} = \begin{bmatrix} -1 & -1 & 1\\ 0 & 1 & 1\\ 1 & 0 & 1 \end{bmatrix}\begin{bmatrix} -\frac{1}{2} & 0 & 0\\ 0 & -\frac{1}{2} & 0\\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} -\frac{1}{3} & -\frac{1}{3} & \frac{2}{3}\\ -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3}\\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{bmatrix}\] Now that we have converted to Jordan Canonical Form, it is extremely easy to compute $\hat{T}^n$
It is an important fact that for any matrix $K$ with Jordan decomposition $SJ_kS^{-1}$ , we have that $K^n = S(J_k)^nS^{-1}$ Using this fact, we aim to find the general solution for the problem. $J^n = \begin{bmatrix} \left(-\frac{1}{2}\right)^n & 0 & 0\\ 0 & \left(-\frac{1}{2}\right)^n & 0\\ 0 & 0 & 1 \end{bmatrix}$ , and using the laws of matrix multiplication, \[SJ^n = \begin{bmatrix} (-1)^{n+1}\left(\frac{1}{2}\right)^n & (-1)^{n+1}\left(\frac{1}{2}\right)^n & 1\\ 0 & \left(-\frac{1}{2}\right)^n & 1\\ \left(-\frac{1}{2}\right)^n & 0 & 1 \end{bmatrix}\] So finally, our final, generalized transition matrix after any number of steps $n$ is: \[\hat{T}^n = \begin{bmatrix} \frac{1}{3} - \frac{1}{3}(-1)^{1+n}(2)^{1-n} & \frac{1}{3} + \frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)} & \frac{1}{3} + \frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)}\\ \frac{1}{3} + \frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)} & \frac{1}{3} - \frac{1}{3}(-1)^{1+n}(2)^{1-n} & \frac{1}{3} + \frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)}\\ \frac{1}{3} + \frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)} & \frac{1}{3} + \frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)} & \frac{1}{3} - \frac{1}{3}(-1)^{1+n}(2)^{1-n} \end{bmatrix}\]
For the sake of this problem, we seek the top left element, which is $\frac{1}{3} - \frac{1}{3}(-1)^{1+n}(2)^{1-n}$ . Substituting $n=10$ readily gives the probability of the bug reaching it's starting position within 10 moves to be $\frac{171}{512} \implies m+n = \boxed{683}$ | null | 683 |
640607f324d1dfa6f0bb460824f86809 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_13 | A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$ | This method does not rigorously get the answer, but it works. As the bug makes more and more moves, the probability of it going back to the origin approaches closer and closer to 1/3. Therefore, after 10 moves, the probability gets close to $341.33/1024$ . We can either round up or down. If we round down, we see $341/1024$ cannot be reduced any further and because the only answers on the AIME are below 1000, this cannot be the right answer. However, if we round up, $342/1024$ can be reduced to $171/512$ where the sum 171+512= $\boxed{683}$ is an accepted answer. | null | 683 |
640607f324d1dfa6f0bb460824f86809 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_13 | A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$ | Start of with any vertex, say $A$ . Denote $a_n$ the number of paths ending where it started. Then notice that for a path to end in the vertex you started on, you must have for the $(n-1)$ case that of the $2^{n-1}$ total paths, we must take away the paths that end in the $(n-1)$ -st term where it started off. (Because imagine on the $(n-1)$ move that it landed on vertex $A$ . Then if we wanted to find the number of length $n$ paths that end at $A$ , we would be able to find that path, because on the $(n-1)$ -st move it landed at $A$ . You can't go from $A$ to $A$ ). Hence we have the recursion $a_n=2^{n-1}-a_{n-1}$ , with $a_3 = 2$ . Now reiterating gives us $a_{10} = 342$ , so that the probability is $\frac{342}{2^{10}} = \frac{171}{512}$ . So we have $171 + 512 = \boxed{683}$ .
~th1nq3r | null | 683 |
2fc351b8914778e7ccd2c2481dd76415 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_14 | Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon can be written in the form $m\sqrt {n},$ where $m$ and $n$ are positive integers and n is not divisible by the square of any prime. Find $m + n.$ | The y-coordinate of $F$ must be $4$ . All other cases yield non-convex and/or degenerate hexagons, which violate the problem statement.
Letting $F = (f,4)$ , and knowing that $\angle FAB = 120^\circ$ , we can use rewrite $F$ using complex numbers: $f + 4 i = (b + 2 i)\left(e^{i(2 \pi / 3)}\right) = (b + 2 i)\left(-1/2 + \frac{\sqrt{3}}{2} i\right) = -\frac{b}{2}-\sqrt{3}+\left(\frac{b\sqrt{3}}{2}-1\right)i$ . We solve for $b$ and $f$ and find that $F = \left(-\frac{8}{\sqrt{3}}, 4\right)$ and that $B = \left(\frac{10}{\sqrt{3}}, 2\right)$
The area of the hexagon can then be found as the sum of the areas of two congruent triangles ( $EFA$ and $BCD$ , with height $8$ and base $\frac{8}{\sqrt{3}}$ ) and a parallelogram ( $ABDE$ , with height $8$ and base $\frac{10}{\sqrt{3}}$ ).
$A = 2 \times \frac{1}{2} \times 8 \times \frac{8}{\sqrt{3}} + 8 \times \frac{10}{\sqrt{3}} = \frac{144}{\sqrt{3}} = 48\sqrt{3}$
Thus, $m+n = \boxed{051}$ | null | 051 |
2fc351b8914778e7ccd2c2481dd76415 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_14 | Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon can be written in the form $m\sqrt {n},$ where $m$ and $n$ are positive integers and n is not divisible by the square of any prime. Find $m + n.$ | [asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),S);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); [/asy] From this image, we can see that the y-coordinate of F is 4, and from this, we can gather that the coordinates of E, D, and C, respectively, are 8, 10, and 6.
[asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),SE);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); xaxis("$x$");yaxis("$y$"); pair b=foot((10/sqrt(3),2),(0,0),(10,0)); pair f=foot((-8/sqrt(3),4),(0,0),(-10,0)); draw(b--(10/sqrt(3),2),dotted); draw(f--(-8/sqrt(3),4),dotted); label("$\theta$",(0,0),7*dir((0,0)--(10/sqrt(3),2)+(4*sqrt(21)/3,0))); [/asy]
Let the angle between the $x$ -axis and segment $AB$ be $\theta$ , as shown above. Thus, as $\angle FAB=120^\circ$ , the angle between the $x$ -axis and segment $AF$ is $60-\theta$ , so $\sin{(60-\theta)}=2\sin{\theta}$ . Expanding, we have
Isolating $\sin{\theta}$ we see that $\frac{\sqrt{3}\cos{\theta}}{2}=\frac{5\sin{\theta}}{2}$ , or $\cos{\theta}=\frac{5}{\sqrt{3}}\sin{\theta}$ . Using the fact that $\sin^2{\theta}+\cos^2{\theta}=1$ , we have $\frac{28}{3}\sin^2{\theta}=1$ , or $\sin{\theta}=\sqrt{\frac{3}{28}}$ . Letting the side length of the hexagon be $y$ , we have $\frac{2}{y}=\sqrt{\frac{3}{28}}$ . After simplification we find that that $y=\frac{4\sqrt{21}}{3}$
In particular, note that by the Pythagorean theorem, $b^2+2^2=y^2$ , hence $b=10\sqrt{3}/3$ . Also, if $C=(c,6)$ , then $y^2=BC^2=4^2+(c-b)^2$ , hence $c-b=8\sqrt{3}/3,$ and thus $c=18\sqrt{3}/3$ . Using similar methods (or symmetry), we determine that $D=(10\sqrt{3}/3,10)$ $E=(0,8)$ , and $F=(-8\sqrt{3}/3,4)$ . By the Shoelace theorem, \[[ABCDEF]=\frac12\left|\begin{array}{cc} 0&0\\ 10\sqrt{3}/3&2\\ 18\sqrt{3}/3&6\\ 10\sqrt{3}/3&10\\ 0&8\\ -8\sqrt{3}/3&4\\ 0&0\\ \end{array}\right|=\frac12|60+180+80-36-60-(-64)|\sqrt{3}/3=48\sqrt{3}.\]
Hence the answer is $\boxed{51}$ | null | 51 |
2fc351b8914778e7ccd2c2481dd76415 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_14 | Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon can be written in the form $m\sqrt {n},$ where $m$ and $n$ are positive integers and n is not divisible by the square of any prime. Find $m + n.$ | This is similar to solution 2 but faster and easier.
First off we see that the y coordinate of F must be 4, the y coordinate of E must be 8, the y coordinate of D must be 10, and the y coordinate of C must be 6 (from the parallel sides of the hexagon).
We then use the sine sum angle formula to find the x coordinate of B (lets call it $x$ ): $2\cdot\cos(120)+x\sin(120)=\frac{x\sqrt3}{2}-1=4\rightarrow x=\frac{10\sqrt3}{3}$ .
Now that we know $x$ we can find the x coordinate of F in multiple ways, including using the cosine sum angle formula or using the fact that triangle AFE is isosceles and AE is on the y axis. Either way, we find that the x coordinate of F is $-\frac{8\sqrt3}{3}$ .
Now, divide ABCDEF into two congruent triangles and a parallelogram: AFE, BCD, and ABDE. The areas of AFE and BCD are each $\frac12\cdot\frac{8\sqrt3}{3}\cdot8=\frac{32\sqrt3}{3}$ . The area of ABDE is $\frac12\cdot8\cdot\frac{10\sqrt3}{3}=\frac{80\sqrt3}3$ .
The total area of the hexagon is $2\cdot\frac{32\sqrt3}3+\frac{80\sqrt3}3=\frac{144\sqrt3}{3}=48\sqrt3\rightarrow48+3=\boxed{051}$ | null | 051 |
2fc351b8914778e7ccd2c2481dd76415 | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_14 | Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon can be written in the form $m\sqrt {n},$ where $m$ and $n$ are positive integers and n is not divisible by the square of any prime. Find $m + n.$ | [asy] size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),SE);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); xaxis("$x$");yaxis("$y$"); pair b=foot((10/sqrt(3),2),(0,0),(10,0)); pair f=foot((-8/sqrt(3),4),(0,0),(-10,0)); draw(b--(10/sqrt(3),2),dotted); draw(f--(-8/sqrt(3),4),dotted); label("$\theta$",(0,0),7*dir((0,0)--(10/sqrt(3),2)+(4*sqrt(21)/3,0))); [/asy]
First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solutions. We can draw a rectangle around the hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A $x$ and $z$ , respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B $y$ . This tells us that the area of the entire rectangle is $10(x+y+z)$ , since the opposite sides are parallel and thus the length of the rectangle is $4+4+2=10$ . Then,
if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as $6x+8z+2y$ . However, noticing that $x=y$ , the area of ABCDEF can also be expressed as $8(x+z)$ . Now we just need to find $x+z$ . Since $AB=AF$ and $\angle BAF = 120$ degrees, $BF=AB\sqrt{3}$ . However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).
From triangle ABX we have that $AB=\sqrt{4+x^2}$ , so $BF=\sqrt{3x^2+12}$ . Since AB=AF, we can also form the equation $4+x^2=16+z^2$ .
We can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us $BF=\sqrt{4+(x+z)^2}$ . Setting our two values of BF equal and substituting $x^2$ as $12+z^2$ and simplifying, we get the equation $3z^4-16z^2-1024=0$ . Now we can use the quadratic formula to get that $z^2=\frac{64}{3}$ or $-18$ , so $z^2=\frac{64}{3}$ . Plugging this value back into the equation $x^2=12+z^2$ , we get that $x^2=\frac{100}{3}$ . Now we get that $x+z$ is $6\sqrt{3}$ , so the area of the hexagon is $8 \cdot 6\sqrt{3}=48\sqrt{3}$ , so the answer is $48+3=\boxed{051}$ | null | 051 |
1b11be1f60d95d3bceb00a56feb6268d | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_15 | Let \[P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).\] Let $z_{1},z_{2},\ldots,z_{r}$ be the distinct zeros of $P(x),$ and let $z_{k}^{2} = a_{k} + b_{k}i$ for $k = 1,2,\ldots,r,$ where $a_{k}$ and $b_{k}$ are real numbers. Let
where $m, n,$ and $p$ are integers and $p$ is not divisible by the square of any prime. Find $m + n + p.$ | This can be factored as:
\[P(x) = x\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right)^2\]
Note that $\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1$ .
So the roots of $x^{23} + x^{22} + \cdots + x^2 + x + 1$ are exactly all $24$ -th complex roots of $1$ , except for the root $x=1$
Let $\omega=\cos \frac{360^\circ}{24} + i\sin \frac{360^\circ}{24}$ . Then the distinct zeros of $P$ are $0,\omega,\omega^2,\dots,\omega^{23}$
We can clearly ignore the root $x=0$ as it does not contribute to the value that we need to compute.
The squares of the other roots are $\omega^2,~\omega^4,~\dots,~\omega^{24}=1,~\omega^{26}=\omega^2,~\dots,~\omega^{46}=\omega^{22}$
Hence we need to compute the following sum:
\[R = \sum_{k = 1}^{23} \left|\, \sin \left( k\cdot \frac{360^\circ}{12} \right) \right|\]
Using basic properties of the sine function, we can simplify this to
\[R = 4 \cdot \sum_{k = 1}^{5} \sin \left( k\cdot \frac{360^\circ}{12} \right)\]
The five-element sum is just $\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \sin 120^\circ + \sin 150^\circ$ .
We know that $\sin 30^\circ = \sin 150^\circ = \frac 12$ $\sin 60^\circ = \sin 120^\circ = \frac{\sqrt 3}2$ , and $\sin 90^\circ = 1$ .
Hence our sum evaluates to:
\[R = 4 \cdot \left( 2\cdot \frac 12 + 2\cdot \frac{\sqrt 3}2 + 1 \right) = 8 + 4\sqrt 3\]
Therefore the answer is $8+4+3 = \boxed{015}$ | null | 015 |
1b11be1f60d95d3bceb00a56feb6268d | https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_15 | Let \[P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).\] Let $z_{1},z_{2},\ldots,z_{r}$ be the distinct zeros of $P(x),$ and let $z_{k}^{2} = a_{k} + b_{k}i$ for $k = 1,2,\ldots,r,$ where $a_{k}$ and $b_{k}$ are real numbers. Let
where $m, n,$ and $p$ are integers and $p$ is not divisible by the square of any prime. Find $m + n + p.$ | As in Solution 1, we find that the roots of $P(x)$ we care about are the 24th roots of unity except $1$ . Therefore, the squares of these roots are the 12th roots of unity. In particular, every 12th root of unity is counted twice, except for $1$ , which is only counted once.
The possible imaginary parts of the 12th roots of unity are $0$ $\pm\frac{1}{2}$ $\pm\frac{\sqrt{3}}{2}$ , and $\pm 1$ . We can disregard $0$ because it doesn't affect the sum.
$8$ squares of roots have an imaginary part of $\pm\frac{1}{2}$ $8$ squares of roots have an imaginary part of $\pm\frac{\sqrt{3}}{2}$ , and $4$ squares of roots have an imaginary part of $\pm 1$ . Therefore, the sum equals $8\left(\frac{1}{2}\right) + 8\left(\frac{\sqrt{3}}{2}\right) + 4(1) = 8 + 4\sqrt{3}$
The answer is $8+4+3=\boxed{015}$ | null | 015 |
0b6e61c0bc39bf2ca7cc2f3b2e7b81a7 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_1 | Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Using complementary counting, we count all of the license plates that do not have the desired property. To not be a palindrome, the first and third characters of each string must be different. Therefore, there are $10\cdot 10\cdot 9$ three-digit non-palindromes, and there are $26\cdot 26\cdot 25$ three-letter non-palindromes. As there are $10^3\cdot 26^3$ total three-letter three-digit arrangements, the probability that a license plate does not have the desired property is $\frac{10\cdot 10\cdot 9\cdot 26\cdot 26\cdot 25}{10^3\cdot 26^3}=\frac{45}{52}$ . We subtract this from 1 to get $1-\frac{45}{52}=\frac{7}{52}$ as our probability. Therefore, our answer is $7+52=\boxed{059}$ | null | 059 |
0b6e61c0bc39bf2ca7cc2f3b2e7b81a7 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_1 | Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Note that we can pick the first and second letters/numbers freely with one choice left for the last letter/number for there to be a palindrome. Thus, the probability of no palindrome is \[\frac{25}{26}\cdot \frac{9}{10}=\frac{45}{52}\] thus we have $1-\frac{45}{52}=\frac{7}{52}$ so our answer is $7+52 = \boxed{059}.$ | null | 059 |
05b1a449efc6af4a6384b428929c7dcd | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_2 | The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\dfrac{1}{2}(\sqrt{p}-q)$ where $p$ and $q$ are positive integers. Find $p+q$ | Let the radius of the circles be $r$ . The longer dimension of the rectangle can be written as $14r$ , and by the Pythagorean Theorem , we find that the shorter dimension is $2r\left(\sqrt{3}+1\right)$
Therefore, $\frac{14r}{2r\left(\sqrt{3}+1\right)}= \frac{7}{\sqrt{3} + 1} \cdot \left[\frac{\sqrt{3}-1}{\sqrt{3}-1}\right] = \frac{1}{2}\left(7\sqrt{3} - 7\right) = \frac{1}{2}\left(\sqrt{p}-q\right)$ . Thus we have $p=147$ and $q=7$ , so $p+q=\boxed{154}$ | null | 154 |
05b1a449efc6af4a6384b428929c7dcd | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_2 | The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\dfrac{1}{2}(\sqrt{p}-q)$ where $p$ and $q$ are positive integers. Find $p+q$ | Since we only care about the ratio between the longer side and shorter side, we can set the longer side to $14$ . So, this means that each of the radii is $1$ . Now, we connect the radii of three circles such that they form an equilateral triangle with side length 4. Obviously, the height this triangle is $2\sqrt{3}$ , and the shorter side if the triangle is therefore $2\sqrt{3}+2$ and we use simplification similar to as showed above, and we reach the result $\frac{1}{2} \cdot (\sqrt{147}-7)$ and the final answer is $147+7 = \boxed{154}$ | null | 154 |
4f1f756e87d115d82d0b042c971a2f92 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_3 | Jane is 25 years old. Dick is older than Jane. In $n$ years, where $n$ is a positive integer, Dick's age and Jane's age will both be two-digit numbers and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let $d$ be Dick's present age. How many ordered pairs of positive integers $(d,n)$ are possible? | Let Jane's age $n$ years from now be $10a+b$ , and let Dick's age be $10b+a$ . If $10b+a>10a+b$ , then $b>a$ . The possible pairs of $a,b$ are:
That makes 36. But $10a+b>25$ , so we subtract all the extraneous pairs: $(1,2), (1,3), (2,3), (1,4), (2,4), (1,5), (2,5), (1,6), (1,7), (1,8),$ and $(1,9)$ $36-11=\boxed{025}$ | null | 025 |
4f1f756e87d115d82d0b042c971a2f92 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_3 | Jane is 25 years old. Dick is older than Jane. In $n$ years, where $n$ is a positive integer, Dick's age and Jane's age will both be two-digit numbers and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let $d$ be Dick's present age. How many ordered pairs of positive integers $(d,n)$ are possible? | Start by assuming that $n < 5$ (essentially, Jane is in the 20s when their ages are 'reverses' of each other). Then we get the pairs \[(61,1),(70,2),(79,3),(88,4).\] Repeating this for the 30s gives \[(34,9),(43,10),(52,11),(61,12),(70,13),(79,14).\] From here, it's pretty clear that every decade we go up we get $(d,n+11)$ as a pair. Since both ages must always be two-digit numbers, we can show that each decade after the 30s, we get 1 fewer option. Therefore, our answer is $4+6+5+\dots+2+1=4+21=\boxed{025}.$ | null | 025 |
dab39563e84f06a65e57b168b56e516d | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_5 | Let $A_1,A_2,A_3,\cdots,A_{12}$ be the vertices of a regular dodecagon. How many distinct squares in the plane of the dodecagon have at least two vertices in the set $\{A_1,A_2,A_3,\cdots,A_{12}\} ?$ | Proceed as above to initially get 198 squares (with overcounting). Then note that any square with all four vertices on the dodecagon has to have three sides "between" each vertex, giving us a total of three squares. However, we counted these squares with all four of their sides plus both of their diagonals, meaning we counted them 6 times. Therefore, our answer is $198-3(6-1)=198-15=\boxed{183}.$ | null | 183 |
c1cc49406c814ef09e02e43fae54c48d | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_6 | The solutions to the system of equations
are $(x_1,y_1)$ and $(x_2,y_2)$ . Find $\log_{30}\left(x_1y_1x_2y_2\right)$ | Let $A=\log_{225}x$ and let $B=\log_{64}y$
From the first equation: $A+B=4 \Rightarrow B = 4-A$
Plugging this into the second equation yields $\frac{1}{A}-\frac{1}{B}=\frac{1}{A}-\frac{1}{4-A}=1 \Rightarrow A = 3\pm\sqrt{5}$ and thus, $B=1\pm\sqrt{5}$
So, $\log_{225}(x_1x_2)=\log_{225}(x_1)+\log_{225}(x_2)=(3+\sqrt{5})+(3-\sqrt{5})=6$ $\Rightarrow x_1x_2=225^6=15^{12}$
And $\log_{64}(y_1y_2)=\log_{64}(y_1)+\log_{64}(y_2)=(1+\sqrt{5})+(1-\sqrt{5})=2$ $\Rightarrow y_1y_2=64^2=2^{12}$
Thus, $\log_{30}\left(x_1y_1x_2y_2\right) = \log_{30}\left(15^{12}\cdot2^{12} \right) = \log_{30}\left(30^{12} \right) = \boxed{012}$ | null | 012 |
ccfb129f7a9a10a9fcc89ac1a111e60a | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_7 | The Binomial Expansion is valid for exponents that are not integers. That is, for all real numbers $x,y$ and $r$ with $|x|>|y|$
\[(x+y)^r=x^r+rx^{r-1}y+\dfrac{r(r-1)}{2}x^{r-2}y^2+\dfrac{r(r-1)(r-2)}{3!}x^{r-3}y^3 \cdots\]
What are the first three digits to the right of the decimal point in the decimal representation of $(10^{2002}+1)^{\frac{10}{7}}$ | $1^n$ will always be 1, so we can ignore those terms, and using the definition ( $2002 / 7 = 286$ ):
\[(10^{2002} + 1)^{\frac {10}7} = 10^{2860}+\dfrac{10}{7}10^{858}+\dfrac{15}{49}10^{-1144}+\cdots\]
Since the exponent of the $10$ goes down extremely fast, it suffices to consider the first few terms. Also, the $10^{2860}$ term will not affect the digits after the decimal, so we need to find the first three digits after the decimal in
\[\dfrac{10}{7}10^{858}\]
(The remainder after this term is positive by the Remainder Estimation Theorem ). Since the repeating decimal of $\dfrac{10}{7}$ repeats every 6 digits, we can cut out a lot of 6's from $858$ to reduce the problem to finding the first three digits after the decimal of
$\dfrac{10}{7}$
That is the same as $1+\dfrac{3}{7}$ , and the first three digits after $\dfrac{3}{7}$ are $\boxed{428}$ | null | 428 |
2de9a60766d30325390901fe54908d20 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_8 | Find the smallest integer $k$ for which the conditions
(1) $a_1,a_2,a_3\cdots$ is a nondecreasing sequence of positive integers
(2) $a_n=a_{n-1}+a_{n-2}$ for all $n>2$
(3) $a_9=k$
are satisfied by more than one sequence. | From $(2)$ $a_9=$ $a_8+a_7=2a_7+a_6=3a_6+2a_5=5a_5+3a_4=8a_4+5a_3=13a_3+8a_2=21a_2+13a_1$ $=k$
Suppose that $a_1=x_0$ is the smallest possible value for $a_1$ that yields a good sequence, and $a_2=y_0$ in this sequence. So, $13x_0+21y_0=k$
Since $\gcd(13,21)=1$ , the next smallest possible value for $a_1$ that yields a good sequence is $a_1=x_0+21$ . Then, $a_2=y_0-13$
By $(1)$ $a_2 \ge a_1 \Rightarrow y_0-13 \ge x_0+21 \Rightarrow y_0 \ge x_0+34 \ge 35$ . So the smallest value of $k$ is attained when $(x_0,y_0)=(1,35)$ which yields $(a_1,a_2)=(1,35)$ or $(22,22)$
Thus, $k=13(1)+21(35)=\boxed{748}$ is the smallest possible value of $k$ | null | 748 |
9f680b5c5f362f6af275d6726052416c | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_9 | Harold, Tanya, and Ulysses paint a very long picket fence.
Call the positive integer $100h+10t+u$ paintable when the triple $(h,t,u)$ of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers. | Note that it is impossible for any of $h,t,u$ to be $1$ , since then each picket will have been painted one time, and then some will be painted more than once.
$h$ cannot be $2$ , or that will result in painting the third picket twice. If $h=3$ , then $t$ may not equal anything not divisible by $3$ , and the same for $u$ . Now for fourth and fifth pickets to be painted, $t$ and $u$ must be $3$ as well. This configuration works, so $333$ is paintable.
If $h$ is $4$ , then $t$ must be even. The same for $u$ , except that it can't be $2 \mod 4$ . Thus $u$ is $0 \mod 4$ and $t$ is $2 \mod 4$ . Since this is all $\mod 4$ $t$ must be $2$ and $u$ must be $4$ , in order for $5,6$ to be paint-able. Thus $424$ is paintable.
$h$ cannot be greater than $4$ , since if that were the case then the answer would be greater than $999$ , which would be impossible for the AIME.
Thus the sum of all paintable numbers is $\boxed{757}$ | null | 757 |
9f680b5c5f362f6af275d6726052416c | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_9 | Harold, Tanya, and Ulysses paint a very long picket fence.
Call the positive integer $100h+10t+u$ paintable when the triple $(h,t,u)$ of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers. | Again, note that $h,t,u \neq 1$ . The three conditions state that no picket number $n$ may satisfy any two of the conditions: $n \equiv 1 \pmod{h},\ n \equiv 2 \pmod{t},\ n \equiv 3 \pmod{u}$ . By the Chinese Remainder Theorem , the greatest common divisor of any pair of the three numbers $\{h,t,u\}$ cannot be $1$ (since otherwise without loss of generality consider $\text{gcd}\,(h,t) = 1$ ; then there will be a common solution $\pmod{h \times t}$ ).
Now for $4$ to be paint-able, we require either $h = 3$ or $t=2$ , but not both.
Thus the answer is $333+424 = \boxed{757}$ | null | 757 |
9f680b5c5f362f6af275d6726052416c | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_9 | Harold, Tanya, and Ulysses paint a very long picket fence.
Call the positive integer $100h+10t+u$ paintable when the triple $(h,t,u)$ of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers. | The three conditions state that no picket number $n$ may satisfy any two of the conditions: $n \equiv 1 \pmod{h},\ n \equiv 2 \pmod{t},\ n \equiv 3 \pmod{u}$ . Note that the smallest number, $min \{ h,t,u \},$ divides the other $2$ , and the next smallest divide the largest number, otherwise there is a common solution by the Chinese Remainder Theorem . It is also a necessary condition so that it paints exactly once. Note that the smallest number can't be at least $5$ , otherwise not all picket will be painted. We are left with few cases (we could also exclude $1$ as the possibility) which could be done quickly. Thus the answer is $333+424 = \boxed{757}$ | null | 757 |
b60f8854d64280ef152f278ab0932698 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_10 | In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$ , and $\overline{AD}$ bisects angle $CAB$ . Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $AE=3$ and $AF=10$ . Given that $EB=9$ and $FC=27$ , find the integer closest to the area of quadrilateral $DCFG$ | By the Pythagorean Theorem, $BC=35$ . Letting $BD=x$ we can use the Angle Bisector Theorem on triangle $ABC$ to get $x/12=(35-x)/37$ , and solving gives $BD=60/7$ and $DC=185/7$
The area of triangle $AGF$ is $10/3$ that of triangle $AEG$ , since they share a common side and angle, so the area of triangle $AGF$ is $10/13$ the area of triangle $AEF$
Since the area of a triangle is $\frac{ab\sin{C}}2$ , the area of $AEF$ is $525/37$ and the area of $AGF$ is $5250/481$
The area of triangle $ABD$ is $360/7$ , and the area of the entire triangle $ABC$ is $210$ . Subtracting the areas of $ABD$ and $AGF$ from $210$ and finding the closest integer gives $\boxed{148}$ as the answer. | null | 148 |
b60f8854d64280ef152f278ab0932698 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_10 | In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$ , and $\overline{AD}$ bisects angle $CAB$ . Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $AE=3$ and $AF=10$ . Given that $EB=9$ and $FC=27$ , find the integer closest to the area of quadrilateral $DCFG$ | By the Pythagorean Theorem, $BC=35$ . From now on, every number we get will be rounded to two decimal points. Using law of cosines on AEF, EF is around 9.46. Using the angle bisector theorem, GF and DC are 7.28 and (185/7), respectively. We also know the lengths of EG and BD, so a quick Stewart's Theorem for triangles ABC and AEF gives lengths 3.76 and 14.75 for AG and AD, respectively. Now we have all segments of triangles AGF and ADC. Joy! It's time for some Heron's Formula. This gives area 10.95 for triangle AGF and 158.68 for triangle ADC. Subtracting gives us 147.73. Round to the nearest whole number and we get $\boxed{148}$ | null | 148 |
b60f8854d64280ef152f278ab0932698 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_10 | In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$ , and $\overline{AD}$ bisects angle $CAB$ . Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $AE=3$ and $AF=10$ . Given that $EB=9$ and $FC=27$ , find the integer closest to the area of quadrilateral $DCFG$ | By the Pythagorean Theorem, $BC=35$ . By the Angle Bisector Theorem $BD = 60/7$ and $DC = 185/7$ . We can find the coordinates of F, and use that the find the equation of line EF. Then, we can find the coordinates of G. Triangle $ADC = 1110/7$ and we can find the area triangle $AGF$ with the shoelace theorem, so subtracting that from $ADC$ gives us $\boxed{148}$ as the closest integer. | null | 148 |
b60f8854d64280ef152f278ab0932698 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_10 | In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$ , and $\overline{AD}$ bisects angle $CAB$ . Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $AE=3$ and $AF=10$ . Given that $EB=9$ and $FC=27$ , find the integer closest to the area of quadrilateral $DCFG$ | By the Pythagorean Theorem, $BC = 35$ , and by the Angle Bisector Theorem $BD = 60/7$ and $DC = 185/7$ . Draw a perpendicular from $F$ to $\overline{AE}$ . Let the intersection of $F$ and $\overline{AE}$ be $H$ . triangle $AHF$ is similar to $ABC$ by $AA$ similarity. thus, $AF/AC = HF/BC$ . We find that $HF = 350/37$ , so the area of $AEF = 525/37$ . The area of triangle $AGF$ is $10/3$ that of triangle $AEG$ , since they share a common side and angle, so the area of triangle $AGF$ is $10/13$ the area of triangle $AEF$ , so the area of $AGF = 5250/481$ . The area of triangle $ADC$ is $1110/7$ , since the base is $185/7$ and the height is $12$ . Thus, the area of $DCFG$ equals the area of $ADC - AGF$ , or rounded to the nearest integer, $\boxed{148}$ | null | 148 |
39f62288b7cfe482f37233f534c78c69 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_11 | Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12$ . A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P$ , which is 7 units from $\overline{BG}$ and 5 units from $\overline{BC}$ . The beam continues to be reflected off the faces of the cube. The length of the light path from the time it leaves point $A$ until it next reaches a vertex of the cube is given by $m\sqrt{n}$ , where $m$ and $n$ are integers and $n$ is not divisible by the square of any prime. Find $m+n$ | When a light beam reflects off a surface, the path is like that of a ball bouncing. Picture that, and also imagine X, Y, and Z coordinates for the cube vertices. The coordinates will all involve 0's and 12's only, so that means that the X, Y, and Z distance traveled by the light must all be divisible by 12. Since the light's Y changes by 5 and the X changes by 7 (the Z changes by 12, don't worry about that), and 5 and 7 are relatively prime to 12, the light must make 12 reflections onto the XY plane or the face parallel to the XY plane.
In each reflection, the distance traveled by the light is $\sqrt{ (12^2) + (5^2) + (7^2) }$ $\sqrt{218}$ . This happens 12 times, so the total distance is $12\sqrt{218}$ $m=12$ and $n=218$ , so therefore, the answer is $m+n=\boxed{230}$ | null | 230 |
39f62288b7cfe482f37233f534c78c69 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_11 | Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12$ . A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P$ , which is 7 units from $\overline{BG}$ and 5 units from $\overline{BC}$ . The beam continues to be reflected off the faces of the cube. The length of the light path from the time it leaves point $A$ until it next reaches a vertex of the cube is given by $m\sqrt{n}$ , where $m$ and $n$ are integers and $n$ is not divisible by the square of any prime. Find $m+n$ | We can use a similar trick as with reflections in 2D: Imagine that the entire space is divided into cubes identical to the one we have. Now let's follow two photons of light that start in $A$ at the same time: one of them will reflect as given in the problem statement, the second will simply fly straight through all cubes. It can easily be seen that at any moment in time the photons are in exactly the same position relative to the cubes they are inside at the moment. In other words, we can take the cube with the first photon, translate it and flip if necessary, to get the cube with the other photon.
It follows that both photons will hit a vertex at the same time, and at this moment they will have travelled the same distance.
Now, the path of the second photon is simply a half-line given by the vector $(12,7,5)$ . That is, the points visited by the photon are of the form $(12t,7t,5t)$ for $t\geq 0$ . We are looking for the smallest $t$ such that all three coordinates are integer multiples of $12$ (which is the length of the side of the cube).
Clearly $t$ must be an integer. As $7$ and $12$ are relatively prime, the smallest solution is $t=12$ . At this moment the second photon will be at the coordinates $(12\cdot 12,7\cdot 12,5\cdot 12)$
Then the distance it travelled is $\sqrt{ (12\cdot 12)^2 + (7\cdot 12)^2 + (5\cdot 12)^2 } = 12\sqrt{12^2 + 7^2 + 5^2}=12\sqrt{218}$ .
And as the factorization of $218$ is $218=2\cdot 109$ , we have $m=12$ and $n=218$ , hence $m+n=\boxed{230}$ | null | 230 |
5a98ceef868cbaf1121707691e9f7223 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_12 | Let $F(z)=\dfrac{z+i}{z-i}$ for all complex numbers $z\neq i$ , and let $z_n=F(z_{n-1})$ for all positive integers $n$ . Given that $z_0=\dfrac{1}{137}+i$ and $z_{2002}=a+bi$ , where $a$ and $b$ are real numbers, find $a+b$ | Iterating $F$ we get:
\begin{align*} F(z) &= \frac{z+i}{z-i}\\ F(F(z)) &= \frac{\frac{z+i}{z-i}+i}{\frac{z+i}{z-i}-i} = \frac{(z+i)+i(z-i)}{(z+i)-i(z-i)}= \frac{z+i+zi+1}{z+i-zi-1}= \frac{(z+1)(i+1)}{(z-1)(1-i)}\\ &= \frac{(z+1)(i+1)^2}{(z-1)(1^2+1^2)}= \frac{(z+1)(2i)}{(z-1)(2)}= \frac{z+1}{z-1}i\\ F(F(F(z))) &= \frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i} = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)} = \frac{2z}{2} = z. \end{align*}
From this, it follows that $z_{k+3} = z_k$ , for all $k$ . Thus $z_{2002} = z_{3\cdot 667+1} = z_1 = \frac{z_0+i}{z_0-i} = \frac{(\frac{1}{137}+i)+i}{(\frac{1}{137}+i)-i}= \frac{\frac{1}{137}+2i}{\frac{1}{137}}= 1+274i.$
Thus $a+b = 1+274 = \boxed{275}$ | null | 275 |
df8fa405ae19db64ec4d0517e7f02155 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_13 | In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths $18$ and $27$ , respectively, and $AB=24$ . Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$ . The area of triangle $AFB$ is $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$ | Applying Stewart's Theorem to medians $AD, CE$ , we have:
Substituting the first equation into the second and simplification yields $24^2 = 2\left(3AC^2 + 2 \cdot 24^2 - 4 \cdot 18^2\right)- 4 \cdot 27^2$ $\Longrightarrow AC = \sqrt{2^5 \cdot 3 + 2 \cdot 3^5 + 2^4 \cdot 3^3 - 2^7 \cdot 3} = 3\sqrt{70}$
By the Power of a Point Theorem on $E$ , we get $EF = \frac{12^2}{27} = \frac{16}{3}$ . The Law of Cosines on $\triangle ACE$ gives
Hence $\sin \angle AEC = \sqrt{1 - \cos^2 \angle AEC} = \frac{\sqrt{55}}{8}$ . Because $\triangle AEF, BEF$ have the same height and equal bases, they have the same area, and $[ABF] = 2[AEF] = 2 \cdot \frac 12 \cdot AE \cdot EF \sin \angle AEF = 12 \cdot \frac{16}{3} \cdot \frac{\sqrt{55}}{8} = 8\sqrt{55}$ , and the answer is $8 + 55 = \boxed{063}$ | null | 063 |
df8fa405ae19db64ec4d0517e7f02155 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_13 | In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths $18$ and $27$ , respectively, and $AB=24$ . Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$ . The area of triangle $AFB$ is $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$ | Let $AD$ and $CE$ intersect at $P$ . Since medians split one another in a 2:1 ratio, we have
This gives isosceles $APE$ and thus an easy area calculation. After extending the altitude to $PE$ and using the fact that it is also a median, we find
Using Power of a Point, we have
By Same Height Different Base,
Solving gives
and
Thus, our answer is $8+55=\boxed{063}$ | null | 063 |
df8fa405ae19db64ec4d0517e7f02155 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_13 | In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths $18$ and $27$ , respectively, and $AB=24$ . Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$ . The area of triangle $AFB$ is $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$ | Use the same diagram as in Solution 1. Call the centroid $P$ . It should be clear that $PE=9$ , and likewise $AP=12$ $AE=12$ . Then, $\sin \angle AEP = \frac{\sqrt{55}}{8}$ . Power of a Point on $E$ gives $FE=\frac{16}{3}$ , and the area of $AFB$ is $AE * EF* \sin \angle AEP$ , which is twice the area of $AEF$ or $FEB$ (they have the same area because of equal base and height), giving $8\sqrt{55}$ for an answer of $\boxed{063}$ | null | 063 |
df8fa405ae19db64ec4d0517e7f02155 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_13 | In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths $18$ and $27$ , respectively, and $AB=24$ . Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$ . The area of triangle $AFB$ is $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$ | Note that, as above, it is quite easy to get that $\sin \angle AEP = \frac{\sqrt{55}}{8}$ (equate Heron's and $\frac{1}{2}ab\sin C$ to find this). Now note that $\angle FEA = \angle BEC$ because they are vertical angles, $\angle FAE = \angle ECB$ , and $\angle EFA = \angle ABC$ (the latter two are derived from the inscribed angle theorem). Therefore $\Delta AEF ~ \Delta CEB$ and so $FE = \frac{144}{27}$ and $\sin \angle FEA = \frac{\sqrt{55}}{8}$ so the area of $\Delta BFA$ is $8\sqrt{55}$ giving us $\boxed{063}$ as our answer. (One may just get the area via triangle similarity too--this is if you are tired by the end of test and just want to bash some stuff out--it may also serve as a useful check). | null | 063 |
df8fa405ae19db64ec4d0517e7f02155 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_13 | In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths $18$ and $27$ , respectively, and $AB=24$ . Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$ . The area of triangle $AFB$ is $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$ | Apply barycentric coordinates on $\triangle ABC$ . We know that $D=\left(0, \frac{1}{2}, \frac{1}{2}\right), E=\left(\frac{1}{2}, \frac{1}{2}, 0\right)$ . We can now get the displacement vectors $\overrightarrow{AD} = \left(1, -\frac{1}{2}, -\frac{1}{2}\right)$ and $\overrightarrow{CE}=\left(-\frac{1}{2}, -\frac{1}{2}, 1\right)$ . Now, applying the distance formula and simplifying gives us the two equations \begin{align*} 2b^2+2c^2-^2&=1296 \\ 2a^2+2b^2-c^2&=2916. \\ \end{align*} Substituting $c=24$ and solving with algebra now gives $a=6\sqrt{31}, b=3\sqrt{70}$ . Now we can find $F$ . Note that $CE$ can be parameterized as $(1:1:t)$ , so plugging into the circumcircle equation and solving for $t$ gives $t=\frac{-c^2}{a^2+b^2}$ so $F=(a^2+b^2:a^2+b^2:-c^2)$ . Plugging in for $a,b$ gives us $F=(1746:1746:-576)$ . Thus, by the area formula, we have \[\frac{[AFB]}{[ABC]}= \left|\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \frac{97}{162} & \frac{97}{162} & -\frac{16}{81} \end{matrix}\right|=\frac{16}{81}.\] By Heron's Formula, we have $[ABC]=\frac{81\sqrt{55}}{2}$ which immediately gives $[AFB]=8\sqrt{55}$ from our ratio, extracting $\boxed{63}$ | null | 63 |
cdda823c51dd852ff9bcadfbe9c83a67 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_15 | Polyhedron $ABCDEFG$ has six faces. Face $ABCD$ is a square with $AB = 12;$ face $ABFG$ is a trapezoid with $\overline{AB}$ parallel to $\overline{GF},$ $BF = AG = 8,$ and $GF = 6;$ and face $CDE$ has $CE = DE = 14.$ The other three faces are $ADEG, BCEF,$ and $EFG.$ The distance from $E$ to face $ABCD$ is 12. Given that $EG^2 = p - q\sqrt {r},$ where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p + q + r.$ | We let $A$ be the origin, or $(0,0,0)$ $B = (0,0,12)$ , and $D = (12,0,0)$ . Draw the perpendiculars from F and G to AB, and let their intersections be X and Y, respectively. By symmetry, $FX = GY = \frac{12-6}2 = 3$ , so $G = (a,b,3)$ , where a and b are variables.
We can now calculate the coordinates of E. Drawing the perpendicular from E to CD and letting the intersection be Z, we have $CZ = DZ = 6$ and $EZ = 4\sqrt{10}$ . Therefore, the x coordinate of $E$ is $12-\sqrt{(4\sqrt{10})^2-12^2}=12-\sqrt{16}=12-4=8$ , so $E = (8,12,6)$
We also know that $A,D,E,$ and $G$ are coplanar, so they all lie on the plane $z = Ax+By+C$ . Since $(0,0,0)$ is on it, then $C = 0$ . Also, since $(12,0,0)$ is contained, then $A = 0$ . Finally, since $(8,12,6)$ is on the plane, then $B = \frac 12$ . Therefore, $b = 6$ . Since $GA = 8$ , then $a^2+6^2+3^2=8^2$ , or $a = \pm \sqrt{19}$ . Therefore, the two permissible values of $EG^2$ are $(8 \pm \sqrt{19})^2+6^2+3^2 = 128 \pm 16\sqrt{19}$ . The only one that satisfies the conditions of the problem is $128 - 16\sqrt{19}$ , from which the answer is $128+16+19=\boxed{163}$ | null | 163 |
7d4056927620bfe181f1cf8b823a3a30 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_1 | Given that \begin{eqnarray*}&(1)& x\text{ and }y\text{ are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \qquad \\ &(2)& y\text{ is the number formed by reversing the digits of }x\text{; and}\\ &(3)& z=|x-y|. \end{eqnarray*}
How many distinct values of $z$ are possible? | We express the numbers as $x=100a+10b+c$ and $y=100c+10b+a$ . From this, we have \begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\&=&|99a-99c|\\&=&99|a-c|\\ \end{eqnarray*} Because $a$ and $c$ are digits, and $a$ and $c$ are both between 1 and 9 (from condition 1), there are $\boxed{009}$ possible values (since all digits except $9$ can be expressed this way). | null | 009 |
16e1cebc94ec4cedd4528a22bd24968e | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_3 | It is given that $\log_{6}a + \log_{6}b + \log_{6}c = 6,$ where $a,$ $b,$ and $c$ are positive integers that form an increasing geometric sequence and $b - a$ is the square of an integer. Find $a + b + c.$ | $abc=6^6$ . Since they form an increasing geometric sequence, $b$ is the geometric mean of the product $abc$ $b=\sqrt[3]{abc}=6^2=36$
Since $b-a$ is the square of an integer, we can find a few values of $a$ that work: $11, 20, 27, 32,$ and $35$ . Out of these, the only value of $a$ that works is $a=27$ , from which we can deduce that $c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36=48$
Thus, $a+b+c=27+36+48=\boxed{111}$ | null | 111 |
16e1cebc94ec4cedd4528a22bd24968e | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_3 | It is given that $\log_{6}a + \log_{6}b + \log_{6}c = 6,$ where $a,$ $b,$ and $c$ are positive integers that form an increasing geometric sequence and $b - a$ is the square of an integer. Find $a + b + c.$ | Let $r$ be the common ratio of the geometric sequence. Since it is increasing, that means that $b = ar$ , and $c = ar^2$ . Simplifying the logarithm, we get $\log_6(a^3*r^3) = 6$ . Therefore, $a^3*r^3 = 6^6$ . Taking the cube root of both sides, we see that $ar = 6^2 = 36$ . Now since $ar = b$ , that means $b = 36$ . Using the trial and error shown in solution 1, we get $a = 27$ , and $r = \frac{4}{3}$ . Now, $27*r^2= c = 48$ . Therefore, the answer is $27+36+48 = \boxed{111}$ | null | 111 |
da91fa0bc8d18efd7fb475bf088cd581 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_4 | Patio blocks that are hexagons $1$ unit on a side are used to outline a garden by placing the blocks edge to edge with $n$ on each side. The diagram indicates the path of blocks around the garden when $n=5$
AIME 2002 II Problem 4.gif
If $n=202$ , then the area of the garden enclosed by the path, not including the path itself, is $m\left(\sqrt3/2\right)$ square units, where $m$ is a positive integer. Find the remainder when $m$ is divided by $1000$ | When $n>1$ , the path of blocks has $6(n-1)$ blocks total in it. When $n=1$ , there is just one lonely block. Thus, the area of the garden enclosed by the path when $n=202$ is
\[(1+6+12+18+\cdots +1200)A=(1+6(1+2+3...+200))A\]
where $A$ is the area of one block. Then, because $n(n+1)/2$ is equal to the sum of the first $n$ integers:
\[(1+6(1+2+3...+200))=(1+6((200)(201)/2))A=120601A\]
Since $A=\dfrac{3\sqrt{3}}{2}$ , the area of the garden is
\[120601\cdot \dfrac{3\sqrt{3}}{2}=\dfrac{361803\sqrt{3}}{2}\]
$m=361803$ $\dfrac{m}{1000}=361$ Remainder $\boxed{803}$ | null | 803 |
32ba39a714336f8b97705aa3c5835f79 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_7 | It is known that, for all positive integers $k$
Find the smallest positive integer $k$ such that $1^2+2^2+3^2+\ldots+k^2$ is a multiple of $200$ | $\frac{k(k+1)(2k+1)}{6}$ is a multiple of $200$ if $k(k+1)(2k+1)$ is a multiple of $1200 = 2^4 \cdot 3 \cdot 5^2$ .
So $16,3,25|k(k+1)(2k+1)$
Since $2k+1$ is always odd, and only one of $k$ and $k+1$ is even, either $k, k+1 \equiv 0 \pmod{16}$
Thus, $k \equiv 0, 15 \pmod{16}$
If $k \equiv 0 \pmod{3}$ , then $3|k$ . If $k \equiv 1 \pmod{3}$ , then $3|2k+1$ . If $k \equiv 2 \pmod{3}$ , then $3|k+1$
Thus, there are no restrictions on $k$ in $\pmod{3}$
It is easy to see that only one of $k$ $k+1$ , and $2k+1$ is divisible by $5$ . So either $k, k+1, 2k+1 \equiv 0 \pmod{25}$
Thus, $k \equiv 0, 24, 12 \pmod{25}$
From the Chinese Remainder Theorem $k \equiv 0, 112, 224, 175, 287, 399 \pmod{400}$ . Thus, the smallest positive integer $k$ is $\boxed{112}$ | null | 112 |
e7e12b0556b2fccdd692fd0e631112a3 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_8 | Find the least positive integer $k$ for which the equation $\left\lfloor\frac{2002}{n}\right\rfloor=k$ has no integer solutions for $n$ . (The notation $\lfloor x\rfloor$ means the greatest integer less than or equal to $x$ .) | Note that if $\frac{2002}n - \frac{2002}{n+1}\leq 1$ , then either $\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor$ ,
or $\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor+1$ . Either way, we won't skip any natural numbers.
The greatest $n$ such that $\frac{2002}n - \frac{2002}{n+1} > 1$ is $n=44$ . (The inequality simplifies to $n(n+1)<2002$ , which is easy to solve by trial, as the solution is obviously $\simeq \sqrt{2002}$ .)
We can now compute: \[\left\lfloor\frac{2002}{45}\right\rfloor=44\] \[\left\lfloor\frac{2002}{44}\right\rfloor=45\] \[\left\lfloor\frac{2002}{43}\right\rfloor=46\] \[\left\lfloor\frac{2002}{42}\right\rfloor=47\] \[\left\lfloor\frac{2002}{41}\right\rfloor=48\] \[\left\lfloor\frac{2002}{40}\right\rfloor=50\]
From the observation above (and the fact that $\left\lfloor\frac{2002}{2002}\right\rfloor=1$ ) we know that all integers between $1$ and $44$ will be achieved for some values of $n$ . Similarly, for $n<40$ we obviously have $\left\lfloor\frac{2002}{n}\right\rfloor > 50$
Hence the least positive integer $k$ for which the equation $\left\lfloor\frac{2002}{n}\right\rfloor=k$ has no integer solutions for $n$ is $\boxed{049}$ | null | 049 |
e7e12b0556b2fccdd692fd0e631112a3 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_8 | Find the least positive integer $k$ for which the equation $\left\lfloor\frac{2002}{n}\right\rfloor=k$ has no integer solutions for $n$ . (The notation $\lfloor x\rfloor$ means the greatest integer less than or equal to $x$ .) | Rewriting the given information and simplifying it a bit, we have \begin{align*} k \le \frac{2002}{n} < k+1 &\implies \frac{1}{k} \ge \frac{n}{2002} > \frac{1}{k+1}. \\ &\implies \frac{2002}{k} \ge n > \frac{2002}{k+1}. \end{align*}
Now note that in order for there to be no integer solutions to $n,$ we must have $\left\lfloor \frac{2002}{k} \right\rfloor = \left\lfloor \frac{2002}{k+1} \right\rfloor.$ We seek the smallest such $k.$ A bit of experimentation yields that $k=49$ is the smallest solution, as for $k=49,$ it is true that $\left\lfloor \frac{2002}{49} \right\rfloor = \left\lfloor \frac{2002}{50} \right\rfloor = 40.$ Furthermore, $k=49$ is the smallest such case. (If unsure, we could check if the result holds for $k=48,$ and as it turns out, it doesn't.) Therefore, the answer is $\boxed{049}.$ | null | 049 |
e7e12b0556b2fccdd692fd0e631112a3 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_8 | Find the least positive integer $k$ for which the equation $\left\lfloor\frac{2002}{n}\right\rfloor=k$ has no integer solutions for $n$ . (The notation $\lfloor x\rfloor$ means the greatest integer less than or equal to $x$ .) | In this solution we use inductive reasoning and a lot of trial and error. Depending on how accurately you can estimate, the solution will come quicker or slower.
Using values of $k$ as $1, 2, 3, 4,$ and $5,$ we can find the corresponding values of $n$ relatively easily. For $k = 1$ $n$ is in the range $[2002-1002]$ ; for $k = 2$ $n$ is the the range $[1001-668]$ , etc: $3, [667,501]; 4, [500-401]; 5, [400-334]$ . For any positive integer $k, n$ is in a range of $\left\lfloor \frac{2002}{k} \right\rfloor -\left\lceil \frac{2002}{k+1} \right\rceil$
Now we try testing $k = 1002$ to get a better understanding of what our solution will look like. Obviously, there will be no solution for $n$ , but we are more interested in how the range will compute to. Using the formula we got above, the range will be $1-2$ . Testing any integer $k$ from $1002-2000$ will result in the same range. Also, notice that each and every one of them have no solution for $n$ . Testing $1001$ gives a range of $2-2$ , and $2002$ gives $1-1$ . They each have a solution for $n$ , and their range is only one value. Therefore, we can assume with relative safety that the integer $k$ we want is the lowest integer that follows this equation
\[\left\lfloor\frac{2002}{k}\right\rfloor + 1 = \left\lceil \frac{2002}{k+1}\right\rceil\]
Now we can easily guess and check starting from $k = 1$ . After a few tests it's not difficult to estimate a few jumps, and it took me only a few minutes to realize the answer was somewhere in the forties (You could also use the fact that $45^2=2025$ ). Then it's just a matter of checking them until we get $\boxed{049}$ .
Alternatively, you could use the equation above and proceed with one of the other two solutions listed. | null | 049 |
5f339a502832f1a6f2d4bd083aa67b5c | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_10 | While finding the sine of a certain angle, an absent-minded professor failed to notice that his calculator was not in the correct angular mode. He was lucky to get the right answer. The two least positive real values of $x$ for which the sine of $x$ degrees is the same as the sine of $x$ radians are $\frac{m\pi}{n-\pi}$ and $\frac{p\pi}{q+\pi}$ , where $m$ $n$ $p$ , and $q$ are positive integers. Find $m+n+p+q$ | Note that $x$ degrees is equal to $\frac{\pi x}{180}$ radians. Also, for $\alpha \in \left[0 , \frac{\pi}{2} \right]$ , the two least positive angles $\theta > \alpha$ such that $\sin{\theta} = \sin{\alpha}$ are $\theta = \pi-\alpha$ , and $\theta = 2\pi + \alpha$
Clearly $x > \frac{\pi x}{180}$ for positive real values of $x$
$\theta = \pi-\alpha$ yields: $x = \pi - \frac{\pi x}{180} \Rightarrow x = \frac{180\pi}{180+\pi} \Rightarrow (p,q) = (180,180)$
$\theta = 2\pi + \alpha$ yields: $x = 2\pi + \frac{\pi x}{180} \Rightarrow x = \frac{360\pi}{180-\pi} \Rightarrow (m,n) = (360,180)$
So, $m+n+p+q = \boxed{900}$ | null | 900 |
e723b9e81fdb60811e7abc27bf6778a1 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_11 | Two distinct, real, infinite geometric series each have a sum of $1$ and have the same second term. The third term of one of the series is $1/8$ , and the second term of both series can be written in the form $\frac{\sqrt{m}-n}p$ , where $m$ $n$ , and $p$ are positive integers and $m$ is not divisible by the square of any prime. Find $100m+10n+p$ | Let the second term of each series be $x$ . Then, the common ratio is $\frac{1}{8x}$ , and the first term is $8x^2$
So, the sum is $\frac{8x^2}{1-\frac{1}{8x}}=1$ . Thus, $64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}$
The only solution in the appropriate form is $x = \frac{\sqrt{5}-1}{8}$ . Therefore, $100m+10n+p = \boxed{518}$ | null | 518 |
e723b9e81fdb60811e7abc27bf6778a1 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_11 | Two distinct, real, infinite geometric series each have a sum of $1$ and have the same second term. The third term of one of the series is $1/8$ , and the second term of both series can be written in the form $\frac{\sqrt{m}-n}p$ , where $m$ $n$ , and $p$ are positive integers and $m$ is not divisible by the square of any prime. Find $100m+10n+p$ | Let's ignore the "two distinct, real, infinite geometric series" part for now and focus on what it means to be a geometric series.
Let the first term of the series with the third term equal to $\frac18$ be $a,$ and the common ratio be $r.$ Then, we get that $\frac{a}{1-r} = 1 \implies a = 1-r,$ and $ar^2 = \frac18 \implies (1-r)(r^2) = \frac18.$
We see that this cubic is equivalent to $r^3 - r^2 + \frac18 = 0.$ Through experimenting, we find that one of the solutions is $r = \frac12.$ Using synthetic division leads to the quadratic $4x^2 - 2x - 1 = 0.$ This has roots $\dfrac{2 \pm \sqrt{4 - 4(4)(-1)}}{8},$ or, when reduced, $\dfrac{1 \pm \sqrt{5}}{4}.$
It becomes clear that the two geometric series have common ratio $\frac{1 + \sqrt{5}}{4}$ and $\frac{1 - \sqrt{5}}{4}.$ Let $\frac{1 + \sqrt{5}}{4}$ be the ratio that we are inspecting. We see that in this case, $a = \dfrac{3 - \sqrt{5}}{4}.$
Since the second term in the series is $ar,$ we compute this and have that \[ar = \left(\dfrac{3 - \sqrt{5}}{4} \right)\left(\dfrac{1+\sqrt{5}}{4}\right) = \dfrac{\sqrt{5} - 1}{8},\] for our answer of $100 \cdot 5 + 1 \cdot 10 + 8 = \boxed{518}.$ | null | 518 |
2d4a7d7ed46c97734cef74717df85b96 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_12 | A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$ $q$ $r$ , and $s$ are primes, and $a$ $b$ , and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$ | We graph the $10$ shots on a grid. Suppose that a made shot is represented by a step of $(0,1)$ , and a missed shot is represented by $(1,0)$ . Then the basketball player's shots can be represented by the number of paths from $(0,0)$ to $(6,4)$ that always stay below the line $y=\frac{2x}{3}$ . We can find the number of such paths using a Pascal's Triangle type method below, computing the number of paths to each point that only move right and up. [asy] size(150); for (int i=0;i<7;++i) {draw((i,0)--(i,4));} for (int i=0;i<5;++i) {draw((0,i)--(6,i));} draw((0,0)--(6,4),dashed); label("$1$",(0,0),SE,fontsize(8)); label("$1$",(1,0),SE,fontsize(8)); label("$1$",(2,0),SE,fontsize(8)); label("$1$",(2,1),SE,fontsize(8)); label("$1$",(3,0),SE,fontsize(8)); label("$2$",(3,1),SE,fontsize(8)); label("$2$",(3,2),SE,fontsize(8)); label("$1$",(4,0),SE,fontsize(8)); label("$3$",(4,1),SE,fontsize(8)); label("$5$",(4,2),SE,fontsize(8)); label("$1$",(5,0),SE,fontsize(8)); label("$4$",(5,1),SE,fontsize(8)); label("$9$",(5,2),SE,fontsize(8)); label("$9$",(5,3),SE,fontsize(8)); label("$1$",(6,0),SE,fontsize(8)); label("$5$",(6,1),SE,fontsize(8)); label("$14$",(6,2),SE,fontsize(8)); label("$23$",(6,3),SE,fontsize(8)); label("$23$",(6,4),SE,fontsize(8)); [/asy] Therefore, there are $23$ ways to shoot $4$ makes and $6$ misses under the given conditions. The probability of each possible sequence occurring is $(.4)^4(.6)^6$ . Hence the desired probability is \[\frac{23\cdot 2^4\cdot 3^6}{5^{10}},\] and the answer is $(23+2+3+5)(4+6+10)=\boxed{660}$ | null | 660 |
2d4a7d7ed46c97734cef74717df85b96 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_12 | A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$ $q$ $r$ , and $s$ are primes, and $a$ $b$ , and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$ | The first restriction is that $a_{10} = .4$ , meaning that the player gets exactly 4 out of 10 baskets. The second restriction is $a_n\le.4$ . This means that the player may never have a shooting average over 40%. Thus, the first and second shots must fail, since $\frac{1}{1}$ and $\frac{1}{2}$ are both over $.4$ , but the player may make the third basket, since $\frac{1}{3} \le .4$ In other words, the earliest the first basket may be made is attempt 3. Using similar reasoning, the earliest the second basket may be made is attempt 5, the earliest the third basket may be made is attempt 8, and the earliest the fourth basket may be made is attempt 10.
Using X to represent a basket and O to represent a failure, this 'earliest' solution may be represented as:
OOXOXOOXOX
To simplify counting, note that the first, second, and tenth shots are predetermined. The first two shots must fail, and the last shot must succeed. Thus, only slots 3-9 need to be counted, and can be abbreviated as follows:
XOXOOXO
The problem may be separated into five cases, since the first shot may be made on attempt 3, 4, 5, 6, or 7. The easiest way to count the problem is to remember that each X may slide to the right, but NOT to the left.
First shot made on attempt 3:
XOXOOXO
XOXOOOX
XOOXOXO
XOOXOOX
XOOOXXO
XOOOXOX
XOOOOXX
Total - 7
First shot made on attempt 4:
Note that all that needs to be done is change each line in the prior case from starting with "XO....." to "OX.....".
Total - 7
First shot made on attempt 5:
OOXXOXO
OOXXOOX
OOXOXXO
OOXOXOX
OOXOOXX
Total - 5
First shot made on attempt 6:
OOOXXXO
OOOXXOX
OOOXOXX
Total - 3
First shot made on attempt 7:
OOOOXXX
Total - 1
The total number of ways the player may satisfy the requirements is $7+7+5+3+1=23$
The chance of hitting any individual combination (say, for example, OOOOOOXXXX) is $\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4$
Thus, the chance of hitting any of these 23 combinations is $23\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4 = \frac{23\cdot3^6\cdot2^4}{5^{10}}$
Thus, the final answer is $(23+3+2+5)(6+4+10)=\boxed{660}$ | null | 660 |
2d4a7d7ed46c97734cef74717df85b96 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_12 | A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$ $q$ $r$ , and $s$ are primes, and $a$ $b$ , and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$ | Note $a_{10}=.4$ . Therefore the player made 4 shots out of 10. He must make the 10th shot, because if he doesn't, then $a_9=\frac{4}{9}>.4$ . Since $a_n\leq .4$ for all $n$ less than 11, we know that $a_1=a_2=0$ . Now we must look at the 3rd through 9th shot.
Now let's take a look at those un-determined shots. Let's put them into groups: the 3rd, 4th, and 5th shots in group A, and the 6th, 7th, 8th, and 9th shots in group B. The total number of shots made in groups A and B must be 3, since the player makes the 10th shot. We cannot have all three shots made in group A, since $a_5\leq .4$ . Therefore we can have two shots made, one shot made, or no shots made in group A.
Case 1: Group A contains no made shots.
The number of ways this can happen in group A is 1. Now we must arrange the shots in group B accordingly. There are four ways to arrange them total, and all of them work. There are $\textbf{4}$ possibilities here.
Case 2: Group A contains one made shot.
The number of ways this could happen in group A is 3. Now we must arrange the shots in group B accordingly. There are six ways to arrange them total, but the arrangement "hit hit miss miss" fails, because that would mean $a_7=\frac{3}{7}>.4$ . All the rest work. Therefore there are $3\cdot5=\textbf{15}$ possibilities here.
Case 3: Group A contains two made shots.
The number of ways this could happen in group A is 2 (hit hit miss doesn't work but the rest do). Now we must arrange the shots in group B accordingly. Note hit miss miss miss and miss hit miss miss fail. Therefore there are only 2 ways to do this, and there are $2\cdot 2=\textbf{4}$ total possibilities for this case.
Taking all these cases into account, we find that there are $4+15+4=23$ ways to have $a_{10} = .4$ and $a_n\leq .4$ . Each of these has a probability of $.4^4\cdot.6^6=\frac{2^4\cdot 3^6}{5^{10}}$ . Therefore the probability that we have $a_{10} = .4$ and $a_n\leq .4$ is $\frac{23\cdot2^4\cdot3^6}{5^{10}}$ . Now we are asked to find the product of the sum of the primes and the sum of the exponents, which is $(23+2+3+5)(4+6+10)=33\cdot20=\boxed{660}$ | null | 660 |
a374ccd51c096e16dcb62edac8d8aa5e | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_13 | In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | First draw $\overline{CP}$ and extend it so that it meets with $\overline{AB}$ at point $X$
[asy] size(10cm); pair A,B,C,D,E,X,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); X=(4.3636,0); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(X); dot(P); dot(Q); dot(R); label("$A$",A,WSW); label("$B$",B,ESE); label("$C$",C,NNW); label("$D$",D,NE); label("$E$",E,WNW); label("$X$",X,SSE); label("$P$",P,NNE); label("$Q$",Q,SSW); label("$R$",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); draw(C--X); [/asy]
We have that $[ABC]=\frac{1}{2}\cdot AC \cdot BC\sin{C}=\frac{1}{2}\cdot 4\cdot {7}\sin{C}=14\sin{C}$
By Ceva's, \[3\cdot{\frac{2}{5}}\cdot{\frac{BX}{AX}}=1\implies BX=\frac{5\cdot AX}{6}\] That means that \[\frac{11\cdot {AX}}{6}=8\implies AX=\frac{48}{11} \ \text{and} \ BX=\frac{40}{11}\]
Now we apply mass points. Assume WLOG that $W_{A}=1$ . That means that
\[W_{C}=3, W_{B}=\frac{6}{5}, W_{X}=\frac{11}{5}, W_{D}=\frac{21}{5}, W_{E}=4, W_{P}=\frac{26}{5}\]
Notice now that $\triangle{PBQ}$ is similar to $\triangle{EBA}$ . Therefore,
\[\frac{PQ}{EA}=\frac{PB}{EB}\implies \frac{PQ}{3}=\frac{10}{13}\implies PQ=\frac{30}{13}\]
Also, $\triangle{PRA}$ is similar to $\triangle{DBA}$ . Therefore,
\[\frac{PA}{DA}=\frac{PR}{DB}\implies \frac{21}{26}=\frac{PR}{5}\implies PR=\frac{105}{26}\]
Because $\triangle{PQR}$ is similar to $\triangle{CAB}$ $\angle{C}=\angle{P}$
As a result, $[PQR]=\frac{1}{2}\cdot PQ \cdot PR \sin{C}=\frac{1}{2}\cdot \frac{30}{13}\cdot \frac{105}{26}\sin{P}=\frac{1575}{338}\sin{C}$
Therefore, \[\frac{[PQR]}{[ABC]}=\frac{\frac{1575}{338}\sin{C}}{14\sin{C}}=\frac{225}{676}\implies 225+676=\boxed{901}\] | null | 901 |
a374ccd51c096e16dcb62edac8d8aa5e | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_13 | In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | [asy] size(10cm); pair A,B,C,D,E,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); dot(Q); dot(R); label("$A$",A,WSW); label("$B$",B,ESE); label("$C$",C,NNW); label("$D$",D,NE); label("$E$",E,WNW); label("$P$",P,NNE); label("$Q$",Q,SSW); label("$R$",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); [/asy] Use the mass of point. Denoting the mass of $C=15,B=6,A=5,D=21,E=20$ , we can see that the mass of Q is $26$ , hence we know that $\frac{BP}{PE}=\frac{10}{3}$ , now we can find that $\frac{PQ}{AE}=\frac{10}{3}$ which implies $PQ=\frac{30}{13}$ , it is obvious that $\triangle{PQR}$ is similar to $\triangle{ACB}$ so we need to find the ration between PQ and AC, which is easy, it is $\frac{15}{26}$ , so our final answer is $\left( \frac{15}{26} \right)^2= \frac{225}{676}$ which is $\boxed{901}$ . ~bluesoul | null | 901 |
dcc430ba28f8ea9b16d4a1c2b581b3c6 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_14 | The perimeter of triangle $APM$ is $152$ , and the angle $PAM$ is a right angle . A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$ . Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$ | Let the circle intersect $\overline{PM}$ at $B$ . Then note $\triangle OPB$ and $\triangle MPA$ are similar. Also note that $AM = BM$ by power of a point . Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have \[\frac{19}{AM} = \frac{152-2AM-19+19}{152} = \frac{152-2AM}{152}\] Solving, $AM = 38$ . So the ratio of the side lengths of the triangles is 2. Therefore, \[\frac{PB+38}{OP}= 2 \text{ and } \frac{OP+19}{PB} = 2\] so $2OP = PB+38$ and $2PB = OP+19.$ Substituting for $PB$ , we see that $4OP-76 = OP+19$ , so $OP = \frac{95}3$ and the answer is $\boxed{098}$ | null | 098 |
dcc430ba28f8ea9b16d4a1c2b581b3c6 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_14 | The perimeter of triangle $APM$ is $152$ , and the angle $PAM$ is a right angle . A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$ . Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$ | Reflect triangle $PAM$ across line $AP$ , creating an isoceles triangle. Let $x$ be the distance from the top of the circle to point $P$ , with $x + 38$ as $AP$ . Given the perimeter is 152, subtracting the altitude yields the semiperimeter $s$ of the isoceles triangle, as $114 - x$ . The area of the isoceles triangle is:
$[PAM] = r \cdot s$
$[PAM] = 19 \cdot (114 - x)$
Now use similarity, draw perpendicular from $O$ to $PM$ , name the new point $D$ . Triangle $PDO$ is similar to triangle $PAM$ , by AA Similarity. Equating the legs, we get:
$\frac{\sqrt{x}}{19} = \frac{\sqrt{x + 38}}{AM}$
Solving for $AM$ , it yields $19 \cdot \sqrt{\frac{x + 38}{x}}$
$19 \cdot (114 - x) = AM \cdot AP = 19 \cdot (x + 38) \cdot \sqrt{\frac{x + 38}{x}}$
The $x^3$ cancels, yielding a quadratic. Solving yields $x = \frac{38}{3}$ .
Add $19$ to find $OP$ , yielding $\frac{95}{3}$ or $\boxed{098}$ | null | 098 |
dcc430ba28f8ea9b16d4a1c2b581b3c6 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_14 | The perimeter of triangle $APM$ is $152$ , and the angle $PAM$ is a right angle . A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$ . Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$ | Let the foot of the perpendicular from $O$ to $PM$ be $D;$ now $OD=19.$ Also let $AM=x$ and $PM=y.$ This means that $OP=\frac{y}{x}\cdot 19$ , since $O$ is on the angle bisector of $\angle M.$
We have that $\tan(\angle AMO)=\frac{19}{x},$ so \[\tan(\angle M)=\tan (2\cdot \angle AMO)=\frac{38x}{x^{2}-361}.\]
However $\tan(\angle M)=\frac{PA}{AM}=\frac{PO+OA}{AM}=\frac{\frac{y}{x}\cdot 19 + 19}{x}$ , so \[\frac{38x}{x^{2}-361}=19\cdot \frac{\frac{y}{x}+1}{x}\] \[\frac{2x^{2}}{x^{2}-361}=\frac{y}{x}+1\] \[\frac{x^{2}+361}{x^{2}-361}=\frac{y}{x}.\] \[x\cdot \frac{x^{2}+361}{x^{2}-361}=y\]
We now use the fact that the perimeter of $\triangle PAM$ is $152$ \[PO+OA+AM+MP=152\] \[\frac{y}{x}\cdot 19+19+x+y=152\] \[19\left(\frac{x^{2}+361}{x^{2}-361}\right)+x\cdot \left(\frac{x^{2}+361}{x^{2}-361}\right)+x+19=152\] \[(x+19)\left(\frac{x^{2}+361}{x^{2}-361}+\frac{x^{2}-361}{x^{2}-361}\right)=152\] \[\frac{2x^{2}}{x-19}=152\] \[x^{2}-76x+19\cdot 76=0.\] This quadratic factors as $(x-38)^{2}=0,$ so $x=38$ , and \[\frac{y}{x}=\frac{38^{2}+361}{38^{2}-361}=\frac{5}{3}\] \[OP=\frac{y}{x}\cdot 19=\frac{95}{3}\to \boxed{98.}\] | null | 98. |
92b8e5e9cbfc989f56adb2dd42aa88b6 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_15 | Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$ , and the product of the radii is $68$ . The x-axis and the line $y = mx$ , where $m > 0$ , are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$ , where $a$ $b$ , and $c$ are positive integers, $b$ is not divisible by the square of any prime, and $a$ and $c$ are relatively prime. Find $a + b + c$ | Let the smaller angle between the $x$ -axis and the line $y=mx$ be $\theta$ . Note that the centers of the two circles lie on the angle bisector of the angle between the $x$ -axis and the line $y=mx$ . Also note that if $(x,y)$ is on said angle bisector, we have that $\frac{y}{x}=\tan{\frac{\theta}{2}}$ . Let $\tan{\frac{\theta}{2}}=m_1$ , for convenience. Therefore if $(x,y)$ is on the angle bisector, then $x=\frac{y}{m_1}$ . Now let the centers of the two relevant circles be $(a/m_1 , a)$ and $(b/m_1 , b)$ for some positive reals $a$ and $b$ . These two circles are tangent to the $x$ -axis, so the radii of the circles are $a$ and $b$ respectively. We know that the point $(9,6)$ is a point on both circles, so we have that
\[(9-\frac{a}{m_1})^2+(6-a)^2=a^2\]
\[(9-\frac{b}{m_1})^2+(6-b)^2=b^2\]
Expanding these and manipulating terms gives
\[\frac{1}{m_1^2}a^2-[(18/m_1)+12]a+117=0\]
\[\frac{1}{m_1^2}b^2-[(18/m_1)+12]b+117=0\]
It follows that $a$ and $b$ are the roots of the quadratic
\[\frac{1}{m_1^2}x^2-[(18/m_1)+12]x+117=0\]
It follows from Vieta's Formulas that the product of the roots of this quadratic is $117m_1^2$ , but we were also given that the product of the radii was 68. Therefore $68=117m_1^2$ , or $m_1^2=\frac{68}{117}$ . Note that the half-angle formula for tangents is
\[\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}\]
Therefore
\[\frac{68}{117}=\frac{1-\cos{\theta}}{1+\cos{\theta}}\]
Solving for $\cos{\theta}$ gives that $\cos{\theta}=\frac{49}{185}$ . It then follows that $\sin{\theta}=\sqrt{1-\cos^2{\theta}}=\frac{12\sqrt{221}}{185}$
It then follows that $m=\tan{\theta}=\frac{12\sqrt{221}}{49}$ . Therefore $a=12$ $b=221$ , and $c=49$ . The desired answer is then $12+221+49=\boxed{282}$ | null | 282 |
92b8e5e9cbfc989f56adb2dd42aa88b6 | https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_15 | Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$ , and the product of the radii is $68$ . The x-axis and the line $y = mx$ , where $m > 0$ , are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$ , where $a$ $b$ , and $c$ are positive integers, $b$ is not divisible by the square of any prime, and $a$ and $c$ are relatively prime. Find $a + b + c$ | Let the centers of $C_1$ and $C_2$ be $A$ and $B$ , respectively, and let the point $(9, 6)$ be $P$
Because both $C_1$ and $C_2$ are tangent to the x-axis, and both of them pass through $P$ , both $A$ and $B$ must be equidistant from $P$ and the x-axis. Therefore, they must both be on the parabola with $P$ as the focus and the x-axis as the directrix. Since the coordinates of $P$ is $(9, 6)$ , we see that this parabola is the graph of the function \[y=\frac{1}{12}(x-9)^2+3=\frac{1}{12}x^2-\frac{3}{2}x+\frac{39}{4}.\]
Let $AB$ be $y=kx$ . Because $C_1$ and $C_2$ are both tangent to the x-axis, the y-coordinates of $A$ and $B$ are $r_1$ and $r_2$ , respectively, so the x-coordinates of $A$ and $B$ are $\frac{r_1}{k}$ and $\frac{r_2}{k}$ . But since $A$ and $B$ are also on the graph of the function $y=\frac{1}{12}x^2-\frac{3}{2}x+\frac{39}{4}$ , the x-coordinates of $A$ and $B$ are also the roots of the equation $kx=\frac{1}{12}x^2-\frac{3}{2}x+\frac{39}{4}$ , and by Vieta's Formulas, their product is $\frac{\frac{39}{4}}{\frac{1}{12}}=117$ . So we have $\frac{r_1}{k}\cdot \frac{r_2}{k}=117$
We are also given that $r_1r_2=68$ , so $k^2=\frac{r_1r_2}{117}=\frac{68}{117}$ , which means that $k=\sqrt{\frac{68}{117}}$ . Note that the line $AB$ is the angle bisector of the angle between the line $y=mx$ and the x-axis. Therefore, we apply the double-angle formula for tangents and get \[m=\frac{2k}{1-k^2}=\frac{2\sqrt{\frac{68}{117}}}{1-\frac{68}{117}}=\frac{12\sqrt{221}}{49}.\] Thus, the answer is $12+221+49=\boxed{282}$ | null | 282 |
3b493f3c5c9dcbb77a9a5a7d1796124c | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_1 | Find the sum of all positive two-digit integers that are divisible by each of their digits. | Let our number be $10a + b$ $a,b \neq 0$ . Then we have two conditions: $10a + b \equiv 10a \equiv 0 \pmod{b}$ and $10a + b \equiv b \pmod{a}$ , or $a$ divides into $b$ and $b$ divides into $10a$ . Thus $b = a, 2a,$ or $5a$ (note that if $b = 10a$ , then $b$ would not be a digit).
If we ignore the case $b = 0$ as we have been doing so far, then the sum is $495 + 120 + 15 = \boxed{630}$ | null | 630 |
3b493f3c5c9dcbb77a9a5a7d1796124c | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_1 | Find the sum of all positive two-digit integers that are divisible by each of their digits. | Using casework, we can list out all of these numbers: \[11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.\] | null | 630 |
3b493f3c5c9dcbb77a9a5a7d1796124c | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_1 | Find the sum of all positive two-digit integers that are divisible by each of their digits. | In this solution, we will do casework on the ones digit.
Before we start, let's make some variables. Let $a$ be the ones digit, and $b$ be the tens digit. Let $n$ equal our number. Our number can be expressed as $10b+a$ . We can easily see that $b|a$ , since $b|n$ , and $b|10b$ . Therefore, $b|(n-10b)$ .
Now, let's start with the casework.
Case 1: $a=1$ Since $b|a$ $b=1$ . From this, we get that $n=11$ satisfies the condition.
Case 2: $a=2$ We either have $b=1$ , or $b=2$ . From this, we get that $n=12$ and $n=22$ satisfy the condition.
Case 3: $a=3$ We have $b=3$ . From this, we get that $n=33$ satisfies the condition. Note that $b=1$ was not included because $3$ does not divide $13$
Case 4: $a=4$ We either have $b=2$ or $b=4$ . From this, we get that $n=24$ and $n=44$ satisfy the condition. $b=1$ was not included for similar reasons as last time.
Case 5: $a=5$ We either have $b=1$ or $b=5$ . From this, we get that $n=15$ and $n=55$ satisfy the condition.
Continuing with this process up to $a=9$ , we get that $n$ could be $11, 12, 22, 33, 24, 44, 15, 55, 36, 66, 77, 48, 88, 99$ . Summing, we get that the answer is $\boxed{630}$ . A clever way to sum would be to group the multiples of $11$ together to get $11+22+\dots+99=(45)(11)=495$ , and then add the remaining $12+24+15+36+48=135$ | null | 630 |
b45c0eee8860e0ae35f1b23ae85b14c1 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_2 | A finite set $\mathcal{S}$ of distinct real numbers has the following properties: the mean of $\mathcal{S}\cup\{1\}$ is $13$ less than the mean of $\mathcal{S}$ , and the mean of $\mathcal{S}\cup\{2001\}$ is $27$ more than the mean of $\mathcal{S}$ . Find the mean of $\mathcal{S}$ | Let $x$ be the mean of $\mathcal{S}$ . Let $a$ be the number of elements in $\mathcal{S}$ .
Then, the given tells us that $\frac{ax+1}{a+1}=x-13$ and $\frac{ax+2001}{a+1}=x+27$ . Subtracting, we have \begin{align*}\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1} \Longrightarrow \frac{2000}{a+1}=40 \Longrightarrow a=49\end{align*} We plug that into our very first formula, and get: \begin{align*}\frac{49x+1}{50}&=x-13 \\ 49x+1&=50x-650 \\ x&=\boxed{651} | null | 651 |
b45c0eee8860e0ae35f1b23ae85b14c1 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_2 | A finite set $\mathcal{S}$ of distinct real numbers has the following properties: the mean of $\mathcal{S}\cup\{1\}$ is $13$ less than the mean of $\mathcal{S}$ , and the mean of $\mathcal{S}\cup\{2001\}$ is $27$ more than the mean of $\mathcal{S}$ . Find the mean of $\mathcal{S}$ | Since this is a weighted average problem, the mean of $S$ is $\frac{13}{27}$ as far from $1$ as it is from $2001$ . Thus, the mean of $S$ is $1 + \frac{13}{13 + 27}(2001 - 1) = \boxed{651}$ | null | 651 |
12c94fcd406b1d7a552618d6f33b3195 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_3 | Find the sum of the roots , real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$ , given that there are no multiple roots. | From Vieta's formulas , in a polynomial of the form $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0$ , then the sum of the roots is $\frac{-a_{n-1}}{a_n}$
From the Binomial Theorem , the first term of $\left(\frac 12-x\right)^{2001}$ is $-x^{2001}$ , but $x^{2001}+-x^{2001}=0$ , so the term with the largest degree is $x^{2000}$ . So we need the coefficient of that term, as well as the coefficient of $x^{1999}$
\begin{align*}\binom{2001}{1} \cdot (-x)^{2000} \cdot \left(\frac{1}{2}\right)^1&=\frac{2001x^{2000}}{2}\\ \binom{2001}{2} \cdot (-x)^{1999} \cdot \left(\frac{1}{2}\right)^2 &=\frac{-x^{1999}*2001*2000}{8}=-2001 \cdot 250x^{1999} \end{align*}
Applying Vieta's formulas, we find that the sum of the roots is $-\frac{-2001 \cdot 250}{\frac{2001}{2}}=250 \cdot 2=\boxed{500}$ | null | 500 |
12c94fcd406b1d7a552618d6f33b3195 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_3 | Find the sum of the roots , real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$ , given that there are no multiple roots. | We find that the given equation has a $2000^{\text{th}}$ degree polynomial. Note that there are no multiple roots. Thus, if $\frac{1}{2} - x$ is a root, $x$ is also a root. Thus, we pair up $1000$ pairs of roots that sum to $\frac{1}{2}$ to get a sum of $\boxed{500}$ | null | 500 |
12c94fcd406b1d7a552618d6f33b3195 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_3 | Find the sum of the roots , real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$ , given that there are no multiple roots. | Note that if $r$ is a root, then $\frac{1}{2}-r$ is a root and they sum up to $\frac{1}{2}.$ We make the substitution $y=x-\frac{1}{4}$ so \[(\frac{1}{4}+y)^{2001}+(\frac{1}{4}-y)^{2001}=0.\] Expanding gives \[2\cdot\frac{1}{4}\cdot\binom{2001}{1}y^{2000}-0y^{1999}+\cdots\] so by Vieta, the sum of the roots of $y$ is 0. Since $x$ has a degree of 2000, then $x$ has 2000 roots so the sum of the roots is \[2000(\sum_{n=1}^{2000} y+\frac{1}{4})=2000(0+\frac{1}{4})=\boxed{500}.\] | null | 500 |
a17a091bda72f61cbdaffc72cd4f4227 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_4 | In triangle $ABC$ , angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$ , and $AT=24$ . The area of triangle $ABC$ can be written in the form $a+b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$ | After chasing angles, $\angle ATC=75^{\circ}$ and $\angle TCA=75^{\circ}$ , meaning $\triangle TAC$ is an isosceles triangle and $AC=24$
Using law of sines on $\triangle ABC$ , we can create the following equation:
$\frac{24}{\sin(\angle ABC)}$ $=$ $\frac{BC}{\sin(\angle BAC)}$
$\angle ABC=45^{\circ}$ and $\angle BAC=60^{\circ}$ , so $BC = 12\sqrt{6}$
We can then use the Law of Sines area formula $\frac{1}{2} \cdot BC \cdot AC \cdot \sin(\angle BCA)$ to find the area of the triangle.
$\sin(75)$ can be found through the sin addition formula.
$\sin(75)$ $=$ $\frac{\sqrt{6} + \sqrt{2}}{4}$
Therefore, the area of the triangle is $\frac{\sqrt{6} + \sqrt{2}}{4}$ $\cdot$ $24$ $\cdot$ $12\sqrt{6}$ $\cdot$ $\frac{1}{2}$
$72\sqrt{3} + 216$
$72 + 3 + 216 =$ $\boxed{291}$ | null | 291 |
a17a091bda72f61cbdaffc72cd4f4227 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_4 | In triangle $ABC$ , angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$ , and $AT=24$ . The area of triangle $ABC$ can be written in the form $a+b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$ | First, draw a good diagram.
We realize that $\angle C = 75^\circ$ , and $\angle CAT = 30^\circ$ . Therefore, $\angle CTA = 75^\circ$ as well, making $\triangle CAT$ an isosceles triangle. $AT$ and $AC$ are congruent, so $AC=24$ . We now drop an altitude from $C$ , and call the foot this altitude point $D$
By 30-60-90 triangles, $AD=12$ and $CD=12\sqrt{3}$
We also notice that $\triangle CDB$ is an isosceles right triangle. $CD$ is congruent to $BD$ , which makes $BD=12\sqrt{3}$ . The base $AB$ is $12+12\sqrt{3}$ , and the altitude $CD=12\sqrt{3}$ . We can easily find that the area of triangle $ABC$ is $216+72\sqrt{3}$ , so $a+b+c=\boxed{291}$ | null | 291 |
2e0e16f594b1609ed140721ef3e527e2 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_5 | An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Denote the vertices of the triangle $A,B,$ and $C,$ where $B$ is in quadrant 4 and $C$ is in quadrant $3.$
Note that the slope of $\overline{AC}$ is $\tan 60^\circ = \sqrt {3}.$ Hence, the equation of the line containing $\overline{AC}$ is \[y = x\sqrt {3} + 1.\] This will intersect the ellipse when \begin{eqnarray*}4 = x^{2} + 4y^{2} & = & x^{2} + 4(x\sqrt {3} + 1)^{2} \\ & = & x^{2} + 4(3x^{2} + 2x\sqrt {3} + 1) \implies x(13x+8\sqrt 3)=0\implies x = \frac { - 8\sqrt {3}}{13}. \end{eqnarray*} We ignore the $x=0$ solution because it is not in quadrant 3.
Since the triangle is symmetric with respect to the y-axis, the coordinates of $B$ and $C$ are now $\left(\frac {8\sqrt {3}}{13},y_{0}\right)$ and $\left(\frac { - 8\sqrt {3}}{13},y_{0}\right),$ respectively, for some value of $y_{0}.$
It is clear that the value of $y_{0}$ is irrelevant to the length of $BC$ . Our answer is \[BC = 2*\frac {8\sqrt {3}}{13}=\sqrt {4\left(\frac {8\sqrt {3}}{13}\right)^{2}} = \sqrt {\frac {768}{169}}\implies m + n = \boxed{937}.\] | null | 937 |
2e0e16f594b1609ed140721ef3e527e2 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_5 | An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Solving for $y$ in terms of $x$ gives $y=\sqrt{4-x^2}/2$ , so the two other points of the triangle are $(x,\sqrt{4-x^2}/2)$ and $(-x,\sqrt{4-x^2}/2)$ , which are a distance of $2x$ apart. Thus $2x$ equals the distance between $(x,\sqrt{4-x^2}/2)$ and $(0,1)$ , so by the distance formula we have
\[2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}.\]
Squaring both sides and simplifying through algebra yields $x^2=192/169$ , so $2x=\sqrt{768/169}$ and the answer is $\boxed{937}$ | null | 937 |
2e0e16f594b1609ed140721ef3e527e2 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_5 | An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Since the altitude goes along the $y$ axis, this means that the base is a horizontal line, which means that the endpoints of the base are $(x,y)$ and $(-x,y)$ , and WLOG, we can say that $x$ is positive.
Now, since all sides of an equilateral triangle are the same, we can do this (distance from one of the endpoints of the base to the vertex and the length of the base):
$\sqrt{x^2 + (y-1)^2} = 2x$
Square both sides,
$x^2 + (y-1)^2 = 4x^2\implies (y-1)^2 = 3x^2$
Now, with the equation of the ellipse: $x^2 + 4y^2 = 4$
$x^2 = 4-4y^2$
$3x^2 = 12-12y^2$
Substituting,
$12-12y^2 = y^2 - 2y +1$
Moving stuff around and solving:
$y = \frac{-11}{13}, 1$
The second is found to be extraneous, so, when we go back and figure out $x$ and then $2x$ (which is the side length), we find it to be:
$\sqrt{\frac{768}{169}}$
and so we get the desired answer of $\boxed{937}$ | null | 937 |
2e0e16f594b1609ed140721ef3e527e2 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_5 | An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Denote $(0,1)$ as vertex $A,$ $B$ as the vertex to the left of the $y$ -axis and $C$ as the vertex to the right of the $y$ -axis. Let $D$ be the intersection of $BC$ and the $y$ -axis.
Let $x_0$ be the $x$ -coordinate of $C.$ This implies \[C=\left(x_0 , \sqrt{\frac{4-x_0^2}{4}}\right)\] and \[B=\left(-x_0 , \sqrt{\frac{4-x_0^2}{4}}\right).\] Note that $BC=2x_0$ and \[\frac{BC}{\sqrt3}=AD=1-\sqrt{\frac{4-x_0^2}{4}}.\] This yields \[\frac{2x_0}{\sqrt3}=1-\sqrt{\frac{4-x_0^2}{4}}.\] Re-arranging and squaring, we have \[\frac{4-x_0^2}{4}=\frac{4x_0^2}{3}-\frac{4x_0}{\sqrt3} +1.\] Simplifying and solving for $x_0$ , we have \[x_0=\frac{48}{13\sqrt 3}.\] As the length of each side is $2x_0,$ our desired length is \[4x_0^2=\frac{768}{169}\] which means our desired answer is \[768+169=\boxed{937}\] | null | 937 |
2e0e16f594b1609ed140721ef3e527e2 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_5 | An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Notice that $x^2+4y^2=4$ can be rewritten as $(x)^2+(2y)^2=2^2$ . The points of the triangle are $(0, 1)$ $(-x, 1-x\sqrt{3})$ , and $(x, 1-x\sqrt{3})$ . When plugging the second coordinate into the equation, we get $x^2+4-8x\sqrt{3}+12x^2=4$ , which equals $13x^2-8x\sqrt{3}=0$ .
This yields $x(13x-8\sqrt{3})=0$ . Obviously x can't be 0, so $x=\frac{8\sqrt{3}}{13}$ . The side length of the equilateral triangle is twice of this, so $\frac{16\sqrt{3}}{13}$ .
This can be rewritten as $\sqrt{\frac{256\cdot3}{169}}=\sqrt{\frac{768}{169}}$ $768+169=\boxed{937}$ .
~ MC413551 | null | 937 |
2e0e16f594b1609ed140721ef3e527e2 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_5 | An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Consider the transformation $(x,y)$ to $(x/2, y).$ This sends the ellipse to the unit circle. If we let $n$ be one-fourth of the side length of the triangle, the equilateral triangle is sent to an isosceles triangle with side lengths $2n, n\sqrt{13}, n\sqrt{13}.$ Let the triangle be $ABC$ such that $AB=AC.$ Let the foot of the altitude from A be $X.$ Then $BX=n,$ and $AX=2n\sqrt{3}.$ Let $C$ be a point such that $AC$ is a diameter of the unit circle. Then $XC=2-2n\sqrt{3}.$ Using power of a point on X, \[n^2=2n\sqrt{3}(2-2n\sqrt{3})\] Simplifying gets us to $13n^2=4n\sqrt{3}.$ Then, $n=\dfrac{4\sqrt{3}}{13}$ which means the side length is $\dfrac{16\sqrt{3}}{13}=\sqrt{\dfrac{768}{169}}.$ Thus, the answer is $768+169=\boxed{937}.$ | null | 937 |
07a8a9cbd118d02c9767ffa55d59e521 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_6 | A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$ | Recast the problem entirely as a block-walking problem. Call the respective dice $a, b, c, d$ . In the diagram below, the lowest $y$ -coordinate at each of $a$ $b$ $c$ , and $d$ corresponds to the value of the roll.
AIME01IN6.png
The red path corresponds to the sequence of rolls $2, 3, 5, 5$ . This establishes a bijection between valid dice roll sequences and block walking paths.
The solution to this problem is therefore $\dfrac{\binom{9}{4}}{6^4} = {\dfrac{7}{72}}$ . So the answer is $\boxed{079}$ | null | 079 |
07a8a9cbd118d02c9767ffa55d59e521 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_6 | A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$ | Let $a, b, c,$ and $d$ be the results of rolling the four dice respectively. We have the range $1\leq a\leq b\leq c\leq d\leq 6$ , and there are $6^4=1296$ total outcomes from rolling the dice. To transfer the inequality into a strictly increasing inequality, we can transform it into $1\leq a<b+1<c+2<d+3\leq 9$ . Now, lets suppose $a'=a, b'=b+1, c'=c+2,$ and $d'=d+3$ . Then, $1\leq a'<b'<c'<d' \leq 9$ . Clearly, there are $\binom{9}{4}=126$ values that satisfy this equation, so our answer is $\dfrac{\binom{9}{4}}{6^4} = {\dfrac{7}{72}}\implies \boxed{079}$ | null | 079 |
07a8a9cbd118d02c9767ffa55d59e521 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_6 | A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$ | If we take any combination of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for four numbers from $\{1,2,3,4,5,6\}$ . We can visualize this as taking the four dice and splitting them into 6 slots (each slot representing one of {1,2,3,4,5,6}), or dividing them amongst 5 separators. Thus, there are ${9\choose4} = 126$ outcomes of four dice. The solution is therefore $\frac{126}{6^4} = \frac{7}{72}$ , and $7 + 72 = \boxed{079}$ | null | 079 |
07a8a9cbd118d02c9767ffa55d59e521 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_6 | A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$ | Call the dice rolls $a, b, c, d$ . The difference between the $a$ and $d$ distinguishes the number of possible rolls there are.
Continuing, we see that the sum is equal to $\sum_{i = 0}^{5}{i+2\choose2}(6 - i) = 6 + 15 + 24 + 30 + 30 + 21 = 126$ . The requested probability is $\frac{126}{6^4} = \frac{7}{72}$ and our answer is $\boxed{79}$ | null | 79 |
07a8a9cbd118d02c9767ffa55d59e521 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_6 | A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$ | The dice rolls can be in the form
where A, B, C, D are some possible value of the dice rolls. (These forms are not keeping track of whether or not the dice are in ascending order, just the possible outcomes.)
There are a total of $6^4$ possible dice rolls.
Thus,
Therefore, our answer is $\boxed{079}$ | null | 079 |
07a8a9cbd118d02c9767ffa55d59e521 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_6 | A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$ | In a manner similar to the above solution, instead consider breaking it down into two sets of two dice rolls. The first subset must have a maximum value which is $\le$ the minimum value of the second subset.
Thus, the number of combinations is $\sum_{i=1}^6 i \cdot \left(\frac{(7 - i)(8 - i)}{2}\right) = 21 + 30 + 30 + 24 + 15 + 6 = 126$ , and the probability again is $\frac7{72}$ , giving $m+n=\boxed{079}$ | null | 079 |
07a8a9cbd118d02c9767ffa55d59e521 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_6 | A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$ | Lets try casework and observe the cases. Notice that if the last roll is a $1$ , then the only dice rolls may be $1-1-1-1$ , which is only $1$ possibility. Observe that if the last roll is $2$ , then there are $4 = 1 + 3$ possibilities. When the last roll is a $3$ , there are $10 = 1 + 3 + 6$ possibilities. Notice when the last roll is $n$ , the number of cases is the sum of the first $n$ positive triangular numbers. This is easily provable when observing the numbers of possibilities after assigning a value to the last and second to last rolls.
So there are a total of $1 + 1 + 3 + 1 + 3 + 6 + 1 + 3 + 6 + 10 + 1 + 3 + 6 + 10 + 15 + 1 + 3 + 6 + 10 + 15 + 21 = 126$ possibilities. So the probability is $\frac{126}{6^4} = \frac{7}{72}$ and $7 + 72 = \boxed{079}$ . ~skyscraper | null | 079 |
07a8a9cbd118d02c9767ffa55d59e521 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_6 | A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$ | This is equivalent to picking a four-element sequence of $\{1, 2, 3, 4, 5, 6\}$ with repetition. Notice that once the sequence is picked, there is one and only one way to order these so that they form a sequence of rolls satisfying our conditions.
Now count the number of such four-element sequences, let $a$ be the number of $1$ s in the sequence, $b$ be the number of $2$ s, $c$ $3$ s, $d$ $4$ s, $e$ $5$ s, and $f$ $6$ s. Now we see that we must have \[a + b + c + d + e + f = 4\] with $a, b, c, d, e, f$ being nonnegative integers since there are a total of $4$ numbers picked. The number of solutions to this is $\dbinom{9}{4},$ so our total number is equal to $\dfrac{\binom{9}{4}}{6^4} = \dfrac{7}{72},$ making our answer $\boxed{079}.$ | null | 079 |
07a8a9cbd118d02c9767ffa55d59e521 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_6 | A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$ | Let the rolls be $a,b,c$ and $d\newline$ let $z=a-1, e=b-a, f=c-b, g=d-c, h=6-d\newline$ $z+e+f+g+h=5\newline$ This equation has $C(5+5-1, 5-1)=126$ integer solutions $\newline$ $126/1296=7/72\newline$ $7+72=\boxed{79}$ ~ryanbear | null | 79 |
07a8a9cbd118d02c9767ffa55d59e521 | https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_6 | A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$ | Let's say the four dice values are all different. These can only be arranged in one way to satisfy our conditions, so there are $\binom{6}{4}=15$ ways. If there are three different values, there are $\binom{6}{3}$ to choose the numbers and $\binom{3}{1}$ to choose which number will have the repeat, so $20\cdot3=60$ ways. If there are two different values, there are $\binom{6}{2}$ ways to choose the numbers. If there are two of each, then there is one way. If there are three of one and one of another, then there are $\binom{2}{1}$ ways. Therefore $15\cdot(1+2)=45$ . Now if all the numbers are the same, there are $\binom{6}{1}=6$ ways. Altogether we have $15+60+45+6=126$ ways. $\frac{126}{6^4}=\frac{21}{216}=\frac{7}{72}$ $7+72=\boxed{79}$ .
~MC413551 | null | 79 |