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33e99f18e99cdf0a415426283897830e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | Rewrite the set after mod 3 as above
1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0
All 0s can be omitted
Case 1
No 1 No 2 $1$
Case 2 $222$ $20$
Case 3 $222222$ $1$
Case 4 $12$ $6*6=36$
Case 5 $12222$ $6*15=90$
Case 6 $1122$ $15*15=225$
Case 7 $1122222$ $15*6=90$
Case 8 $111$ $20$
Case 9 $111222$ $20*20=400$
Case 10 $111222222$ $20$
Case 11 $11112$ $15*6=90$
Case 12 $11112222$ $15*15=225$
Case 13 $1111122$ $6*15=90$
Case 14 $1111122222$ $6*6=36$
Case 15 $111111$ $1$
Case 16 $111111222$ $20$
Case 17 $111111222222$ $1$
Total $1+20+1+36+90+225+90+20+400+20+90+225+90+36+1+20+1=484+360+450+72=1366$
$P=\frac{1366}{2^{12}}=\frac{683}{2^{11}}$
ANS= $\boxed{683}$ | null | 683 |
33e99f18e99cdf0a415426283897830e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | Notice that six numbers are $0\pmod3$ , six are $1\pmod3$ , and six are $2\pmod3$ . Having numbers $0\pmod3$ will not change the remainder when $s(T)$ is divided by $3$ , so we can choose any number of these in our subset. We ignore these for now. The number of numbers that are $1\pmod3$ , minus the number of numbers that are $2\pmod3$ , must be a multiple of $3$ , possibly zero or negative. We can now split into cases based on how many numbers that are $1\pmod3$ are in the set.
CASE 1- $0$ $3$ , or $6$ integers: There can be $0$ $3$ , or $6$ integers that are $2\pmod3$ . We can choose these in \[\left(\binom60+\binom63+\binom66\right)\cdot\left(\binom60+\binom63+\binom66\right)=(1+20+1)^2=484.\]
CASE 2- $1$ or $4$ integers: There can be $2$ or $5$ integers that are $2\pmod3$ . We can choose these in \[\left(\binom61+\binom64\right)\cdot\left(\binom62+\binom65\right)=(6+15)^2=441.\]
CASE 3- $2$ or $5$ integers: There can be $1$ or $4$ integers that are $2\pmod3$ . We can choose these in \[\left(\binom62+\binom65\right)\cdot\left(\binom61+\binom64\right)=(15+6)^2=441.\]
Adding these up, we get that there are $1366$ ways to choose the numbers such that their sum is a multiple of three. Putting back in the possibility that there can be multiples of $3$ in our set, we have that there are \[1366\cdot\left(\binom60+\binom61+\binom62+\binom63+\binom64+\binom65+\binom66\right)=1366\cdot2^6\] subsets $T$ with a sum that is a multiple of $3$ . Since there are $2^{18}$ total subsets, the probability is \[\frac{1366\cdot2^6}{2^{18}}=\frac{683}{2^{11}},\] so the answer is $\boxed{683}$ | null | 683 |
33e99f18e99cdf0a415426283897830e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | We use generating functions. Each element of $U$ has two choices that occur with equal probability--either it is in the set or out of the set. Therefore, given $n\in U$ , the probability generating function is \[\frac{1}{2}+\frac{1}{2}x^n.\] Therefore, in the generating function \[\frac{1}{2^{18}}(1+x)(1+x^2)(1+x^3)\cdots (1+x^{18}),\] the coefficient of $x^k$ represents the probability of obtaining a sum of $k$ . We wish to find the sum of the coefficients of all terms of the form $x^{3k}$ (where $k \in \mathbb{Z}_{\geq0}$ ). If $\omega=e^{2\pi i/3}$ is a cube root of unity, then it is well know that for a polynomial $P(x)$ \[\frac{P(1)+P(\omega)+P(\omega^2)}{3}\] will yield the sum of the coefficients of the terms of the form $x^{3k}$ . (This is known as the third Roots of Unity Filter.) Then we find \begin{align*} \frac{1}{2^{18}}(1+1)(1+1^2)(1+1^3)\cdots (1+1^{18})&=1\\\frac{1}{2^{18}}(1+\omega)(1+\omega^2)(1+\omega^3)\cdots (1+\omega^{18})&=\frac{1}{2^{12}}\\\frac{1}{2^{18}}(1+\omega^2)(1+\omega^4)(1+\omega^6)\cdots (1+\omega^{36})&=\frac{1}{2^{12}}. \end{align*} To evaluate the last two products, we utilized the facts that $\omega^3=1$ and $1+\omega+\omega^2=0$ . Therefore, the desired probability is \[\frac{1+1/2^{12}+1/2^{12}}{3}=\frac{683}{2^{11}}.\] Thus the answer is $\boxed{683}$ | null | 683 |
33e99f18e99cdf0a415426283897830e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | Define a set that "works" to be a set for which the sum of the terms is $0$ mod $3$ . The given set mod $3$ is \[\{1,2,0,1,2,0,1,2,0,1,2,0,1,2,0,1,2,0\}.\] Let $w(N)$ be the number of subsets of the first $N$ terms of this set that "work."
Consider just the first $3$ terms: \[\{1,2,0\}.\] There are $2^3=8$ total subsets, and $w(3)=4$ (the subsets are $\emptyset, \{0\}, \{1,2\},$ and $\{1,2,0\}$ ). Now consider the first $6$ terms: \[\{1,2,0,1,2,0\}.\] To find $w(6)$ , we set aside the last $3$ terms as a "reserve" that we can pair with subsets of the first $3$ terms (which we looked at earlier).
First, all $2^3$ subsets of the first $3$ terms can either be paired with a $1$ or a $2$ (or nothing) from the "reserve" terms so that they "work," creating $2^3$ subsets that "work" already. But for each of these, we have the option to add a $0$ from the reserve, so we now have $2\cdot2^3$ subsets that "work." For each of the $w(3)=4$ subsets of the first $3$ terms that "work", we can also add on a $\{1,2\}$ or a $\{1,2,0\}$ from the reserves, creating $2w(3)$ new subsets that "work." And that covers all of them. With all of this information, we can write $w(6)$ as \[w(6)=2\cdot2^3+2w(3)=2(2^3+w(3)).\] The same process can be used to calculate $w(9)$ then $w(12)$ etc. so we can generalize it to \[w(N)=2(2^{N-3}+w(N-3)).\] Thus, we calculate $w(18)$ with this formula: \[w(18)=2( 2^{15} + 2( 2^{12} + 2( 2^9 + 2( 2^6 + 2( 2^3 + 4 ) ) ) ) )=87424.\] Because there are $2^{18}$ total possible subsets, the desired probability is \[\frac{w(18)}{2^{18}}=\frac{87424}{2^{18}}=\frac{683}{2048},\] so the answer is $\boxed{683}.$ | null | 683 |
33e99f18e99cdf0a415426283897830e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | Try smaller cases and find a pattern. Using similar casework as in Solution 1, we can easily find the desired probability if $U$ is of a small size.
If $U = \{ 1,2,0\} \pmod 3$ , then $4$ out of $8$ subsets work, for a probability of $\tfrac12$
If $U = \{ 1,2,0,1,2,0\} \pmod 3$ , then $24$ out of $64$ subsets work, for a probability of $\tfrac38$
If $U = \{ 1,2,0,1,2,0,1,2,0\} \pmod 3$ , then $176$ out of $512$ subsets work, for a probability of $\tfrac{11}{32}$
Let $a_n$ be the numerator of the desired probability if $U$ is of size $3n$ . Then $a_1 = 1, a_2 = 3,$ and $a_3 = 11$ . Noticing that the denominators are multiplied by $4$ each time, we can conclude that the pattern of the numerators is $a_n = 4a_{n-1} - 1$ , so $a_6 = \boxed{683}$ | null | 683 |
33e99f18e99cdf0a415426283897830e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | The total number of subsets is simply $2^{18}.$ Now we need to find the number of subsets that have a sum divisible by $3.$
Ignore the 6 numbers in the list that are divisible by 3. We look only at the number of subsets of $\{1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17\}$ then multiply by $2^6$ at the end. This is because adding a multiple of $3$ to the sum will not change whether it is divisible by $3$ or not and for each of the $6$ multiples, we have two options of whether to add it to the subset or not.
Now, let $f(3k)$ be the number of subsets of $\{1, 2, 4, 5, \dots, 3k-2, 3k-1\}$ that have a sum divisible by $3$ . There are $2k$ numbers in the set. Let us look at the first $2k - 2$ numbers: all subsets of the first $2k - 2$ numbers will have a residue $0, 1,$ or $2$ mod $3.$ If it is $0,$ add both $3k-2$ and $3k-1$ to the subset. If it is $1$ add just $3k-1$ to the set, and if it is $2$ add just $3k-2.$ We don't have to compute all three cases separately; since there is a 1 to 1 correspondence, we only need the sum of the three cases (which we know is $2^{2k-2}$ ).
Now, this counts all the subsets except for those that include neither $3k-1$ nor $3k-2.$ However, this is just $f(3k - 3).$ Thus, $f(3k) = 2^{2k-2} + f(3k-3)$ and the base case is $f(0) = 1.$ We have,
$f(18) = 2^{10} + f(15) = 2^{10} + 2^{8} + f(12) = \dots = 2^{10} + 2^8 + 2^6 + 2^4 + 2^2 + 2^0 + f(0).$ $=2^{10} + 2^8 + \dots + 2^2 + 2.$
Multiplying this by $2^6$ we have, \[\frac{2^7(1 + 2^1 + 2^3 + 2^5 + 2^7 + 2^9)}{2^{18}} .\] The $2^7$ cancels out with the denominator. However, the second term in the product is obviously odd and so does not cancel further with the denominator, which is just a power of $2.$ Thus, we want to find $1 + 2 + 2^3 + 2^5 + 2^7 + 2^9,$ which equals $\boxed{683}.$ | null | 683 |
33e99f18e99cdf0a415426283897830e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_12 | For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$ , let $s(T)$ be the sum of the elements of $T$ , with $s(\emptyset)$ defined to be $0$ . If $T$ is chosen at random among all subsets of $U$ , the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | The total number of subsets is $\sum_{i=0}^{18}\tbinom{18}{i}=2^{18}$ . If $s(T)\equiv 0\bmod{3}$ , the sum of the elements divides 3. We can rewrite the set as 6 0s, 6 1s, and 6 2s. We can ignore the zeros for now, since they won't influence the sum so we focus on each configuration of the 6 1s and 6 2s such that the sum is divisible by 3 and then multiply by $\sum_{i=0}^{6}\tbinom{6}{i}=2^6$ at the end. We proceed with casework on the number of values that are equivalent to $2\pmod{3}$ as follows:
Case 1: There are zero elements:
Then, there are only configurations of the values congruent to 1 mod 3. There are $\tbinom{6}{0}+\tbinom{6}{3}+\tbinom{6}{6}$ ways to assign values in the set that are congruent to 1 mod 3 such that the sum of those values divides 3. Therefore there are $22$ configurations for this case.
Case 2: There is one element:
There are $6$ ways to choose which element is included in the subset and $\tbinom{6}{1}+\tbinom{6}{4}$ ways to assign values to the numbers congruent to 1 mod 3 in the set. Therefore there are $6\cdot21=126$ configurations for this case.
Case 3: There are two elements:
There are $\tbinom{6}{2}=12$ ways to choose the 2 elements in the set that are congruent to 2 mod 3 and $\tbinom{6}{2}+\tbinom{6}{5}$ possible ways to assign values to the numbers congruent to 1 mod 3. Therefore there are $15\cdot21=315$ configurations for this case.
Case 4: There are three elements:
There are $\tbinom{6}{3}=20$ ways to choose three elements in the set that are congruent to 2 mod 3. There are $\tbinom{6}{0}+\tbinom{6}{3}+\tbinom{6}{6}$ ways to assign values to the numbers congruent to 1 mod 3. Therefore, there are $20\cdot22=440$ configurations for this case.
Case 5, Case 6, and Case 7 are symmetric to Case 3, Case 2, and Case 1 respectively so we can multiply by 2 for those cases.
Putting all the cases together we obtain $2(22+126+315) + 440=1366$ . Multiplying by $2^6$ gives $2^6\cdot1366=2^7\cdot683$ . Since $n=2^{18}$ $\frac{m}{n}=\frac{683}{2^{11}}$ . Therefore, $m=\boxed{683}$ | null | 683 |
66e18f4de28e0e9635ab2ecc93d5d3c4 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_13 | Let $\triangle ABC$ have side lengths $AB=30$ $BC=32$ , and $AC=34$ . Point $X$ lies in the interior of $\overline{BC}$ , and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$ , respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$ | First note that \[\angle I_1AI_2 = \angle I_1AX + \angle XAI_2 = \frac{\angle BAX}2 + \frac{\angle CAX}2 = \frac{\angle A}2\] is a constant not depending on $X$ , so by $[AI_1I_2] = \tfrac12(AI_1)(AI_2)\sin\angle I_1AI_2$ it suffices to minimize $(AI_1)(AI_2)$ . Let $a = BC$ $b = AC$ $c = AB$ , and $\alpha = \angle AXB$ . Remark that \[\angle AI_1B = 180^\circ - (\angle I_1AB + \angle I_1BA) = 180^\circ - \tfrac12(180^\circ - \alpha) = 90^\circ + \tfrac\alpha 2.\] Applying the Law of Sines to $\triangle ABI_1$ gives \[\frac{AI_1}{AB} = \frac{\sin\angle ABI_1}{\sin\angle AI_1B}\qquad\Rightarrow\qquad AI_1 = \frac{c\sin\frac B2}{\cos\frac\alpha 2}.\] Analogously one can derive $AI_2 = \tfrac{b\sin\frac C2}{\sin\frac\alpha 2}$ , and so \[[AI_1I_2] = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{2\cos\frac\alpha 2\sin\frac\alpha 2} = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{\sin\alpha}\geq bc\sin\frac A2 \sin\frac B2\sin\frac C2,\] with equality when $\alpha = 90^\circ$ , that is, when $X$ is the foot of the perpendicular from $A$ to $\overline{BC}$ . In this case the desired area is $bc\sin\tfrac A2\sin\tfrac B2\sin\tfrac C2$ . To make this feasible to compute, note that \[\sin\frac A2=\sqrt{\frac{1-\cos A}2}=\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}2} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}}.\] Applying similar logic to $\sin \tfrac B2$ and $\sin\tfrac C2$ and simplifying yields a final answer of \begin{align*}bc\sin\frac A2\sin\frac B2\sin\frac C2&=bc\cdot\dfrac{(a-b+c)(b-c+a)(c-a+b)}{8abc}\\&=\dfrac{(30-32+34)(32-34+30)(34-30+32)}{8\cdot 32}=\boxed{126} | null | 126 |
66e18f4de28e0e9635ab2ecc93d5d3c4 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_13 | Let $\triangle ABC$ have side lengths $AB=30$ $BC=32$ , and $AC=34$ . Point $X$ lies in the interior of $\overline{BC}$ , and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$ , respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$ | It's clear that $\angle I_{1}AI_{2}=\frac{1}{2}\angle BAX+\frac{1}{2}\angle CAX=\frac{1}{2}\angle A$ . Thus \begin{align*} AI_{1}I_{2}&=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\angle I_{1}AI_{2} \\ &=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\left(\frac{1}{2}\angle A\right).\end{align*} By the Law of Sines on $\triangle AI_{1}B$ \[\frac{AI_{1}}{\sin\left(\frac{1}{2}\angle B\right)}=\frac{AB}{\sin\angle AI_{1}B}.\] Similarly, \[\frac{AI_{2}}{\sin\left(\frac{1}{2}\angle C\right)}=\frac{AC}{\sin\angle AI_{2}C}.\] It is well known that \[\angle AI_{1}B=90+\frac{1}{2}\angle AXB~~~\text{and}~~~\angle AI_{2}C=90+\frac{1}{2}\angle AXC.\] Denote $\alpha=\frac{1}{2}\angle AXB$ and $\theta=\frac{1}{2}\angle AXC$ , with $\alpha+\theta=90^{\circ}$ . Thus $\sin\alpha=\cos\theta$ and \begin{align*}\frac{AI_{1}}{\sin\left(\frac{1}{2}\angle B\right)}=\frac{AB}{\sin\angle AI_{1}B}\Longrightarrow\frac{AB}{\sin\left(90^{\circ}+\alpha\right)}\Longrightarrow\frac{AB}{\cos\alpha} \\\frac{AI_{2}}{\sin\left(\frac{1}{2}\angle C\right)}=\frac{AC}{\sin\angle AI_{2}C}\Longrightarrow\frac{AC}{\sin\left(90^{\circ}+\theta\right)}\Longrightarrow\frac{AC}{\cos\theta}\Longrightarrow\frac{AC}{\sin\alpha}.\end{align*} Thus \[AI_{1}=\frac{AB\sin\left(\frac{1}{2}\angle B\right)}{\cos\alpha}~~~\text{and}~~~AI_{2}=\frac{AC\sin\left(\frac{1}{2}\angle C\right)}{\sin\alpha}\] so \begin{align*}[AI_{1}I_{2}]&=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\left(\frac{1}{2}\angle A\right) \\ &=\frac{AB\sin\left(\frac{1}{2}\angle B\right)\cdot AC\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right)}{2\sin\alpha\cos\alpha} \\ &=\frac{AB\sin\left(\frac{1}{2}\angle B\right)\cdot AC\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right)}{\sin(2\alpha)}.\end{align*} We intend to minimize this expression, which is equivalent to maximizing $\sin(2\alpha)$ , and that occurs when $\alpha=45^{\circ}$ , or $\angle AXB=90^{\circ}$ . Ergo, $X$ is the foot of the altitude from $A$ to $\overline{BC}$ . In that case, we intend to compute \[AB\cdot AC\cdot\sin\left(\frac{1}{2}\angle B\right)\cdot\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right).\] Recall that \[\sin\left(\frac{1}{2}\angle B\right)=\sqrt{\frac{1-\cos\angle B}{2}}\] and similarly for angles $C$ and $A$ . Applying the Law of Cosines to each angle of $\triangle ABC$ gives \begin{align*}\angle B&:\cos\angle B=\frac{30^{2}+32^{2}-34^{2}}{2\cdot 30\cdot 32}=\frac{2}{5} \\ \angle C&:\cos\angle C=\frac{32^{2}+34^{2}-30^{2}}{2\cdot 32\cdot 34}=\frac{10}{17} \\ \angle A&:\cos\angle A=\frac{30^{2}+34^{2}-32^{2}}{2\cdot 30\cdot 34}=\frac{43}{85}.\end{align*} Thus \begin{align*}\sin\left(\frac{1}{2}\angle B\right)=\sqrt{\frac{1-\tfrac{2}{5}}{2}}=\sqrt{\frac{3}{10}} \\ \sin\left(\frac{1}{2}\angle C\right)=\sqrt{\frac{1-\tfrac{10}{17}}{2}}=\sqrt{\frac{7}{34}} \\ \sin\left(\frac{1}{2}\angle A\right)=\sqrt{\frac{1-\tfrac{43}{85}}{2}}=\sqrt{\frac{21}{85}}.\end{align*} Thus the answer is \begin{align*} & 30\cdot 34\cdot\sqrt{\frac{3}{10}\cdot\frac{7}{34}\cdot\frac{21}{85}} \\ =&~30\cdot 34\cdot\sqrt{\frac{3}{2\cdot 5}\cdot\frac{7}{2\cdot 17}\cdot\frac{3\cdot 7}{5\cdot 17}} \\ =&~30\cdot 34\cdot\frac{3\cdot 7}{2\cdot 5\cdot 17} \\ =&~(2\cdot 3\cdot 5)\cdot(2\cdot 17)\cdot\frac{3\cdot 7}{2\cdot 5\cdot 17} \\ =&~2\cdot 3^{2}\cdot 7 \\ =&~\boxed{126} | null | 126 |
66e18f4de28e0e9635ab2ecc93d5d3c4 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_13 | Let $\triangle ABC$ have side lengths $AB=30$ $BC=32$ , and $AC=34$ . Point $X$ lies in the interior of $\overline{BC}$ , and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$ , respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$ | First, instead of using angles to find $[AI_1I_2]$ , let's try to find the area of other, simpler figures, and subtract that from $[ABC]$ . However, to do this, we need to be able to figure out the length of the inradii, and so, we need to find $AX$
To minimize $[AI_1I_2]$ , intuitively, we should try to minimize the length of $AX$ , since, after using the $rs=A$ formula for the area of a triangle, we'll be able to minimize the inradii lengths, and thus, eventually minimize the area of $[AI_1I_2]$ . (Proof needed here).
We need to minimize $AX$ . Let $AX=d$ $BX=s$ , and $CX=32-s$ . After an application of Stewart's Theorem, we will get that \[d=\sqrt{s^2-24s+900}\] To minimize this quadratic, $s=12$ whereby we conclude that $d=6\sqrt{21}$
From here, draw perpendiculars down from $I_1$ and $I_2$ to $AB$ and $AC$ respectively, and label the foot of these perpendiculars $D$ and $E$ respectively. After, draw the inradii from $I_1$ to $BX$ , and from $I_2$ to $CX$ , and draw in $I_1I_2$
Label the foot of the inradii to $BX$ and $CX$ $F$ and $G$ , respectively. From here, we see that to find $[AI_1I_2]$ , we need to find $[ABC]$ , and subtract off the sum of $[DBCEI_2I_1], [ADI_1],$ and $[AEI_2]$
$[DBCEI_2I_1]$ can be found by finding the area of two quadrilaterals $[DBFI_1]+[ECGI_2]$ as well as the area of a trapezoid $[FGI_2I_1]$ . If we let the inradius of $ABX$ be $r_1$ and if we let the inradius of $ACX$ be $r_2$ , we'll find, after an application of basic geometry and careful calculations on paper, that $[DBCEI_2I_1]=13r_1+19r_2$
The area of two triangles can be found in a similar fashion, however, we must use $XYZ$ substitution to solve for $AD$ as well as $AE$ . After doing this, we'll get a similar sum in terms of $r_1$ and $r_2$ for the area of those two triangles which is equal to \[\frac{(9+3\sqrt{21})(r_1)}{2} + \frac{(7+3\sqrt{21})(r_2)}{2}.\]
Now we're set. Summing up the area of the Hexagon and the two triangles and simplifying, we get that the formula for $[AI_1I_2]$ is just \[[ABC]-\left(\frac{(35+3\sqrt{21})(r_1)}{2}+\frac{(45+3r_2\sqrt{21})(r_2)}{2}\right).\]
Using Heron's formula, $[ABC]=96\sqrt{21}$ . Solving for $r_1$ and $r_2$ using Heron's in $ABX$ and $ACX$ , we get that $r_1=3\sqrt{21}-9$ and $r_2=3\sqrt{21}-7$ . From here, we just have to plug into our above equation and solve.
Doing so gets us that the minimum area of $AI_1I_2=\boxed{126}.$ | null | 126 |
0e66f08bf6c3efc73d0f4020b081b636 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_14 | Let $SP_1P_2P_3EP_4P_5$ be a heptagon. A frog starts jumping at vertex $S$ . From any vertex of the heptagon except $E$ , the frog may jump to either of the two adjacent vertices. When it reaches vertex $E$ , the frog stops and stays there. Find the number of distinct sequences of jumps of no more than $12$ jumps that end at $E$ | This is easily solved by recursion/dynamic programming. To simplify the problem somewhat, let us imagine the seven vertices on a line $E \leftrightarrow P_4 \leftrightarrow P_5 \leftrightarrow S \leftrightarrow P_1 \leftrightarrow P_2 \leftrightarrow P_3 \leftrightarrow E$ . We can count the number of left/right (L/R) paths of length $\le 11$ that start at $S$ and end at either $P_4$ or $P_3$
We can visualize the paths using the common grid counting method by starting at the origin $(0,0)$ , so that a right (R) move corresponds to moving 1 in the positive $x$ direction, and a left (L) move corresponds to moving 1 in the positive $y$ direction. Because we don't want to move more than 2 units left or more than 3 units right, our path must not cross the lines $y = x+2$ or $y = x-3$ . Letting $p(x,y)$ be the number of such paths from $(0,0)$ to $(x,y)$ under these constraints, we have the following base cases:
$p(x,0) = \begin{cases} 1 & x \le 3 \\ 0 & x > 3 \end{cases} \qquad p(0,y) = \begin{cases} 1 & y \le 2 \\ 0 & y > 2 \end{cases}$
and recursive step $p(x,y) = p(x-1,y) + p(x,y-1)$ for $x,y \ge 1$
The filled in grid will look something like this, where the lower-left $1$ corresponds to the origin:
$\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline 0 & 0 & 0 & 0 & \textbf{89} & & & \\ \hline 0 & 0 & 0 & \textbf{28} & 89 & & & \\ \hline 0 & 0 & \textbf{9} & 28 & 61 & 108 & 155 & \textbf{155} \\ \hline 0 & \textbf{3} & 9 & 19 & 33 & 47 & \textbf{47} & 0 \\ \hline \textbf{1} & 3 & 6 & 10 & 14 & \textbf{14} & 0 & 0 \\ \hline 1 & 2 & 3 & 4 & \textbf{4} & 0 & 0 & 0 \\ \hline 1 & 1 & 1 & \textbf{1} & 0 & 0 & 0 & 0 \\ \hline \end{tabular}$
The bolded numbers on the top diagonal represent the number of paths from $S$ to $P_4$ in 2, 4, 6, 8, 10 moves, and the numbers on the bottom diagonal represent the number of paths from $S$ to $P_3$ in 3, 5, 7, 9, 11 moves. We don't care about the blank entries or entries above the line $x+y = 11$ . The total number of ways is $1+3+9+28+89+1+4+14+47+155 = \boxed{351}$ | null | 351 |
0e66f08bf6c3efc73d0f4020b081b636 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_14 | Let $SP_1P_2P_3EP_4P_5$ be a heptagon. A frog starts jumping at vertex $S$ . From any vertex of the heptagon except $E$ , the frog may jump to either of the two adjacent vertices. When it reaches vertex $E$ , the frog stops and stays there. Find the number of distinct sequences of jumps of no more than $12$ jumps that end at $E$ | Let $E_n$ denotes the number of sequences with length $n$ that ends at $E$ . Define similarly for the other vertices. We seek for a recursive formula for $E_n$ \begin{align*} E_n&=P_{3_{n-1}}+P_{4_{n-1}} \\ &=P_{2_{n-2}}+P_{5_{n-2}} \\ &=P_{1_{n-3}}+P_{3_{n-3}}+S_{n-3}+P_{4_{n-3}} \\ &=(P_{3_{n-3}}+P_{4_{n-3}})+S_{n-3}+P_{1_{n-3}} \\ &=E_{n-2}+S_{n-3}+P_{1_{n-3}} \\ &=E_{n-2}+P_{1_{n-4}}+P_{5_{n-4}}+S_{n-4}+P_{2_{n-4}} \\ &=E_{n-2}+(S_{n-4}+P_{1_{n-4}})+P_{5_{n-4}}+P_{2_{n-4}} \\ &=E_{n-2}+(E_{n-1}-E_{n-3})+S_{n-5}+P_{4_{n-5}}+P_{1_{n-5}}+P_{3_{n-5}} \\ &=E_{n-2}+(E_{n-1}-E_{n-3})+(S_{n-5}+P_{1_{n-5}})+(P_{4_{n-5}}+P_{3_{n-5}}) \\ &=E_{n-2}+(E_{n-1}-E_{n-3})+(E_{n-2}-E_{n-4})+E_{n-4} \\ &=E_{n-1}+2E_{n-2}-E_{n-3} \\ \end{align*} Computing a few terms we have $E_0=0$ $E_1=0$ $E_2=0$ $E_3=1$ , and $E_4=1$
Using the formula yields $E_5=3$ $E_6=4$ $E_7=9$ $E_8=14$ $E_9=28$ $E_{10}=47$ $E_{11}=89$ , and $E_{12}=155$
Finally adding yields $\sum_{k=0}^{12}E_k=\boxed{351}$ | null | 351 |
84110ddc425b669da7f9d67eaa1df52a | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_15 | David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$ , which can each be inscribed in a circle with radius $1$ . Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$ , and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\tfrac{2}{3}$ $\sin\varphi_B=\tfrac{3}{5}$ , and $\sin\varphi_C=\tfrac{6}{7}$ . All three quadrilaterals have the same area $K$ , which can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Suppose our four sides lengths cut out arc lengths of $2a$ $2b$ $2c$ , and $2d$ , where $a+b+c+d=180^\circ$ . Then, we only have to consider which arc is opposite $2a$ . These are our three cases, so \[\varphi_A=a+c\] \[\varphi_B=a+b\] \[\varphi_C=a+d\] Our first case involves quadrilateral $ABCD$ with $\overarc{AB}=2a$ $\overarc{BC}=2b$ $\overarc{CD}=2c$ , and $\overarc{DA}=2d$
Then, by Law of Sines, $AC=2\sin\left(\frac{\overarc{ABC}}{2}\right)=2\sin(a+b)$ and $BD=2\sin\left(\frac{\overarc{BCD}}{2}\right)=2\sin(a+d)$ . Therefore,
\[K=\frac{1}{2}\cdot AC\cdot BD\cdot \sin(\varphi_A)=2\sin\varphi_A\sin\varphi_B\sin\varphi_C=\frac{24}{35},\] so our answer is $24+35=\boxed{059}$ | null | 059 |
84110ddc425b669da7f9d67eaa1df52a | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_15 | David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$ , which can each be inscribed in a circle with radius $1$ . Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$ , and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\tfrac{2}{3}$ $\sin\varphi_B=\tfrac{3}{5}$ , and $\sin\varphi_C=\tfrac{6}{7}$ . All three quadrilaterals have the same area $K$ , which can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Suppose the four side lengths of the quadrilateral cut out arc lengths of $2a$ $2b$ $2c$ , and $2d$ $a+b+c+d=180^\circ$ .
Therefore, without losing generality,
\[\varphi_A=a+b\] \[\varphi_B=b+c\] \[\varphi_C=a+c\]
$(1)+(3)-(2)$ $(1)+(2)-(3)$ , and $(2)+(3)-(1)$ yields
\[2a=\varphi_A+\varphi_C-\varphi_B\] \[2b=\varphi_A+\varphi_B-\varphi_C\] \[2c=\varphi_B+\varphi_C-\varphi_A\]
Because $2d=360^\circ-2a-2b-2c,$ Therefore,
\[2d=360^\circ-\varphi_A-\varphi_B-\varphi_C\]
Using the sum-to-product identities , our area of the quadrilateral $K$ then would be
\begin{align*} K&=\frac{1}{2}(\sin(2a)+\sin(2b)+\sin(2c)+\sin(2d))\\ &=\frac{1}{2}(\sin(\varphi_A+\varphi_B-\varphi_C)+\sin(\varphi_B+\varphi_C-\varphi_A)+\sin(\varphi_C+\varphi_A-\varphi_B)-\sin(\varphi_A+\varphi_B+\varphi_C))\\ &=\frac{1}{2}(2\sin\varphi_B\cos(\varphi_A-\varphi_C)-2\sin\varphi_B\cos(\varphi_A+\varphi_C))\\ &=\frac{1}{2}\cdot2\cdot2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\ &=2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\ &=\frac{24}{35}\\ \end{align*}
Therefore, our answer is $24+35=\boxed{059}$ | null | 059 |
84110ddc425b669da7f9d67eaa1df52a | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_15 | David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$ , which can each be inscribed in a circle with radius $1$ . Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$ , and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\tfrac{2}{3}$ $\sin\varphi_B=\tfrac{3}{5}$ , and $\sin\varphi_C=\tfrac{6}{7}$ . All three quadrilaterals have the same area $K$ , which can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let the four stick lengths be $a$ $b$ $c$ , and $d$ . WLOG, let’s say that quadrilateral $A$ has sides $a$ and $d$ opposite each other, quadrilateral $B$ has sides $b$ and $d$ opposite each other, and quadrilateral $C$ has sides $c$ and $d$ opposite each other. The area of a convex quadrilateral can be written as $\frac{1}{2} d_1 d_2 \sin{\theta}$ , where $d_1$ and $d_2$ are the lengths of the diagonals of the quadrilateral and $\theta$ is the angle formed by the intersection of $d_1$ and $d_2$ . By Ptolemy's theorem $d_1 d_2 = ad+bc$ for quadrilateral $A$ , so, defining $K_A$ as the area of $A$ \[K_A = \frac{1}{2} (ad+bc)\sin{\varphi_A}\] Similarly, for quadrilaterals $B$ and $C$ \[K_B = \frac{1}{2} (bd+ac)\sin{\varphi_B}\] and \[K_C = \frac{1}{2} (cd+ab)\sin{\varphi_C}\] Multiplying the three equations and rearranging, we see that \[K_A K_B K_C = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\sin{\varphi_A}\sin{\varphi_B}\sin{\varphi_B}\] \[K^3 = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\left(\frac{2}{3}\right) \left(\frac{3}{5}\right) \left(\frac{6}{7}\right)\] \[\frac{70}{3}K^3 = (ab+cd)(ac+bd)(ad+bc)\] The circumradius $R$ of a cyclic quadrilateral with side lengths $a$ $b$ $c$ , and $d$ and area $K$ can be computed as $R = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4K}$ .
Inserting what we know, \[1 = \frac{\sqrt{\frac{70}{3}K^3}}{4K}\quad \Rightarrow \quad 16K^2 = \frac{70}{3}K^3\quad \Rightarrow \quad \frac{24}{35} = K\] So our answer is $24 + 35 = \boxed{059}$ | null | 059 |
84110ddc425b669da7f9d67eaa1df52a | https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_15 | David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$ , which can each be inscribed in a circle with radius $1$ . Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$ , and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\tfrac{2}{3}$ $\sin\varphi_B=\tfrac{3}{5}$ , and $\sin\varphi_C=\tfrac{6}{7}$ . All three quadrilaterals have the same area $K$ , which can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let the sides of the quadrilaterals be $a,b,c,$ and $d$ in some order such that $A$ has $a$ opposite of $c$ $B$ has $a$ opposite of $b$ , and $C$ has $a$ opposite of $d$ . Then, let the diagonals of $A$ be $e$ and $f$ . Similarly to solution $2$ , we get that $\tfrac{2}{3}(ac+bd)=\tfrac{3}{5}(ab+cd)=\tfrac{6}{7}(ad+bc)=2K$ , but this is also equal to $2\cdot\tfrac{eab+ecd}{4(1)}=2\cdot\tfrac{fad+fbc}{4(1)}$ using the area formula for a triangle using the circumradius and the sides, so $\tfrac{e(ab+cd)}{2}=\tfrac{3}{5}(ab+cd)$ and $\tfrac{f(ad+bc)}{2}=\tfrac{6}{7}(ad+bc)$ . Solving for $e$ and $f$ , we get that $e=\tfrac{6}{5}$ and $f=\tfrac{12}{7}$ , but $K=\tfrac{1}{2}\cdot\tfrac{2}{3}\cdot{}ef$ , similarly to solution $2$ , so $K=\tfrac{24}{35}$ and the answer is $24+35=\boxed{059}$ | null | 059 |
f25de092f1d8a467236e582cac44853a | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_1 | Points $A$ $B$ , and $C$ lie in that order along a straight path where the distance from $A$ to $C$ is $1800$ meters. Ina runs twice as fast as Eve, and Paul runs twice as fast as Ina. The three runners start running at the same time with Ina starting at $A$ and running toward $C$ , Paul starting at $B$ and running toward $C$ , and Eve starting at $C$ and running toward $A$ . When Paul meets Eve, he turns around and runs toward $A$ . Paul and Ina both arrive at $B$ at the same time. Find the number of meters from $A$ to $B$ | We know that in the same amount of time given, Ina will run twice the distance of Eve, and Paul would run quadruple the distance of Eve. Let's consider the time it takes for Paul to meet Eve: Paul would've run 4 times the distance of Eve, which we can denote as $d$ . Thus, the distance between $B$ and $C$ is $4d+d=5d$ . In that given time, Ina would've run twice the distance $d$ , or $2d$ units.
Now, when Paul starts running back towards $A$ , the same amount of time would pass since he will meet Ina at his starting point. Thus, we know that he travels another $4d$ units and Ina travels another $2d$ units.
Therefore, drawing out the diagram, we find that $2d+2d+4d+d=9d=1800 \implies d=200$ , and distance between $A$ and $B$ is the distance Ina traveled, or $4d=4(200)=\boxed{800}$ | null | 800 |
f25de092f1d8a467236e582cac44853a | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_1 | Points $A$ $B$ , and $C$ lie in that order along a straight path where the distance from $A$ to $C$ is $1800$ meters. Ina runs twice as fast as Eve, and Paul runs twice as fast as Ina. The three runners start running at the same time with Ina starting at $A$ and running toward $C$ , Paul starting at $B$ and running toward $C$ , and Eve starting at $C$ and running toward $A$ . When Paul meets Eve, he turns around and runs toward $A$ . Paul and Ina both arrive at $B$ at the same time. Find the number of meters from $A$ to $B$ | Let $x$ be the distance from $A$ to $B$ . Then the distance from $B$ to $C$ is $1800-x$ . Since Eve is the slowest, we can call her speed $v$ , so that Ina's speed is $2v$ and Paul's speed is $4v$
For Paul and Eve to meet, they must cover a total distance of $1800-x$ which takes them a time of $\frac{1800-x}{4v+v}$ . Paul must run the same distance back to $B$ , so his total time is $\frac{2(1800-x)}{5v}$
For Ina to reach $B$ , she must run a distance of $x$ at a speed of $2v$ , taking her a time of $\frac{x}{2v}$
Since Paul and Ina reach $B$ at the same time, we know that $\frac{2(1800-x)}{5v} = \frac{x}{2v}$ (notice that $v$ cancels out on both sides). Solving the equation gives $x = \boxed{800}$ | null | 800 |
23de272b9020150bb0726405b3e5d488 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_2 | Let $a_{0} = 2$ $a_{1} = 5$ , and $a_{2} = 8$ , and for $n > 2$ define $a_{n}$ recursively to be the remainder when $4$ $a_{n-1}$ $+$ $a_{n-2}$ $+$ $a_{n-3}$ ) is divided by $11$ . Find $a_{2018} \cdot a_{2020} \cdot a_{2022}$ | When given a sequence problem, one good thing to do is to check if the sequence repeats itself or if there is a pattern.
After computing more values of the sequence, it can be observed that the sequence repeats itself every 10 terms starting at $a_{0}$
$a_{0} = 2$ $a_{1} = 5$ $a_{2} = 8$ $a_{3} = 5$ $a_{4} = 6$ $a_{5} = 10$ $a_{6} = 7$ $a_{7} = 4$ $a_{8} = 7$ $a_{9} = 6$ $a_{10} = 2$ $a_{11} = 5$ $a_{12} = 8$ $a_{13} = 5$
We can simplify the expression we need to solve to $a_{8}\cdot a_{10} \cdot a_{2}$
Our answer is $7 \cdot 2 \cdot 8$ $= \boxed{112}$ | null | 112 |
23de272b9020150bb0726405b3e5d488 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_2 | Let $a_{0} = 2$ $a_{1} = 5$ , and $a_{2} = 8$ , and for $n > 2$ define $a_{n}$ recursively to be the remainder when $4$ $a_{n-1}$ $+$ $a_{n-2}$ $+$ $a_{n-3}$ ) is divided by $11$ . Find $a_{2018} \cdot a_{2020} \cdot a_{2022}$ | Notice that the characteristic polynomial of this is $x^3-4x^2-4x-4\equiv 0\pmod{11}$
Then since $x\equiv1$ is a root, using Vieta's formula, the other two roots $r,s$ satisfy $r+s\equiv3$ and $rs\equiv4$
Let $r=7+d$ and $s=7-d$
We have $49-d^2\equiv4$ so $d\equiv1$ . We found that the three roots of the characteristic polynomial are $1,6,8$
Now we want to express $a_n$ in an explicit form as $a(1^n)+b(6^n)+c(8^n)\pmod{11}$
Plugging in $n=0,1,2$ we get
$(*)$ $a+b+c\equiv2,$
$(**)$ $a+6b+8c\equiv5,$
$(***)$ $a+3b+9c\equiv8$
$\frac{(***)-(*)}{2}$ $\implies b+4c\equiv3$ and $(***)-(**)$ $\implies -3b+c\equiv3$
so $a\equiv6,$ $b\equiv1,$ and $c\equiv6$
Hence, $a_n\equiv 6+(6^n)+6(8^n)\equiv(2)^{-n\pmod{10}}+(2)^{3n-1\pmod{10}}-5\pmod{11}$
Therefore
$a_{2018}\equiv4+8-5=7$
$a_{2020}\equiv1+6-5=2$
$a_{2022}\equiv3+10-5=8$
And the answer is $7\times2\times8=\boxed{112}$ | null | 112 |
ea61f96c7ab2746a629215cc84ba453e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_3 | Find the sum of all positive integers $b < 1000$ such that the base- $b$ integer $36_{b}$ is a perfect square and the base- $b$ integer $27_{b}$ is a perfect cube. | The first step is to convert $36_{b}$ and $27_{b}$ into base-10 numbers. Then, we can write \[36_{b} = 3b + 6\] and \[27_{b} = 2b + 7\] . It should also be noted that $8 \leq b < 1000$
Because there are less perfect cubes than perfect squares for the restriction we are given on $b$ , it is best to list out all the perfect cubes. Since the maximum $b$ can be is 1000 and $2$ $1000 + 7 = 2007$ , we can list all the perfect cubes less than 2007.
Now, $2b + 7$ must be one of \[3^3, 4^3, ... , 12^3\] . However, $2b + 7$ will always be odd, so we can eliminate the cubes of the even numbers and change our list of potential cubes to \[3^3, 5^3, 7^3, 9^3\text{, and }11^3\]
Because $3b + 6$ is a perfect square and is clearly divisible by 3, it must be divisible by 9, so $b + 2$ is divisible by 3. Thus the cube, which is \[2b + 7 = 2(b + 2) + 3\] , must also be divisible by 3. Therefore, the only cubes that $2b + 7$ could potentially be now are $3^3$ and $9^3$
We need to test both of these cubes to make sure $3b + 6$ is a perfect square.
$\textbf{Case 1:}$ If we set \[3^3 = 2b + 7\] so \[b = 10\] . If we plug this value of b into $3b + 6$ , the expression equals $36$ , which is indeed a perfect square.
$\textbf{Case 2:}$ If we set \[9^3 = 2b + 7\] so \[b = 361\] . If we plug this value of b into $3b + 6$ , the expression equals $1089$ , which is $33^2$
We have proven that both $b = 10$ and $b = 361$ are the only solutions, so \[10 + 361 = \boxed{371}\] | null | 371 |
ea61f96c7ab2746a629215cc84ba453e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_3 | Find the sum of all positive integers $b < 1000$ such that the base- $b$ integer $36_{b}$ is a perfect square and the base- $b$ integer $27_{b}$ is a perfect cube. | The conditions are: \[3b+6 = n^2\] \[2b+7 = m^3\] We can see $n$ is multiple is 3, so let $n=3k$ , then $b= 3k^2-2$ . Substitute $b$ into second condition and we get $m^3=3(2k^2+1)$ . Now we know $m$ is both a multiple of 3 and odd. Also, $m$ must be smaller than 13 for $b$ to be smaller than 1000. So the only two possible values for $m$ are 3 and 9. Test and they both work. The final answer is $10 + 361 =$ $\boxed{371}$ .
-Mathdummy | null | 371 |
ea61f96c7ab2746a629215cc84ba453e | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_3 | Find the sum of all positive integers $b < 1000$ such that the base- $b$ integer $36_{b}$ is a perfect square and the base- $b$ integer $27_{b}$ is a perfect cube. | As shown above, let \[3b+6 = n^2\] \[2b+7 = m^3\] such that \[6b+12=2n^2\] \[6b+21=3m^3\]
Subtracting the equations we have \[3m^3-2n^2=9 \implies 3m^3-2n^2-9=0.\]
We know that $m$ and $n$ both have to be integers, because then the base wouldn't be an integer. Furthermore, any integer solution $m$ must divide $9$ by the Rational Root Theorem.
We can instantly know $m \neq -9,-3,-1,1$ since those will have negative solutions.
When $m=3$ we have $n=6$ , so then $b=10$
When $m=9$ we have $n=33$ , so then $b=361$
Therefore, the sum of all possible values of $b$ is \[10+361=\boxed{371}.\] | null | 371 |
9e716310bae0c2d788a5b2d9eff1270d | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_4 | In equiangular octagon $CAROLINE$ $CA = RO = LI = NE =$ $\sqrt{2}$ and $AR = OL = IN = EC = 1$ . The self-intersecting octagon $CORNELIA$ encloses six non-overlapping triangular regions. Let $K$ be the area enclosed by $CORNELIA$ , that is, the total area of the six triangular regions. Then $K =$ $\dfrac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Find $a + b$ | We can draw $CORNELIA$ and introduce some points.
2018 AIME II Problem 4.png
The diagram is essentially a 3x3 grid where each of the 9 squares making up the grid have a side length of 1.
In order to find the area of $CORNELIA$ , we need to find 4 times the area of $\bigtriangleup$ $ACY$ and 2 times the area of $\bigtriangleup$ $YZW$
Using similar triangles $\bigtriangleup$ $ARW$ and $\bigtriangleup$ $YZW$ (We look at their heights), $YZ$ $=$ $\frac{1}{3}$ . Therefore, the area of $\bigtriangleup$ $YZW$ is $\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{1}{2}$ $=$ $\frac{1}{12}$
Since $YZ$ $=$ $\frac{1}{3}$ and $XY = ZQ$ $XY$ $=$ $\frac{1}{3}$ and $CY$ $=$ $\frac{4}{3}$
Therefore, the area of $\bigtriangleup$ $ACY$ is $\frac{4}{3}\cdot$ $1$ $\cdot$ $\frac{1}{2}$ $=$ $\frac{2}{3}$
Our final answer is $\frac{1}{12}$ $\cdot$ $2$ $+$ $\frac{2}{3}$ $\cdot$ $4$ $=$ $\frac{17}{6}$
$17 + 6 =$ $\boxed{023}$ | null | 023 |
29a3931a6c989b9cfee0a5d4c70619bd | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_5 | Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$ | The First (pun intended) thing to notice is that $xy$ and $zx$ have a similar structure, but not exactly conjugates, but instead once you take out the magnitudes of both, simply multiples of a root of unity. It turns out that root of unity is $e^{\frac{3\pi i}{2}}$ . Anyway this results in getting that $\left(\frac{-3i}{10}\right)y=z$ . Then substitute this into $yz$ to get, after some calculation, that $y=10e^{\frac{5\pi i}{4}}\sqrt{2}$ and $z=-3e^{\frac{7\pi i}{4}}\sqrt{2}$ . Then plug $z$ into $zx$ , you could do the same thing with $xy$ but $zx$ looks like it's easier due to it being smaller. Anyway you get $x=20+12i$ . Then add all three up, it turns out easier than it seems because for $z$ and $y$ the $\sqrt{2}$ disappears after you expand the root of unity (e raised to a specific power). Long story short, you get $x=20+12i, y=-3+3i, z=-10-10i \implies x+y+z=7+5i \implies a^2+b^2=\boxed{074}$ | null | 074 |
29a3931a6c989b9cfee0a5d4c70619bd | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_5 | Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$ | First we evaluate the magnitudes. $|xy|=80\sqrt{17}$ $|yz|=60$ , and $|zx|=24\sqrt{17}$ . Therefore, $|x^2y^2z^2|=17\cdot80\cdot60\cdot24$ , or $|xyz|=240\sqrt{34}$ . Divide to find that $|z|=3\sqrt{2}$ $|x|=4\sqrt{34}$ , and $|y|=10\sqrt{2}$ [asy] draw((0,0)--(4,0)); dot((4,0),red); draw((0,0)--(-4,0)); draw((0,0)--(0,-4)); draw((0,0)--(-4,1)); dot((-4,1),red); draw((0,0)--(-1,-4)); dot((-1,-4),red); draw((0,0)--(4,4),red); draw((0,0)--(4,-4),red); [/asy] This allows us to see that the argument of $y$ is $\frac{\pi}{4}$ , and the argument of $z$ is $-\frac{\pi}{4}$ . We need to convert the polar form to a standard form. Simple trig identities show $y=10+10i$ and $z=3-3i$ . More division is needed to find what $x$ is. \[x=-20-12i\] \[x+y+z=-7-5i\] \[(-7)^2+(-5)^2=\boxed{74}\] \[QED\blacksquare\] Written by a1b2 | null | 74 |
29a3931a6c989b9cfee0a5d4c70619bd | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_5 | Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$ | Solve this system the way you would if the RHS of all equations were real. Multiply the first and 3rd equations out and then factor out $60$ to find $x^2$ , then use standard techniques that are used to evaluate square roots of irrationals. let \[x = c+di\] , then you get \[c^2 - d^2 = 256\] and \[2cd = 480\] Solve to get $x$ as $20+12i$ and $-20-12i$ . Both will give us the same answer, so use the positive one. Divide $-80-320i$ by $x$ , and you get $10+10i$ as $y$ . This means that $z$ is a multiple of $1-i$ to get a real product, so you find $z$ is $3-3i$ . Now, add the real and imaginary parts separately to get $-7-5i$ , and calculate $a^2 + b^2$ to get $\boxed{74}$ .
~minor latex improvements done by jske25 and jdong2006 | null | 74 |
29a3931a6c989b9cfee0a5d4c70619bd | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_5 | Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$ | Dividing the first equation by the second equation given, we find that $\frac{xy}{yz}=\frac{x}{z}=\frac{-80-320i}{60}=-\frac{4}{3}-\frac{16}{3}i \implies x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)$ . Substituting this into the third equation, we get $z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-1-4i}=3\cdot \frac{(-24+6i)(-1+4i)}{1+16}=3\cdot \frac{-102i}{17}=-18i$ . Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of $y$ is the negative of that of $z$ , and their magnitudes multiply to $60$ . Thus, we have $z=\sqrt{-18i}=3-3i$ and $3\sqrt{2}\cdot |y|=60 \implies |y|=10\sqrt{2} \implies y=10+10i$ . To find $x$ , we can use the previous substitution we made to find that $x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)=-\frac{4}{3}\cdot (3-3i)(1+4i)=-4(1-i)(1+4i)=-4(5+3i)=-20-12i$ .
Therefore, $x+y+z=(-20+10+3)+(-12+10-3)i=-7-5i \implies a^2+b^2=(-7)^2+(-5)^2=49+25=\boxed{74}$ | null | 74 |
29a3931a6c989b9cfee0a5d4c70619bd | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_5 | Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$ | We observe that by multiplying $xy,$ $yz,$ and $zx,$ we get $(xyz)^2=(-80-320i)(60)(-96+24i).$ Next, we divide $(xyz)^2$ by $(yz)^2$ to
get $x^2.$ We have $x^2=\frac{(-80-320i)(60)(-96+24i)}{3600}=256+480i.$ We can write $x$ in the form of $a+bi,$ so we get $(a+bi)^2=256+480i.$ Then, $a^2-b^2+2abi=256+480i,$ $a^2-b^2=256,$ and $2ab=480.$ Solving this system of equations is relatively
simple. We have two cases, $a=20, b=12,$ and $a=-20, b=-12.$ Case 1: $a=20, b=12,$ so $x=20+12i.$ We solve for $y$ and $z$ by plugging in $x$ to the two equations. We see $y=\frac{-80-320i}{20+12i}=-10-10i$ and $z=\frac{-96+24i}{20+12i}=-3+3i.$ $x+y+z=7+5i,$ so $a=7$ and $b=5.$ Solving, we end up with $7^2+5^2=\boxed{074}$ as our answer.
Case 2: $a=-20, b=-12,$ so $x=-20-12i.$ Again, we solve for $y$ and $z.$ We find $y=\frac{-80-320i}{-20-12i}=10+10i,$ $z=\frac{-96+24i}{-20-12i}=3-3i,$ so $x+y+z=-7-5i.$ We again have $(-7)^2+(-5)^2=\boxed{074}.$ Solution by Airplane50 | null | 074 |
29a3931a6c989b9cfee0a5d4c70619bd | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_5 | Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$ | According to the Euler's Theory, we can rewrite $x$ $y$ and $z$ as \[x=r_{1}e^{i{\theta}_1}\] \[y=r_{2}e^{i{\theta}_2}\] \[x=r_{3}e^{i{\theta}_3}\] As a result, \[|xy|=r_{1}r_{2}=\sqrt{80^2+320^2}=80\sqrt{17}\] \[|yz|=r_{2}r_{3}=60\] \[|xz|=r_{1}r_{3}=\sqrt{96^2+24^2}=24\sqrt{17}\] Also, it is clear that \[yz=r_{2}e^{i{\theta}_2}r_{3}e^{i{\theta}_3}=|yz|e^{i({\theta}_2+{\theta}_3)}=|yz|=60\] So ${\theta}_2+{\theta}_3=0$ , or \[{\theta}_2=-{\theta}_3\] Also, we have \[xy=-80\sqrt{17}e^{i\arctan{4}}\] \[yz=60\] \[xz=-24\sqrt{17}e^{i\arctan{-\frac{1}{4}}}\] So now we have $r_{1}r_{2}=80\sqrt{17}$ $r_{2}r_{3}=60$ $r_{1}r_{3}=24\sqrt{17}$ ${\theta}_1+{\theta}_2=\arctan{4}$ and ${\theta}_1-{\theta}_2=\arctan {-\frac{1}{4}}$ . Solve these above, we get \[r_{1}=4\sqrt{34}\] \[r_{2}=10\sqrt{2}\] \[r_{3}=3\sqrt{2}\] \[{\theta}_2=\frac{\arctan{4}-\arctan{-\frac{1}{4}}}{2}=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}\] So we can get \[y=r_{2}e^{i{\theta}_2}=10\sqrt{2}e^{i\frac{\pi}{4}}=10+10i\] \[z=r_{3}e^{i{\theta}_3}=r_{3}e^{-i{\theta}_2}=3\sqrt{2}e^{-i\frac{\pi}{4}}=3-3i\] Use $xy=-80-320i$ we can find that \[x=-20-12i\] So \[x+y+z=-20-12i+10+10i+3-3i=-7-5i\] So we have $a=-7$ and $b=-5$
As a result, we finally get \[a^2+b^2=(-7)^2+(-5)^2=\boxed{074}\] | null | 074 |
29a3931a6c989b9cfee0a5d4c70619bd | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_5 | Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$ | We can turn the expression $x+y+z$ into $\sqrt{x^2+y^2+z^2+2xy+2yz+2xz}$ , and this would allow us to plug in the values after some computations.
Based off of the given products, we have
$xy^2z=60(-80-320i)$
$xyz^2=60(-96+24i)$
$x^2yz=(-96+24i)(-80-320i)$
Dividing by the given products, we have
$y^2=\frac{60(-80-320i)}{-96+24i}$
$z^2=\frac{60(-96+24i)}{-80-320i}$
$x^2=\frac{(-96+24i)(-80-320i)}{60}$
Simplifying, we get that this expression becomes $\sqrt{24+70i}$ . This equals $\pm{(7+5i)}$ , so the answer is $7^2+5^2=\boxed{74}$ | null | 74 |
29a3931a6c989b9cfee0a5d4c70619bd | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_5 | Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$ | Multiplying $xy \cdot yz \cdot zx = (xyz)^2$ we obtain $60 \cdot 960(16+30i)$ (too lazy to do $60 \cdot 960$ , you don't need to). Taking the square root, we get $240\sqrt{16+30i}$ . Letting $(a+bi)^2=16+30i,$ we have $a^2+2abi-b^2=16+30i.$ Thus, $(a+b)(a-b)=16,$ and $2ab=30.$ Guessing and checking, we get $a+bi=5+3i$ . Therefore, $xyz=240(5+3i).$ Dividing this by each of the equations provided in the original problem, we get $x=20+12i,y=-10-10i,$ and $z=-3+3i$ $20+12i-10-10i-3+3i=7+5i$ . Finally, $7^2+5^2=\boxed{074}.$ | null | 074 |
cfeaf5d9b1523fae2579281c8ee06fe7 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_6 | A real number $a$ is chosen randomly and uniformly from the interval $[-20, 18]$ . The probability that the roots of the polynomial
$x^4 + 2ax^3 + (2a - 2)x^2 + (-4a + 3)x - 2$
are all real can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | The polynomial we are given is rather complicated, so we could use Rational Root Theorem to turn the given polynomial into a degree-2 polynomial. With Rational Root Theorem, $x = 1, -1, 2, -2$ are all possible rational roots. Upon plugging these roots into the polynomial, $x = -2$ and $x = 1$ make the polynomial equal 0 and thus, they are roots that we can factor out.
The polynomial becomes:
$(x - 1)(x + 2)(x^2 + (2a - 1)x + 1)$
Since we know $1$ and $-2$ are real numbers, we only need to focus on the quadratic.
We should set the discriminant of the quadratic greater than or equal to 0.
$(2a - 1)^2 - 4 \geq 0$
This simplifies to:
$a \geq \dfrac{3}{2}$
or
$a \leq -\dfrac{1}{2}$
This means that the interval $\left(-\dfrac{1}{2}, \dfrac{3}{2}\right)$ is the "bad" interval. The length of the interval where $a$ can be chosen from is 38 units long, while the bad interval is 2 units long. Therefore, the "good" interval is 36 units long.
$\dfrac{36}{38} = \dfrac{18}{19}$
$18 + 19 = \boxed{037}$ | null | 037 |
9041e66e69ce0e40646b5ef22df9f5d0 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_7 | Triangle $ABC$ has side lengths $AB = 9$ $BC =$ $5\sqrt{3}$ , and $AC = 12$ . Points $A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B$ are on segment $\overline{AB}$ with $P_{k}$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, ..., 2449$ , and points $A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C$ are on segment $\overline{AC}$ with $Q_{k}$ between $Q_{k-1}$ and $Q_{k+1}$ for $k = 1, 2, ..., 2449$ . Furthermore, each segment $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2449$ , is parallel to $\overline{BC}$ . The segments cut the triangle into $2450$ regions, consisting of $2449$ trapezoids and $1$ triangle. Each of the $2450$ regions has the same area. Find the number of segments $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2450$ , that have rational length. | For each $k$ between $2$ and $2450$ , the area of the trapezoid with $\overline{P_kQ_k}$ as its bottom base is the difference between the areas of two triangles, both similar to $\triangle{ABC}$ . Let $d_k$ be the length of segment $\overline{P_kQ_k}$ . The area of the trapezoid with bases $\overline{P_{k-1}Q_{k-1}}$ and $P_kQ_k$ is $\left(\frac{d_k}{5\sqrt{3}}\right)^2 - \left(\frac{d_{k-1}}{5\sqrt{3}}\right)^2 = \frac{d_k^2-d_{k-1}^2}{75}$ times the area of $\triangle{ABC}$ . (This logic also applies to the topmost triangle if we notice that $d_0 = 0$ .) However, we also know that the area of each shape is $\frac{1}{2450}$ times the area of $\triangle{ABC}$ . We then have $\frac{d_k^2-d_{k-1}^2}{75} = \frac{1}{2450}$ . Simplifying, $d_k^2-d_{k-1}^2 = \frac{3}{98}$ . However, we know that $d_0^2 = 0$ , so $d_1^2 = \frac{3}{98}$ , and in general, $d_k^2 = \frac{3k}{98}$ and $d_k = \frac{\sqrt{\frac{3k}{2}}}{7}$ . The smallest $k$ that gives a rational $d_k$ is $6$ , so $d_k$ is rational if and only if $k = 6n^2$ for some integer $n$ .The largest $n$ such that $6n^2$ is less than $2450$ is $20$ , so $k$ has $\boxed{020}$ possible values. | null | 020 |
9041e66e69ce0e40646b5ef22df9f5d0 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_7 | Triangle $ABC$ has side lengths $AB = 9$ $BC =$ $5\sqrt{3}$ , and $AC = 12$ . Points $A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B$ are on segment $\overline{AB}$ with $P_{k}$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, ..., 2449$ , and points $A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C$ are on segment $\overline{AC}$ with $Q_{k}$ between $Q_{k-1}$ and $Q_{k+1}$ for $k = 1, 2, ..., 2449$ . Furthermore, each segment $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2449$ , is parallel to $\overline{BC}$ . The segments cut the triangle into $2450$ regions, consisting of $2449$ trapezoids and $1$ triangle. Each of the $2450$ regions has the same area. Find the number of segments $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2450$ , that have rational length. | We have that there are $2449$ trapezoids and $1$ triangle of equal area, with that one triangle being $AP_1Q_1$ . Notice, if we "stack" the trapezoids on top of $\bigtriangleup AP_1Q_1$ the way they already are, we'd create a similar triangle, all of which are similar to $\bigtriangleup ABC$ , and since the trapezoids and $\bigtriangleup AP_1Q_1$ have equal area, each of these similar triangles $AP_kQ_k$ have area $\frac{k}{2450}\left[ ABC\right]$ , and so $\frac{\left[ AP_kQ_k\right]}{\left[ABC\right]}=\frac{k}{2450}$ . We want the ratio of the side lengths $P_kQ_k:BC$ . Since area is a 2-dimensional unit of measurement, and side lengths are 1-dimensional, the ratio is simply the square root of the areas, or \[\frac{P_kQ_k}{BC}=\sqrt{\frac{k}{2450}}\] \[\implies P_kQ_k=BC\cdot \sqrt{\frac{k}{2450}}=5\sqrt{3}\cdot\sqrt{\frac{k}{2450}}=\frac{1}{7}\cdot \sqrt{\frac{3k}{2}}=\frac{3}{7}\sqrt{\frac{k}{6}}\] \[\implies k=6n^2<2450\] \[\implies 0<n\leq 20\] so there are $\boxed{020}$ solutions. | null | 020 |
9041e66e69ce0e40646b5ef22df9f5d0 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_7 | Triangle $ABC$ has side lengths $AB = 9$ $BC =$ $5\sqrt{3}$ , and $AC = 12$ . Points $A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B$ are on segment $\overline{AB}$ with $P_{k}$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, ..., 2449$ , and points $A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C$ are on segment $\overline{AC}$ with $Q_{k}$ between $Q_{k-1}$ and $Q_{k+1}$ for $k = 1, 2, ..., 2449$ . Furthermore, each segment $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2449$ , is parallel to $\overline{BC}$ . The segments cut the triangle into $2450$ regions, consisting of $2449$ trapezoids and $1$ triangle. Each of the $2450$ regions has the same area. Find the number of segments $\overline{P_{k}Q_{k}}$ $k = 1, 2, ..., 2450$ , that have rational length. | Let $T_1$ stand for $AP_1Q_1$ , and $T_k = AP_kQ_k$ . All triangles $T$ are similar by AA. Let the area of $T_1$ be $x$ . The next trapezoid will also have an area of $x$ , as given. Therefore, $T_k$ has an area of $kx$ . The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore, $P_k Q_k=P_1 Q_1\cdot \sqrt{k}$ , and the same if $Q$ is substituted for $P$ throughout. We want the side $P_k Q_k$ to be rational. Setting up proportions: \[5\sqrt{3}Â : \sqrt{2450}=35\sqrt{2}\] \[\sqrt{6}Â : 14\] which shows that $P_1 Q_1=\frac{\sqrt{6}}{14}$ . In order for $\sqrt{k} P_1 Q_1$ to be rational, $\sqrt{k}$ must be some rational multiple of $\sqrt{6}$ . This is achieved at $\sqrt{k}=\sqrt{6}, 2\sqrt{6}, \ldots, 20\sqrt{6}$ . We end there as $21\sqrt{6}=\sqrt{2646}$ . There are 20 numbers from 1 to 20, so there are $\boxed{020}$ solutions. | null | 020 |
de6440dd971dd8c3e52c8f7f9885aa32 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_8 | A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$ , the frog can jump to any of the points $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , or $(x, y + 2)$ . Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$ | We solve this problem by working backwards. Notice, the only points the frog can be on to jump to $(4,4)$ in one move are $(2,4),(3,4),(4,2),$ and $(4,3)$ . This applies to any other point, thus we can work our way from $(0,0)$ to $(4,4)$ , recording down the number of ways to get to each point recursively.
$(0,0): 1$
$(1,0)=(0,1)=1$
$(2,0)=(0, 2)=2$
$(3,0)=(0, 3)=3$
$(4,0)=(0, 4)=5$
$(1,1)=2$ $(1,2)=(2,1)=5$ $(1,3)=(3,1)=10$ $(1,4)=(4,1)= 20$
$(2,2)=14, (2,3)=(3,2)=32, (2,4)=(4,2)=71$
$(3,3)=84, (3,4)=(4,3)=207$
$(4,4)=2\cdot \left( (4,2)+(4,3)\right) = 2\cdot \left( 207+71\right)=2\cdot 278=\boxed{556}$ | null | 556 |
de6440dd971dd8c3e52c8f7f9885aa32 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_8 | A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$ , the frog can jump to any of the points $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , or $(x, y + 2)$ . Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$ | We'll refer to the moves $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , and $(x, y + 2)$ as $R_1$ $R_2$ $U_1$ , and $U_2$ , respectively. Then the possible sequences of moves that will take the frog from $(0,0)$ to $(4,4)$ are all the permutations of $U_1U_1U_1U_1R_1R_1R_1R_1$ $U_2U_1U_1R_1R_1R_1R_1$ $U_1U_1U_1U_1R_2R_1R_1$ $U_2U_1U_1R_2R_1R_1$ $U_2U_2R_1R_1R_1R_1$ $U_1U_1U_1U_1R_2R_2$ $U_2U_2R_2R_1R_1$ $U_2U_1U_1R_2R_2$ , and $U_2U_2R_2R_2$ . We can reduce the number of cases using symmetry.
Case 1: $U_1U_1U_1U_1R_1R_1R_1R_1$
There are $\frac{8!}{4!4!} = 70$ possibilities for this case.
Case 2: $U_2U_1U_1R_1R_1R_1R_1$ or $U_1U_1U_1U_1R_2R_1R_1$
There are $2 \cdot \frac{7!}{4!2!} = 210$ possibilities for this case.
Case 3: $U_2U_1U_1R_2R_1R_1$
There are $\frac{6!}{2!2!} = 180$ possibilities for this case.
Case 4: $U_2U_2R_1R_1R_1R_1$ or $U_1U_1U_1U_1R_2R_2$
There are $2 \cdot \frac{6!}{2!4!} = 30$ possibilities for this case.
Case 5: $U_2U_2R_2R_1R_1$ or $U_2U_1U_1R_2R_2$
There are $2 \cdot \frac{5!}{2!2!} = 60$ possibilities for this case.
Case 6: $U_2U_2R_2R_2$
There are $\frac{4!}{2!2!} = 6$ possibilities for this case.
Adding up all these cases gives us $70+210+180+30+60+6=\boxed{556}$ ways. | null | 556 |
de6440dd971dd8c3e52c8f7f9885aa32 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_8 | A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$ , the frog can jump to any of the points $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , or $(x, y + 2)$ . Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$ | Mark the total number of distinct sequences of jumps for the frog to reach the point $(x,y)$ as $\varphi (x,y)$ . Consider for each point $(x,y)$ in the first quadrant, there are only $4$ possible points in the first quadrant for frog to reach point $(x,y)$ , and these $4$ points are \[(x-1,y); (x-2,y); (x,y-1); (x,y-2)\] . As a result, the way to count $\varphi (x,y)$ is \[\varphi (x,y)=\varphi (x-1,y)+\varphi (x-2,y)+\varphi (x,y-1)+\varphi (x,y-2)\]
Also, for special cases, \[\varphi (0,y)=\varphi (0,y-1)+\varphi (0,y-2)\]
\[\varphi (x,0)=\varphi (x-1,0)+\varphi (x-2,0)\]
\[\varphi (x,1)=\varphi (x-1,1)+\varphi (x-2,1)+\varphi (x,0)\]
\[\varphi (1,y)=\varphi (1,y-1)+\varphi (1,y-2)+\varphi (0,y)\]
\[\varphi (1,1)=\varphi (1,0)+\varphi (0,1)\]
Start with $\varphi (0,0)=1$ , use this method and draw the figure below, we can finally get \[\varphi (4,4)=556\] (In order to make the LaTeX thing more beautiful to look at, I put $0$ to make every number $3$ digits)
\[005-020-071-207-\boxed{556}\] \[003-010-032-084-207\] \[002-005-014-032-071\] \[001-002-005-010-020\] \[001-001-002-003-005\] | null | 556 |
de6440dd971dd8c3e52c8f7f9885aa32 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_8 | A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$ , the frog can jump to any of the points $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , or $(x, y + 2)$ . Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$ | Casework Solution:
x-distribution: 1-1-1-1 (1 way to order)
y-distribution: 1-1-1-1 (1 way to order) $\dbinom{8}{4} = 70$ ways total
x-distribution: 1-1-1-1 (1 way to order)
y-distribution: 1-1-2 (3 ways to order) $\dbinom{7}{3} \times 3= 105$ ways total
x-distribution: 1-1-1-1 (1 way to order)
y-distribution: 2-2 (1 way to order) $\dbinom{6}{4} = 15$ ways total
x-distribution: 1-1-2 (3 ways to order)
y-distribution: 1-1-1-1 (1 way to order) $\dbinom{7}{3} \times 3= 105$ ways total
x-distribution: 1-1-2 (3 ways to order)
y-distribution: 1-1-2 (3 ways to order) $\dbinom{6}{3} \times 9= 180$ ways total
x-distribution: 1-1-2 (3 ways to order)
y-distribution: 2-2 (1 way to order) $\dbinom{5}{3} \times 3 = 30$ ways total
x-distribution: 2-2 (1 way to order)
y-distribution: 1-1-1-1 (1 way to order) $\dbinom{6}{4} = 15$ ways total
x-distribution: 2-2 (1 way to order)
y-distribution: 1-1-2 (3 ways to order) $\dbinom{5}{3} \times 3 = 30$ ways total
x-distribution: 2-2 (1 way to order)
y-distribution: 2-2 (1 way to order) $\dbinom{4}{2} = 6$ ways total
$6+30+15+105+180+70+30+15+105=\boxed{556}$ -fidgetboss_4000 | null | 556 |
6dc854619ced590aca33dcdab022d2f6 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_9 | Octagon $ABCDEFGH$ with side lengths $AB = CD = EF = GH = 10$ and $BC = DE = FG = HA = 11$ is formed by removing 6-8-10 triangles from the corners of a $23$ $\times$ $27$ rectangle with side $\overline{AH}$ on a short side of the rectangle, as shown. Let $J$ be the midpoint of $\overline{AH}$ , and partition the octagon into 7 triangles by drawing segments $\overline{JB}$ $\overline{JC}$ $\overline{JD}$ $\overline{JE}$ $\overline{JF}$ , and $\overline{JG}$ . Find the area of the convex polygon whose vertices are the centroids of these 7 triangles.
[asy] unitsize(6); pair P = (0, 0), Q = (0, 23), R = (27, 23), SS = (27, 0); pair A = (0, 6), B = (8, 0), C = (19, 0), D = (27, 6), EE = (27, 17), F = (19, 23), G = (8, 23), J = (0, 23/2), H = (0, 17); draw(P--Q--R--SS--cycle); draw(J--B); draw(J--C); draw(J--D); draw(J--EE); draw(J--F); draw(J--G); draw(A--B); draw(H--G); real dark = 0.6; filldraw(A--B--P--cycle, gray(dark)); filldraw(H--G--Q--cycle, gray(dark)); filldraw(F--EE--R--cycle, gray(dark)); filldraw(D--C--SS--cycle, gray(dark)); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(J); dot(H); defaultpen(fontsize(10pt)); real r = 1.3; label("$A$", A, W*r); label("$B$", B, S*r); label("$C$", C, S*r); label("$D$", D, E*r); label("$E$", EE, E*r); label("$F$", F, N*r); label("$G$", G, N*r); label("$H$", H, W*r); label("$J$", J, W*r); [/asy] | We represent octagon $ABCDEFGH$ in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that $A=(0,-6), B=(8, 0), C=(19,0), D=(27, -6), E=(27, -17), F=(19, -23), G=(8, -23), H=(0, -17), J=(0, -\frac{23}{2})$ . Recall that the centroid is $\frac{1}{3}$ way up each median in the triangle. Thus, we can find the centroids easily by finding the midpoint of the side opposite of point $J$ . Furthermore, we can take advantage of the reflective symmetry across the line parallel to $BC$ going through $J$ by dealing with less coordinates and ommiting the $\frac{1}{2}$ in the shoelace formula.
By doing some basic algebra, we find that the coordinates of the centroids of $\bigtriangleup JAB, \bigtriangleup JBC, \bigtriangleup JCD, \bigtriangleup JDE$ are $\left(\frac{8}{3}, -\frac{35}{6}\right), \left(9, -\frac{23}{6}\right), \left(\frac{46}{3}, -\frac{35}{6}\right),$ and $\left(18, -\frac{23}{2}\right)$ , respectively. We'll have to throw in the projection of the centroid of $\bigtriangleup JAB$ to the line of reflection to apply shoelace, and that point is $\left( \frac{8}{3}, -\frac{23}{2}\right)$
Finally, applying Shoelace, we get: $\left|\left(\frac{8}{3}\cdot (-\frac{23}{6})+9\cdot (-\frac{35}{6})+\frac{46}{3}\cdot (-\frac{23}{2})+18\cdot (\frac{-23}{2})+\frac{8}{3}\cdot (-\frac{35}{6})\right) - \left((-\frac{35}{6}\cdot 9) +\\(-\frac{23}{6}\cdot \frac{46}{3})+ (-\frac{35}{6}\cdot 18)+(\frac{-23}{2}\cdot \frac{8}{3})+(-\frac{23}{2}\cdot \frac{8}{3})\right)\right|$ $=\left|\left(-\frac{92}{9}-\frac{105}{2}-\frac{529}{3}-207-\frac{140}{9}\right)-\left(-\frac{105}{2}-\frac{529}{9}-105-\frac{92}{3}-\frac{92}{3}\right)\right|$ $=\left|-\frac{232}{9}-\frac{1373}{6}-207+\frac{529}{9}+\frac{184}{3}+105+\frac{105}{2}\right|$ $=\left|\frac{297}{9}-\frac{690}{6}-102\right|=\left| 33-115-102\right|=\left|-184\right|=\boxed{184}$ | null | 184 |
6dc854619ced590aca33dcdab022d2f6 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_9 | Octagon $ABCDEFGH$ with side lengths $AB = CD = EF = GH = 10$ and $BC = DE = FG = HA = 11$ is formed by removing 6-8-10 triangles from the corners of a $23$ $\times$ $27$ rectangle with side $\overline{AH}$ on a short side of the rectangle, as shown. Let $J$ be the midpoint of $\overline{AH}$ , and partition the octagon into 7 triangles by drawing segments $\overline{JB}$ $\overline{JC}$ $\overline{JD}$ $\overline{JE}$ $\overline{JF}$ , and $\overline{JG}$ . Find the area of the convex polygon whose vertices are the centroids of these 7 triangles.
[asy] unitsize(6); pair P = (0, 0), Q = (0, 23), R = (27, 23), SS = (27, 0); pair A = (0, 6), B = (8, 0), C = (19, 0), D = (27, 6), EE = (27, 17), F = (19, 23), G = (8, 23), J = (0, 23/2), H = (0, 17); draw(P--Q--R--SS--cycle); draw(J--B); draw(J--C); draw(J--D); draw(J--EE); draw(J--F); draw(J--G); draw(A--B); draw(H--G); real dark = 0.6; filldraw(A--B--P--cycle, gray(dark)); filldraw(H--G--Q--cycle, gray(dark)); filldraw(F--EE--R--cycle, gray(dark)); filldraw(D--C--SS--cycle, gray(dark)); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(J); dot(H); defaultpen(fontsize(10pt)); real r = 1.3; label("$A$", A, W*r); label("$B$", B, S*r); label("$C$", C, S*r); label("$D$", D, E*r); label("$E$", EE, E*r); label("$F$", F, N*r); label("$G$", G, N*r); label("$H$", H, W*r); label("$J$", J, W*r); [/asy] | Draw the heptagon whose vertices are the midpoints of octagon $ABCDEFGH$ except $J$ .
We have a homothety since:
1. $J$ passes through corresponding vertices of the two heptagons.
2. By centroid properties, our ratio between the sidelengths is $\frac{2}{3},$ and their area ratio is hence $\frac{4}{9}.$
Compute the area of the large heptagon by dividing into two congruent trapezoids and a triangle. The area of each trapezoid is \[= \frac{1}{2}(17+23)\cdot \frac{19}{2}=190.\] The area of each triangle is \[= \frac{1}{2}\cdot 17\cdot 4=34.\]
Hence, the area of the large heptagon is \[2\cdot 190+34=414.\] Then, from our homothety, the area of the required heptagon is \[\frac{4}{9}\cdot 414=\boxed{184}.\] ~novus677 | null | 184 |
ff9e88338bf49a31a107a5199593b536 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_10 | Find the number of functions $f(x)$ from $\{1, 2, 3, 4, 5\}$ to $\{1, 2, 3, 4, 5\}$ that satisfy $f(f(x)) = f(f(f(x)))$ for all $x$ in $\{1, 2, 3, 4, 5\}$ | We perform casework on the number of fixed points (the number of points where $f(x) = x$ ). To better visualize this, use the grid from Solution 1.
Case 1: 5 fixed points
Case 2: 4 fixed points
Case 3: 3 fixed points
Case 4: 2 fixed points
Case 5: 1 fixed point
Therefore, the answer is $1+20+150+380+205 = \boxed{756}$ | null | 756 |
e2b096df9581c31f516af2b1e148f51b | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11 | Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ | If the first number is $6$ , then there are no restrictions. There are $5!$ , or $120$ ways to place the other $5$ numbers.
If the first number is $5$ $6$ can go in four places, and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.
If the first number is $4$ , ....
4 6 _ _ _ _ $\implies$ 24 ways
4 _ 6 _ _ _ $\implies$ 24 ways
4 _ _ 6 _ _ $\implies$ 24 ways
4 _ _ _ 6 _ $\implies$ 5 must go between $4$ and $6$ , so there are $3 \cdot 3! = 18$ ways.
$24 + 24 + 24 + 18 = 90$ ways if 4 is first.
If the first number is $3$ , ....
3 6 _ _ _ _ $\implies$ 24 ways
3 _ 6 _ _ _ $\implies$ 24 ways
3 1 _ 6 _ _ $\implies$ 4 ways
3 2 _ 6 _ _ $\implies$ 4 ways
3 4 _ 6 _ _ $\implies$ 6 ways
3 5 _ 6 _ _ $\implies$ 6 ways
3 5 _ _ 6 _ $\implies$ 6 ways
3 _ 5 _ 6 _ $\implies$ 6 ways
3 _ _ 5 6 _ $\implies$ 4 ways
$24 + 24 + 4 + 4 + 6 + 6 + 6 + 6 + 4 = 84$ ways
If the first number is $2$ , ....
2 6 _ _ _ _ $\implies$ 24 ways
2 _ 6 _ _ _ $\implies$ 18 ways
2 3 _ 6 _ _ $\implies$ 4 ways
2 4 _ 6 _ _ $\implies$ 6 ways
2 5 _ 6 _ _ $\implies$ 6 ways
2 5 _ _ 6 _ $\implies$ 6 ways
2 _ 5 _ 6 _ $\implies$ 4 ways
2 4 _ 5 6 _ $\implies$ 2 ways
2 3 4 5 6 1 $\implies$ 1 way
$24 + 18 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71$ ways
Grand Total : $120 + 96 + 90 + 84 + 71 = \boxed{461}$ | null | 461 |
e2b096df9581c31f516af2b1e148f51b | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11 | Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ | If $6$ is the first number, then there are no restrictions. There are $5!$ , or $120$ ways to place the other $5$ numbers.
If $6$ is the second number, then the first number can be $2, 3, 4,$ or $5$ , and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.
If $6$ is the third number, then we cannot have the following:
1 _ 6 _ _ _ $\implies$ 24 ways
2 1 6 _ _ _ $\implies$ 6 ways
$120 - 24 - 6 = 90$ ways
If $6$ is the fourth number, then we cannot have the following:
1 _ _ 6 _ _ $\implies$ 24 ways
2 1 _ 6 _ _ $\implies$ 6 ways
2 3 1 6 _ _ $\implies$ 2 ways
3 1 2 6 _ _ $\implies$ 2 ways
3 2 1 6 _ _ $\implies$ 2 ways
$120 - 24 - 6 - 2 - 2 - 2 = 84$ ways
If $6$ is the fifth number, then we cannot have the following:
_ _ _ _ 6 5 $\implies$ 24 ways
1 5 _ _ 6 _ $\implies$ 6 ways
1 _ 5 _ 6 _ $\implies$ 6 ways
2 1 5 _ 6 _ $\implies$ 2 ways
1 _ _ 5 6 _ $\implies$ 6 ways
2 1 _ 5 6 _ $\implies$ 2 ways
2 3 1 5 6 4, 3 1 2 5 6 4, 3 2 1 5 6 4 $\implies$ 3 ways
$120 - 24 - 6 - 6 - 2 - 6 - 2 - 3 = 71$ ways
Grand Total : $120 + 96 + 90 + 84 + 71 = \boxed{461}$ | null | 461 |
e2b096df9581c31f516af2b1e148f51b | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11 | Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ | Note the condition in the problem is equivalent to the following condition: for each $k$ with $1 \le k \le 5$ , the first $k$ terms is not a permutation $(1, 2, \ldots, k)$ (since it would mean there must be some integer $x$ in the first $k$ terms such that $x \not \in \{1, 2, \ldots, k\} \implies x > k$ ). Then, let $a_n$ denote the number of permutations of $(1, 2, \ldots, n)$ satisfying the condition in the problem. We use complementary counting to find $a_n$ . Notice that in order to not satisfy the condition in the problem, there are $n-1$ cases: the first $1 \le k \le n-1$ (we don't include $k = n$ since the condition in the problem only holds up to $n-1$ ) terms are a permutation of $(1, 2, \ldots, k)$ , and for all $k+1 \le i \le n-1$ , the condition in the problem still holds. Then, for each of these cases, there are $k!$ ways to arrange the first $k$ terms, and then $a_{n-k}$ ways to arrange the $k + 1$ to $n$ terms (assume by contradiction that terms from $k+1$ to $i$ is a permutation of $(k+1, k+2, \ldots, i)$ . Then, since the first $k$ terms are already a permutation of $(1, 2, \ldots, k)$ , the first $i$ terms form a permutation of $(1, 2, \ldots, i)$ . This contradicts our assumption that for all $k+1 \le i \le n-1$ , the condition still holds. Thus, we can only include the $a_{n-k}$ permutations of these terms). Now, summing the cases up and subtracting from $n!$ , we have: $a_n = n! - \sum_{k=1}^{n-1} a_{n-k} k!$ . From this recursion, we derive that $a_1 = 1$ $a_2 = 1$ $a_3 = 3$ $a_4 = 13$ $a_5 = 71$ , and finally $a_6 = \boxed{461}$ | null | 461 |
e2b096df9581c31f516af2b1e148f51b | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11 | Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ | Let $A_i$ be the set of permutations such that there is no number greater than $i$ in the first $i$ places. Note that $\bigcap^{k}_{i=0}{A_{b_i}}=\prod^k_{i=1}{(b_i-b_{i-1})!}$ for all $1\le b_0 < b_1\cdots < b_k \le 5$ and that the set of restricted permutations is $A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5$
We will compute the cardinality of this set with PIE. \begin{align*} &|A_1| + |A_2| + |A_3| + |A_4| + |A_5|\\ = &120 + 48 + 36 + 48 + 120 = 372\\ \\ &|A_1 \cap A_2| + |A_1 \cap A_3| + |A_1 \cap A_4| + |A_1 \cap A_5| + |A_2 \cap A_3|\\ + &|A_2 \cap A_4| + |A_2 \cap A_5| + |A_3 \cap A_4| + |A_3 \cap A_5| + |A_4 \cap A_5|\\=&24+12+12+24+12+8+12+12+12+24=152\\ \\ &|A_1 \cap A_2 \cap A_3| + |A_1 \cap A_2 \cap A_4| + |A_1 \cap A_2 \cap A_5| + |A_1 \cap A_3 \cap A_4| + |A_1 \cap A_3 \cap A_5|\\ +& |A_1 \cap A_4 \cap A_5| + |A_2 \cap A_3 \cap A_4| + |A_2 \cap A_3 \cap A_5| + |A_2 \cap A_4 \cap A_5| + |A_3 \cap A_4 \cap A_5|\\=&6 + 4 + 6 + 4 + 4 + 6 + 4 + 4 + 4 + 6 = 48\\ \\ &|A_1 \cap A_2 \cap A_3 \cap A_4| + |A_1 \cap A_2 \cap A_3 \cap A_5| + |A_1 \cap A_2 \cap A_4 \cap A_5| + |A_1 \cap A_3 \cap A_4 \cap A_5| + |A_2 \cap A_3 \cap A_4 \cap A_5|\\=&2 + 2 + 2 + 2 + 2 = 10\\ \\ &|A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5| = 1 \end{align*} To finish, $720 - 372 + 152 - 48 + 10 - 1 = \boxed{461}$ | null | 461 |
e2b096df9581c31f516af2b1e148f51b | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11 | Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ | Define the function $f(p,q)$ as the amount of permutations with maximum digit $q$ and string length $p$ that satisfy the condition within bounds. For example, $f(4,5)$ would be the amount of ways to make a string with length $4$ with the highest digit being $5$ . We wish to obtain $f(6,6)=f(5,6)$
To generate recursion, consider how we would get to $f(p,q)$ from $f(p-1,a)$ for all $a$ such that $p\le{a}\le6$ . We could either jump from the old maximum $a$ to the new $q$ by concatenating the old string and the new digit $q$ , or one could retain the maximum, in which case $a=q$ . To retain the maximum, one would have to pick a new available digit not exceeding $q$
In the first case, there is only one way to pick the new digit, namely picking $q$ . For the second case, there are $q-p+1$ digits left to choose, because there are $q$ digits between 1 and $q$ total and there are $p-1$ digits already chosen below or equal to $q$ . Thus, $f(p,q)=[\sum^{q-1}_{n=p}f(p-1,n)] + (q-p+1)f(p-1,q)$ . Now that we have the recursive function, we can start evaluating the values of $f(p,q)$ until we get to $f(6,6)=f(5,6)$
\[f(2,3)=3, f(2,4)=5, f(2,5)=7, f(2,6)=9\] \[f(3,4)=13, f(3,5)=29, f(3,6)=51\] \[f(4,5)=71, f(4,6)=195\] \[f(5,6)=461\] Our requested answer is thus $\boxed{461}$ ~sigma | null | 461 |
e2b096df9581c31f516af2b1e148f51b | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11 | Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ | We can also solve this problem by counting the number of permutations that do NOT satisfy the given conditions; namely, these permutations must satisfy the condition that none of the first $k$ terms is greater than $k$ for $1\leq$ $k$ $\leq$ $5$ . We can further simplify this method by approaching it through casework on the first $k$ terms.
Case 1: None of the first one terms is greater than 1
The first term must obviously be one. Since there are five spaces left for numbers, there are a total of $5!=120$ permutations for this case.
Case 2: None of the first two terms is greater than 2
The first two terms must be 1 and 2 in some order. However, we already counted all cases starting with 1 in the previous case, so all of the permutations in this case must begin with $12\cdots$ . Since there are four spaces left, there are a total of $4!=24$ permutations for this case.
Case 3: None of the first three terms is greater than 3
The first three terms must be 1, 2, and 3 in some order. However, the cases beginning with 1__ and 21_ have already been accounted for. There are now $3!-3 = 3$ ways to order the first three numbers of the permutation, and $3!$ ways to order the last three numbers, for a total of $3\times6 = 18$ permutations.
Case 4: None of the first four terms is greater than 4
You can see where the pattern is going - the first four terms must be 1, 2, 3, and 4 in some order. All cases starting with 1 (there are $3!=6$ ), the cases starting with 21 (there are $2!=2$ ), and the 3 cases from case 3 (there are $3\times 1! = 3$ ) have been accounted for, so there are a total of $(4!-6-2-3)2!=26$ permutations for this case.
Case 5: None of the first five terms is greater than 5
This is perhaps the hardest case to work with, simply because there are so many subcases, so keeping track is crucial here. Obviously, the first five terms must be 1, 2, 3, 4, and 5, meaning there are 120 ways to order them. Now, we count the permutations we have already counted in previous cases. $4!$ start with 1, $3!$ start with 2, $3\times2!=6$ start with 3, and $13\times1!=13$ start with 4. Subtracting, we get a total of $120-24-6-6-13=71$ permutations.
Now, we subtract the total number of permutations from our cases from the total number of permutations, which is $6!$ $720 - 120 - 24 - 18 - 26 - 71 = \boxed{461}$ | null | 461 |
3ce68a1555ca9aea100f77f9dfa72c3c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | For reference, $2\sqrt{65} \approx 16$ , so $\overline{AD}$ is the longest of the four sides of $ABCD$ . Let $h_1$ be the length of the altitude from $B$ to $\overline{AC}$ , and let $h_2$ be the length of the altitude from $D$ to $\overline{AC}$ . Then, the triangle area equation becomes
\[\frac{h_1}{2}AP + \frac{h_2}{2}CP = \frac{h_1}{2}CP + \frac{h_2}{2}AP \rightarrow \left(h_1 - h_2\right)AP = \left(h_1 - h_2\right)CP \rightarrow AP = CP\]
What an important finding! Note that the opposite sides $\overline{AB}$ and $\overline{CD}$ have equal length, and note that diagonal $\overline{DB}$ bisects diagonal $\overline{AC}$ . This is very similar to what happens if $ABCD$ were a parallelogram with $AB = CD = 10$ , so let's extend $\overline{DB}$ to point $E$ , such that $AECD$ is a parallelogram. In other words, \[AE = CD = 10\] and \[EC = DA = 2\sqrt{65}\] Now, let's examine $\triangle ABE$ . Since $AB = AE = 10$ , the triangle is isosceles, and $\angle ABE \cong \angle AEB$ . Note that in parallelogram $AECD$ $\angle AED$ and $\angle CDE$ are congruent, so $\angle ABE \cong \angle CDE$ and thus \[\text{m}\angle ABD = 180^\circ - \text{m}\angle CDB\] Define $\alpha := \text{m}\angle CDB$ , so $180^\circ - \alpha = \text{m}\angle ABD$
We use the Law of Cosines on $\triangle DAB$ and $\triangle CDB$
\[\left(2\sqrt{65}\right)^2 = 10^2 + BD^2 - 20BD\cos\left(180^\circ - \alpha\right) = 100 + BD^2 + 20BD\cos\alpha\]
\[14^2 = 10^2 + BD^2 - 20BD\cos\alpha\]
Subtracting the second equation from the first yields
\[260 - 196 = 40BD\cos\alpha \rightarrow BD\cos\alpha = \frac{8}{5}\]
This means that dropping an altitude from $B$ to some foot $Q$ on $\overline{CD}$ gives $DQ = \frac{8}{5}$ and therefore $CQ = \frac{42}{5}$ . Seeing that $CQ = \frac{3}{5}\cdot BC$ , we conclude that $\triangle QCB$ is a 3-4-5 right triangle, so $BQ = \frac{56}{5}$ . Then, the area of $\triangle BCD$ is $\frac{1}{2}\cdot 10 \cdot \frac{56}{5} = 56$ . Since $AP = CP$ , points $A$ and $C$ are equidistant from $\overline{BD}$ , so \[\left[\triangle ABD\right] = \left[\triangle CBD\right] = 56\] and hence \[\left[ABCD\right] = 56 + 56 = \boxed{112}\] -kgator | null | 112 |
3ce68a1555ca9aea100f77f9dfa72c3c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | Denote $\angle APB$ by $\alpha$ . Then $\sin(\angle APB)=\sin \alpha = \sin(\angle APD)$ .
Using the formula for the area of a triangle, we get \[\frac{1}{2} (AP\cdot BP+ CP\cdot DP)\sin\alpha=\frac{1}{2}(AP\cdot DP+ CP\cdot BP)\sin\alpha ,\] so \[(AP-CP)(BP-DP)=0\] Hence $AP=CP$ (note that $BP=DP$ makes no difference here).
Now, assume that $AP=CP=x$ $BP=y$ , and $DP=z$ . Using the cosine rule for $\triangle APB$ and $\triangle BPC$ , it is clear that \[x^2+y^2-100=2 xy \cdot \cos{APB}=-(2 xy \cdot \cos{(\pi-CPB)})=-(x^2+y^2-196)\] or \begin{align}x^2+y^2=148\end{align}. Likewise, using the cosine rule for triangles $APD$ and $CPD$ \begin{align}\tag{2}x^2+z^2=180\end{align}. It follows that \begin{align}\tag{3}z^2-y^2=32\end{align}. Since $\sin\alpha=\sqrt{1-\cos^2\alpha}$ \[\sqrt{1-\frac{(x^2+y^2-100)^2}{4x^2y^2}}=\sqrt{1-\frac{(x^2+z^2-260)^2}{4x^2z^2}}\] which simplifies to \[\frac{48^2}{y^2}=\frac{80^2}{z^2} \qquad \Longrightarrow \qquad 5y=3z.\] Plugging this back to equations $(1)$ $(2)$ , and $(3)$ , it can be solved that $x=\sqrt{130},y=3\sqrt{2},z=5\sqrt{2}$ . Then, the area of the quadrilateral is \[x(y+z)\sin\alpha=\sqrt{130}\cdot8\sqrt{2}\cdot\frac{14}{\sqrt{260}}=\boxed{112}\] --Solution by MicGu | null | 112 |
3ce68a1555ca9aea100f77f9dfa72c3c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | As in all other solutions, we can first find that either $AP=CP$ or $BP=DP$ , but it's an AIME problem, we can take $AP=CP$ , and assume the other choice will lead to the same result (which is true).
From $AP=CP$ , we have $[DAP]=[DCP]$ , and $[BAP]=[BCP] \implies [ABD] = [CBD]$ , therefore, \begin{align} \nonumber \frac 12 \cdot AB\cdot AD\sin A &= \frac 12 \cdot BC\cdot CD\sin C \\ \Longrightarrow \hspace{1in} 7\sin C &= \sqrt{65}\sin A \end{align} By Law of Cosines, \begin{align} \nonumber 10^2+14^2-2\cdot 10\cdot 14\cos C &= 10^2+4\cdot 65-2\cdot 10\cdot 2\sqrt{65}\cos A \\ \Longrightarrow \hspace{1in} -\frac 85 - 7\cos C &= \sqrt{65}\cos A \tag{2} \end{align} Square $(1)$ and $(2)$ , and add them, to get \[\left(\frac 85\right)^2 + 2\cdot \frac 85 \cdot 7\cos C + 7^2 = 65\] Solve, $\cos C = 3/5 \implies \sin C = 4/5$ \[[ABCD] = 2[BCD] = BC\cdot CD\cdot \sin C = 14\cdot 10\cdot \frac 45 = \boxed{112}\] -Mathdummy | null | 112 |
3ce68a1555ca9aea100f77f9dfa72c3c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | Either $PA=PC$ or $PD=PB$ . Let $PD=PB=s$ . Applying Stewart's Theorem on $\triangle ABD$ and $\triangle BCD$ , dividing by $2s$ and rearranging, \[\tag{1}CP^2+s^2=148\] \[\tag{2}AP^2+s^2=180\] Applying Stewart on $\triangle CAB$ and $\triangle CAD$ \[\tag{3} 5CP^2=3AP^2\] Substituting equations 1 and 2 into 3 and rearranging, $s=BP=PD\sqrt{130}, CP=3\sqrt{2}, PA=5\sqrt{2}$ . By Law of Cosines on $\triangle APB$ $\cos(\angle APB)=\frac{4\sqrt{65}}{65}$ so $\sin(\angle APB)=\sin(\angle BPC)=\sin(\angle CPD)=\sin(\angle DPA)=\frac{7\sqrt{65}}{65}$ . Using $[\triangle ABC]=\frac{ab\sin(\angle C)}{2}$ to find unknown areas, $[ABCD]=[\triangle APB]+[\triangle BPC]+[\triangle CPD]+[\triangle DPA]=\boxed{112}$ | null | 112 |
3ce68a1555ca9aea100f77f9dfa72c3c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | Now we prove P is the midpoint of $BD$ . Denote the height from $B$ to $AC$ as $h_1$ , height from $D$ to $AC$ as $h_2$ .According to the problem, $AP* h_1 +CP* h_2 =CP* h_1 +AP* h_2$ implies $h_1 (AP-CP)= h_2 (AP-CP), h_1 = h_2$ . Then according to basic congruent triangles we get $BP=DP$ Firstly, denote that $CP=a,BP=b,CP=c,AP=d$ . Applying Stewart theorem, getting that $100c+196b=(b+c)(bc+a^2); 100b+260c=(b+c)(bc+c^2); 100c+196b=100b+260c, 3b=5c$ , denote $b=5x,c=3x$ Applying Stewart Theorem, getting $260a+100a=2a(a^2+25x^2); 196a+100a=2a(9x^2+a^2)$ solve for a, getting $a=\sqrt{130},AP=5\sqrt{2}; CP=3\sqrt{2}$ Now everything is clear, we can find $cos\angle{BPA}=\frac{4}{\sqrt{65}}$ using LOC, $sin\angle{BPA}=\frac{7\sqrt{65}}{65}$ , the whole area is $\sqrt{130}*8\sqrt{2}*\frac{7\sqrt{65}}{65}=\boxed{112}$ | null | 112 |
3ce68a1555ca9aea100f77f9dfa72c3c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | $BP = PD$ as in another solutions.
Let $D'$ be the reflection of $D$ across $C$ .
Let points $E, E',$ and $H$ be the foot of perpendiculars on $AC$ from $D,D'$ , and $B$ respectively. \begin{align*} &\qquad AB = CD = CD', \quad \textrm{and} \quad BH = DE = D'E' \\ \Rightarrow &\qquad \triangle ABH = \triangle CDE = \triangle CD'E' \\ \Rightarrow &\qquad \angle BAC = \angle ACD' \\ \Rightarrow &\qquad \triangle ABC = \triangle AD'C \\ \Rightarrow &\qquad BC = AD'. \end{align*} The area of quadrilateral $ABCD$ is equal to the area of triangle $ADD'$ with sides $AD' = 14, AD = 2\sqrt{65}, DD' = 2 \cdot 10 = 20$ .
The semiperimeter is $s = 17 + \sqrt{65},$ the area \[[ADD'] = \sqrt {(17 + \sqrt{65}) (17 - \sqrt{65})(3 + \sqrt{65})(\sqrt{65}-3)} = \sqrt{(289 – 65)\cdot(65-9)} =\sqrt{56 \cdot 4 \cdot 56} = \boxed{112}.\] | null | 112 |
3ce68a1555ca9aea100f77f9dfa72c3c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | Use your favorite method to get that $P$ is the midpoint of one of the two diagonals (suppose it's the midpoint of $\overline{AC}$ ). From here, let $x=AP=PC, y=BP, z=PD, a=\cos\theta$ where $\theta$ is the angle that the diagonals make. Then we have a system of four equations:
\begin{align*} x^2+y^2+2xya &= 100 \\ z^2+x^2+2xza &= 100 \\ x^2+y^2-2xya &= 196 \\ x^2+z^2-2xza &= 260 \\ \end{align*}
From these equations we get that \begin{align*} xya &= -24 \\ xza &= -40 \\ x^2+y^2-48 &= 10 \\ x^2+z^2-80 &= 10 \end{align*}
From here we can see that $\frac{z}{y}={5}{3}, z^2-y^2=32,$ so $z=5\sqrt{2}, y=3\sqrt{2}.$ Furthermore, this implies $x=\sqrt{130}$ and $xa=-4\sqrt{2},$ which implies $a=\cos\theta=\frac{4}{\sqrt{65}}.$ Then note that the area of the quadrilateral is \[\frac{1}{2}\sin\theta (xy+xz+xz+xy)=\sin\theta (\sqrt{130}\cdot 3\sqrt{2}+\sqrt{130} \cdot 5\sqrt{2})=7\cdot (3\cdot 2+5\cdot 2)=7(6+10)=7\cdot 16=\boxed{112}.\] | null | 112 |
3ce68a1555ca9aea100f77f9dfa72c3c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | Note that $\angle APB = \angle CPD = 180-\angle APD = 180-\angle BPC.$ (All angles are in degrees)
Since $\sin(\theta)=\sin(180-\theta),$ we can use sine area formula to get the following(after some simplifying steps): \begin{align*} BP \times AP + CP \times DP = BP \times PC + AP \times PD. \end{align*} For convenience, let $AP=a, BP=b, CP=c,DP=d.$ The above equation simplifies to: \begin{align*} ab + cd = bc + ad \\ab-ad+cd-bc=0 \\a(b-d)-c(b-d)=0 \\(a-c)(b-d)=0 \end{align*} From here, we see that $a=c$ or $b=d$ . Without loss of generality, let $a=c$ . Since triangles $ABP$ and $CDP$ are obviously not congruent, we see that one triangle is obtuse and the other one is acute.(Refer to the diagram) However, if we drop perpendiculars from $B$ to $AC$ and $D$ to $AC$ , we do get congruent triangles. If the foot of the perpendicular from $B$ is $M$ , and the foot of the perpendicular from $D$ is $N$ , then right triangle $BMP$ is congruent to right triangle $DNP$ .
From here, we see that the altitudes of triangles $ABC$ and $ADC$ to $AC$ are equal. Since they share base $AC$ , their areas are equal. We can use Heron's formula. To not have any fractions, let $AC=2x.$ \begin{align*} \sqrt{(12+x)(12-x)(x+2)(x-2)}=\sqrt{5+\sqrt{65}+x)(5+\sqrt{65}-x)(5-\sqrt{65}+x)(\sqrt{65}-5+x)} \end{align*} Even though this looks bad at first, it actually isn't too complicated to simplify. Expanding the differences of squares and simplifying completely, we get $x^2=32.$ Plugging this $x$ back into the Heron's formula, we get that the area of $ABC$ (or $ADC$ ) is $56$ . Since these triangles have equal area, the area of the quadrilateral is $2 \times 56 = \boxed{112}$ , and we are done. $\blacksquare$ | null | 112 |
3ce68a1555ca9aea100f77f9dfa72c3c | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_12 | Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$ $BC = 14$ , and $AD = 2\sqrt{65}$ . Assume that the diagonals of $ABCD$ intersect at point $P$ , and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$ . Find the area of quadrilateral $ABCD$ | Use any method to derive that $P$ is the midpoint of $A$ and $C$ . Now, either construct a parallelogram with as shown in the diagram to the right, or use law of sines. We will take the parallelogram route here, but the same thing can be done with law of sines on triangles $\triangle \textnormal{ABP}$ and $\triangle \textnormal{CPD}$ . Reflect $D$ across $P$ to get $D'$ . Since $CD = AD' = AB = 10$ $\triangle \textnormal{ABD'}$ is isosceles. Thus, $\angle AD'B = \angle ABD'$ , and because $ADCD'$ is a parallelogram (since $AP = PC$ and $DP = PD'$ ), $\angle AD'B = \angle BDC = \angle ABD'$ . So, $\angle ABD = 180 - \angle ABD' = 180 - \angle BDC$ . Now, apply law of cosines on $\triangle \textnormal{ABD}$ and $\triangle \textnormal{CDB}$ . We get:
\begin{align}
100 + BD^2 - 20BD \cos{\angle ABD} &= 100 + BD^2 - 20 BD \cos {(180 - \angle BDC)} = \\
100 + BD^2 + 20 BD \cos{\angle BDC} &= 260 \\
&\textnormal{and} \\
100 + BD^2 - 20 BD \cos{\angle BDC} &= 196 \\
\textnormal{summing }&\textnormal{and simplifying,} \\
BD &= 8\sqrt{2}
\end{align}
Then, applying law of cosines on $\triangle \textnormal{BCD}$ again, we obtain \[100 + 196 - 280 \cos{\angle BCD} = BD^2 = 128 \implies \cos{\angle BCD} = \frac{3}{5} \implies \sin{\angle BCD} = \frac{4}{5}\] Since $AP = PD$ $[ABD] = [BCD] \implies [ABCD] = [ABD] + [BCD] = 2[BCD]$ . Thus, $[ABCD] = 2[BCD] = 2 \cdot \frac{1}{2} \cdot 10 \cdot 14 \sin{\angle BCD} = 140 \cdot \frac{4}{5} = \boxed{112}$ | null | 112 |
28c3818bb0eebd518a05f951d51140d7 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $P_n$ be the probability of getting consecutive $1,2,3$ rolls in $n$ rolls and not rolling $1,2,3$ prior to the nth roll.
Let $x = P_3+P_5+...=1-(P_4+P_6+..)$ . Following Solution 2, one can see that \[P_{n+1}=P_{n}-\frac{P_{n-2}}{6^3}\] for all positive integers $n \ge 5$ . Summing for $n=5,7,...$ gives \[(1-x)-\frac{1}{6^3}=x-\frac{1}{6^3}-\frac{x}{6^3}\] \[\implies x = \frac{m}{n} = \frac{216}{431} \implies m+n=216+431= \boxed{647}\] | null | 647 |
28c3818bb0eebd518a05f951d51140d7 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $P_{odd}=\frac{m}{n}$ , with the subscript indicating an odd number of rolls. Then $P_{even}=1-\frac{m}{n}$
The ratio of $\frac{P_{odd}}{P_{even}}$ is just $\frac{P_{odd}}{1-P_{odd}}$
We see that $P_{odd}$ is the sum of $P_3$ $P_5$ $P_7$ ,... , while $P_{even}$ is the sum of $P_4$ $P_6$ $P_8$ ,... .
$P_3$ , the probability of getting rolls of 1-2-3 in exactly 3 rolls, is obviously $\frac{1}{216}$
We set this probability of $P_3$ aside, meaning we totally remove the chance of getting 1-2-3 in 3 rolls. Now the ratio of $P_4+P_6+P_8+...$ to $P_5+P_7+P_9+...$ should be equal to the ratio of $\frac{P_{odd}}{P_{even}}$ , because in this case the 1st roll no longer matters, so we can disregard the very existence of it in counting how many times of rolls, and thus, 4 rolls, 6 rolls, 8 rolls... would become an odd number of rolls (while 5 rolls, 7 rolls, 9 rolls... would become even number of rolls).
Notice $P_4+P_6+P_8+...=P_{even}$ , and also $P_5+P_7+P_9+...=P_{odd}-P_3=P_{odd}-\frac{1}{216}$
So we have $\frac{P_{even}}{P_{odd}-\frac{1}{216}}=\frac{P_{odd}}{P_{even}}$
Finally, we get $P_{odd}=\frac{m}{n}=\frac{216}{431}$ .
Therefore, $m+n = \boxed{647}$ | null | 647 |
28c3818bb0eebd518a05f951d51140d7 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Call the probability you win on a certain toss $f_n$ , where $n$ is the toss number.
Obviously, since the sequence has length 3, $f_1=0$ and $f_2=0$ .
Additionally, $f_3=\left(\frac{1}{6}\right)^3$ . We can call this value $x$ , to keep our further equations looking clean.
We can now write our general form for $f$ as $f_n=x\left(1-\sum_{i=1}^{n-3}f_i\right)$ . This factors the probability of the last 3 rolls being 1-2-3, and the important probability that the sequence has not been rolled in the past (because then the game would already be over).
Note that $\sum_{i=1}^{\infty}f_i=1$ since you'll win at some point.
An intermediate step here is figuring out $f_n-f_{n+1}$ . This is equal to $x\left(1-\sum_{i=1}^{n-3}f_i\right)-x\left(1-\sum_{i=1}^{n-2}f_i\right)=x\left(\sum_{i=1}^{n-2}f_i-\sum_{i=1}^{n-3}f_i\right)=xf_{n-2}$ .
Adding up all the differences, i.e. $\sum_{i=2}^{\infty}(f_{2n-1}-f_{2n})$ will give us the amount by which the odds probability exceeds the even probability. Since they sum to 1, that means the odds probability will be half of the difference above one-half. Subbing in our earlier result from the intermediate step, the odd probability is equal to $\frac{1}{2}+\frac{1}{2}x\sum_{i=2}^{\infty}f_{2n-3}$ .
Another way to find the odd probability is simply summing it up, which turns out to be $\sum_{i=1}^{\infty}f_{2n-1}$ . Note the infinite sums in both expressions are equal; let's call it $P$ . Equating them gives $\frac{1}{2}+\frac{1}{2}xP=P$ , or $P=\frac{1}{2-x}$ .
Finally, substituting $x=\frac{1}{216}$ , we find that $P=\frac{216}{431}$ , giving us a final answer of $216 + 431 = \boxed{647}$ | null | 647 |
28c3818bb0eebd518a05f951d51140d7 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $S(n)$ be the number of strings of length $n$ containing the digits $1$ through $6$ that do not contain the string $123$ . Then we have $S(n) = 6 \cdot S(n-1) - S(n-3)$ because we can add any digit to end of a string with length $n-1$ but we have to subtract all the strings that end in $123$ . We rewrite this as \begin{align*} S(n) &= 6 \cdot S(n-1) - S(n-3) \\ &= 6 \cdot (6 \cdot S(n-2) - S(n-4)) - (6 \cdot (S(n-4) - S(n-6)) \\ &= 36 \cdot S(n-2) - 12 \cdot S(n-4) + S(n-6) \end{align*} We wish to compute $P=\sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}}$ since the last three rolls are $123$ for the game to end. Summing over the recursion, we obtain \[\sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}} =36 \cdot \sum_{n=0}^\infty \frac{S(2n-2)}{6^{2n+3}} - 12 \cdot \sum_{n=0}^\infty \frac{S(2n-4)}{6^{2n+3}}+ \sum_{n=0}^\infty \frac{S(2n-6)}{6^{2n+3}}\] Now shift the indices so that the inside term is the same: \begin{align*} \sum_{n=3}^\infty \frac{S(2n)}{6^{2n+3}} &= \frac{36}{6^2} \cdot \sum_{n=2}^\infty \frac{S(2n)}{6^{2n+3}} - \frac{12}{6^4} \cdot \sum_{n=1}^\infty \frac{S(2n)}{6^{2n+3}} + \frac{1}{6^6} \cdot \sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}} \\ \left(P - \frac{S(0)}{6^3} - \frac{S(2)}{6^5} -\frac{S(4)}{6^7} \right) &= \frac{36}{6^2} \cdot \left( P - \frac{S(0)}{6^3} - \frac{S(2)}{6^5}\right) - \frac{12}{6^4} \cdot \left( P - \frac{S(0)}{6^3} \right) + \frac{1}{6^6} \cdot P \end{align*} Note that $S(0) = 1, S(2) = 36$ and $S(4) = 6^4 - 2 \cdot 6 = 1284$ . Therefore, \begin{align*} \left(P - \frac{1}{6^3} - \frac{36}{6^5} -\frac{1284}{6^7} \right) = \frac{36}{6^2} \cdot \left( P - \frac{1}{6^3} - \frac{36}{6^5}\right) - \frac{12}{6^4} \cdot \left( P - \frac{1}{6^3} \right) + \frac{1}{6^6} \cdot P \end{align*} Solving for $P$ , we obtain $P = \frac{216}{431} \implies m+n = \boxed{647}$ | null | 647 |
28c3818bb0eebd518a05f951d51140d7 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $A=\frac{1}{6} \begin{bmatrix} 5 & 1 & 0 & 0 \\ 4 & 1 & 1 & 0 \\ 4 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$ $A$ is a transition matrix for the prefix of 1-2-3 matched so far. The state corresponding to a complete match has no outgoing probability mass. The probability that we roll the dice exactly $k$ times is $(A^k)_{1,4}$ . Thus the probability that we roll the dice an odd number of times is $1-\left(\sum_{k=0}^\infty A^{2k}\right)_{1,4} = 1-\left((I - A^2)^{-1}\right)_{1,4} = \frac{216}{431}$ . Thus the answer is $216+431=\boxed{647}$ | null | 647 |
28c3818bb0eebd518a05f951d51140d7 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Consider it as a contest of Odd and Even. Let $P_o$ and $P_e$ be probability that Odd and Even wins, respectively. If we consider every 3 rolls as an atomic action, then we can have a simple solution. If the rolls is 1-2-3, Odd wins; otherwise, Odd and Even switch the odds of winning. Therefore, we have \[P_o = \frac{1}{216} + \frac{215}{216}P_e\] Plug in $P_e = 1 - P_o$ and we can easily solve for $Po=\frac{216}{431}$
$\boxed{647}$ | null | 647 |
28c3818bb0eebd518a05f951d51140d7 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $P_o$ $P_e$ be the winning probabilities respectively. We call Odd "in position" when a new sequence of 1-2-3 starts at odd position, and likewise, call Even is "in position" when a new sequence starts at even position.
Now, consider the situation when the first roll is $1.$ The conditional probability for Odd or Even to eventually win out depends on whose is in position. So let's denote by $P_o(1), P_e(1)$ the probabilities of Odd and Even winning out, respectively, both when Odd is in position. Remember that the probabilities simply switch if Even is in position. Similarly, after 1-2 is rolled, we denote by $P_o(2), P_e(2)$ the conditional probabilities of Odd and Even winning out, when Odd is in position.
Consider the first roll. If it's not a 1, the sequence restarts, but Even is now in position; if it's a 1, then Odd's winning probability becomes $P_o(1)$ . So, \[P_o = \frac{1}{6}P_o(1) + \frac{5}{6}P_e\] In the next roll, there are 3 outcomes. If the roll is 2, then Odd's winning probability becomes $P_o(2)$ ; if the roll is 1, then we stay in the sequence, but Even is now in position, so the probability of Odd winning now becomes $P_e(1)$ ; if the rolls is any other number, then the sequence restarts, and Odd is still in position. So, \[P_o(1) = \frac{1}{6}P_o(2) + \frac{1}{6}P_e(1) + \frac{4}{6}P_o\] In the next roll after a 1-2 sequence, there are 3 outcomes. If the roll is a 3, Odd wins; if it's a 1, we go back to the state when 1 is just rolled, and Odd is in position; if it's any other number, then the sequence restarts, and Even is in position. So, \[P_o(2) = \frac{1}{6} + \frac{1}{6}P_o(1) + \frac{4}{6}P_e\]
Plug in $P_e = 1-P_o$ and $P_e(1) = 1 - P_o(1)$ , we have a 3-equation linear system which is not hard to solve. The final answer is $Po=\frac{216}{431}$ $\boxed{647}$ . (We want P_o because if it starts odd, it will also end odd) | null | 647 |
28c3818bb0eebd518a05f951d51140d7 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_13 | Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Take a block of any three possible rolls. You have $6^3$ different possibilities for a block and $6^3 - 1$ different possibilities to have a block without 1-2-3. This gives two probabilities of $\frac{6^3-1}{6^3}$ and $\frac{1}{6^3}$ , respectively. To get an odd number of total rolls, we need an even number of blocks without a 1-2-3. Then, we can sum up the probabilities up to infinity to see the total probability of getting an even number of rolls. \[\frac{1}{6^3}\sum_{k=0}^{\infty}\left(\frac{6^3-1}{6^3}\right)^{2k}\] \[=\frac{\frac{1}{6^3}}{1-\left(\frac{6^3-1}{6^3}\right)^2}\] Using difference of squares on the bottom half we can reduce it to $\left(1-\frac{6^3-1}{6^3}\right)\left(1+\frac{6^3-1}{6^3}\right)=\frac{2\times 6^3 -1}{(6^3)^2}$ . Plugging this back in to the equation we get- \[\frac{\frac{1}{6^3}}{\frac{2\times 6^3 -1}{(6^3)^2}} = \frac{6^3}{2\times 6^3 - 1} = \frac{216}{431}\] Then, our answer is $\boxed{647}$ -Mathiscool109 | null | 647 |
0d01b82c26e82e434fa59fda8363e253 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_14 | The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$ . Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$ . Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$ . Assume that $AP = 3$ $PB = 4$ $AC = 8$ , and $AQ = \dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let the sides $\overline{AB}$ and $\overline{AC}$ be tangent to $\omega$ at $Z$ and $W$ , respectively. Let $\alpha = \angle BAX$ and $\beta = \angle AXC$ . Because $\overline{PQ}$ and $\overline{BC}$ are both tangent to $\omega$ and $\angle YXC$ and $\angle QYX$ subtend the same arc of $\omega$ , it follows that $\angle AYP = \angle QYX = \angle YXC = \beta$ . By equal tangents, $PZ = PY$ . Applying the Law of Sines to $\triangle APY$ yields \[\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.\] Similarly, applying the Law of Sines to $\triangle ABX$ gives \[\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.\] It follows that \[2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,\] implying $AZ = \tfrac{21}5$ . Applying the same argument to $\triangle AQY$ yields \[2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),\] from which $AQ = \tfrac{168}{59}$ . The requested sum is $168 + 59 = \boxed{227}$ | null | 227 |
0d01b82c26e82e434fa59fda8363e253 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_14 | The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$ . Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$ . Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$ . Assume that $AP = 3$ $PB = 4$ $AC = 8$ , and $AQ = \dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let the incircle of $ABC$ be tangent to $AB$ and $AC$ at $Z$ and $W$ . By Brianchon's theorem on tangential hexagons $QWCBZP$ and $PYQCXB$ , we know that $ZW,CP,BQ$ and $XY$ are concurrent at a point $O$ . Let $PQ \cap BC = M$ . Then by La Hire's $A$ lies on the polar of $M$ so $M$ lies on the polar of $A$ . Therefore, $ZW$ also passes through $M$ . Then projecting through $M$ , we have \[-1 = (A,O;Y,X) \stackrel{M}{=} (A,Z;P,B) \stackrel{M}{=} (A,W;Q,C).\] Therefore, $\frac{AP \cdot ZB}{MP \cdot AB} = 1 \implies \frac{3 \cdot ZB}{ZP \cdot 7} = 1$ . Since $ZB+ZP=4$ we know that $ZP = \frac{6}{5}$ and $ZB = \frac{14}{5}$ . Therefore, $AW = AZ = \frac{21}{5}$ and $WC = 8 - \frac{21}{5} = \frac{19}{5}$ . Since $(A,W;Q,C) = -1$ , we also have $\frac{AQ \cdot WC}{NQ \cdot AC} = 1 \implies \frac{AQ \cdot \tfrac{19}{5}}{(\tfrac{21}{5} - AQ) \cdot 8} = 1$ . Solving for $AQ$ , we obtain $AQ = \frac{168}{59} \implies m+n = \boxed{227}$ .
😃
-Vfire | null | 227 |
0d01b82c26e82e434fa59fda8363e253 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_14 | The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$ . Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$ . Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$ . Assume that $AP = 3$ $PB = 4$ $AC = 8$ , and $AQ = \dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let the center of the incircle of $\triangle ABC$ be $O$ . Link $OY$ and $OX$ . Then we have $\angle OYP=\angle OXB=90^{\circ}$
$\because$ $OY=OX$
$\therefore$ $\angle OYX=\angle OXY$
$\therefore$ $\angle PYX=\angle YXB$
$\therefore$ $\sin \angle PYX=\sin \angle YXB=\sin \angle YXC=\sin \angle PYA$
Let the incircle of $ABC$ be tangent to $AB$ and $AC$ at $M$ and $N$ , let $MP=YP=x$ and $NQ=YQ=y$
Use Law of Sine in $\triangle APY$ and $\triangle AXB$ , we have
$\frac{\sin \angle PAY}{PY}=\frac{\sin \angle PYA}{PA}$
$\frac{\sin \angle BAX}{BX}=\frac{\sin \angle AXB}{AB}$
therefore we have
$\frac{3}{x}=\frac{7}{4-x}$
Solve this equation, we have $x=\frac{6}{5}$
As a result, $MB=4-x=\frac{14}{5}=BX$ $AM=x+3=\frac{21}{5}=AN$ $NC=8-AN=\frac{19}{5}=XC$ $AQ=\frac{21}{5}-y$ $PQ=\frac{6}{5}+y$
So, $BC=\frac{14}{5}+\frac{19}{5}=\frac{33}{5}$
Use Law of Cosine in $\triangle BAC$ and $\triangle PAQ$ , we have
$\cos \angle BAC=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}$
$\cos \angle PAQ=\frac{AP^2+AQ^2-PQ^2}{2\cdot AP\cdot AQ}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}$
And we have
$\cos \angle BAC=\cos \angle PAQ$
So
$\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}$
Solve this equation, we have $y=\frac{399}{295}=QN$
As a result, $AQ=AN-QN=\frac{21}{5}-\frac{399}{295}=\frac{168}{59}$
So, the final answer of this question is $168+59=\boxed{227}$ | null | 227 |
4a2eb40a0f009f84dde0d2be362753d5 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_15 | Find the number of functions $f$ from $\{0, 1, 2, 3, 4, 5, 6\}$ to the integers such that $f(0) = 0$ $f(6) = 12$ , and \[|x - y| \leq |f(x) - f(y)| \leq 3|x - y|\] for all $x$ and $y$ in $\{0, 1, 2, 3, 4, 5, 6\}$ | Because $f(n)$ and $f(n+1)$ can differ by at most 3 for $n=0,1,2,3,4,5$ $f$ can decrease at most once. This is because if $f$ decreases $2$ times (or more) then we get a contradiction: $12 = f(6) \le 3\cdot 4 + (-1)\cdot 2 = 10$
If $f$ never decreases, then $f(n+1)-f(n)\in \{1,2,3\}$ for all $n$ . Let $a, b$ , and $c$ denote the number of times this difference is $1, 2$ , and $3$ , respectively. Then $a+b+c=6$ and $a+2b+3c = 12$ . Subtracting the first equation from the second yields $b+2c=6$ , so $(b,c)=(6,0), (4,1), (2,2)$ , or $(0,3)$ . These yield $a=0,1,2$ , or $3$ , respectively, so the number of possibilities in this case is \[\binom{6}{0,6,0}+\binom{6}{1,4,1}+\binom{6}{2,2,2}+\binom{6}{3,0,3}=141.\] If $f$ decreases from $f(0)$ to $f(1)$ or from $f(5)$ to $f(6)$ , then $f(2)$ or $f(4)$ , respectively, is determined. The only solutions to $a+b+c=4$ and $a+2b+3c=10$ are $(a,b,c)=(1,0,3)$ and $(a,b,c)=(0,2,2)$ , so the number of functions is \[2\left[\binom{4}{1,0,3}+\binom{4}{0,2,2}\right]=20.\] Finally, suppose that $f(n+1)<f(n)$ for some $n=1,2,3,4$ . Note that the condition $|(n+1)-(n-1)|\le |f(n+1)-f(n-1)|$ implies that $f(n+1)-f(n-1)\ge 2$ , so it must be that $f(n)-f(n+1)=1$ and \[f(n+2)-f(n+1)=f(n)-f(n-1)=3.\] This means that $f(n-1)$ and $f(n+2)$ are uniquely determined by the value of $f(n)$ , and, in particular, that $f(n+2)-f(n-1)=5$ . As a result, there are three more values of $f$ to determine, and they must provide a total increase of $7$ . The only ways to do this are either to have two differences of $3$ and one difference of $1$ , which can be arranged in $3$ ways, or to have one difference of $3$ and two differences of $2$ , which can be arranged in $3$ ways. Thus for each of the $4$ possibilities for $n$ , there are $6$ ways to arrange the increases, giving a total of $24$ ways.
The total number of functions is $141+20+24=\boxed{185}$ | null | 185 |
aca10aa4623333946fa1fd2141d19a18 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_1 | Fifteen distinct points are designated on $\triangle ABC$ : the 3 vertices $A$ $B$ , and $C$ $3$ other points on side $\overline{AB}$ $4$ other points on side $\overline{BC}$ ; and $5$ other points on side $\overline{CA}$ . Find the number of triangles with positive area whose vertices are among these $15$ points. | Every triangle is uniquely determined by 3 points. There are $\binom{15}{3}=455$ ways to choose 3 points, but that counts the degenerate triangles formed by choosing three points on a line. There are $\binom{5}{3}$ invalid cases on segment $AB$ $\binom{6}{3}$ invalid cases on segment $BC$ , and $\binom{7}{3}$ invalid cases on segment $CA$ for a total of $65$ invalid cases. The answer is thus $455-65=\boxed{390}$ | null | 390 |
85f23e58ee2d13aa19a4d5dbe17c4eef | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_2 | When each of $702$ $787$ , and $855$ is divided by the positive integer $m$ , the remainder is always the positive integer $r$ . When each of $412$ $722$ , and $815$ is divided by the positive integer $n$ , the remainder is always the positive integer $s \neq r$ . Find $m+n+r+s$ | Let's work on both parts of the problem separately. First, \[855 \equiv 787 \equiv 702 \equiv r \pmod{m}.\] We take the difference of $855$ and $787$ , and also of $787$ and $702$ . We find that they are $85$ and $68$ , respectively. Since the greatest common divisor of the two differences is $17$ (and the only one besides one), it's safe to assume that $m = 17$
Then, we divide $855$ by $17$ , and it's easy to see that $r = 5$ . Dividing $787$ and $702$ by $17$ also yields remainders of $5$ , which means our work up to here is correct.
Doing the same thing with $815$ $722$ , and $412$ , the differences between $815$ and $722$ and $412$ are $310$ and $93$ , respectively. Since the only common divisor (besides $1$ , of course) is $31$ $n = 31$ . Dividing all $3$ numbers by $31$ yields a remainder of $9$ for each, so $s = 9$ . Thus, $m + n + r + s = 17 + 5 + 31 + 9 = \boxed{062}$ | null | 062 |
85f23e58ee2d13aa19a4d5dbe17c4eef | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_2 | When each of $702$ $787$ , and $855$ is divided by the positive integer $m$ , the remainder is always the positive integer $r$ . When each of $412$ $722$ , and $815$ is divided by the positive integer $n$ , the remainder is always the positive integer $s \neq r$ . Find $m+n+r+s$ | We know that $702 = am + r, 787 = bm + r,$ and $855 = cm+r$ where $a-c$ are integers.
Subtracting the first two, the first and third, and the last two, we get $85 = (b-a)m, 153=(c-a)m,$ and $68=(c-b)m.$
We know that $b-a, c-a$ and $c-b$ must be integers, so all the numbers are divisible by $m.$
Factorizing the numbers, we get $85 = 5 \cdot 17, 153 = 3^2 \cdot 17,$ and $68 = 2^2 \cdot 17.$ We see that all these have a factor of 17, so $m=17.$
Finding the remainder when $702$ is divided by $17,$ we get $n=5.$
Doing the same thing for the next three numbers, we get $17 + 5 + 31 + 9 = \boxed{062}$ | null | 062 |
85f23e58ee2d13aa19a4d5dbe17c4eef | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_2 | When each of $702$ $787$ , and $855$ is divided by the positive integer $m$ , the remainder is always the positive integer $r$ . When each of $412$ $722$ , and $815$ is divided by the positive integer $n$ , the remainder is always the positive integer $s \neq r$ . Find $m+n+r+s$ | As in Solution 1, we are given $855\equiv787\equiv702\equiv r\pmod{m}$ and $815\equiv722\equiv412\equiv s\pmod{n}.$ Tackling the first equation, we can simply look at $855\equiv787\equiv702\pmod{m}$ . We subtract $702$ from each component of the congruency to get $153\equiv85\equiv0\pmod{m}$ . Thus, we know that $153$ and $85$ must both be divisible by $m$ . The only possible $m$ , in this case, become $17$ and $1$ ; obviously, $m\neq1$ , so we know $m=17$ . We go back to the original equation, plug in $m$ , and we find that $r=5$
Similarly, we can subtract out the smallest value in the second congruency, $412$ . We end up with $403\equiv310\equiv0\pmod{n}$ . Again, we find that $n=31$ or $1$ , so $n=31$ . We also find that $s=9$
Thus, our answer is $\boxed{062}$ | null | 062 |
0151a3f9a20a6bcebf44fc8dc8e4c171 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_3 | For a positive integer $n$ , let $d_n$ be the units digit of $1 + 2 + \dots + n$ . Find the remainder when \[\sum_{n=1}^{2017} d_n\] is divided by $1000$ | We see that $d_n$ appears in cycles of $20$ and the cycles are \[1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,\] adding a total of $70$ each cycle.
Since $\left\lfloor\frac{2017}{20}\right\rfloor=100$ , we know that by $2017$ , there have been $100$ cycles and $7000$ has been added. This can be discarded as we're just looking for the last three digits.
Adding up the first $17$ of the cycle of $20$ , we can see that the answer is $\boxed{069}$ | null | 069 |
0ec19092ec060873f0e1b2b9b3613ef0 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_4 | A pyramid has a triangular base with side lengths $20$ $20$ , and $24$ . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$ . The volume of the pyramid 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 the triangular base be $\triangle ABC$ , with $\overline {AB} = 24$ . We find that the altitude to side $\overline {AB}$ is $16$ , so the area of $\triangle ABC$ is $(24*16)/2 = 192$
Let the fourth vertex of the tetrahedron be $P$ , and let the midpoint of $\overline {AB}$ be $M$ . Since $P$ is equidistant from $A$ $B$ , and $C$ , the line through $P$ perpendicular to the plane of $\triangle ABC$ will pass through the circumcenter of $\triangle ABC$ , which we will call $O$ . Note that $O$ is equidistant from each of $A$ $B$ , and $C$ . Then,
\[\overline {OM} + \overline {OC} = \overline {CM} = 16\]
Let $\overline {OM} = d$ . Then $OC=OA=\sqrt{d^2+12^2}.$ Equation $(1)$ \[d + \sqrt {d^2 + 144} = 16\]
Squaring both sides, we have
\[d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256\]
\[2d^2 + 2d\sqrt {d^2+144} = 112\]
\[2d(d + \sqrt {d^2+144}) = 112\]
Substituting with equation $(1)$
\[2d(16) = 112\]
\[d = 7/2\]
We now find that $\sqrt{d^2 + 144} = 25/2$
Let the distance $\overline {OP} = h$ . Using the Pythagorean Theorem on triangle $AOP$ $BOP$ , or $COP$ (all three are congruent by SSS):
\[25^2 = h^2 + (25/2)^2\]
\[625 = h^2 + 625/4\]
\[1875/4 = h^2\]
\[25\sqrt {3} / 2 = h\]
Finally, by the formula for volume of a pyramid,
\[V = Bh/3\]
\[V = (192)(25\sqrt{3}/2)/3\] This simplifies to $V = 800\sqrt {3}$ , so $m+n = \boxed{803}$ | null | 803 |
0ec19092ec060873f0e1b2b9b3613ef0 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_4 | A pyramid has a triangular base with side lengths $20$ $20$ , and $24$ . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$ . The volume of the pyramid is $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ | We can place a three dimensional coordinate system on this pyramid. WLOG assume the vertex across from the line that has length $24$ is at the origin, or $(0, 0, 0)$ . Then, the two other vertices can be $(-12, -16, 0)$ and $(12, -16, 0)$ . Let the fourth vertex have coordinates of $(x, y, z)$ . We have the following $3$ equations from the distance formula.
\[x^2+y^2+z^2=625\]
\[(x+12)^2+(y+16)^2+z^2=625\]
\[(x-12)^2+(y+16)^2+z^2=625\]
Adding the last two equations and substituting in the first equation, we get that $y=-\frac{25}{2}$ . If you drew a good diagram, it should be obvious that $x=0$ . Now, solving for $z$ , we get that $z=\frac{25\sqrt{3}}{2}$ . So, the height of the pyramid is $\frac{25\sqrt{3}}{2}$ . The base is equal to the area of the triangle, which is $\frac{1}{2} \cdot 24 \cdot 16 = 192$ . The volume is $\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}$ . Thus, the answer is $800+3 = \boxed{803}$ | null | 803 |
0ec19092ec060873f0e1b2b9b3613ef0 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_4 | A pyramid has a triangular base with side lengths $20$ $20$ , and $24$ . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$ . The volume of the pyramid is $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ | Label the four vertices of the tetrahedron and the midpoint of $\overline {AB}$ , and notice that the area of the base of the tetrahedron, $\triangle ABC$ , equals $192$ , according to Solution 1.
Notice that the altitude of $\triangle CPM$ from $\overline {CM}$ to point $P$ is the height of the tetrahedron. Side $\overline {PM}$ is can be found using the Pythagorean Theorem on $\triangle APM$ , giving us $\overline {PM}=\sqrt{481}.$
Using Heron's Formula, the area of $\triangle CPM$ can be written as \[\sqrt{\frac{41+\sqrt{481}}{2}(\frac{41+\sqrt{481}}{2}-16)(\frac{41+\sqrt{481}}{2}-25)(\frac{41+\sqrt{481}}{2}-\sqrt{481})}\] \[=\frac{\sqrt{(41+\sqrt{481})(9+\sqrt{481})(-9+\sqrt{481})(41-\sqrt{481})}}{4}\]
Notice that both $(41+\sqrt{481})(41-\sqrt{481})$ and $(9+\sqrt{481})(-9+\sqrt{481})$ can be rewritten as differences of squares; thus, the expression can be written as \[\frac{\sqrt{(41^2-481)(481-9^2)}}{4}=\frac{\sqrt{480000}}{4}=100\sqrt{3}.\]
From this, we can determine the height of both $\triangle CPM$ and tetrahedron $ABCP$ to be $\frac{100\sqrt{3}}{8}$ ; therefore, the volume of the tetrahedron equals $\frac{100\sqrt{3}}{8} \cdot 192=800\sqrt{3}$ ; thus, $m+n=800+3=\boxed{803}.$ | null | 803 |
0ec19092ec060873f0e1b2b9b3613ef0 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_4 | A pyramid has a triangular base with side lengths $20$ $20$ , and $24$ . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$ . The volume of the pyramid is $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ | Notation is shown on diagram. \[AM = MB = c = 12, AC = BC = b = 20,\] \[DA = DB = DC = a = 25.\] \[CM = x + y = \sqrt{b^2-c^2} = 16,\] \[x^2 - y^2 = CD^2 – DM^2 = CD^2 – (BD^2 – BM^2) = c^2 = 144,\] \[x – y = \frac{x^2 – y^2}{x+y} = \frac {c^2} {16} = 9,\] \[x = \frac {16 + 9}{2} = \frac {a}{2},\] \[h = \sqrt{a^2 -\frac{ a^2}{4}} = a \frac {\sqrt{3}}{2},\] \[V = \frac{h\cdot CM \cdot c}{3}= \frac{16\cdot 25 \sqrt{3} \cdot 12}{3} = 800 \sqrt{3} \implies \boxed{803}.\] vladimir.shelomovskii@gmail.com, vvsss | null | 803 |
3fbb53572c9bd3a252b19c76a4968cde | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_5 | A rational number written in base eight is $\underline{ab} . \underline{cd}$ , where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$ . Find the base-ten number $\underline{abc}$ | First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base eight that have equal tens and ones digits in base 12.
$11_{12}=15_8$
$22_{12}=32_8$
$33_{12}=47_8$
$44_{12}=64_8$
$55_{12}=101_8$
We stop because we only can have two-digit numbers in base 8 and 101 is not a 2 digit number.
Compare the ones places to check if they are equal. We find that they are equal if $b=2$ or $b=4$ .
Evaluating the places to the right side of the decimal point gives us $22.23_{12}$ or $44.46_{12}$ .
When the numbers are converted into base 8, we get $32.14_8$ and $64.30_8$ . Since $d\neq0$ , the first value is correct. Compiling the necessary digits leaves us a final answer of $\boxed{321}$ | null | 321 |
3fbb53572c9bd3a252b19c76a4968cde | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_5 | A rational number written in base eight is $\underline{ab} . \underline{cd}$ , where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$ . Find the base-ten number $\underline{abc}$ | The parts before and after the decimal points must be equal. Therefore $8a + b = 12b + b$ and $c/8 + d/64 = b/12 + a/144$ . Simplifying the first equation gives $a = (3/2)b$ . Plugging this into the second equation gives $3b/32 = c/8 + d/64$ . Multiplying both sides by 64 gives $6b = 8c + d$ $a$ and $b$ are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using $a = 3/2b$ $(a,b) = (3,2)$ or $(6,4)$ . Testing these gives that $(6,4)$ doesn't work, and $(3,2)$ gives $a = 3, b = 2, c = 1$ , and $d = 4$ . Therefore $abc = \boxed{321}$ | null | 321 |
3fbb53572c9bd3a252b19c76a4968cde | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_5 | A rational number written in base eight is $\underline{ab} . \underline{cd}$ , where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$ . Find the base-ten number $\underline{abc}$ | Converting to base $10$ we get
$4604a+72c+9d=6960b$
Since $72c$ and $9d$ are much smaller than the other two terms, dividing by $100$ and approximating we get
$46a=70b$
Writing out the first few values of $a$ and $b$ , the first possible tuple is
$a=3, b=2, c=1, d=4$
and the second possible tuple is
$a=6, b=4, c=3, d=0$
Note that $d$ can not be $0$ , therefore the answer is $\boxed{321}$ | null | 321 |
3fbb53572c9bd3a252b19c76a4968cde | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_5 | A rational number written in base eight is $\underline{ab} . \underline{cd}$ , where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$ . Find the base-ten number $\underline{abc}$ | In the problem, we are given that \[8a+b+\dfrac c8+\dfrac d{64}=12b+b+\dfrac b{12}+\dfrac a{144}.\] We multiply by the LCM of the denominators, which is $576$ to get \[4608a+576b+72c+9d=6912b+576b+48b+4a.\] We then group like terms and factor to get \[4604a+72c+9d=48b(144+1)=145\cdot48b=24\cdot290b.\]
Observe that the coefficients of $a$ $c$ , and $b$ are all divisible by $4$ . Therefore, we know that $d$ must also be divisible by $4$ to compensate. (To observe this, one could rearrange the terms to see $9d=24\cdot290b-4604a-72c=4(6\cdot290b-1151a-18c)$ . At this point, it is obvious that $9d$ must be divisible by $4$ , so $d$ must be divisible by $4$ .) Thus, we let $d=4y$ to see $4604a+72c+9\cdot4y=24\cdot290b$ . We divide by $4$ in the equation to get \[1151a+18c+9y=6\cdot290b.\]
Observe that the coefficients of $c$ $y$ , and $b$ are all divisible by $3$ . By similar reasoning, $a$ must be divisible by $3$ , so we let $a=3x$ . Substituting and dividing by $3$ , we get \[1151x+6c+3y=2\cdot290b.\] We observe that this can no longer be reduced by similar means.
We know that $0<a,b,c<8\implies0<3x,b,4y<8$ . We examine $0<4y<8$ ; this becomes $0<y<2$ . It is apparent that $y=1\implies d=4$ . However, the problem does not even ask for $d$ , so it may appear that this find is meaningless. However, we substitute our value of $y$ in to get \[1151x+6c+3=580b.\]
We know that $x$ is either $1$ or $2$ , since $0<3x<8$ . We take our equation modulo $6$ to find \[5x+3\equiv4b\pmod{6}\implies3\equiv4b+x\pmod{6}.\] If $x=2$ , then $1\equiv4b\pmod{6}$ - this is equivalent to saying that $1+6n=4b\implies 1=4b-6n$ . However, $4b-6n$ is always even, and $1$ is odd. Thus, this case is not possible, so $x=1\implies a=3$ . We know that $x=1$ ; thus, $3\equiv4b+1\pmod{6}\implies2\equiv4b\pmod{6}\implies1\equiv2b\pmod{3}$ . Obviously, $b=2$ and $b=5$ work; however, if $b=5$ , then the RHS of our original equation ( $2\cdot290b$ ) is much too large to be equal to the LHS (the maximum possible value of the LHS is $1151\cdot1+6\cdot7+3\cdot1$ , which is less than $1200$ while the RHS would become $2\cdot290\cdot5=2900$ ), so we have $b=2$
We recall our equation of $1151x+6c+3=580b$ . Plugging in what we know, we have $1151\cdot1+6c+3=580\cdot2\implies6c=6\implies c=1$
Therefore, $\overline{abc}=\boxed{321}$ | null | 321 |
e254c54d86a87692e2ae5f672f669f86 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_6 | A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure $x$ . Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\frac{14}{25}$ . Find the difference between the largest and smallest possible values of $x$ | The probability that the chord doesn't intersect the triangle is $\frac{11}{25}$ . The only way this can happen is if the two points are chosen on the same arc between two of the triangle vertices. The probability that a point is chosen on one of the arcs opposite one of the base angles is $\frac{2x}{360}=\frac{x}{180}$ (this comes from the Central Angle Theorem, which states that the central angle from two points on a circle is always twice the inscribed angle from those two points), and the probability that a point is chosen on the arc between the two base angles is $\frac{180-2x}{180}$ . Therefore, we can write \[2\left(\frac{x}{180}\right)^2+\left(\frac{180-2x}{180}\right)^2=\frac{11}{25}\] This simplifies to \[x^2-120x+3024=0\]
(Note that the simplification is quite tedious)
Which factors as \[(x-84)(x-36)=0\] So $x=84, 36$ . The difference between these is $\boxed{048}$ | null | 048 |
e254c54d86a87692e2ae5f672f669f86 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_6 | A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure $x$ . Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\frac{14}{25}$ . Find the difference between the largest and smallest possible values of $x$ | Because we know that we have an isosceles triangle with angles of $x$ (and we know that x is an inscribed angle), that means that the arc that is intercepted by this angle is $2x$ . We form this same conclusion for the other angle $x$ , and $180-2x$ . Therefore we get $3$ arcs, namely, $2x$ $2x$ , and $360-4x$ . To have the chords intersect the triangle, we need the two points selected (to make a chord) to be on completely different arcs. An important idea to understand is that order matters in this case, so we have the equation $2$ $\frac{2x}{360}$ $\frac{2x}{360}$ $2$ $2$ $\frac{2x}{360}$ $\frac{360-4x}{360}$ $\frac{14}{25}$ which using trivial algebra gives you $x^2-120x+3024$ and factoring gives you $(x-84)(x-36)$ and so your answer is $\boxed{048}$ .
~jske25 | null | 048 |
e254c54d86a87692e2ae5f672f669f86 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_6 | A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure $x$ . Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\frac{14}{25}$ . Find the difference between the largest and smallest possible values of $x$ | After constructing the circumscribed circle, realize that the only time when the chord does not intersect the circle is when our $2$ points fall on only one arc formed by the sides of the triangle. Thus, lets call our isosceles triangle $ABC$ , where $AB=BC$ . Thus, the arcs formed by $BC$ and $AB$ can be called $a$ , and the arc formed by $AC$ is called $b$ . So, we can create the following system
\[2a+b=1\] \[2a^2+4ab=\frac{14}{25}\] \[2a^2+b^2=\frac{11}{25}\]
Notice we are denoting $a$ and $b$ as our probabilities, which we will be converting to degrees later. The 2 remaining systems can be calculated by using our rule about intersecting arcs and chords. So, after some hairy algebra we get:
\[a=\frac{1}{5}\] if \[b=\frac{3}{5}\] \[a=\frac{7}{15}\] if \[b=\frac{1}{15}\]
From here we find the absolute difference by doing $\frac{7}{15}-\frac{1}{5} = \frac{4}{15}$ . Converting to degrees, since the angles of a triangle add up to $180^o$ , we find that $\frac{4}{15} \cdot 180 =\boxed{048}$ , which is our answer. | null | 048 |
3ad06ae0abad68f43335d56d49600079 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_7 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ | Let $c=6-(a+b)$ , and note that $\binom{6}{a + b}=\binom{6}{c}$ . The problem thus asks for the sum $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$ . Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots by considering each column separately, which is equal to $\binom{18}{6}=18564$ . Therefore, the answer is $\boxed{564}$ | null | 564 |
3ad06ae0abad68f43335d56d49600079 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_7 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ | Alternatively, one can note that we can consider groups where $a+b$ is constant, say $c$ . Fix any value of $c$ . Then the sum of all of the values of $T(a,b)$ such that $a+b=c$ is $\binom{6}{a+b} \sum_{a+b=c} \binom{6}{a}\binom{6}{b}$ which by Vandermonde's is $\binom{6}{a+b}\binom{12}{a+b}$ . Remember, that expression is the resulting sum for a fixed $a+b$ . So, for $a+b\le 6$ , we want $\sum_{c=0}^{6} \binom{6}{c}\binom{12}{c}$ . This is (by Vandermonde's or committee forming) $\binom{18}{6} = 18564 \implies \boxed{564}$ ~ firebolt360 | null | 564 |
3ad06ae0abad68f43335d56d49600079 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_7 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ | Treating $a+b$ as $n$ , this problem asks for \[\sum_{n=0}^{6} \left[\binom{6}{n}\sum_{m=0}^{n}\left[\binom{6}{m}\binom{6}{n-m}\right]\right].\] But \[\sum_{m=0}^{n} \left[\binom{6}{m} \binom{6}{n-m}\right]\] can be computed through the following combinatorial argument. Choosing $n$ elements from a set of size $12$ is the same as splitting the set into two sets of size $6$ and choosing $m$ elements from one, $n-m$ from the other where $0\leq m\leq n$ . The number of ways to perform such a procedure is simply $\binom{12}{n}$ . Therefore, the requested sum is \[\sum_{n=0}^{6} \left[\binom{6}{n} \binom{12}{n}\right] = 18564.\] As such, our answer is $\boxed{564}$ | null | 564 |
3ad06ae0abad68f43335d56d49600079 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_7 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ | Case 1: $a<b$
Subcase 1: $a=0$ \[\binom{6}{0}\binom{6}{1}\binom{6}{1}=36\] \[\binom{6}{0}\binom{6}{2}\binom{6}{2}=225\] \[\binom{6}{0}\binom{6}{3}\binom{6}{3}=400\] \[\binom{6}{0}\binom{6}{4}\binom{6}{4}=225\] \[\binom{6}{0}\binom{6}{5}\binom{6}{5}=36\] \[\binom{6}{0}\binom{6}{6}\binom{6}{6}=1\] \[36+225+400+225+36+1=923\] Subcase 2: $a=1$ \[\binom{6}{1}\binom{6}{2}\binom{6}{3}=1800 \equiv 800 \pmod {1000}\] \[\binom{6}{1}\binom{6}{3}\binom{6}{4}=1800 \equiv 800 \pmod {1000}\] \[\binom{6}{1}\binom{6}{4}\binom{6}{5}=540\] \[\binom{6}{1}\binom{6}{5}\binom{6}{6}=36\] \[800+800+540+36=2176 \equiv 176 \pmod {1000}\] Subcase 3: $a=2$ \[\binom{6}{2}\binom{6}{3}\binom{6}{5}=1800\equiv800\pmod{1000}\] \[\binom{6}{2}\binom{6}{4}\binom{6}{6}=225\] \[800+225=1025\equiv25\pmod{1000}\]
\[923+176+25=1124\equiv124\pmod{1000}\]
Case 2: $b<a$
By just switching $a$ and $b$ in all of the above cases, we will get all of the cases such that $b>a$ is true. Therefore, this case is also $124\pmod{1000}$
Case 3: $a=b$ \[\binom{6}{0}\binom{6}{0}\binom{6}{0}=1\] \[\binom{6}{1}\binom{6}{1}\binom{6}{2}=540\] \[\binom{6}{2}\binom{6}{2}\binom{6}{4}=3375\equiv375\pmod{1000}\] \[\binom{6}{3}\binom{6}{3}\binom{6}{6}=400\] \[1+540+375+400=1316\equiv316\pmod{1000}\]
\[316+124+124=\boxed{564}\] | null | 564 |
3ad06ae0abad68f43335d56d49600079 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_7 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ | We begin as in solution 1 to rewrite the sum as $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$ .
Consider the polynomial $P(x)=\left(\binom{6}{0} + \binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \cdot \cdot \cdot + \binom{6}{6}x^6\right)^3$ .
We can see the sum we wish to compute is just the coefficient of the $x^6$ term. However $P(x)=((1+x)^6)^3=(1+x)^{18}$ . Therefore, the coefficient of the $x^6$ term is just $\binom{18}{6} = 18564$ so the answer is $\boxed{564}$ | null | 564 |
3ad06ae0abad68f43335d56d49600079 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_7 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ | Let $c=6-(a+b)$ . Then $\binom{6}{a+b}=\binom{6}{c}$ , and $a+b+c=6$ . The problem thus asks for \[\sum_{a+b+c=6}\binom{6}{a}\binom{6}{b}\binom{6}{c} \pmod {1000}.\] Suppose we have $6$ red balls, $6$ green balls, and $6$ blue balls lined up in a row, and we want to choose $6$ balls from this set of $18$ balls by considering each color separately. Over all possible selections of $6$ balls from this set, there are always a nonnegative number of balls in each color group. The answer is $\binom{18}{6} \pmod {1000}=18\boxed{564}$ | null | 564 |
3ad06ae0abad68f43335d56d49600079 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_7 | For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$ | Since $\binom{6}{n}=\binom{6}{6-n}$ , we can rewrite $T(a,b)$ as $\binom{6}{a}\binom{6}{b}\binom{6}{6-(a+b)}$ . Consider the number of ways to choose a committee of 6 people from a group of 6 democrats, 6 republicans, and 6 independents. We can first pick $a$ democrats, then pick $b$ republicans, provided that $a+b \leq 6$ . Then we can pick the remaining $6-(a+b)$ people from the independents. But this is just $T(a,b)$ , so the sum of all $T(a,b)$ is equal to the number of ways to choose this committee.
On the other hand, we can simply pick any 6 people from the $6+6+6=18$ total politicians in the group. Clearly, there are $\binom{18}{6}$ ways to do this. So the desired quantity is equal to $\binom{18}{6}$ . We can then compute (routinely) the last 3 digits of $\binom{18}{6}$ as $\boxed{564}$ | null | 564 |
f2ebcdf2a13127e59a552357903db8d9 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_8 | Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Noting that $\angle OQP$ and $\angle ORP$ are right angles, we realize that we can draw a semicircle with diameter $\overline{OP}$ and points $Q$ and $R$ on the semicircle. Since the radius of the semicircle is $100$ , if $\overline{QR} \leq 100$ , then $\overarc{QR}$ must be less than or equal to $60^{\circ}$
This simplifies the problem greatly. Since the degree measure of an angle on a circle is simply half the degree measure of its subtended arc, the problem is simply asking:
Given $a, b$ such that $0<a, b<75$ , what is the probability that $|a-b| \leq 30$ ?
Through simple geometric probability, we get that $P = \frac{16}{25}$
The answer is $16+25=\boxed{041}$ | null | 041 |
f2ebcdf2a13127e59a552357903db8d9 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_8 | Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Put $\triangle POQ$ and $\triangle POR$ with $O$ on the origin and the triangles on the $1^{st}$ quadrant.
The coordinates of $Q$ and $P$ is $(200 \cos^{2}a,200 \cos a\sin a )$ $(200\cos^{2}b,200\cos(b)\sin b)$ . So $PQ^{2}$ $(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2}$ , which we want to be less then $100^{2}$ .
So $(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2} <= 100^{2}$ \[(\cos^{2} a - \cos^{2} b)^{2} +(\cos a \sin a - \cos b \sin b)^{2} \le \frac{1}{4}\] \[\cos^{4} a + \cos^{4} b - 2\cos^{2} a \cos^{2} b +\cos^{2}a \sin^{2} a + \cos^{2} b \sin^{2} b - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}\] \[\cos^{2} a(\cos^{2} a + \sin^{2} a)+\cos^{2} b(\cos^{2} b+\sin^{2} b) - 2\cos^{2} a \cos^{2} b- 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}\] \[\cos^{2} a(1-\cos^{2} b)+\cos^{2} b(1-\cos^{2} a) - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}\] \[(\cos a\sin b)^{2} +(\cos b\sin a)^{2} - 2 (\cos a \sin b)(\cos b \sin a)\le \frac{1}{4}\] \[(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4}\] \[\sin^{2} (b-a) \le \frac{1}{4}\] So we want $-\frac{1}{2} \le \sin (b-a) \le \frac{1}{2}$ , which is equivalent to $-30 \le b-a \le 30$ or $150 \le b-a \le 210$ . The second inequality is impossible so we only consider what the first inequality does to our $75$ by $75$ box in the $ab$ plane. This cuts off two isosceles right triangles from opposite corners with side lengths $45$ from the $75$ by $75$ box. Hence the probability is $1-\frac{45^2}{75^2} = 1- \frac{9}{25}=\frac{16}{25}$ and the answer is $16+25 = \boxed{41}$ | null | 41 |
f2ebcdf2a13127e59a552357903db8d9 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_8 | Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | [asy] pair O, P, Q, R; draw(circle(O, 10)); O = (10, 0); P = (-10, 0); Q = (10*cos(pi/3), 10*sin(pi/3)); R = (10*cos(5*pi/6), 10*sin(5*pi/6)); dot(Q); dot(O); dot(P); dot(R); draw(P--O--Q--P--R--O); draw(Q--R, red); label("$O$", O, 2*E); label("$P$", P, 2*W); label("$Q$", Q, NE); label("$R$", R, NW); label("$200$", (0,0), 2*S); label("$x$", (Q+R)/2, N); draw(rightanglemark(O, Q, P, 38)); draw(rightanglemark(O, R, P, 38)); [/asy] Let $QR=x.$ Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: $OQ = 200 \cos a, PQ = 200 \sin a, PR = 200 \sin b, OR = 200 \cos b.$ Now observe that quadrilateral $OQRP$ is a cyclic quadrilateral . Thus, we are able to apply Ptolemy's Theorem to it: \[200 x + (200 \cos a) (200 \sin b) = (200 \sin a) (200 \cos b),\] \[x + 200 (\cos a \sin b) = 200 (\sin a \cos b),\] \[x = 200(\sin a \cos b - \sin b \cos a),\] \[x = 200 \sin(a-b).\] We want $|x| \le 100$ (the absolute value comes from the fact that $a$ is not necessarily greater than $b,$ so we cannot assume that $Q$ is to the right of $R$ as in the diagram), so we substitute: \[|200 \sin(a-b)| \le 100,\] \[|\sin(a-b)| \le \frac{1}{2},\] \[|a-b| \le 30 ^\circ,\] \[-30 \le a-b \le 30.\] By simple geometric probability (see Solution 2 for complete explanation), $\frac{m}{n} = 1 - \frac{2025}{5625} = 1 - \frac{9}{25} = \frac{16}{25},$ so $m+n = \boxed{041}.$ | null | 041 |
f2ebcdf2a13127e59a552357903db8d9 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_8 | Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Impose a coordinate system as follows:
Let the midpoint of $\overline{OP}$ be the origin, and let $\overline{OP}$ be the x-axis. We construct a circle with center at the origin with radius 100. Since $\angle OQP$ and $\angle ORP$ are both right angles, points $Q$ and $R$ are on our circle. Place $Q$ and $R$ in the first quadrant of the Cartesian Plane. Suppose we construct $Q'$ and $R'$ such that they are clockwise rotations of $Q$ and $R$ , respectively by an angle of $2b$ degrees. Thus, we see that $\overline{QR}=100\sqrt{2}\sqrt{\cos(2|a-b|)}$ . We want this quantity to be less than $100$ . This happens when $\cos(2|a-b|) \ge 1/2,$ or when $|a-b|\le 30^{\circ}$ . The probability that the last inequality is satisfied is $16/25$ . Therefore, the probability that $QR$ is less than $100$ is $16/25$ . Hence, $m+n=\boxed{41}$ | null | 41 |
f2ebcdf2a13127e59a552357903db8d9 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_8 | Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | WLOG, let $b\ge a$ . It does not actually matter, but it is necessary for this particular setup. It should be apparent that $\Delta RAQ\sim\Delta OAP$ . We write the equation
\[\dfrac{RA}{AO}=\dfrac{RQ}{OP}.\]
If we examine right triangle $\Delta ROA$ , we can see that $\sin(b-a)=\dfrac{RA}{AO}$ . Also, we are given $OP=200$ , so now we have
\[\sin(b-a)=\dfrac{QR}{200}.\]
We want $QR$ to be less than or equal to $100$ ; this is equivalent to $\dfrac{QR}{200}\le\dfrac12.$ We solve from there:
\begin{align*}
\dfrac{QR}{200}&\le\dfrac12 \\
\sin(b-a)&\le\dfrac12 \\
\arcsin(\sin(b-a))&\le\arcsin\left(\dfrac12\right) \\
b-a&\le30^\circ. \\
\end{align*}
(Notice that if $a>b$ , then this would become $a-b\le30^circ.$ As in Solution 1, we can write $|a-b|\le30$ .) One can now proceed as in Solution 1, but let us tackle the geometric probability for completeness.
We now have transformed this problem into another problem asking for the probability of two uniformly, randomly, and independently chosen real numbers between $0$ and $75$ being no more than $30$ from each other.
If the first number (let this be $x$ ) is between $30$ and $45$ , then the other number can be from $x-30$ to $x+30$ - a range of $60$ . Thus, the probability that this contributes is $\dfrac{45-30}{75}\cdot\dfrac{60}{75}=\dfrac4{25}$
If $x$ is between $0$ and $30$ or $45$ to $75$ (these two cases are equivalent), the chance is the same as that of the average value since the ranges are uniform. For $x=15$ (the average), the second number can be from $0$ to $15+30=45$ - a range of $45$ . The total range is $(30-0)+(75-45)=30+30=60$ . Thus, this case contributes $\dfrac{45}{75}\cdot\dfrac{60}{75}=\dfrac{12}{25}$
Adding the two, we get $\dfrac{16}{25}$ for an answer of $16+25=\boxed{041}$ | null | 041 |
3eafd85dc85bb5b34d71f2af315b8122 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_9 | Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$ | Writing out the recursive statement for $a_n, a_{n-1}, \dots, a_{10}$ and summing them gives \[a_n+\dots+a_{10}=100(a_{n-1}+\dots+a_{10})+n+\dots+10\] Which simplifies to \[a_n=99(a_{n-1}+\dots+a_{10})+\frac{1}{2}(n+10)(n-9)\] Therefore, $a_n$ is divisible by 99 if and only if $\frac{1}{2}(n+10)(n-9)$ is divisible by 99, so $(n+10)(n-9)$ needs to be divisible by 9 and 11. Assume that $n+10$ is a multiple of 11. Writing out a few terms, $n=12, 23, 34, 45$ , we see that $n=45$ is the smallest $n$ that works in this case. Next, assume that $n-9$ is a multiple of 11. Writing out a few terms, $n=20, 31, 42, 53$ , we see that $n=53$ is the smallest $n$ that works in this case. The smallest $n$ is $\boxed{045}$ | null | 045 |
3eafd85dc85bb5b34d71f2af315b8122 | https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_9 | Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$ | \[a_n \equiv a_{n-1} + n \pmod {99}\] By looking at the first few terms, we can see that \[a_n \equiv 10+11+12+ \dots + n \pmod {99}\] This implies \[a_n \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99}\] Since $a_n \equiv 0 \pmod {99}$ , we can rewrite the equivalence, and simplify \[0 \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99}\] \[0 \equiv n(n+1) - 90 \pmod {99}\] \[0 \equiv 4n^2+4n+36 \pmod {99}\] \[0 \equiv (2n+1)^2+35 \pmod {99}\] \[64 \equiv (2n+1)^2 \pmod {99}\] The only squares that are congruent to $64 \pmod {99}$ are $(\pm 8)^2$ and $(\pm 19)^2$ , so \[2n+1 \equiv -8, 8, 19, \text{or } {-19} \pmod {99}\] $2n+1 \equiv -8 \pmod {99}$ yields $n=45$ as the smallest integer solution.
$2n+1 \equiv 8 \pmod {99}$ yields $n=53$ as the smallest integer solution.
$2n+1 \equiv -19 \pmod {99}$ yields $n=89$ as the smallest integer solution.
$2n+1 \equiv 19 \pmod {99}$ yields $n=9$ as the smallest integer solution. However, $n$ must be greater than $10$
The smallest positive integer solution greater than $10$ is $n=\boxed{045}$ | null | 045 |