Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$

Solution: (Complementary Counting with probability)
Notice that m can only equal 4, 5, or 6, and 4 and 6 are symmetric.
WLOG let $m=4$
1. There is a $\frac{15}{28}$ chance that exactly one of 1, 2, 3 is in the same row with 4.
There are 3 ways to select which of the smaller numbers will get in the row, and then 5
ways to select the number larger than 4.
$\frac{\dbinom{3}{1}\cdot\dbinom{5}{1}}{\dbinom{8}{2}} = \frac{15}{28}$
2. There is a $\frac{2}{5}$ chance that the other two smaller numbers end up in the same row.
There are 2 ways to select the row that the two smaller number are in, and then $\dbinom{3}{2}$ ways
to place the smaller numbers in the row.
$\frac{\dbinom{2}{1}\cdot\dbinom{3}{2}}{\dbinom{6}{2}} = \frac{2}{5}$
$9!(1-2*\frac{15}{28}*\frac{2}{5})=362880*\frac{4}{7}=207\boxed{360}$

Answer:  360
Problem: Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$

Solution: We will make sure to multiply by $3!$ in the end to account for all the possible permutation of the rows.
WLOG, let $5$ be present in the Row # $1$
Notice that $5$ MUST be placed with a number lower than it and a number higher than it.
This happens in $4\cdot4$ ways. You can permutate Row # $1$ in $3!$ ways.
Now, take a look at Row $2$ and Row $3$
Because there are $6$ numbers to choose from now, you can assign #'s to Row's #2&3 in
$\frac{\binom{6}{3}\cdot\binom{3}{3}}{2}$ ways. There are $3!\cdot3!$ ways to permute the numbers in the individual Rows.
Hence, our answer is $3!(4\cdot4\cdot3!\cdot{10}\cdot{3!}\cdot{3!})=3!(16\cdot60\cdot36)=3!(34560)\implies{207\boxed{360}$

Answer:  360
Problem: Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$

Solution: We see that if one of the medians is 5, then there are two remaining numbers greater than 5 and two less than 5, so it follows that $m=5$ . There are 3 ways to choose which row to have 5 in, $4\cdot 4=16$ ways to choose the other two numbers in that row, $3!=6$ ways to arrange the numbers in that row, and $6!=720$ ways for the remaining numbers, for our answer is $3\cdot 16\cdot 6\cdot 720=207\boxed{360}$ .
-Stormersyle

Answer:  360
Problem: Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$

Solution: We take the grid, and we do a bunch of stuff with it. First, we sort each row, smallest on the left, largest on the right, then we arrange these 3 rows such that the middle #s are increasing, from top to bottom. Thus, we get that the cell in the very center of the grid must be 5. (We need to multiply by 1296 at the end)
Let S mean a number that is < 5, L mean a number > 5.
Bobby.jpg
In the 2 blank corners, one can be S, and the other one has to be L.
If the top right corner is S, there are 16 ways, otherwise, there are 144 ways. (This is left as an exercise to the reader).
Thus, there are 160 * 1296 total configurations, which gets us an answer of $207\boxed{360}$

Answer:  360
Problem: Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$

Solution: We note that if $5$ is a median of one of the rows, then $m=5$ . First, focus on the row with $5$ in it. There are $4^2$ ways to choose the other numbers in that row and then $3!$ ways to order it. Now, clearly, there are $6!$ ways to put the other $6$ numbers into the remaining slots so $Q=6!\cdot3!\cdot4^2\cdot3=207360$ . Hence, our answer is $\boxed{360}$

Answer:  360
Problem: Call a set $S$ product-free if there do not exist $a, b, c \in S$ (not necessarily distinct) such that $a b = c$ . For example, the empty set and the set $\{16, 20\}$ are product-free, whereas the sets $\{4, 16\}$ and $\{2, 8, 16\}$ are not product-free. Find the number of product-free subsets of the set $\{1, 2, 3, 4,..., 7, 8, 9, 10\}$

Solution: We shall solve this problem by doing casework on the lowest element of the subset. Note that the number $1$ cannot be in the subset because $1*1=1$ . Let $S$ be a product-free set. If the lowest element of $S$ is $2$ , we consider the set $\{3, 6, 9\}$ . We see that 5 of these subsets can be a subset of $S$ $\{3\}$ $\{6\}$ $\{9\}$ $\{6, 9\}$ , and the empty set). Now consider the set $\{5, 10\}$ . We see that 3 of these subsets can be a subset of $S$ $\{5\}$ $\{10\}$ , and the empty set). Note that $4$ cannot be an element of $S$ , because $2$ is. Now consider the set $\{7, 8\}$ . All four of these subsets can be a subset of $S$ . So if the smallest element of $S$ is $2$ , there are $5*3*4=60$ possible such sets.
If the smallest element of $S$ is $3$ , the only restriction we have is that $9$ is not in $S$ . This leaves us $2^6=64$ such sets.
If the smallest element of $S$ is not $2$ or $3$ , then $S$ can be any subset of $\{4, 5, 6, 7, 8, 9, 10\}$ , including the empty set. This gives us $2^7=128$ such subsets.
So our answer is $60+64+128=\boxed{252}$

Answer:  252
Problem: Call a set $S$ product-free if there do not exist $a, b, c \in S$ (not necessarily distinct) such that $a b = c$ . For example, the empty set and the set $\{16, 20\}$ are product-free, whereas the sets $\{4, 16\}$ and $\{2, 8, 16\}$ are not product-free. Find the number of product-free subsets of the set $\{1, 2, 3, 4,..., 7, 8, 9, 10\}$

Solution: We will consider the $2^9 = 512$ subsets that do not contain 1. A subset is product-free if and only if it does not contain one of the groups $\{2, 4\}, \{3, 9\}, \{2, 3, 6\},$ or $\{2, 5, 10\}$ . There are $2^7$ subsets that contain 2 and 4 since each of the seven elements other than 2 and 4 can either be in the subset or not. Similarly, there are $2^7$ subsets that contain 3 and 9, $2^6$ subsets that contain 2, 3, and 6, and $2^6$ subsets that contain 2, 5, and 10. The number of sets that contain one of the groups is: \[2^7 + 2^7 + 2^6 + 2^6 = 384\] For sets that contain two of the groups, we have: \[2^5 + 2^5 + 2^5 + 2^5 + 2^4 + 2^4 = 160\] For sets that contain three of the groups, we have: \[2^4 + 2^3 + 2^3 + 2^3 = 40\] For sets that contain all of the groups, we have: \[2^2 = 4\] By the principle of inclusion and exclusion, the number of product-free subsets is \[512 - 384 + 160 - 40 + 4 = \boxed{252}\]

Answer:  252
Problem: Call a set $S$ product-free if there do not exist $a, b, c \in S$ (not necessarily distinct) such that $a b = c$ . For example, the empty set and the set $\{16, 20\}$ are product-free, whereas the sets $\{4, 16\}$ and $\{2, 8, 16\}$ are not product-free. Find the number of product-free subsets of the set $\{1, 2, 3, 4,..., 7, 8, 9, 10\}$

Solution: Let $X$ be a product-free subset, and note that 1 is not in $x$ . We consider four cases:
1.) both 2 and 3 are not in $X$ . Then there are $2^7=128$ possible subsets for this case.
2.) 2 is in $X$ , but 3 is not. Then 4 in not in $X$ , so there are $2^6=64$ subsets; however, there is a $\frac{1}{4}$ chance that 5 and 10 are both in $X$ , so there are $64\cdot \frac{3}{4}=48$ subsets for this case.
3.) 2 is not in $X$ , but 3 is. Then, 9 is not in $X$ , so there are $2^6=64$ subsets for this case.
4.) 2 and 3 are both in $X$ . Then, 4, 6, and 9 are not in $X$ , so there are $2^4=16$ total subsets; however, there is a $\frac{1}{4}$ chance that 5 and 10 are both in $X$ , so there are $16\cdot \frac{3}{4}=12$ subsets for this case.
Hence our answer is $128+48+64+12=\boxed{252}$ . -Stormersyle

Answer:  252
Problem: Call a set $S$ product-free if there do not exist $a, b, c \in S$ (not necessarily distinct) such that $a b = c$ . For example, the empty set and the set $\{16, 20\}$ are product-free, whereas the sets $\{4, 16\}$ and $\{2, 8, 16\}$ are not product-free. Find the number of product-free subsets of the set $\{1, 2, 3, 4,..., 7, 8, 9, 10\}$

Solution: Note that if any element $s\geq6$ makes an invalid set, it must be $c$ of $ab=c$ . Otherwise, $ab\geq12>10$ so no $ab=c$ will suffice. Therefore, whether or not an element $s\geq6$ depends only on the previous elements $1\leq\omega\leq5$
First, if a set is product-free, it must never contain an element $\omega=1$ or $1\cdot1=1$
We do casework on the subset of $S\in{1,2,3,4,5}=s_1$ to determine $S\in{6,7,8,9,10}=s_2$ (There are only 12 to cover so we just list them out):
When it is empty, there are clearly no restrictions for $s_2$ so there are $2^5=32$ cases.
$s_1={2}, s_1={4}, s_1={5}$ has no restrictions $s_2$ so we get $2^5\cdot3=96$ total cases.
$s_1=3$ only cannot contain 9 so there are $2^4=16$ cases.
$s_1={2, 5}; s_1={3, 4}; s_1={3, 5}$ each have one restriction in $s_2$ so there are $2^4\cdot3=48$ ways.
$s_1={2, 3}$ has 2 restrictions (6 and 9) so there are $2^3=8$ ways.
$s_1={4, 5}$ has no restrictions so there are $2^5=32$ ways.
$s_1={2, 3, 5}$ has 3 restrictions (6, 9, and 10) so there are $2^2=4$ ways.
$s_1={3, 4, 5}$ has 1 restriction so there are $2^4=16$ cases.
Summing the numbers, we get $32+96+16+48+8+32+4+16=\boxed{252}$

Answer:  252
Problem: For every $m \geq 2$ , let $Q(m)$ be the least positive integer with the following property: For every $n \geq Q(m)$ , there is always a perfect cube $k^3$ in the range $n < k^3 \leq m \cdot n$ . Find the remainder when \[\sum_{m = 2}^{2017} Q(m)\] is divided by 1000.

Solution: We claim that $Q(m) = 1$ when $m \ge 8$
When $m \ge 8$ , for every $n \ge Q(m) = 1$ , we need to prove there exists an integer $k$ , such that $n < k^3 \le m \cdot n$
Thats because $\sqrt[3]{m \cdot n} - \sqrt[3]{n} \ge 2\sqrt[3]{n} - \sqrt[3]{n} = \sqrt[3]{n} \ge 1$ , so k exists between $\sqrt[3]{m \cdot n}$ and $\sqrt[3]{n}$
$\sqrt[3]{n} < k \le \sqrt[3]{m \cdot n}$
We can then hand evaluate $Q(m)$ for $m = 2,3,4,5,6,7$ , and get $Q(2) = 32$ $Q(3) = 9$ , and all the others equal 2.
There are a total of 2010 integers from 8 to 2017.
\[\sum_{m = 2}^{2017} Q(m) \equiv \sum_{m = 2}^{7} Q(m) + 2010 \equiv 32+9+2+2+2+2+10 = \boxed{059} \mod 1000\]

Answer:  059
Problem: Let $a > 1$ and $x > 1$ satisfy $\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128$ and $\log_a(\log_a x) = 256$ . Find the remainder when $x$ is divided by $1000$

Solution: The first condition implies
\[a^{128} = \log_a\log_a 2 + \log_a 24 - 128\]
\[128+a^{128} = \log_a\log_a 2^{24}\]
\[a^{a^{128}a^{a^{128}}} = 2^{24}\]
\[\left(a^{a^{128}}\right)^{\left(a^{a^{128}}\right)} = 2^{24} = 8^8\]
So $a^{a^{128}} = 8$
Putting each side to the power of $128$
\[\left(a^{128}\right)^{\left(a^{128}\right)} = 8^{128} = 64^{64},\]
so $a^{128} = 64 \implies a = 2^{\frac{3}{64}}$ . Specifically,
\[\log_a(x) = \frac{\log_2(x)}{\log_2(a)} = \frac{64}{3}\log_2(x)\]
so we have that
\[256 = \log_a(\log_a(x)) = \frac{64}{3}\log_2\left(\frac{64}{3}\log_2(x)\right)\]
\[12 = \log_2\left(\frac{64}{3}\log_2(x)\right)\]
\[2^{12} = \frac{64}{3}\log_2(x)\]
\[192 = \log_2(x)\]
\[x = 2^{192}\]
We only wish to find $x\bmod 1000$ . To do this, we note that $x\equiv 0\bmod 8$ and now, by the Chinese Remainder Theorem, wish only to find $x\bmod 125$ . By Euler's Totient Theorem
\[2^{\phi(125)} = 2^{100} \equiv 1\bmod 125\]
so
\[2^{192} \equiv \frac{1}{2^8} \equiv \frac{1}{256} \equiv \frac{1}{6} \bmod 125\]
so we only need to find the inverse of $6 \bmod 125$ . It is easy to realize that $6\cdot 21 = 126 \equiv 1\bmod 125$ , so
\[x\equiv 21\bmod 125, x\equiv 0\bmod 8.\]
Using Chinese Remainder Theorem , we get that $x\equiv \boxed{896}\bmod 1000$ , finishing the solution.

Answer:  896
Problem: Let $a > 1$ and $x > 1$ satisfy $\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128$ and $\log_a(\log_a x) = 256$ . Find the remainder when $x$ is divided by $1000$

Solution: If you've found $x$ but you don't know that much number theory.
Note $192 = 3 * 2^6$ , so what we can do is take $2^3$ and keep squaring it (mod 1000).
\[2^3 = 8\] \[2^6 = 8*8 = 64\] \[2^{12} = 64*64 \equiv 96\bmod 1000\] \[2^{24} \equiv 96*96 \equiv 216\bmod 1000\] \[2^{48} \equiv 216*216 \equiv 656\bmod 1000\] \[2^{96} \equiv 656*656 \equiv 336\bmod 1000\] \[2^{192} \equiv 336*336 \equiv \boxed{896}\bmod 1000\]

Answer:  896
Problem: The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy]

Solution: Lemma: If $x,y$ satisfy $px+qy=1$ , then the minimal value of $\sqrt{x^2+y^2}$ is $\frac{1}{\sqrt{p^2+q^2}}$
Proof: Recall that the distance between the point $(x_0,y_0)$ and the line $px+qy+r = 0$ is given by $\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}$ . In particular, the distance between the origin and any point $(x,y)$ on the line $px+qy=1$ is at least $\frac{1}{\sqrt{p^2+q^2}}$
Let the vertices of the right triangle be $(0,0),(5,0),(0,2\sqrt{3}),$ and let $(a,0),(0,b)$ be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is $\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)$ . This point must lie on the hypotenuse $\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1$ , i.e. $a,b$ must satisfy \[\frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,\] which can be simplified to \[\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.\]
By the lemma, the minimal value of $\sqrt{a^2+b^2}$ is \[\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},\] so the minimal area of the equilateral triangle is \[\frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},\] and hence the answer is $75+3+67=\boxed{145}$

Answer:  145
Problem: The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy]

Solution: Let $AB=2\sqrt{3}, BC=5$ $D$ lies on $BC$ $F$ lies on $AB$ and $E$ lies on $AC$
Set $D$ as the origin, $BD=a,BF=b$ $F$ can be expressed as $-a+bi$ in argand plane, the distance of $CD$ is $5-a$
We know that $(-a+bi)\cdot\cos(-\frac{\pi}{3})=(-\frac{a+\sqrt{3}b}{2}+\frac{\sqrt{3}a+b}{2}i)$ . We know that the slope of $AC$ is $-\frac{2\sqrt{3}}{5}$ , we have that $\frac{5-a-(-\frac{a+\sqrt{3}b}{2})}{(\sqrt{3}a+b)/2}=\frac{5}{2\sqrt{3}}$ , after computation, we have $11b+7\sqrt{3}a=20\sqrt{3}$
Now the rest is easy with C-S inequality, $(a^2+b^2)(147+121)\geq (7\sqrt{3}a+11a)^2, a^2+b^2\geq \frac{300}{67}$ so the smallest area is $\frac{\sqrt{3}}{4}\cdot \frac{300}{67}=\frac{75\sqrt{3}}{67}$ , and the answer is $\boxed{145}$

Answer:  145
Problem: The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy]

Solution: Let $\triangle ABC$ be the right triangle with sides $AB = x$ $AC = y$ , and $BC = z$ and right angle at $A$
Let an equilateral triangle touch $AB$ $AC$ , and $BC$ at $D$ $E$ , and $F$ respectively, having side lengths of $c$
Now, call $AD$ as $a$ and $AE$ as $b$ . Thus, $DB = x-a$ and $EC = y-b$
By Law of Sines on triangles $\triangle DBF$ and $ECF$
$BF = \frac{z(a\sqrt{3}+b)} {2y}$ and $CF = \frac{z(a+b\sqrt{3})} {2x}$
Summing,
$BF+CF =  \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z$
Now substituting $AB = x = 2\sqrt{3}$ $AC = y = 5$ , and $BC = \sqrt{37}$ and solving, $\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1$
We seek to minimize $[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}$
This is equivalent to minimizing $a^2+b^2$
Using the lemma from solution 1, we conclude that $\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}$
Thus, $[DEF] = \frac{75\sqrt{3}}{67}$ and our final answer is $\boxed{145}$

Answer:  145
Problem: The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy]

Solution: We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are $5$ and $2\sqrt{3}i$ , respectively. Now let the vertex of the equilateral triangle on the real axis be $a$ and let the vertex of the equilateral triangle on the imaginary axis be $bi$ . Then, the third vertex of the equilateral triangle is given by: \[(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{b}{2})i\]
For this to be on the hypotenuse of the right triangle, we also have the following: \[\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}\]
Note that the area of the equilateral triangle is given by $\frac{\sqrt{3}(a^2+b^2)}{4}$ , so we seek to minimize $a^2+b^2$ . This can be done by using the Cauchy Schwarz Inequality on the relation we derived above: \[1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}\]
Thus, the minimum we seek is simply $\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}$ , so the desired answer is $\boxed{145}$

Answer:  145
Problem: The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy]

Solution: We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be $a$ and the point on the imaginary axis be $bi$ . Then, we see that $(a-bi)\left(\text{cis}\frac{\pi}{3}\right)+bi=(a-bi)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+bi=\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b\right)+i\left(\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right).$ Now we switch back to Cartesian coordinates. The equation of the hypotenuse is $y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.$ This means that the point $\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)$ is on the line. Plugging the numbers in, we have $\frac{\sqrt{3}}{2}a+\frac{1}{2}b=-\frac{\sqrt{3}}{5}a-\frac{3}{5}b+2\sqrt{3} \implies 7\sqrt{3}a+11b=20\sqrt{3}.$ Now, we note that the side length of the equilateral triangle is $a^2+b^2$ so it suffices to minimize that. By Cauchy-Schwarz, we have $(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.$ Thus, the area of the smallest triangle is $\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}$ so our desired answer is $\boxed{145}$

Answer:  145
Problem: The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$
[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy]

Solution: Let the right triangle's lower-left point be at $O(0,0)$ . Notice the 2 other points will determine a unique equilateral triangle. Let 2 points be on the $x$ -axis ( $B$ ) and the $y$ -axis ( $A$ ) and label them $(b, 0)$ and $(0, a)$ respectively.  The third point ( $C$ ) will then be located on the hypotenuse.  We proceed to find the third point's coordinates in terms of $a$ and $b$
1. Find the slope of $AB$ and take the negative reciprocal of it to find the slope of the line containing $C$ .  Notice the line contains the midpoint of $AB$ so we can then have an equation of the line.
2. Let $AB=x.$ For $ABC$ to be an equilateral triangle, the altitude from $C$ to $AB$ must be $\frac{x\sqrt{3}}{2}.$
We then have two equations and two variables, so we can solve for $C$ 's coordinates.
We can find $C(\frac{a+b\sqrt{3}}{2}), (\frac{b+a\sqrt{3}}{2}).$ Also, note that $C$ must be on the hypotenuse of the triangle $\frac{x}{5}+\frac{y}{2\sqrt{3}}=1.$ We can plug in $x$ and $y$ as the coordinates of $C$ ,  which simplifies to
\[11b+7\sqrt{3}a=20\sqrt{3}.\]
We aim to minimize the side length of the triangle, which is $\sqrt{a^2+b^2}.$ Applying the Cauchy inequality gives us
\[(a^2+b^2)(7\sqrt{3}^2+11^2)\geq (11b+7\sqrt3a)^2 = 1200\]
From which we obtain $\sqrt{a^2+b^2} \geq \sqrt{\frac{300}{67}}.$ Thus, the area of the triangle = $\frac{75\sqrt{3}}{67}$ which leads to the answer $75+3+67=\boxed{145}.$

Answer:  145