Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $x,y,$ and $z$ be real numbers satisfying the system \begin{align*} \log_2(xyz-3+\log_5 x)&=5,\\ \log_3(xyz-3+\log_5 y)&=4,\\ \log_4(xyz-3+\log_5 z)&=4.\\ \end{align*} Find the value of $|\log_5 x|+|\log_5 y|+|\log_5 z|$

Solution: First, we get rid of logs by taking powers: $xyz-3+\log_5 x=2^{5}=32$ $xyz-3+\log_5 y=3^{4}=81$ , and $(xyz-3+\log_5 z)=4^{4}=256$ . Adding all the equations up and using the $\log {xy}=\log {x}+\log{y}$ property, we have $3xyz+\log_5{xyz} = 378$ , so we have $xyz=125$ . Solving for $x,y,z$ by substituting $125$ for $xyz$ in each equation, we get $\log_5 x=-90, \log_5 y=-41, \log_5 z=134$ , so adding all the absolute values we have $90+41+134=\boxed{265}$

Answer:  265
Problem: An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$ faces of the box contains exactly $20$ green cubes, exactly $25$ yellow cubes, and some red cubes. Find the smallest possible volume of the box.

Solution: By counting the number of green cubes $2$ different ways, we have $12a=20b$ , or $a=\dfrac{5}{3} b$ . Notice that there are only $3$ possible colors for unit cubes, so for each of the $1 \times b \times c$ layers, there are $bc-21$ yellow cubes, and similarly there are $ac-45$ red cubes in each of the $1 \times a \times c$ layers. Therefore, we have $a(bc-21)=25b$ and $b(ac-45)=9a$ . We check a few small values of $a,b$ and solve for $c$ , checking $(a,b)=(5,3)$ gives $c=12$ with a volume of $180$ $(a,b)=(10,6)$ gives $c=6$ with a volume of $360$ , and $(a,b)=(15,9)$ gives $c=4$ , with a volume of $540$ . Any higher $(a,b)$ will $ab>180$ , so therefore, the minimum volume is $\boxed{180}$

Answer:  180
Problem: An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$ faces of the box contains exactly $20$ green cubes, exactly $25$ yellow cubes, and some red cubes. Find the smallest possible volume of the box.

Solution: The total number of green cubes is given by $12a=20b\Longrightarrow a=\frac{5}{3}b$
Let $r$ be the number of red cubes on each one of the $b$ layers then the total number of red cubes is $9a=br$ . Substitute $a=\frac{5}{3}b$ gives $r=15$
Repeating the procedure on the number of yellow cubes $y$ on each of the $a$ layers gives $y=15$
Therefore $bc=9+12+15=36$ and $ac=15+20+25=60$ . Multiplying yields $abc^2=2160$
Since $abc^2$ is fixed, $abc$ is minimized when $c$ is maximized, which occurs when $a$ $b$ are minimized (since each of $ac$ $bc$ is fixed). Thus $(a,b,c)=(3,5,12)\Longrightarrow abc=\boxed{180}$

Answer:  180
Problem: Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$

Solution: Do note that by counting the area in 2 ways, the first altitude is $x = \frac{ab}{c}$ . By similar triangles, the common ratio is $\rho = \frac{a}{c}$ for each height, so by the geometric series formula, we have \begin{align}     6p=\frac{x}{1-\rho} = \frac{ab}{c-a}. \end{align} Writing $p=a+b+c$ and clearing denominators, we get \[13a=6p .\] Thus $p=13q$ $a=6q$ , and $b+c=7q$ , i.e. $c=7q-b$ . Plugging these into $(1)$ , we get $78q(q-b)=6bq$ , i.e., $14b=13q$ . Thus $q=14r$ and $p=182r$ $b=13r$ $a=84r$ $c=85r$ . Taking $r=1$ (since $a,b,c$ are relatively prime) we get $p=\boxed{182}$

Answer:  182
Problem: Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$

Solution: Note that by counting the area in 2 ways, the first altitude is $\dfrac{ab}{c}$ . By similar triangles, the common ratio is $\dfrac{a}{c}$ for reach height, so by the geometric series formula, we have $6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}$ . Multiplying by the denominator and expanding, the equation becomes $\dfrac{ab}{c}=6a+6b+6c-\dfrac{6a^2}{c}-\dfrac{6ab}{c}-6a$ . Cancelling $6a$ and multiplying by $c$ yields $ab=6bc+6c^2-6a^2-6ab$ , so $7ab = 6bc+6b^2$ and $7a=6b+6c$ . Checking for Pythagorean triples gives $13,84,$ and $85$ , so $p=13+84+85=\boxed{182}$

Answer:  182
Problem: Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$

Solution: We start by splitting the sum of all $C_{n-2}C_{n-1}$ into two parts: those where $n-2$ is odd and those where $n-2$ is even.
First consider the sum of the lengths of the segments for which $n-2$ is odd for each $n\geq2$ . The perimeters of these triangles can be expressed using $p$ and ratios that result because of similar triangles. Considering triangles where $n-2$ is odd, we find that the perimeter for each such $n$ is $p\left(\frac{C_{n-1}C_{n}}{C_{0}B}\right)$ . Thus,
$p\sum_{n=1}^{\infty}\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B$
Simplifying,
$\sum_{n=1}^{\infty}C_{2n-1}C_{2n}=6C_{0}B + \frac{(C_{0}B)^2}{p}=C_{0}B\left(6+\frac{C_{0}B}{p}\right)$ .     (1)
Continuing with a similar process for the sum of the lengths of the segments for which $n-2$ is even,
$p\sum_{n=1}^{\infty}\frac{C_{2n-2}C_{2n-1}}{C_{0}B}=6p+C_{0}A+AB=7p-C_{0}B$
Simplifying,
$\sum_{n=1}^{\infty}C_{2n-2}C_{2n-1}=C_{0}B\left(7-\frac{C_{0}B}{p}\right)$ .     (2)
Adding (1) and (2) together, we find that
$6p=13C_{0}B \Rightarrow p=\frac{13C_{0}B}{6}=C_{0}B+C_{0}A+AB \Rightarrow \frac{7C_{0}B}{6}=C_{0}A+AB \Rightarrow 7C_{0}B=6C_{0}A + 6AB$
Setting $a=C_{0}B$ $b=C_{0}A$ , and $c=AB$ , we can now proceed as in Shaddoll's solution, and our answer is $p=13+84+85=\boxed{182}$

Answer:  182
Problem: Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$

Solution: [asy] size(10cm);  // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B);  // Labels label("$A$", A, SE); label("$B$", B, NW); label("$C_0$", C0, SW); label("$C_1$", C1, NE); label("$C_2$", C2, W); label("$C_3$", C3, NE); label("$C_4$", C4, W); label("$a$", (B+C0)/2, W); label("$b$", (A+C0)/2, S); label("$c$", (A+B)/2, NE);   // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2, red); draw(C2--C3--C4--C5--C6--C7--C8); draw(C0--C2, blue); [/asy]
Let $a = BC_0$ $b = AC_0$ , and $c = AB$ .
Note that the total length of the red segments in the figure above is equal to the length of the blue segment times $\frac{a+c}{b}$
The desired sum is equal to the total length of the infinite path $C_0 C_1 C_2 C_3 \cdots$ , shown in red in the figure below.
Since each of the triangles $\triangle C_0 C_1 C_2, \triangle C_2 C_3 C_4, \dots$ on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the blue segment times $\frac{a+c}{b}$ .
In other words, we have that $a\left(\frac{a+c}{b}\right) = 6p$
Guessing and checking Pythagorean triples reveals that $a = 84$ $b=13$ $c = 85$ , and $p = a + b + c = \boxed{182}$ satisfies this equation.

Answer:  182
Problem: Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$

Solution: This solution proceeds from $\frac{\frac{ab}{c}}{1-\frac{a}{c}} = \frac{\frac{ab}{c}}{\frac{c-a}{c}} = \frac{ab}{c-a} = 6(a+b+c)$ . Note the general from for a primitive pythagorean triple, $m^2-n^2, 2mn, m^2+n^2$ and after substitution, letting $a = m^2-n^2, b = 2mn, c = m^2+n^2$ into the previous equation simplifies down very nicely into $m = 13n$ . Thus $a = 168n^2, b = 26n^2, c = 170n^2$ . Since we know all three side lengths are relatively prime, we must divide each by 2 and let n = 1 giving $a = 84, b = 13, c = 85$ yielding $p = a + b + c = \boxed{182}$

Answer:  182
Problem: Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$

Solution: For this problem, first notice that its an infinite geometric series of $6(a+b+c)=\frac{ab}{c-b}$ if $c$ is the hypotenuse. WLOG $a<b$ , we can generalize a pythagorean triple of $x^2-y^2, 2xy, x^2+y^2$ . Let $b=2xy$ , then this generalization gives $6(a+b+c)(c-b)=ab$ \[(x^2-y^2)2xy=6(2x^2+2xy)(x-y)^2\] \[(x+y)xy=6(x^2+xy)(x-y)\] \[xy=6x(x-y)\] \[7xy=6x^2\] \[y=\frac{6}{7}x\]
Now this is just clear. Let $x=7m$ and $y=6m$ for $m$ to be a positive integer, the pythagorean triple is $13-84-85$ which yields $\boxed{182}$

Answer:  182
Problem: For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Note that all the coefficients of odd-powered terms is an odd number of odd degree terms multiplied together, and all coefficients of even-powered terms have an even number of odd degree terms multiplied together. Since every odd degree term is negative, and every even degree term is positive, the sum is just equal to $Q(-1)=P(-1)^{5}=\left( \dfrac{3}{2}\right)^{5}=\dfrac{243}{32}$ , so the desired answer is $243+32=\boxed{275}$

Answer:  275
Problem: For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: We are looking for the sum of the absolute values of the coefficients of $Q(x)$ . By defining $P'(x) = 1 + \frac{1}{3}x+\frac{1}{6}x^2$ , and defining $Q'(x) = P'(x)P'(x^3)P'(x^5)P'(x^7)P'(x^9)$ , we have made it so that all coefficients in $Q'(x)$ are just the positive/absolute values of the coefficients of $Q(x)$ . .
To find the sum of the absolute values of the coefficients of $Q(x)$ , we can just take the sum of the coefficients of $Q'(x)$ . This sum is equal to \[Q'(1) = P'(1)P'(1)P'(1)P'(1)P'(1) = \left(1+\frac{1}{3}+\frac{1}{6}\right)^5 = \frac{243}{32},\]
so our answer is $243+32 = \boxed{275}$

Answer:  275
Problem: For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Multiply $P(x)P(x^3)$ and notice that the odd degree terms have a negative coefficient. Observing that this is probably true for all polynomials like this (including $P(x)P(x^3)P(x^5)P(x^7)P(x^9)$ ), we plug in $-1$ to get $\frac{243}{32} \implies \boxed{275}$

Answer:  275
Problem: Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} || \overline{EF}$ . The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$ . Find the difference between the largest and smallest positive integer values for the area of $IJKL$

Solution: Letting $AI=a$ and $IB=b$ , we have \[IJ^{2}=a^{2}+b^{2} \geq 1008\] by AM-GM inequality . Also, since $EFGH||ABCD$ , the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and $2$ adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since \[2016=12^{2} \cdot 14\] we have the maximum area is \[2016 \cdot \dfrac{11}{12} = 1848\] (the areas of the squares from largest to smallest are $12^{2} \cdot 14, 11 \cdot 12 \cdot 14, 11^{2} \cdot 14$ forming a geometric progression).
The minimum area is $1008$ (every square is half the area of the square whose sides its vertices touch), so the desired answer is \[1848-1008=\boxed{840}\]

Answer:  840
Problem: Find the number of sets $\{a,b,c\}$ of three distinct positive integers with the property that the product of $a,b,$ and $c$ is equal to the product of $11,21,31,41,51,61$

Solution: Note that the prime factorization of the product is $3^{2}\cdot 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61$ . Ignoring overcounting, by stars and bars there are $6$ ways to choose how to distribute the factors of $3$ , and $3$ ways to distribute the factors of the other primes, so we have $3^{6} \cdot 6$ ways. However, some sets have $2$ numbers that are the same, namely the ones in the form $1,1,x$ and $3,3,x$ , which are each counted $3$ times, and each other set is counted $6$ times, so the desired answer is $\dfrac{729 \cdot 6-6}{6} = \boxed{728}$

Answer:  728
Problem: Find the number of sets $\{a,b,c\}$ of three distinct positive integers with the property that the product of $a,b,$ and $c$ is equal to the product of $11,21,31,41,51,61$

Solution: Again, notice that the prime factors of the product are $3, 3, 7, 11, 17, 31, 41, 61$ . In this problem, we are asked to partition this set of distinct(ish) factors into three smaller indistinct sets. To do this, we can use Stirling numbers of the second kind, but Stirling numbers of the second kind would assume no empty parts, which isn't what we want. However, this is easy to fix. Denote Stirling numbers of the second kind with $S(n,k)$ . We may start at the situation when all three sets are nonempty. Then, the number of partitions is simply $S(8,3)$ . However, we are overcounting, since every time the two threes are in different sets, (unless they are both individually in their own sets), each one is double counted. To fix this, we can see that they were in different sets $S(8, 3) - S(7,3)$ times by complementary counting, but one of these times they were in their own individual sets, so the total overcount is $\dfrac{S(8,3) - S(7,3) - 1}{2}$ . Similarly, we can do the cases for if two of the sets are nonempty and one of the sets are nonempty, (but in those last two cases the threes cannot be individually in their own sets). We then find that the "answer" is given by: \[S(8,3) - \dfrac{S(8,3) - S(7,3) - 1}{2} + S(8,2) - \dfrac{S(8,2) - S(7,2)}{2}+ S(8,1) - \dfrac{S(8,1) - S(7,1)}{2}\] \[= \dfrac{S(8,3)+S(8,2) + S(8,1) + S(7,3) + S(7,2) + S(7,1) + 1}{2}\] Drawing out the Stirling number triangle and evaluating yields $730$ for the last expression. However, throughout the entire solution, we ignored the fact that the three numbers $a$ $b$ , and $c$ needed to be distinct. However, this is easy to fix, since there are only two sets in which they are not, namely $(1,1,x)$ and $(3,3,x)$ . Thus, the actual answer is $730 - 2 = \boxed{728}$

Answer:  728
Problem: The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$ . There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$ . Find $c_k$

Solution: Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for $b_2$ . When we get to $b_2=9$ and $a_2=91$ , we have $a_4=271$ and $b_4=729$ , which works, therefore, the answer is $b_3+a_3=81+181=\boxed{262}$

Answer:  262
Problem: The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$ . There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$ . Find $c_k$

Solution: Using the same reasoning ( $100$ isn't very big), we can guess which terms will work. The first case is $k=3$ , so we assume the second and fourth terms of $c$ are $100$ and $1000$ . We let $r$ be the common ratio of the geometric sequence and write the arithmetic relationships in terms of $r$
The common difference is $100-r - 1$ , and so we can equate: $2(99-r)+100-r=1000-r^3$ . Moving all the terms to one side and the constants to the other yields
$r^3-3r = 702$ , or $r(r^2-3) = 702$ . Simply listing out the factors of $702$ shows that the only factor $3$ less than a square that works is $78$ . Thus $r=9$ and we solve from there to get $\boxed{262}$

Answer:  262
Problem: The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$ . There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$ . Find $c_k$

Solution: The reason for bashing in this context can also be justified by the fact 100 isn't very big.
Let the common difference for the arithmetic sequence be $a$ , and the common ratio for the geometric sequence be $b$ . The sequences are now $1, a+1, 2a+1, \ldots$ , and $1, b, b^2, \ldots$ . We can now write the given two equations as the following:
$1+(k-2)a+b^{k-2} = 100$
$1+ka+b^k = 1000$
Take the difference between the two equations to get $2a+(b^2-1)b^{k-2} = 900$ . Since 900 is divisible by 4, we can tell $a$ is even and $b$ is odd. Let $a=2m$ $b=2n+1$ , where $m$ and $n$ are positive integers. Substitute variables and divide by 4 to get:
$m+(n+1)(n)(2n+1)^{k-2} = 225$
Because very small integers for $n$ yield very big results, we can bash through all cases of $n$ . Here, we set an upper bound for $n$ by setting $k$ as 3. After trying values, we find that $n\leq 4$ , so $b=9, 7, 5, 3$ . Testing out $b=9$ yields the correct answer of $\boxed{262}$ . Note that even if this answer were associated with another b value like $b=3$ , the value of $k$ can still only be 3 for all of the cases.

Answer:  262
Problem: Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: [asy] import cse5; pathpen = black; pointpen = black; pointfontsize = 9; size(8cm);  pair A = origin, B = (13,0), P = (4,0), Q = (7,0), 	T = B + 5 dir(220), C = IP(circumcircle(A,B,T),Line(T,Q,-0.1,10)), 	S = IP(circumcircle(A,B,C),Line(C,P,-0.1,10));  Drawing(A--B--C--cycle); D(circumcircle(A,B,C),rgb(0,0.6,1)); DrawPathArray(C--S^^C--T,rgb(1,0.4,0.1)); DrawPathArray(A--S^^B--T,rgb(0,0.4,0)); D(S--T,rgb(1,0.2,0.4));  D("A",A,dir(215)); D("B",B,dir(330)); D("P",P,dir(240)); D("Q",Q,dir(240)); D("T",T,dir(290)); D("C",C,dir(120)); D("S",S,dir(250));  MP("4",(A+P)/2,dir(90)); MP("3",(P+Q)/2,dir(90)); MP("6",(Q+B)/2,dir(90)); MP("5",(B+T)/2,dir(140)); MP("7",(A+S)/2,dir(40)); [/asy] Let $\angle ACP=\alpha$ $\angle PCQ=\beta$ , and $\angle QCB=\gamma$ . Note that since $\triangle ACQ\sim\triangle TBQ$ we have $\tfrac{AC}{CQ}=\tfrac56$ , so by the Ratio Lemma \[\dfrac{AP}{PQ}=\dfrac{AC}{CQ}\cdot\dfrac{\sin\alpha}{\sin\beta}\quad\implies\quad \dfrac{\sin\alpha}{\sin\beta}=\dfrac{24}{15}.\] Similarly, we can deduce $\tfrac{PC}{CB}=\tfrac47$ and hence $\tfrac{\sin\beta}{\sin\gamma}=\tfrac{21}{24}$
Now Law of Sines on $\triangle ACS$ $\triangle SCT$ , and $\triangle TCB$ yields \[\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}=\dfrac{TB}{\sin\gamma}.\] Hence \[\dfrac{ST^2}{\sin^2\beta}=\dfrac{TB\cdot AS}{\sin\alpha\sin\gamma},\] so \[TS^2=TB\cdot AS\left(\dfrac{\sin\beta}{\sin\alpha}\dfrac{\sin\beta}{\sin\gamma}\right)=\dfrac{15\cdot 21}{24^2}\cdot 5\cdot 7=\dfrac{35^2}{8^2}.\] Hence $ST=\tfrac{35}8$ and the requested answer is $35+8=\boxed{43}$

Answer:  43
Problem: Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Projecting through $C$ we have \[\frac{3}{4}\times \frac{13}{6}=(A,Q;P,B)\stackrel{C}{=}(A,T;S,B)=\frac{ST}{7}\times \frac{13}{5}\] which easily gives $ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{043}$

Answer:  043
Problem: Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: By Ptolemy's Theorem applied to quadrilateral $ASTB$ , we find \[5\cdot 7+13\cdot ST=AT\cdot BS.\] Therefore, in order to find $ST$ , it suffices to find $AT\cdot BS$ . We do this using similar triangles, which can be found by using Power of a Point theorem.
As $\triangle APS\sim \triangle CPB$ , we find \[\frac{4}{PC}=\frac{7}{BC}.\] Therefore, $\frac{BC}{PC}=\frac{7}{4}$
As $\triangle BQT\sim\triangle CQA$ , we find \[\frac{6}{CQ}=\frac{5}{AC}.\] Therefore, $\frac{AC}{CQ}=\frac{5}{6}$
As $\triangle ATQ\sim\triangle CBQ$ , we find \[\frac{AT}{BC}=\frac{7}{CQ}.\] Therefore, $AT=\frac{7\cdot BC}{CQ}$
As $\triangle BPS\sim \triangle CPA$ , we find \[\frac{9}{PC}=\frac{BS}{AC}.\] Therefore, $BS=\frac{9\cdot AC}{PC}$ . Thus we find \[AT\cdot BS=\left(\frac{7\cdot BC}{CQ}\right)\left(\frac{9\cdot AC}{PC}\right).\] But now we can substitute in our previously found values for $\frac{BC}{PC}$ and $\frac{AC}{CQ}$ , finding \[AT\cdot BS=63\cdot \frac{7}{4}\cdot \frac{5}{6}=\frac{21\cdot 35}{8}.\] Substituting this into our original expression from Ptolemy's Theorem, we find \begin{align*}35+13ST&=\frac{21\cdot 35}{8}\\13ST&=\frac{13\cdot 35}{8}\\ST&=\frac{35}{8}.\end{align*} Thus the answer is $\boxed{43}$

Answer:  43