Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$

Solution: [asy] pair A,B,C,D,P; A=(-3,3); B=(3,3); C=(3,-3); D=(-3,-3); draw(A--B--C--D--cycle); label(A,"$A$",NW); label(B,"$B$",NE); label(C,"$C$",SE); label(D,"$D$",SW); draw(circle((0,0),4.24264068712)); P=(-1,4.12310562562); label(P,"$P$", NW); pen k=red+dashed; draw(P--A,k); draw(P--B,k); draw(P--C,k); draw(P--D,k); dot(P); [/asy] Let $P=(a,b)$ on the upper quarter of the circle, and let $k$ be the side length of the square. Hence, we want to find $k^2$ . Let the center of the circle be $(0,0)$ .
The two equations would thus become: \[\left(\left(a+\dfrac{k}2\right)^2+\left(b-\dfrac{k}2\right)^2\right)\left(\left(a-\dfrac{k}2\right)^2+\left(b+\dfrac{k}2\right)^2\right)=56^2\] \[\left(\left(a-\dfrac{k}2\right)^2+\left(b-\dfrac{k}2\right)^2\right)\left(\left(a+\dfrac{k}2\right)^2+\left(b+\dfrac{k}2\right)^2\right)=90^2\] Now, let $m=\left(a+\dfrac{k}2\right)^2$ $n=\left(a-\dfrac{k}2\right)^2$ $o=\left(b+\dfrac{k}2\right)^2$ , and $p=\left(b-\dfrac{k}2\right)^2$ . Our equations now change to $(m+p)(n+o)=56^2=mn+op+mo+pn$ and $(n+p)(m+o)=90^2=mn+op+no+pm$ . Subtracting the first from the second, we have $pm+no-mo-pn=p(m-n)-o(m-n)=(m-n)(p-o)=34\cdot146$ . Substituting back in and expanding, we have $2ak\cdot-2bk=34\cdot146$ , so $abk^2=-17\cdot73$ . We now have one of our terms we need ( $k^2$ ). Therefore, we only need to find $ab$ to find $k^2$ .
We now write the equation of the circle, which point $P$ satisfies: \[a^2+b^2=\left(\dfrac{k\sqrt{2}}{2}\right)^2=\dfrac{k^2}2\] We can expand the second equation, yielding \[\left(a^2+b^2+\dfrac{k^2}2+(ak+bk)\right)\left(a^2+b^2+\dfrac{k^2}2-(ak+bk)\right)=(k^2+k(a+b))(k^2-k(a+b))=8100.\] Now, with difference of squares, we get $k^4-k^2\cdot(a+b)^2=k^2(k^2-(a+b)^2)=8100$ . We can add $2abk^2=-17\cdot73\cdot2=-2482$ to this equation, which we can factor into $k^2(k^2-(a+b)^2+2ab)=k^2(k^2-(a^2+b^2))=8100-2482$ . We realize that $a^2+b^2$ is the same as the equation of the circle, so we plug its equation in: $k^2\left(k^2-\dfrac{k^2}2\right)=5618$ . We can combine like terms to get $k^2\cdot\dfrac{k^2}2=5618$ , so $(k^2)^2=11236$ .
Since the answer is an integer, we know $11236$ is a perfect square. Since it is even, it is divisible by $4$ , so we can factor $11236=2^2\cdot2809$ . With some testing with approximations and last-digit methods, we can find that $53^2=2809$ . Therefore, taking the square root, we find that $k^2$ , the area of square $ABCD$ , is $2\cdot53=\boxed{106}$

Answer:  106
Problem: Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$

Solution: WLOG, let $P$ be on minor arc $AD.$ Draw in $AP$ $BP$ $CP$ $DP$ and let $\angle ABP = x.$ We can see, by the inscribed angle theorem, that $\angle APB = \angle ACB = 45$ , and $\angle CPD = \angle CAD = 45.$ Then, $\angle PAB = 135-x$ $\angle PCD = \angle PAD = (135-x)-90 = 45-x$ , and $\angle PDC = 90+x.$ Letting $(PA, PB, PC, PD, AB) = (a,b,c,d,s)$ , we can use the law of sines on triangles $PAB$ and $PCD$ to get \[s\sqrt{2} = \frac{a}{\sin(x)} = \frac{b}{\sin(135-x)} = \frac{c}{\sin(90+x)} = \frac{d}{\sin(45-x)}.\] Making all the angles in the above equation acute gives \[s\sqrt{2} = \frac{a}{\sin(x)} = \frac{b}{\sin(45+x)} = \frac{c}{\sin(90-x)} = \frac{d}{\sin(45-x)}.\]
Note that we are looking for $s^{2}.$ We are given that $ac = 56$ and $bd = 90.$ This means that $s^{2}\sin(x)\sin(90-x) = 28$ and $s^{2}\sin(45+x)\sin(45-x) = 45.$ However, \[\sin(x)\sin(90-x) = \sin(x)\cos(x) = \frac{\sin(2x)}{2}\] and \[\sin(45+x)\sin(45-x) = \frac{(\cos(x) + \sin(x))(\cos(x) - \sin(x))}{2} = \frac{\cos^{2}(x) - \sin^{2}(x)}{2} = \frac{\cos(2x)}{2}.\] Therefore, $s^{2}\sin(2x) = 56$ and $s^{2}\cos(2x) = 90.$ Therefore, by the Pythagorean Identity, \[s^{2} = \sqrt{(s^{2}\sin(2x))^{2} + (s^{2}\cos(2x))^{2}} = \sqrt{56^{2} + 90^{2}} = \boxed{106}.\]

Answer:  106
Problem: Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$

Solution: Similar to Solution 6, let $P$ be on minor arc $\overarc {AB}$ $r$ and $O$ be the radius and center of the circumcircle respectively, and $\theta = \angle AOP$ . Since $\triangle APC$ is a right triangle, $PA \cdot PC$ equals the hypotenuse, $2r$ , times its altitude, which can be represented as $r \sin \theta$ . Therefore, $2r^2 \sin \theta = 56$ . Applying similar logic to $\triangle BPD$ , we get $2r^2 \sin (90^\circ - \theta) = 2r^2 \cos \theta = 90$
Dividing the two equations, we have \begin{align*} \frac{\sin \theta}{\cos \theta} &= \frac{56}{90} \\ 56 \cos \theta &= 90 \sin \theta \\ (56 \cos \theta)^2 &= (90 \sin \theta)^2. \end{align*} Adding $(56 \sin \theta)^2$ to both sides allows us to get rid of $\cos \theta$ \begin{align*} (56 \cos \theta)^2 + (56 \sin \theta)^2 &= (90 \sin \theta)^2 + (56 \sin \theta)^2 \\ 56^2 &= (90^2 + 56^2)(\sin \theta)^2 \\ \frac{56^2}{90^2 + 56^2} &= (\sin \theta)^2 \\ \frac{28}{53} &= \sin \theta. \end{align*} Therefore, we have $2r^2\left(\frac{28}{53}\right) = 56$ , and since the area of the square can be represented as $2r^2$ , the answer is $56 \cdot \frac{53}{28} = \boxed{106}$

Answer:  106
Problem: Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: At any point in the game, Alice should guess whichever color has come up less frequently thus far (although if both colors have come up equally often, she may guess whichever she likes); using this strategy, her probability of guessing correctly is at least $\frac{1}{2}$ on any given card, as desired.
There are ${6 \choose 3} = 20$ possible orderings of cards, all equally likely (since any of the $6! = 720$ permutations of the cards is equally likely, and each ordering covers $3!^2 = 6^2 = 36$ permutations).
Each of the $10$ orderings that start with red cards corresponds with one that starts with a black card; the problem is symmetrical with respect to red and black cards, so we can, without loss of generality, consider only the orderings that start with red cards.
We then generate a tally table showing whether Alice's guesses are correct for each ordering; for a given card, she guesses correctly if fewer than half the previously shown cards were the same color, guesses incorrectly if more than half were the same color, and guesses correctly with probability $\frac{1}{2}$ if exactly half were the same color.
In this table, $\mid$ denotes a correct guess, $\--$ denotes an incorrect guess, and $/$ denotes a guess with $\frac{1}{2}$ probability of being correct.
Now we sum the tallies across orderings, obtaining $41$ , and finally divide by the number of orderings ( $10$ ) to obtain the expected number of correct guesses, $\frac{41}{10}$ , which yields an answer of $41 + 10 = \boxed{051}.$

Answer:  051
Problem: Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: Denote by $N \left( a, b \right)$ the optimal expected number of cards that Alice guesses correctly, where the number of red and black cards are $a$ and $b$ , respectively.
Thus, for $a, b \geq 1$ , we have \begin{align*} N \left( a, b \right) & = \max \left\{ \frac{a}{a+b} \left( 1 + N \left( a - 1 , b \right) \right) + \frac{b}{a+b} N \left( a , b - 1 \right) , \right. \\ & \hspace{1cm} \left. \frac{a}{a+b} N \left( a - 1 , b \right) + \frac{b}{a+b} \left( 1 + N \left( a , b - 1 \right) \right) \right\} . \end{align*}
For $a = 0$ , Alice always guesses black. So $N \left( 0 , b \right) = b$
For $b = 0$ , Alice always guesses red. So $N \left( a , 0 \right) = a$
To solve this dynamic program, we can also exploit its symmetry that $N \left( a , b \right) = N \left( b , a \right)$
By solving this dynamic program, we get $N \left( 1, 1 \right) = \frac{3}{2}$ $N \left( 1, 2 \right) = \frac{7}{3}$ $N \left( 1 , 3 \right) = \frac{13}{4}$ $N \left( 2 , 2 \right) = \frac{17}{6}$ $N \left( 2 , 3 \right) = \frac{18}{5}$ $N \left( 3, 3 \right) = \frac{41}{10}$
Therefore, the answer is $41 + 10 = \boxed{051}$

Answer:  051
Problem: Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: Denote by $N_{i,j}$ the optimal expected number of cards that Alice guesses correctly, where the number of cards are $i$ and $j \ge i.$
If $i = 0$ then Alice guesses correctly all cards, so $N_{0,j} = j.$
If $j = i$ then Alice guesses next card with probability $\frac {1}{2} \implies N_{i,i} = \frac {1}{2} + N_{i-1,i}.$
If $j = i+1$ then Alice guesses next card with probability $\frac {i+1}{2i+1} \implies N_{i,i+1} = \frac {i+1}{2i+1} (1+ N_{i,i}) + \frac{i}{2i+1}  N_{i-1,i+1}.$
If $j = i+2$ then Alice guesses next card with probability $\frac {i+2}{2i+2} \implies N_{i,i+2} = \frac {i+2}{2i+2} (1+ N_{i,i+1}) + \frac{i}{2i+2}  N_{i-1,i+2}.$
One can find consistently: $N_{1,1} = \frac {1}{2} + N_{0,1} = \frac {3}{2},$ \[N_{1,2} = \frac {2}{3} (1 + N_{1,1}) + \frac {1}{3} N_{0,2} = \frac {7}{3}.\] \[N_{2,2} = \frac {1}{2} + N_{1,2} = \frac {17}{6}.\] \[N_{1,3} = \frac {3}{4} (1 + N_{1,2}) + \frac {1}{4} N_{0,3} = \frac {13}{4}.\] \[N_{2,3} = \frac {3}{5} (1 + N_{2,2}) + \frac {2}{5} N_{1,3} = \frac {18}{5}.\] \[N_{3,3} = \frac {1}{2} + N_{2,3} = \frac {41}{10}.\] Therefore, the answer is $41 + 10 = \boxed{051}$

Answer:  051
Problem: Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: Denote $E_n$ the expected number of cards Alice guesses correctly given $n$ red cards and $n$ black cards. We want to find $E_3$
Alice has a $\frac{1}{2}$ chance of guessing the first card. WLOG assume the first card color is red. For the next card, Alice has a $\frac{3}{5}$ chance of guessing the card if she chooses black; if they guess right, there's one less red and black card, so the expected number of cards Alice guesses from here is $E_2$ . If Alice does not guess correctly (which occurs with probability $\frac{2}{5}$ ), this means that there's 3 black cards and 1 red card left, so Alice should guess black next with a $\frac{3}{4}$ chance of being right. If Alice is wrong (with probability $\frac{1}{4}$ ), there are only 3 black cards left, so Alice can guess these with certainty; if Alice is right, there are 2 blacks and 1 red left, so Alice should again guess black. If Alice is right (with probability $\frac{2}{3}$ ), there is now 1 black and red card each, so the expected number of cards guessed is $E_1$ ; if she is wrong (with probability $\frac{1}{3}$ ), there are 2 black cards left, so Alice can guess these with certainty.
Summing this up into a formula: \[E_3 = \frac{1}{2} + \frac{3}{5} \left(1 + E_2 \right) + \frac{2}{5} \left( \frac{1}{4}(3) + \frac{3}{4}\left(1 + \frac{2}{3}\left(1 + E_1 \right) + \frac{1}{3}(2)\right) \right)\]
We can apply similar logic to compute $E_2$ and get \[E_2 = \frac{1}{2} + \frac{2}{3}(1 + E_1) + \frac{1}{3}(2)\]
To compute $E_1$ , we know that Alice can guess the last card with certainty, and there's a $\frac{1}{2}$ chance they get the first card as well, so $E_1 = \frac{3}{2}$
Thus, $E_2 = \frac{17}{6}$ , and after long computation, we get $E_3 = \frac{41}{10}$ . The requested answer is $41 + 10 = \boxed{51}$

Answer:  51
Problem: Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$

Solution: $n$ can either be $0$ or $1$ (mod $2$ ).
Case 1: $n \equiv 0 \pmod{2}$
Then, $n \equiv 2 \pmod{4}$ , which implies $n \equiv 1 \pmod{3}$ and $n \equiv 4 \pmod{6}$ , and therefore $n \equiv 3 \pmod{5}$ . Using CRT , we obtain $n \equiv 58 \pmod{60}$ , which gives $16$ values for $n$
Case 2: $n \equiv 1 \pmod{2}$
$n$ is then $3 \pmod{4}$ . If $n \equiv 0 \pmod{3}$ $n \equiv 3 \pmod{6}$ , a contradiction. Thus, $n \equiv 2 \pmod{3}$ , which implies $n \equiv 5 \pmod{6}$ $n$ can either be $0 \pmod{5}$ , which implies that $n \equiv 35 \pmod{60}$ by CRT, giving $17$ cases; or $4 \pmod{5}$ , which implies that $n \equiv 59 \pmod{60}$ by CRT, giving $16$ cases.
The total number of extra-distinct numbers is thus $16 + 16 + 17 = \boxed{049}$

Answer:  049
Problem: Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$

Solution: Because the LCM of all of the numbers we are dividing by is $60$ , we know that all of the remainders are $0$ again at $60$ , meaning that we have a cycle that repeats itself every $60$ numbers.
After listing all of the remainders up to $60$ , we find that $35$ $58$ , and $59$ are extra-distinct. So, we have $3$ numbers every $60$ which are extra-distinct. $60\cdot16 = 960$ and $3\cdot16 = 48$ , so we have $48$ extra-distinct numbers in the first $960$ numbers. Because of our pattern, we know that the numbers from $961$ thru $1000$ will have the same remainders as $1$ thru $40$ , so we have $1$ other extra-distinct number ( $35$ ).
$48 + 1 =  \boxed{049}$

Answer:  049
Problem: Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$

Solution: $\textbf{Case 0:  } {\rm Rem} \ \left( n, 6 \right) = 0$
We have ${\rm Rem} \ \left( n, 2 \right) = 0$ . This violates the condition that $n$ is extra-distinct.
Therefore, this case has no solution.
$\textbf{Case 1:  } {\rm Rem} \ \left( n, 6 \right) = 1$
We have ${\rm Rem} \ \left( n, 2 \right) = 1$ . This violates the condition that $n$ is extra-distinct.
Therefore, this case has no solution.
$\textbf{Case 2:  } {\rm Rem} \ \left( n, 6 \right) = 2$
We have ${\rm Rem} \ \left( n, 3 \right) = 2$ . This violates the condition that $n$ is extra-distinct.
Therefore, this case has no solution.
$\textbf{Case 3:  } {\rm Rem} \ \left( n, 6 \right) = 3$
The condition ${\rm Rem} \ \left( n, 6 \right) = 3$ implies ${\rm Rem} \ \left( n, 2 \right) = 1$ ${\rm Rem} \ \left( n, 3 \right) = 0$
Because $n$ is extra-distinct, ${\rm Rem} \ \left( n, l \right)$ for $l \in \left\{ 2, 3, 4 \right\}$ is a permutation of $\left\{ 0, 1 ,2 \right\}$ .
Thus, ${\rm Rem} \ \left( n, 4 \right) = 2$
However, ${\rm Rem} \ \left( n, 4 \right) = 2$ conflicts ${\rm Rem} \ \left( n, 2 \right) = 1$ .
Therefore, this case has no solution.
$\textbf{Case 4:  } {\rm Rem} \ \left( n, 6 \right) = 4$
The condition ${\rm Rem} \ \left( n, 6 \right) = 4$ implies ${\rm Rem} \ \left( n, 2 \right) = 0$ and ${\rm Rem} \ \left( n, 3 \right) = 1$
Because $n$ is extra-distinct, ${\rm Rem} \ \left( n, l \right)$ for $l \in \left\{ 2, 3, 4 , 5 \right\}$ is a permutation of $\left\{ 0, 1 ,2 , 3 \right\}$
Because ${\rm Rem} \ \left( n, 2 \right) = 0$ , we must have ${\rm Rem} \ \left( n, 4 \right) = 2$ . Hence, ${\rm Rem} \ \left( n, 5 \right) = 3$
Hence, $n \equiv -2 \pmod{{\rm lcm} \left( 4, 5 , 6 \right)}$ .
Hence, $n \equiv - 2 \pmod{60}$
We have $1000 = 60 \cdot 16 + 40$ .
Therefore, the number extra-distinct $n$ in this case is 16.
$\textbf{Case 5:  } {\rm Rem} \ \left( n, 6 \right) = 5$
The condition ${\rm Rem} \ \left( n, 6 \right) = 5$ implies ${\rm Rem} \ \left( n, 2 \right) = 1$ and ${\rm Rem} \ \left( n, 3 \right) = 2$
Because $n$ is extra-distinct, ${\rm Rem} \ \left( n, 4 \right)$ and ${\rm Rem} \ \left( n, 5 \right)$ are two distinct numbers in $\left\{ 0, 3, 4 \right\}$ .
Because ${\rm Rem} \ \left( n, 4 \right) \leq 3$ and $n$ is odd, we have ${\rm Rem} \ \left( n, 4 \right) = 3$ .
Hence, ${\rm Rem} \ \left( n, 5 \right) = 0$ or 4.
$\textbf{Case 5.1:  } {\rm Rem} \ \left( n, 6 \right) = 5$ ${\rm Rem} \ \left( n, 4 \right) = 3$ ${\rm Rem} \ \left( n, 5 \right) = 0$
We have $n \equiv 35 \pmod{60}$
We have $1000 = 60 \cdot 16 + 40$ .
Therefore, the number extra-distinct $n$ in this subcase is 17.
$\textbf{Case 5.2:  } {\rm Rem} \ \left( n, 6 \right) = 5$ ${\rm Rem} \ \left( n, 4 \right) = 3$ ${\rm Rem} \ \left( n, 5 \right) = 4$
$n \equiv - 1 \pmod{60}$
We have $1000 = 60 \cdot 16 + 40$ .
Therefore, the number extra-distinct $n$ in this subcase is 16.
Putting all cases together, the total number of extra-distinct $n$ is $16 + 17 + 16 = \boxed{049}$

Answer:  049
Problem: Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$

Solution: This solution refers to the Diagram section.
Define points $O,R,S,$ and $T$ as Solution 1 does. Moreover, let $H$ be the foot of the perpendicular from $P$ to $\overleftrightarrow{CD},$ $M$ be the foot of the perpendicular from $O$ to $\overleftrightarrow{HS},$ and $N$ be the foot of the perpendicular from $O$ to $\overleftrightarrow{RT}.$
We obtain the following diagram: [asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */  size(300); pair A, B, C, D, O, P, R, S, T, H, M, N; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); H = foot(S,C,D); M = foot(O,S,H); N = foot(O,R,T);  fill(O--M--P--cycle,yellow); fill(O--N--P--cycle,green); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C)^^rightanglemark(O,M,P)^^rightanglemark(O,N,P)^^rightanglemark(S,H,D),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T^^D--H^^O--M^^O--N^^O--P); draw(P--R^^P--S^^P--T^^P--H,red+dashed); dot("$A$",A,1.5*dir(225),linewidth(4.5)); dot("$B$",B,1.5*dir(-45),linewidth(4.5)); dot("$C$",C,1.5*dir(45),linewidth(4.5)); dot("$D$",D,1.5*dir(90),linewidth(4.5)); dot("$P$",P,1.5*dir(60),linewidth(4.5)); dot("$R$",R,1.5*dir(135),linewidth(4.5)); dot("$S$",S,1.5*dir(-90),linewidth(4.5)); dot("$T$",T,1.5*dir(-45),linewidth(4.5)); dot("$O$",O,1.5*dir(45),linewidth(4.5)); dot("$H$",H,1.5*dir(90),linewidth(4.5)); dot("$M$",M,1.5*dir(180),linewidth(4.5)); dot("$N$",N,1.5*dir(15),linewidth(4.5));  label("$9$",midpoint(P--R),dir(A-D),red); label("$5$",midpoint(P--S),dir(180),red); label("$16$",midpoint(P--T),dir(A-D),red); [/asy] Note that the diameter of $\odot O$ is $HS=RT=25,$ so $OP=\frac{25}{2}.$ It follows that:
Since $\overline{MO}\parallel\overline{AB}$ and $\overline{ON}\parallel\overline{DA},$ we conclude that $\angle A = \angle MON.$ We apply the Sine of a Sum Formula: \begin{align*} \sin\angle A &= \sin\angle MON \\ &= \sin(\angle MOP + \angle PON) \\ &= \sin\angle MOP \cos\angle PON + \cos\angle MOP \sin\angle PON \\ &= \frac{3}{5}\cdot\frac{24}{25} + \frac{4}{5}\cdot\frac{7}{25} \\ &= \frac{4}{5}. \end{align*} Note that \[\sin\angle A = \frac{HS}{DA},\] from which $\frac{4}{5} = \frac{25}{DA}.$ We solve this equation to get $DA=\frac{125}{4}.$
Finally, the perimeter of $ABCD$ is $4DA = \boxed{125}.$

Answer:  125
Problem: Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$

Solution: Label the points of the rhombus to be $X$ $Y$ $Z$ , and $W$ and the center of the incircle to be $O$ so that $9$ $5$ , and $16$ are the distances from point $P$ to side $ZW$ , side $WX$ , and $XY$ respectively. Through this, we know that the distance from the two pairs of opposite lines of rhombus $XYZW$ is $25$ and circle $O$ has radius $\frac{25}{2}$
Call the feet of the altitudes from $P$ to side $ZW$ , side $WX$ , and side $XY$ to be $A$ $B$ , and $C$ respectively. Additionally, call the feet of the altitudes from $O$ to side $ZW$ , side $WX$ , and side $XY$ to be $D$ $E$ , and $F$ respectively.
Draw a line segment from $P$ to $\overline{OD}$ so that it is perpendicular to $\overline{OD}$ . Notice that this segment length is equal to $AD$ and is $\sqrt{\left(\frac{25}{2}\right)^2-\left(\frac{7}{2}\right)^2}=12$ by Pythagorean Theorem.
Similarly, perform the same operations with perpendicular from $P$ to $\overline{OE}$ to get $BE=10$
By equal tangents, $WD=WE$ . Now, label the length of segment $WA=n$ and $WB=n+2$
Using Pythagorean Theorem again, we get
\begin{align*} WA^2+PA^2&=WB^2+PB^2 \\ n^2+9^2&=(n+2)^2+5^2 \\ n&=13. \end{align*}
Which also gives us $\tan{\angle{OWX}}=\frac{1}{2}$ and $OW=\frac{25\sqrt{5}}{2}$
Since the diagonals of the rhombus intersect at $O$ and are angle bisectors and are also perpendicular to each other, we can get that
\begin{align*} \frac{OX}{OW}&=\tan{\angle{OWX}} \\ OX&=\frac{25\sqrt{5}}{4} \\ WX^2&=OW^2+OX^2 \\ WX&=\frac{125}{4} \\ 4WX&=\boxed{125}

Answer:  125
Problem: Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$

Solution: Denote by $O$ the center of $ABCD$ .
We drop an altitude from $O$ to $AB$ that meets $AB$ at point $H$ .
We drop altitudes from $P$ to $AB$ and $AD$ that meet $AB$ and $AD$ at $E$ and $F$ , respectively.
We denote $\theta = \angle BAC$ .
We denote the side length of $ABCD$ as $d$
Because the distances from $P$ to $BC$ and $AD$ are $16$ and $9$ , respectively, and $BC \parallel AD$ , the distance between each pair of two parallel sides of $ABCD$ is $16 + 9 = 25$ .
Thus, $OH = \frac{25}{2}$ and $d \sin \theta = 25$
We have \begin{align*} \angle BOH & = 90^\circ - \angle HBO \\ & = 90^\circ - \angle HBD \\ & = 90^\circ - \frac{180^\circ - \angle C}{2} \\ & = 90^\circ - \frac{180^\circ - \theta}{2} \\ & = \frac{\theta}{2} . \end{align*}
Thus, $BH = OH \tan \angle BOH = \frac{25}{2} \tan \frac{\theta}{2}$
In $FAEP$ , we have $\overrightarrow{FA} + \overrightarrow{AE} + \overrightarrow{EP} + \overrightarrow{PF} = 0$ .
Thus, \[ AF + AE e^{i \left( \pi - \theta \right)} + EP e^{i \left( \frac{3 \pi}{2} - \theta \right)} - PF i . \]
Taking the imaginary part of this equation and plugging $EP = 5$ and $PF = 9$ into this equation, we get \[ AE = \frac{9 + 5 \cos \theta}{\sin \theta} . \]
We have \begin{align*} OP^2 & = \left( OH - EP \right)^2 + \left( AH - AE \right)^2 \\ & = \left( \frac{25}{2} - 5 \right)^2 + \left( d - \frac{25}{2} \tan \frac{\theta}{2} - \frac{9 + 5 \cos \theta}{\sin \theta} \right) \\ & = \left( \frac{15}{2} \right)^2 + \left( \frac{25}{\sin \theta} - \frac{25}{2} \tan \frac{\theta}{2} - \frac{9 + 5 \cos \theta}{\sin \theta} \right) . \hspace{1cm} (\bigstar) \end{align*}
Because $P$ is on the incircle of $ABCD$ $OP = \frac{25}{2}$ . Plugging this into $(\bigstar)$ , we get the following equation \[ 20 \sin \theta - 15 \cos \theta = 7 . \]
By solving this equation, we get $\sin \theta = \frac{4}{5}$ and $\cos \theta = \frac{3}{5}$ .
Therefore, $d = \frac{25}{\sin \theta} = \frac{125}{4}$
Therefore, the perimeter of $ABCD$ is $4d = \boxed{125}$

Answer:  125