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2025 SSMO Accuracy Round Problems/Problem 10

Problem

Let $ABCDE$ be a convex pentagon with $\angle{BAC} = \angle{CAD} = \angle{DAE}$ and $\angle{ABC} = \angle{ACD} = \angle{ADE}$. Let $BD$ and $CE$ meet at $P$. Given that $BC = 6$, $\sin{\angle{BAC}} = \tfrac{3}{5}$, and $\tfrac{AC}{AB} = 5$, the length of $AP$ can be expressed as $\tfrac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers such that $n$ is square-free. Find $m+n$.

Solution

[asy] import geometry; unitsize(1cm); point A=(0,0); point B=dir(80); point C=2*dir(30); point D=4*dir(-20); point E=8*dir(-70); point P=intersectionpoints(circle(A,B,C), circle(A,D,E))[0]; point O1=circumcenter(A,B,C); point O2=circumcenter(A,D,E); draw(A--P,p=magenta); draw(O1--A--O2--cycle,p=heavygreen); draw(B--D,p=blue); draw(C--E,p=blue); draw(A--B--C--cycle); draw(A--D--C); draw(A--E--D); draw(circle(A,B,C),p=red+dotted); draw(circle(A,D,E),p=red+dotted); label("$A$",A,dir(140)); label("$B$",B,dir(120)); label("$C$",C,dir(50)); label("$D$",D,dir(0)); label("$E$",E,dir(-80)); label("$P$",P,dir(30)); label("$O_1$",O1+(0.1,-0.12),dir(45)); label("$O_2$",O2,dir(-135)); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); dot(O1); dot(O2); [/asy]

The main idea is to consider the transformation $\mathcal{T}$ that sends $\triangle ABC$ to $\triangle ACD$. By the given angle equalities, AA similarity gives us $\triangle ABC \sim \triangle ACD \sim \triangle ADE$. Since $\tfrac{AC}{AB} = 5$, it follows that $\mathcal{T}$ is composed of a rotation by $\angle BAC$ about $A$ followed by a dilation of scale factor $5$ centered at $A$.

We use $\mathcal{T}$ to prove that $P$ is the the second intersection point of $(ABC)$ and $(ADE)$. Notice that $\mathcal{T}$ sends $BD$ to $CE$, so $\angle BPC = \angle BAC$. This implies that $P$ lies on $(ABC)$. Since $\angle DPE = \angle BPC = \angle BAC = \angle DAE$, $P$ also lies on $(ADE)$, as desired.

Next, let $O_1$ and $O_2$ be the centers of $(ABC)$ and $(ADE)$, respectively. We seek the value of $\cos (\angle O_1AO_2)$. Notice that $\mathcal{T}^2$ sends $\triangle ABC$ to $\triangle ADE$, so $\mathcal{T}^2$ sends $O_1$ to $O_2$. (Here, $\mathcal{T}^2$ denotes $\mathcal{T}$ applied twice.) Therefore, $\angle O_1AO_2 = 2\angle BAC$. Then, \begin{align*} \cos (\angle O_1AO_2) &= \cos(2\angle BAC) \\ &= 1 - 2 \sin^2(\angle BAC) \\ &= \frac{7}{25}. \end{align*} By the extended Law of Sines, the radius of $(ABC)$ is $\tfrac{BC}{2\sin (\angle BAC)} = 5$, which means that the radius of $(ADE)$ is $25 \cdot 5 = 125$. By the Law of Cosines on $\triangle O_1AO_2$, we have \begin{align*} (O_1O_2)^2 &= (AO_1)^2 + (AO_2)^2 - 2(AO_1)(AO_2)\cos(\angle O_1AO_2) \\ &= 5^2 + 125^2 - 2(5)(125)(\tfrac{7}{25}) \\ &= 15300, \end{align*} so $O_1O_2 = 30\sqrt{17}$.

Finally, to find $AP$, we compute $[O_1AO_2]$. We have $\sin (\angle O_1AO_2) = \sqrt{1 - \cos^2(\angle O_1AO_2)} = \tfrac{24}{25}$, and by the sine area formula, we obtain $[ABC] = \tfrac{1}{2}(5)(125)(\tfrac{24}{25}) = 300$. Now, since $O_1O_2$ is the perpendicular bisector of $AP$, the length of the altitude from $A$ to $O_1O_2$ is $\tfrac{AP}{2}$. Thus, $[ABC] = \tfrac{1}{2}(30\sqrt{17})(\tfrac{AP}{2}) = 300$. Solving, we find $AP = \tfrac{40}{\sqrt{17}}$, and we extract $40+17 = \boxed{57}$.

~Sedro

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