2025 SSMO Speed Round Problems/Problem 9

Problem

Let $ABC$ be a triangle. The point $P$ lies on side $BC,$ the point $Q$ lies on side $AB,$ and the point $R$ lies on side $AC$ such that $PQ = BQ,$ $CR = PR,$ and $\angle APB < 90^\circ$ . Let $H$ be the foot of the altitude from $A$ to $BC$ . Given that $BP = 3,$ $CP = 5,$ and $[AQPR] = \tfrac{3}{7} \cdot [ABC],$ the value of $BH \cdot CH$ can be expressed in the form $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

$[asy] unitsize(1.3cm); pair A,B,C,P,Q,R,H,X,Y; A=(1.75,3.5); B=(0,0); C=(8,0); P=(3,0); X=(1.5,0); Y=(5.5,0); Q=(1.5,3); R=(5.5,1.4); H=(1.75,0); fill(A--Q--P--R--cycle,palegreen); draw(A--B--C--cycle); draw(Q--P--R); draw(Q--X,red+dashed); draw(R--Y,red+dashed); draw(A--H,dashed); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(300)); label("$X$",(1.4,0),dir(270)); label("$Y$",Y,dir(270)); label("$P$",P,dir(270)); label("$Q$",(1.4,2.9),dir(120)); label("$R$",(5.5,1.5),dir(90)); label("$H$",(1.85,0),dir(270)); draw(rightanglemark(A,H,P,4)); draw(rightanglemark(Q,X,B,4)); draw(rightanglemark(R,Y,P,4)); dot(A); dot(B); dot(C); dot(P); dot(Q); dot(R); dot(X); dot(Y); dot(H); [/asy]$

Let $X$ and $Y$ be the projections of $Q$ and $R$ onto $BC$ , respectively. Since $\triangle BQP$ and $\triangle CRP$ are both isosceles, $X$ and $Y$ are the midpoints of $BP$ and $CP$ , respectively. Let $BH = x$ , $CH = y$ , and $AH = h$ . Note that $x+y = BC = 8$ , and also note that the condition $\angle APB < 90^\circ$ implies $x=BH < BP = 3$ .

Next, observe that $\triangle QBX \sim \triangle ABH$ , and hence $\tfrac{QX}{BX} = \tfrac{AH}{BH}$ . From, this, we get $QX = \tfrac{AH\cdot BX}{BH} = \tfrac{3h}{2x}$ . Analogously, from the similarity $\triangle RCY \sim \triangle ACH$ , we get $RY = \tfrac{5h}{2y}$ . Thus, $[BQP] + [CRP] = \frac{9h}{4x} + \frac{25h}{4y}.$ Now, using the area condition in the problem statement, we have \begin{align*} \frac{9h}{4x} + \frac{25h}{4y} &= [BQP] + [CRP] \\ &= [ABC] - [AQPR] \\ &= \tfrac{4}{7}[ABC] \\ &= \frac{16h}{7}. \end{align*}Cancelling a factor of $h$ from the first and last equations in the above chain and substituting $y=8-x$ , we obtain $\frac{9}{4x} + \frac{25}{4(8-x)} =\frac{16}{7}.$ Solving this equation, we find $x = \tfrac{7}{4}, \tfrac{9}{2}$ . Since $x<3$ , we must have $x = \tfrac{7}{4}$ , which implies $y =\tfrac{25}{7}$ . Hence, $BH \cdot CH = xy = \tfrac{175}{16}$ , and we extract $175+16 = \boxed{191}$ .

~Sedro

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2025 SSMO Speed Round Problems/Problem 9

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