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Difference between revisions of "2025 SSMO Speed Round Problems/Problem 9"

(Created page with "==Problem== Let <math>ABC</math> be a triangle. The point <math>P</math> lies on side <math>BC,</math> the point <math>Q</math> lies on side <math>AB,</math> and the point <m...")
 
(→‎Solution)
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
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<asy>
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unitsize(1.3cm);
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pair A,B,C,P,Q,R,H,X,Y;
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A=(1.75,3.5);
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B=(0,0);
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C=(8,0);
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P=(3,0);
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X=(1.5,0);
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Y=(5.5,0);
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Q=(1.5,3);
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R=(5.5,1.4);
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H=(1.75,0);
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fill(A--Q--P--R--cycle,palegreen);
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draw(A--B--C--cycle);
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draw(Q--P--R);
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draw(Q--X,red+dashed);
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draw(R--Y,red+dashed);
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draw(A--H,dashed);
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label("$A$",A,dir(90));
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label("$B$",B,dir(225));
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label("$C$",C,dir(300));
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label("$X$",(1.4,0),dir(270));
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label("$Y$",Y,dir(270));
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label("$P$",P,dir(270));
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label("$Q$",(1.4,2.9),dir(120));
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label("$R$",(5.5,1.5),dir(90));
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label("$H$",(1.85,0),dir(270));
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draw(rightanglemark(A,H,P,4));
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draw(rightanglemark(Q,X,B,4));
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draw(rightanglemark(R,Y,P,4));
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dot(A);
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dot(B);
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dot(C);
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dot(P);
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dot(Q);
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dot(R);
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dot(X);
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dot(Y);
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dot(H);
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</asy>
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Let <math>X</math> and <math>Y</math> be the projections of <math>Q</math> and <math>R</math> onto <math>BC</math>, respectively. Since <math>\triangle BQP</math> and <math>\triangle CRP</math> are both isosceles, <math>X</math> and <math>Y</math> are the midpoints of <math>BP</math> and <math>CP</math>, respectively. Let <math>BH = x</math>, <math>CH = y</math>, and <math>AH = h</math>. Note that <math>x+y = BC = 8</math>, and also note that the condition <math>\angle APB < 90^\circ</math> implies <math>x=BH < BP = 3</math>.
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Next, observe that <math>\triangle QBX \sim \triangle ABH</math>, and hence <math>\tfrac{QX}{BX} = \tfrac{AH}{BH}</math>. From, this, we get <math>QX = \tfrac{AH\cdot BX}{BH} =  \tfrac{3h}{2x}</math>. Analogously, from the similarity <math>\triangle RCY \sim \triangle ACH</math>, we get <math>RY = \tfrac{5h}{2y}</math>. Thus,<cmath>[BQP] + [CRP] = \frac{9h}{4x} + \frac{25h}{4y}.</cmath>Now, using the area condition in the problem statement, we have
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\begin{align*}
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\frac{9h}{4x} + \frac{25h}{4y} &= [BQP] + [CRP] \\
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&= [ABC] - [AQPR] \\
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&= \tfrac{4}{7}[ABC] \\
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&= \frac{16h}{7}.
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\end{align*}Cancelling a factor of <math>h</math> from the first and last equations in the above chain and substituting <math>y=8-x</math>, we obtain<cmath>\frac{9}{4x} + \frac{25}{4(8-x)}  =\frac{16}{7}.</cmath>Solving this equation, we find <math>x = \tfrac{7}{4}, \tfrac{9}{2}</math>. Since <math>x<3</math>, we must have <math>x = \tfrac{7}{4}</math>, which implies <math>y =\tfrac{25}{7}</math>. Hence, <math>BH \cdot CH = xy = \tfrac{175}{16}</math>, and we extract <math>175+16 = \boxed{191}</math>.
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 +
~Sedro

Latest revision as of 22:45, 10 September 2025

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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