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2025 SSMO Team Round Problems/Problem 11

Problem

Squares $s_1,$ $s_2,$ and $s_3$ with side lengths $9,$ $17,$ and $10,$ respectively, lie inside of $\triangle ABC$ such that:

  • $s_1$ has a side that lies on $\overline{AB},$ $s_2$ has a side that lies on $\overline{BC},$ and $s_3$ has a side that lies on $\overline{CA}$;
  • each square shares exactly one vertex with each of the other two squares.

Find the perimeter of $\triangle ABC$.

Solution

[asy] unitsize(0.1cm); import geometry; pair A,B,C,V_1,W_1,V_2,W_2,V_3,W_3,S_1,S_2,S_3,Ap,Bp,Cp,P; A=(0,0); B=(17,0); C=dir(aSin(sqrt((36)/2*((36)/2-17)*((36)/2-9)*((36)/2-10))*2/17/9))*9; V_1=0.5*A+0.5*C; W_1=rotate(90,A)*V_1; V_2=0.5*C-0.5*B; W_2=rotate(-90,A)*V_2; V_3=0.5*A+0.5*B; W_3=rotate(-90,A)*V_3; S_1=V_1+W_1; S_2=V_2+W_2+B; S_3=V_3+W_3; Ap=intersectionpoint(line(2*V_1+2*W_1,2*W_1),line(2*V_2+2*W_2+B,2*W_2+B)); Bp=intersectionpoint(line(2*V_1+2*W_1,2*W_1),line(2*V_3+2*W_3,2*W_3)); Cp=intersectionpoint(line(2*V_3+2*W_3,2*W_3),line(2*V_2+2*W_2+B,2*W_2+B)); P=intersectionpoint(line(A,Bp),line(B,Cp)); draw(Ap--P,red); draw(Bp--P,red); draw(Cp--P,red); draw(Ap--Bp--Cp--cycle); fill(A--C--2*V_1+2*W_1--2*W_1--cycle,palegreen); draw(A--C--2*V_1+2*W_1--2*W_1--cycle); fill(B--C--2*V_2+2*W_2+B--2*W_2+B--cycle--cycle,palegreen); draw(B--C--2*V_2+2*W_2+B--2*W_2+B--cycle); fill(A--B--2*V_3+2*W_3--2*W_3--cycle,palegreen); draw(A--B--2*V_3+2*W_3--2*W_3--cycle); dot(Ap); dot(Bp); dot(Cp); label("$B$",Bp,dir(180)); label("$C$",Cp,dir(0)); label("$A$",Ap,dir(90)); label("$s_1$",V_1+W_1); label("$s_3$",V_2+W_2+B); label("$s_2$",V_3+W_3); [/asy]

[asy] unitsize(0.6cm); import geometry; pair A,B,C,V_1,W_1,V_2,W_2,V_3,W_3,S_1,S_2,S_3,Ap,Bp,Cp,P,X,Y,Z; A=(0,0); B=(17,0); C=dir(aSin(sqrt((36)/2*((36)/2-17)*((36)/2-9)*((36)/2-10))*2/17/9))*9; V_1=0.5*A+0.5*C; W_1=rotate(90,A)*V_1; V_2=0.5*C-0.5*B; W_2=rotate(-90,A)*V_2; V_3=0.5*A+0.5*B; W_3=rotate(-90,A)*V_3; S_1=V_1+W_1; S_2=V_2+W_2+B; S_3=V_3+W_3; Ap=intersectionpoint(line(2*V_1+2*W_1,2*W_1),line(2*V_2+2*W_2+B,2*W_2+B)); Bp=intersectionpoint(line(2*V_1+2*W_1,2*W_1),line(2*V_3+2*W_3,2*W_3)); Cp=intersectionpoint(line(2*V_3+2*W_3,2*W_3),line(2*V_2+2*W_2+B,2*W_2+B)); P=intersectionpoint(line(A,Bp),line(B,Cp)); fill(0.3*W_1--A--C--2*V_1+0.3*W_1--cycle,palegreen); draw(0.3*W_1--A--C--2*V_1+0.3*W_1); fill(0.3*W_2+B--B--C--2*V_2+0.3*W_2+B--cycle--cycle,palegreen); draw(0.3*W_2+B--B--C--2*V_2+0.3*W_2+B); fill(0.3*W_3--A--B--2*V_3+0.3*W_3--cycle,palegreen); draw(0.3*W_3--A--B--2*V_3+0.3*W_3); draw(P--0.95*A+0.05*Bp,p=red); draw(P--0.95*B+0.05*Cp,p=red); draw(P--0.95*C+0.05*Ap,p=red); label("$D$",C+(0.04,0.06),dir(20)); label("$E$",A+(0.03,0.09),dir(75)); label("$F$",B+(-0.03,-0.05),dir(-135)); label("$P$",P+(0.32,-0.45)); X=foot(P,A,C); Y=foot(P,A,B); Z=foot(P,B,C); label("$X$",X,dir(120)); label("$Y$",Y,dir(-90)); label("$Z$",Z+(0.1,-0.04),dir(35)); draw(P--X,dashed+blue); draw(P--Y,dashed+blue); draw(P--Z,dashed+blue); draw(rightanglemark(P,X,C)); draw(rightanglemark(P,Z,B)); draw(rightanglemark(P,Y,A)); dot(A); dot(B); dot(C); dot(P); dot(X); dot(Y); dot(Z); [/asy]

Let $D$ be the shared vertex of $s_1$ and $s_3$, let $E$ be the shared vertex of $s_1$ and $s_2$, and let $F$ be the shared vertex of $s_2$ and $s_3$. Since $AB\parallel ED$, $EF\parallel BC$, and $FD\parallel CA$, $\triangle ABC$ is the image of $\triangle DEF$ under a homothety with scale factor $k$. Therefore, lines $AD$, $BE$, and $CF$ are concurrent and meet at the center of homothety, which we call $P$.

Let $X$, $Y$, and $Z$ be the projections of $P$ onto $DE$, $EF$, and $FD$, respectively. By homothety, we have\[\frac{PX+9}{PX} = \frac{PY+17}{PY} = \frac{PZ+10}{PZ}=k.\]This implies that $PX:PY:PZ = 9:17:10$, so let $PX=9h$, $PY=17h$, and $PZ=10h$ for some positive real $h$. Note that $k=1+\tfrac{1}{h}$.

By Heron's formula, $[DEF]=36$. However, we also have \begin{align*} [DEF] &= [DPE]+[EPF]+[FPD] \\ &=\tfrac{1}{2}(9\cdot 9h +17\cdot 17h +10\cdot 10h) \\ &= 235h. \end{align*}Thus, $h = \tfrac{36}{235}$ and $k = 1+\tfrac{235}{36} = \tfrac{271}{36}$. Since the perimeter of $\triangle DEF$ is $36$, the perimeter of $\triangle ABC$ is $36k = \boxed{271}$.

~Sedro

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