1994 USAMO Problems/Problem 3

Problem

A convex hexagon $ABCDEF$ is inscribed in a circle such that $AB=CD=EF$ and diagonals $AD,BE$ , and $CF$ are concurrent. Let $P$ be the intersection of $AD$ and $CE$ . Prove that $\frac{CP}{PE}=(\frac{AC}{CE})^2$ .

Solution

Let the diagonals $AD$ , $BE$ , $CF$ meet at $Q$ .

First, let's show that the triangles $\triangle AEC$ and $\triangle QED$ are similar.

$[asy] pair A,B,C,D,E,F,P,Q; A=(-0.96,0.28); B=(-0.352,0.936); C=(0,1); D=(4/5,3/5); E=(4/5,-3/5); F=(0,-1); P=IntersectionPoint(A--D,C--E); Q=IntersectionPoint(A--D,C--F); draw(A--B); draw(B--C); draw(C--D); draw(D--E,green); draw(E--F); draw(F--A); draw(A--C,red); draw(A--Q); draw(A--E,red); draw(B--Q); draw(C--E,red); draw(C--F); draw(Q--E,green); draw(Q--D,green); draw(circle((0,0),1)); label("$A$",A,W); label("$B$",B,N); label("$C$",C,N); label("$D$",D,NE); label("$E$",E,SE); label("$F$",F,S); label("$P$",(0.3,0.8),S); label("$Q$",(-0.15,0.4),S); [/asy]$

$\angle ACE=\angle ADE$ because $A$ , $C$ , $D$ and $E$ all lie on the circle, and $\angle ADE=\angle QDE$ . $\angle AEB=\angle CED$ because $AB=CD$ , and $A$ , $B$ , $C$ , $D$ and $E$ all lie on the circle. Then,

$\angle AEB=\angle CED \rightarrow \angle AEB+\angle BEC=\angle CED+\angle BEC \rightarrow \angle AEC=\angle QED$

Therefore, $\triangle AEC$ and $\triangle QED$ are similar, so $AC/CE=QD/DE$ .

Next, let's show that $\triangle AEC$ and $\triangle CDQ$ are similar.

$[asy] pair A,B,C,D,E,F,P,Q; A=(-0.96,0.28); B=(-0.352,0.936); C=(0,1); D=(4/5,3/5); E=(4/5,-3/5); F=(0,-1); P=IntersectionPoint(A--D,C--E); Q=IntersectionPoint(A--D,C--F); draw(A--B); draw(B--C); draw(C--D,green); draw(D--E); draw(E--F); draw(F--A); draw(A--C,red); draw(A--Q); draw(A--E,red); draw(B--E); draw(C--E,red); draw(Q--F); draw(Q--C,green); draw(Q--D,green); draw(circle((0,0),1)); label("$A$",A,W); label("$B$",B,N); label("$C$",C,N); label("$D$",D,NE); label("$E$",E,SE); label("$F$",F,S); label("$P$",(0.3,0.8),S); label("$Q$",(-0.15,0.4),S); [/asy]$

$\angle AEC=\angle ADC$ because $A$ , $C$ , $D$ and $E$ all lie on the circle, and $\angle ADC=\angle CDQ$ . $\angle EAD=\angle ECD$ because $A$ , $C$ , $D$ and $E$ all lie on the circle. $\angle DAC=\angle ECF$ because $CD=EF$ , and $A$ , $C$ , $D$ , $E$ and $F$ all lie on the circle. Then,

$\angle EAC=\angle EAD+\angle DAC=\angle ECD+\angle ECF=\angle DCQ$

Therefore, $\triangle AEC$ and $\triangle CDQ$ are similar, so $AC/CE=CQ/QD$ .

Lastly, let's show that $\triangle CPQ$ and $\triangle EPD$ are similar.

$[asy] pair A,B,C,D,E,F,P,Q; A=(-0.96,0.28); B=(-0.352,0.936); C=(0,1); D=(4/5,3/5); E=(4/5,-3/5); F=(0,-1); P=IntersectionPoint(A--D,C--E); Q=IntersectionPoint(A--D,C--F); draw(A--B); draw(B--C); draw(C--D); draw(D--E,green); draw(E--F); draw(F--A); draw(A--C); draw(A--Q); draw(A--E); draw(B--E); draw(P--E,green); draw(Q--F); draw(C--P,red); draw(Q--P,red); draw(C--Q,red); draw(D--P,green); draw(circle((0,0),1)); label("$A$",A,W); label("$B$",B,N); label("$C$",C,N); label("$D$",D,NE); label("$E$",E,SE); label("$F$",F,S); label("$P$",(0.3,0.8),S); label("$Q$",(-0.15,0.4),S); [/asy]$

Because $CD=EF$ and $C$ , $D$ , $E$ and $F$ all lie on the circle, $CF$ is parallel to $DE$ . So, $\triangle CPQ$ and $\triangle EPD$ are similar, and $CQ/DE=CP/PE$ .

Putting it all together, $\frac{CP}{PE}=\frac{CQ}{DE}=\frac{AC}{CE}\cdot \frac{QD}{DE}=(\frac{AC}{CE})^2$ .

Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol943.html

1994 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5
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1994 USAMO Problems/Problem 3

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