1994 USAMO Problems/Problem 3

Revision as of 08:02, 19 July 2016 by 1=2 (talk | contribs) (Reconstructed from page template)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

See Also

1994 USAMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS