Difference between revisions of "1994 USAMO Problems/Problem 3"
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==Problem== | ==Problem== | ||
− | A convex hexagon <math>ABCDEF</math> is inscribed in a circle such that <math>AB=CD=EF</math> and diagonals <math>AD,BE</math>, and <math>CF</math> are concurrent. Let <math>P</math> be the intersection of <math>AD</math> and <math>CE</math>. Prove that <math>\frac{CP}{ | + | A convex hexagon <math>ABCDEF</math> is inscribed in a circle such that <math>AB=CD=EF</math> and diagonals <math>AD,BE</math>, and <math>CF</math> are concurrent. Let <math>P</math> be the intersection of <math>AD</math> and <math>CE</math>. Prove that <math>\frac{CP}{PE}=(\frac{AC}{CE})^2</math>. |
==Solution== | ==Solution== | ||
+ | |||
+ | Let the diagonals <math>AD</math>, <math>BE</math>, <math>CF</math> meet at <math>Q</math>. | ||
+ | |||
+ | First, let's show that the triangles <math>\triangle AEC</math> and <math>\triangle QED</math> are similar. | ||
+ | |||
+ | <center> | ||
+ | <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(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | |||
+ | </asy> | ||
+ | </center> | ||
+ | |||
+ | <math>\angle ACE=\angle ADE</math> because <math>A</math>,<math>C</math>,<math>D</math> and <math>E</math> all lie on the circle, and <math>\angle ADE=\angle QDE</math>. <math>\angle AEB=\angle CED</math> because <math>AB=CD</math>, and <math>A</math>,<math>B</math>,<math>C</math>,<math>D</math> and <math>E</math> all lie on the circle. Then, | ||
+ | |||
+ | <math>\angle AEB=\angle CED \rightarrow \angle AEB+\angle BEC=\angle CED+\angle BEC \rightarrow \angle AEC=\angle QED</math> | ||
+ | |||
+ | Therefore, <math>\triangle AEC</math> and <math>\triangle QED</math> are similar, so <math>AC/CE=QD/DE</math>. | ||
+ | |||
+ | Next, let's show that <math>\triangle AEC</math> and <math>\triangle CDQ</math> are similar. | ||
+ | |||
+ | <center> | ||
+ | <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(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | |||
+ | </asy> | ||
+ | </center> | ||
+ | |||
+ | <math>\angle AEC=\angle ADC</math> because <math>A</math>,<math>C</math>,<math>D</math> and <math>E</math> all lie on the circle, and <math>\angle ADC=\angle CDQ</math>. <math>\angle EAD=\angle ECD</math> because <math>A</math>,<math>C</math>,<math>D</math> and <math>E</math> all lie on the circle. <math>\angle DAC=\angle ECF</math> because <math>CD=EF</math>, and <math>A</math>,<math>C</math>,<math>D</math>,<math>E</math> and <math>F</math> all lie on the circle. Then, | ||
+ | |||
+ | <math>\angle EAC=\angle EAD+\angle DAC=\angle ECD+\angle ECF=\angle DCQ</math> | ||
+ | |||
+ | Therefore, <math>\triangle AEC</math> and <math>\triangle CDQ</math> are similar, so <math>AC/CE=CQ/QD</math>. | ||
+ | |||
+ | Lastly, let's show that <math>\triangle CPQ</math> and <math>\triangle EPD</math> are similar. | ||
+ | |||
+ | <center> | ||
+ | <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(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | |||
+ | </asy> | ||
+ | </center> | ||
+ | |||
+ | Because <math>CD=EF</math> and <math>C</math>,<math>D</math>,<math>E</math> and <math>F</math> all lie on the circle, <math>CF</math> is parallel to <math>DE</math>. So, <math>\triangle CPQ</math> and <math>\triangle EPD</math> are similar, and <math>CQ/DE=CP/PE</math>. | ||
+ | |||
+ | Putting it all together, <math>\frac{CP}{PE}=\frac{CQ}{DE}=\frac{AC}{CE}\cdot \frac{QD}{DE}=(\frac{AC}{CE})^2</math>. | ||
+ | |||
+ | Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol943.html | ||
+ | |||
+ | ==See Also== | ||
+ | {{USAMO box|year=1994|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 07:02, 19 July 2016
Problem
A convex hexagon is inscribed in a circle such that and diagonals , and are concurrent. Let be the intersection of and . Prove that .
Solution
Let the diagonals , , meet at .
First, let's show that the triangles and are similar.
because ,, and all lie on the circle, and . because , and ,,, and all lie on the circle. Then,
Therefore, and are similar, so .
Next, let's show that and are similar.
because ,, and all lie on the circle, and . because ,, and all lie on the circle. because , and ,,, and all lie on the circle. Then,
Therefore, and are similar, so .
Lastly, let's show that and are similar.
Because and ,, and all lie on the circle, is parallel to . So, and are similar, and .
Putting it all together, .
Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol943.html
See Also
1994 USAMO (Problems • Resources) | ||
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.