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== |
Revision as of 15:34, 30 May 2014
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, .