# Difference between revisions of "2011 USAJMO Problems/Problem 5"

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+ | Points <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math> lie on a circle <math>\omega</math> and point <math>P</math> lies outside the circle. The given points are such that (i) lines <math>PB</math> and <math>PD</math> are tangent to <math>\omega</math>, (ii) <math>P</math>, <math>A</math>, <math>C</math> are collinear, and (iii) <math>\overline{DE} \parallel \overline{AC}</math>. Prove that <math>\overline{BE}</math> bisects <math>\overline{AC}</math>. | ||

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+ | == Solution == | ||

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Let <math>O</math> be the center of the circle, and let <math>M</math> be the midpoint of <math>AC</math>. Let <math>\omega</math> denote the circle with diameter <math>OP</math>. Since <math>\angle OBP = \angle OMP = \angle ODP = 90^\circ</math>, <math>B</math>, <math>D</math>, and <math>M</math> all lie on <math>\omega</math>. | Let <math>O</math> be the center of the circle, and let <math>M</math> be the midpoint of <math>AC</math>. Let <math>\omega</math> denote the circle with diameter <math>OP</math>. Since <math>\angle OBP = \angle OMP = \angle ODP = 90^\circ</math>, <math>B</math>, <math>D</math>, and <math>M</math> all lie on <math>\omega</math>. | ||

## Revision as of 19:31, 28 April 2011

## Problem

Points , , , , lie on a circle and point lies outside the circle. The given points are such that (i) lines and are tangent to , (ii) , , are collinear, and (iii) . Prove that bisects .

## Solution

Let be the center of the circle, and let be the midpoint of . Let denote the circle with diameter . Since , , , and all lie on .

Since quadrilateral is cyclic, . Triangles and are congruent, so , so . Since and are parallel, lies on .