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

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== Solutions == | == Solutions == | ||

+ | Connet segment PO, and name the interaction of PO and the circle as point M. | ||

+ | Since PB and PD are tangent to the circle, it's easy to see that M is the midpoint of arc BD. | ||

+ | ∠ BOA = 1/2 arc AB + 1/2 arc CE | ||

+ | Since AC // DE, arc AD = arc CE, | ||

+ | thus, ∠ BOA = 1/2 arc AB + 1/2 arc AD = 1/2 arc BD = arc BM = ∠ BOM | ||

+ | Therefore, PBOM is cyclic, ∠ PFO = ∠ OBP = 90°, AF = AC (F is the interaction of BE and AC) | ||

+ | BE bisects AC, proof completed! | ||

==Solution 1== | ==Solution 1== |

## Revision as of 07:37, 27 April 2019

## 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 .

## Solutions

Connet segment PO, and name the interaction of PO and the circle as point M. Since PB and PD are tangent to the circle, it's easy to see that M is the midpoint of arc BD. ∠ BOA = 1/2 arc AB + 1/2 arc CE Since AC // DE, arc AD = arc CE, thus, ∠ BOA = 1/2 arc AB + 1/2 arc AD = 1/2 arc BD = arc BM = ∠ BOM Therefore, PBOM is cyclic, ∠ PFO = ∠ OBP = 90°, AF = AC (F is the interaction of BE and AC) BE bisects AC, proof completed!

## Solution 1

Let be the center of the circle, and let be the intersection of and . Let be and be .

, ,

Thus is a cyclic quadrilateral and and so is the midpoint of chord .

~pandadude

## Solution 2

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 . Because and are parallel, lies on (using Euclid's Parallel Postulate).

## Solution 3

Note that by Lemma 9.9 of EGMO, is a harmonic bundle. We project through onto , Where is the point at infinity for parallel lines and . Thus, we get , and is the midpoint of . ~novus677

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