Difference between revisions of "2011 USAJMO Problems/Problem 5"
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== Solution == | == Solution == | ||
− | Let <math>O</math> be the center of the circle, and let <math>M</math> be the midpoint of <math>AC</math>. Let <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>\theta</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>\theta</math>. |
<asy> | <asy> | ||
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dot("$O$", O, dir(0)); | dot("$O$", O, dir(0)); | ||
dot("$P$", P, W); | dot("$P$", P, W); | ||
− | label("$\ | + | label("$\theta$", (O + P)/2 + abs(O - P)/2*dir(120), NW); |
</asy> | </asy> | ||
Since quadrilateral <math>BOMP</math> is cyclic, <math>\angle BMP = \angle BOP</math>. Triangles <math>BOP</math> and <math>DOP</math> are congruent, so <math>\angle BOP = \angle BOD/2 = \angle BED</math>, so <math>\angle BMP = \angle BED</math>. Since <math>AC</math> and <math>DE</math> are parallel, <math>M</math> lies on <math>BE</math>. | Since quadrilateral <math>BOMP</math> is cyclic, <math>\angle BMP = \angle BOP</math>. Triangles <math>BOP</math> and <math>DOP</math> are congruent, so <math>\angle BOP = \angle BOD/2 = \angle BED</math>, so <math>\angle BMP = \angle BED</math>. Since <math>AC</math> and <math>DE</math> are parallel, <math>M</math> lies on <math>BE</math>. |
Revision as of 14:19, 29 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 .