Difference between revisions of "2011 USAJMO Problems/Problem 5"
(→Solution) |
|||
Line 49: | Line 49: | ||
</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 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>. Thus, <math>AM</math> and <math>DE</math> are parallel. Because <math>AC</math> and <math>DE</math> are parallel, <math>M</math> lies on <math>BE</math> (using Euler's Parallel Postulate). |
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:53, 16 April 2014
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
. Thus,
and
are parallel. Because
and
are parallel,
lies on
(using Euler's Parallel Postulate).
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.