Difference between revisions of "2011 USAJMO Problems/Problem 5"
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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>. Because <math>AC</math> and <math>DE</math> are parallel, <math>M</math> lies on <math>BE</math> (using Euler's Parallel Postulate). | 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>. 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}} | ||
+ | |||
+ | ==Solution 3== | ||
+ | Note that by Lemma 9.9 of EGMO, <math>(A,C;B,D)</math> is a harmonic bundle. We project through <math>E</math> onto <math>\overline{AC}</math>, | ||
+ | <cmath>(A,C;B,D)\stackrel{E}{=}(A,C;M,P_{\infty})=-1</cmath> | ||
+ | Thus, we get <math>\frac{MA}{MC}=-1</math>, and <math>M</math> is the midpoint of <math>AC</math>. ~novus677 |
Revision as of 01:28, 2 June 2018
Contents
[hide]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
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 Euler's Parallel Postulate).
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Solution 3
Note that by Lemma 9.9 of EGMO, is a harmonic bundle. We project through
onto
,
Thus, we get
, and
is the midpoint of
. ~novus677