Difference between revisions of "2011 USAJMO Problems/Problem 5"
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<cmath>-1=(A,C;B,D)\stackrel{E}{=}(A,C;M,P_{\infty})</cmath> | <cmath>-1=(A,C;B,D)\stackrel{E}{=}(A,C;M,P_{\infty})</cmath> | ||
Where <math>P_{\infty}</math> is the point at infinity for parallel lines <math>\overline{DE}</math> and <math>\overline{AC}</math>. Thus, we get <math>\frac{MA}{MC}=-1</math>, and <math>M</math> is the midpoint of <math>AC</math>. ~novus677 | Where <math>P_{\infty}</math> is the point at infinity for parallel lines <math>\overline{DE}</math> and <math>\overline{AC}</math>. Thus, we get <math>\frac{MA}{MC}=-1</math>, and <math>M</math> is the midpoint of <math>AC</math>. ~novus677 | ||
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Revision as of 19:07, 2 June 2018
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).
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
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.