Difference between revisions of "2011 USAJMO Problems/Problem 5"
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Let <math>O</math> be the center of the circle, and let <math>X</math> be the intersection of <math>AC</math> and <math>BE</math>. Let <math>\angle OPA</math> be <math>x</math> and <math>\angle OPD</math> be <math>y</math>. | Let <math>O</math> be the center of the circle, and let <math>X</math> be the intersection of <math>AC</math> and <math>BE</math>. Let <math>\angle OPA</math> be <math>x</math> and <math>\angle OPD</math> be <math>y</math>. | ||
− | <math>\angle OPB = \angle OPD = y</math> | + | <math> \angle OPB = \angle OPD = y </math>, <math> \angle BED = \frac{\angle DOB}{2} = 90-y </math>, |
− | <math>\angle BED = frac{\angle DOB}{2} = 90-y</math> | + | <math> \angle ODE = \angle PDE - 90 = 90-x-y </math> |
− | <math>\angle ODE = \angle PDE - 90 = 90-x-y</math> | + | <math> \angle OBE = \angle PBE - 90 = x = \angle OPA </math> |
− | <math>\angle OBE = \angle PBE - 90 = x = \angle OPA</math> | ||
Thus <math>PBXO</math> is a cyclic quadrilateral and <math>\angle OXP = \angle OBP = 90</math> and so <math>X</math> is the midpoint of chord <math>AC</math>. | Thus <math>PBXO</math> is a cyclic quadrilateral and <math>\angle OXP = \angle OBP = 90</math> and so <math>X</math> is the midpoint of chord <math>AC</math>. |
Revision as of 19:53, 10 May 2018
Contents
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).
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