Difference between revisions of "2011 USAJMO Problems/Problem 5"
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+ | == Problem == | ||
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+ | Points <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math> lie on a circle <math>\omega</math> and point <math>P</math> lies outside the circle. The given points are such that (i) lines <math>PB</math> and <math>PD</math> are tangent to <math>\omega</math>, (ii) <math>P</math>, <math>A</math>, <math>C</math> are collinear, and (iii) <math>\overline{DE} \parallel \overline{AC}</math>. Prove that <math>\overline{BE}</math> bisects <math>\overline{AC}</math>. | ||
+ | |||
+ | == 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>\omega</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>\omega</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>\omega</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>\omega</math>. | ||
Revision as of 18:31, 28 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
.