Difference between revisions of "1994 USAMO Problems/Problem 3"
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==Problem== | ==Problem== | ||
− | A convex hexagon <math>ABCDEF</math> is inscribed in a circle such that <math>AB=CD=EF</math> and diagonals <math>AD,BE</math>, and <math>CF</math> are concurrent. Let <math>P</math> be the intersection of <math>AD</math> and <math>CE</math>. Prove that <math>\frac{CP}{ | + | A convex hexagon <math>ABCDEF</math> is inscribed in a circle such that <math>AB=CD=EF</math> and diagonals <math>AD,BE</math>, and <math>CF</math> are concurrent. Let <math>P</math> be the intersection of <math>AD</math> and <math>CE</math>. Prove that <math>\frac{CP}{PE}=(\frac{AC}{CE})^2</math>. |
==Solution== | ==Solution== |
Revision as of 15:34, 30 May 2014
Problem
A convex hexagon is inscribed in a circle such that
and diagonals
, and
are concurrent. Let
be the intersection of
and
. Prove that
.
Solution
Let the diagonals ,
,
meet at
.
First, let's show that the triangles and
are similar.
because
,
,
and
all lie on the circle, and
.
because
, and
,
,
,
and
all lie on the circle. Then,
Therefore, and
are similar, so
.
Next, let's show that and
are similar.
because
,
,
and
all lie on the circle, and
.
because
,
,
and
all lie on the circle.
because
, and
,
,
,
and
all lie on the circle. Then,
Therefore, and
are similar, so
.
Lastly, let's show that and
are similar.
Because and
,
,
and
all lie on the circle,
is parallel to
. So,
and
are similar, and
.
Putting it all together, .