Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 7"
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− | Then <math>CP-BP= \frac{75\sqrt{26}}{13} - \frac{15\sqrt{26}}{13} = \frac{60\sqrt{26}}{13}</math>, and so <math>a+b+c=60+26+13=\boxed{ | + | Then <math>CP-BP= \frac{75\sqrt{26}}{13} - \frac{15\sqrt{26}}{13} = \frac{60\sqrt{26}}{13}</math>, and so <math>a+b+c=60+26+13=\boxed{099}</math>. |
Latest revision as of 21:38, 3 January 2012
Problem
In trapezoid
and
There is a point
on side
such that the circumcircle of triangle
is tangent to
If
can be expressed in the form
where
are positive integers and
are relatively prime. Find
Solution
Let be the foot of the perpendicular from
to
. Then
. By the Pythagorean Theorem on
,
. Solving yields
, and so
.
Let be the intersection of lines
and
.
by
similarity. Then
, or
, and so
. Also,
, or
, and so
. Therefore,
. By the Power of a Point of
,
. Then
.
From the Law of Cosines of ,
. Solving yields
. Similarly, from the Law of Cosines of
,
. Solving yields
.
Then , and so
.