Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 14"
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− | To find <math>EC</math>, we can proceed by Power of a Point using point <math>D</math> on circle <math>(ABC)</math> to get <math>DE \cdot DC = DB \cdot DA.</math> Since <math>DC=13</math>, <math>DB = 7</math>, and <math>AD = 12</math>, we have <math>DE=\frac{84}{13}.</math> Since <math>CD=13</math>, we have <cmath>EC=CD-DE=\frac{85}{13} \qquad (2).</cmath> | + | To find <math>EC</math>, we can proceed by Power of a Point using point <math>D</math> on circle <math>(ABC)</math> to get <math>DE \cdot DC = DB \cdot DA.</math> Since <math>DC=13</math>, <math>DB = 7</math>, and <math>AD = 12</math>, we have <math>DE=\frac{84}{13}.</math> Since <math>CD=13</math>, we have <cmath>EC=CD-DE=\frac{85}{13} \qquad (2).</cmath> ~samrocksnature |
Revision as of 15:11, 11 July 2021
Problem
Let there be a such that
,
, and
, and let
be a point on
such that
Let the circumcircle of
intersect hypotenuse
at
and
. Let
intersect
at
. If the ratio
can be expressed as
where
and
are relatively prime, find
Solution
We claim that
is the angle bisector of
.
Observe that
, which tells us that
is a
triangle. In cyclic quadrilateral
, we have
and
Since
, we have
. This means that
, or equivalently
, is an angle bisector of
in
.
By the angle bisector theorem and our
We seek the lengths
and
.
To find , we can proceed by Power of a Point using point
on circle
to get
Since
,
, and
, we have
Since
, we have
~samrocksnature