Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 7"
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+ | As above, use Power of a Point to compute <math>CF=2</math> and <math>FB=1</math>. Since triangles <math>AFE</math> and <math>CEF</math> share the same height, <math>\frac{[AEF]}{[CFE]}=\frac{CE}{EA}=5</math>. Similarly, <math>\frac{[CFE]}{[CFD]}=\frac{EF}{FD}</math>. Using Menelaus's Theorem on points <math>E, F, D</math> on the sides of triangle <math>ABC</math>, we see that <cmath>\frac{AE}{EC}\cdot\frac{CF}{FB}\cdot\frac{BD}{DA}=1\implies \frac{AB}{BD}=9.</cmath>Let <math>X=AF\cap CD</math>. Using Ceva's Theorem on points <math>X, E, B</math> lying on the sides of triangle <math>ACD</math>, we find that <cmath>\frac{AE}{EC}\cdot\frac{CX}{XD}\cdot\frac{DB}{BA}=1\implies \frac{CX}{XD}=\frac{9}{5}.</cmath>Then, using Menelaus's Theorem on the points <math>A, F, X</math> on the sides of triangle <math>ECD</math>, we see that <cmath>\frac{EA}{AC}\cdot\frac{CX}{XD}\cdot\frac{DF}{FE}=1\implies \frac{DF}{FE}=\frac{2}{3}.</cmath> Thus, <math>\frac{[CFE]}{[CFD]}=\frac{3}{2},</math> so that <cmath>\frac{[AEF]}{[CFD]}=\frac{[AEF]}{[CFE]}\cdot\frac{[CFE]}{[CFD]}=5\cdot\frac{3}{2}=\frac{15}{2}=\frac{m}{n}\implies m+n=\boxed{17}.</cmath> | ||
− | *[[Mock AIME 1 2006-2007/Problem 8 | Next Problem]] | + | ---- |
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+ | *[[Mock AIME 1 2006-2007 Problems/Problem 6 | Previous Problem]] | ||
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+ | *[[Mock AIME 1 2006-2007 Problems/Problem 8 | Next Problem]] | ||
*[[Mock AIME 1 2006-2007]] | *[[Mock AIME 1 2006-2007]] |
Latest revision as of 16:54, 25 November 2015
Problem
Let have
and
. Point
is such that
and
. Construct point
on segment
such that
.
and
are extended to meet at
. If
where
and
are positive integers, find
(note:
denotes the area of
).
Solution
We can immediately see that quadrilateral is cyclic, since
. We then have, from Power of a Point, that
. In other words,
.
is then 2, and
is 1. We can now use Menelaus on line
with respect to triangle
:
This shows that .
Now let , for some real
. Therefore
, and
. Similarly,
and
. The desired ratio is then
Therefore .
Alternate Solution
As above, use Power of a Point to compute and
. Since triangles
and
share the same height,
. Similarly,
. Using Menelaus's Theorem on points
on the sides of triangle
, we see that
Let
. Using Ceva's Theorem on points
lying on the sides of triangle
, we find that
Then, using Menelaus's Theorem on the points
on the sides of triangle
, we see that
Thus,
so that