Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 7"
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− | {{ | + | We can immediately see that quadrilateral <math>AEFB</math> is cyclic, since <math>\angle AEB=\angle AFB</math>. We then have, from Power of a Point, that <math>CE\cdot CA=CF\cdot CB</math>. In other words, <math>1\cdot 6 = CF\cdot 3</math>. <math>CF</math> is then 2, and <math>BF</math> is 1. We can now use Menelaus on line <math>DF</math> with respect to triangle <math>ABC</math>: |
+ | |||
+ | <cmath>\frac{AE}{EC}\cdot \frac{CF}{FB}\cdot \frac{BD}{DA}=1</cmath> | ||
+ | |||
+ | <cmath>\frac{5}{1}\cdot \frac{2}{1}\cdot \frac{BD}{DA}=1</cmath> | ||
+ | |||
+ | <cmath>\frac{BD}{DA}=\frac{1}{10}</cmath> | ||
+ | |||
+ | This shows that <math>\frac{BA}{BD}=9</math>. | ||
+ | |||
+ | Now let <math>[ABC]=x</math>, for some real <math>x</math>. Therefore <math>[CFA]=\frac{CF}{CB}\cdot [ABC]=\frac{2x}{3}</math>, and <math>[AEF]=\frac{AE}{AC}\cdot [CFA]=\frac{5}{6}\cdot \frac{2x}{3}=\frac{5x}{9}</math>. Similarly, <math>[CBD]=\frac{DB}{AB}\cdot [ABC]=\frac{x}{9}</math> and <math>[CFD]=\frac{CF}{CB}\cdot [CBD]=\frac{2}{3}\cdot \frac{x}{9}=\frac{2x}{27}</math>. The desired ratio is then | ||
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+ | <cmath>\frac{[AEF]}{[CFD]}=\frac{\frac{5x}{9}}{\frac{2x}{27}}=\frac{5\cdot 27}{9\cdot 2}=\frac{15}{2}</cmath> | ||
+ | |||
+ | Therefore <math>m+n=\boxed{017}</math>. | ||
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Revision as of 23:06, 28 December 2011
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 .