Mock AIME 5 2005-2006 Problems/Problem 12

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Problem

Let $ABC$ be a triangle with $AB = 13$, $BC = 14$, and $AC = 15$. Let $D$ be the foot of the altitude from $A$ to $BC$ and $E$ be the point on $BC$ between $D$ and $C$ such that $BD = CE$. Extend $AE$ to meet the circumcircle of $ABC$ at $F$. If the area of triangle $FAC$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.

Solution

Let area of $\triangle XYZ$ be denoted by $[XYZ]$.


By Heron's theorem , We get

$[ABC]=84$

$Or,\frac{AD.BC}{2}=84$

$Or,AD=12$


By pythagoras theorem on $\triangle ABD$ , We get $BD=5=CE$ So, $DE=4$.

Again applying pythagoras theorem on $\triangle AED$ , We get, $AE=\sqrt{160}$

As $\triangle ABC$ and $\triangle AEC$ share same height ,

So, $\frac{[AEC]}{[ABC]}=\frac{EC}{BC}=\frac{5}{14}$

Thus $[AEC]=30$

So, $[ABE]=84-30=54$


Now, $\angle AEB = \angle FEC$ [Vertical angle]

$\angle BAF = \angle BCF$ [shares same chord BF]

$Or, \angle BAE = \angle ECF$

So, $\triangle ABE$ is simillar to $\triangle CFE$

Now, $\frac{[CEF]}{[AEB]}=\frac{CE^2}{AE^2}=\frac{25}{160}=\frac{5}{32}$


So, $[CEF]=\frac{135}{16}$


Now, $[ACF]=[CEF]+[AEC]=\frac{135}{16}+30=\frac{615}{16}$

Thus required answer=$615+16=\fbox{631}$

~by NOOBMASTER_M

Solution

See also

Mock AIME 5 2005-2006 (Problems, Source)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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