Mock AIME 5 2005-2006 Problems/Problem 12

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

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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Mock AIME 5 2005-2006 Problems/Problem 12

Contents

Problem

Solution

Solution

See also