Mock AIME 3 Pre 2005 Problems/Problem 11

Problem

$ABC$ is an acute triangle with perimeter $60$ . $D$ is a point on $\overline{BC}$ . The circumcircles of triangles $ABD$ and $ADC$ intersect $\overline{AC}$ and $\overline{AB}$ at $E$ and $F$ respectively such that $DE = 8$ and $DF = 7$ . If $\angle{EBC} \cong \angle{BCF}$ , then the value of $\frac{AE}{AF}$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Compute $m + n$ .

Solution

Remark that since $ABDE$ is cyclic we have $\angle CED=\angle DBF$ , and similarly $\angle BFD=\angle DCE$ . Therefore by AA similarity $\triangle DBF\sim\triangle DEC$ . Thus there exists a spiral similarity sending $B$ to $E$ and $F$ to $D$ , so by a fundamental theorem of spiral similarity $\triangle BDE\sim\triangle FDC$ . The angle equality condition gives $\angle CFD=\angle EBD=\angle DCF$ , so $\triangle CDF$ is isosceles and $DC=7$ . Similarly, $BD=8$ . Finally, note that the congruent side lengths actually imply $\triangle DBF=\triangle DEC$ , so $EC=BF$ .

Let $x=CE=BF$ and $b=AC$ . Remark that from the perimeter condition $AB=45-b$ . Now from Power of a Point we have the system of two equations $\begin{cases}7\cdot 15&=xb,\\8\cdot 15&=x(45-b).\end{cases}$ Expanding the second equation and rearranging variables gives $45x=8\cdot 15+bx=15^2\implies x=5$ . Back-substitution yields $AC=21$ and consequently $AB=24$ . Thus $AE=16$ and $AF=19$ , so the desired ratio is $\tfrac{16}{19}\implies\boxed{035}$ .

Mock AIME 3 Pre 2005 (Problems, Source)
Preceded by Problem 10	Followed by Problem 12
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Mock AIME 3 Pre 2005 Problems/Problem 11

Problem

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