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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 $\tfrac{16}{19}\implies\boxed{035}$.

Solution 2

Notice that \[\angle{BCF} = \angle{DAF} = \frac{\overarc{DF}}{2}\] and \[\angle{EBC} = \angle{EBD} = \angle{EAD} =\frac{\overarc{DE}}{2}\] Hence, $\angle{BAD} = \angle{CAD}$. Furthermore, through cyclic quadrilaterals, we can find that $\triangle{BFD} \sim \triangle {BCA}$ and $\triangle{ECD} \sim \triangle{BCA}$.


See Also

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