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Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 11"
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Let <math>x=CE=BF</math> and <math>b=AC</math>. Remark that from the perimeter condition <math>AB=45-b</math>. Now from Power of a Point we have the system of two equations <cmath>\begin{cases}7\cdot 15&=xb,\\8\cdot 15&=x(45-b).\end{cases}</cmath> Expanding the second equation and rearranging variables gives <math>45x=8\cdot 15+bx=15^2\implies x=5</math>. Back-substitution yields <math>AC=21</math> and consequently <math>AB=24</math>. Thus <math>AE=16</math> and <math>AF=19</math>, so the desired ratio is <math>\tfrac{16}{19}\implies\boxed{035}</math>.
 
Let <math>x=CE=BF</math> and <math>b=AC</math>. Remark that from the perimeter condition <math>AB=45-b</math>. Now from Power of a Point we have the system of two equations <cmath>\begin{cases}7\cdot 15&=xb,\\8\cdot 15&=x(45-b).\end{cases}</cmath> Expanding the second equation and rearranging variables gives <math>45x=8\cdot 15+bx=15^2\implies x=5</math>. Back-substitution yields <math>AC=21</math> and consequently <math>AB=24</math>. Thus <math>AE=16</math> and <math>AF=19</math>, so the desired ratio is <math>\tfrac{16}{19}\implies\boxed{035}</math>.
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==Solution 2==
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Notice that <math>\angle BCF = \angle DAF = \frac{\arc DF}{2}</math> and <math>\angle EBC = \angle EBD = \angle EAD =\frac{\arc DE}{2}</math>. Hence, <math>\angle BAD = \angle CAD</math>. Furthermore, through cyclic quadrilaterals, we can find that <math>\triangle{BFD} \sim \triangle {BCA}</math> and <math>\triangle{ECD} \sim \triangle{BCA}</math>.
  
 
==See Also==
 
==See Also==
 
{{Mock AIME box|year=Pre 2005|n=3|num-b=10|num-a=12}}
 
{{Mock AIME box|year=Pre 2005|n=3|num-b=10|num-a=12}}

Revision as of 12:46, 30 November 2024

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 $\angle BCF = \angle DAF = \frac{\arc DF}{2}$ (Error compiling LaTeX. Unknown error_msg) and $\angle EBC = \angle EBD = \angle EAD =\frac{\arc DE}{2}$ (Error compiling LaTeX. Unknown error_msg). 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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