Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 7"

m
(Solution)
 
Line 21: Line 21:
 
----
 
----
  
 +
==Alternate Solution==
 +
As above, use Power of a Point to compute <math>CF=2</math> and <math>FB=1</math>. Since triangles <math>AFE</math> and <math>CEF</math> share the same height, <math>\frac{[AEF]}{[CFE]}=\frac{CE}{EA}=5</math>. Similarly, <math>\frac{[CFE]}{[CFD]}=\frac{EF}{FD}</math>. Using Menelaus's Theorem on points <math>E, F, D</math> on the sides of triangle <math>ABC</math>, we see that <cmath>\frac{AE}{EC}\cdot\frac{CF}{FB}\cdot\frac{BD}{DA}=1\implies \frac{AB}{BD}=9.</cmath>Let <math>X=AF\cap CD</math>. Using Ceva's Theorem on points <math>X, E, B</math> lying on the sides of triangle <math>ACD</math>, we find that <cmath>\frac{AE}{EC}\cdot\frac{CX}{XD}\cdot\frac{DB}{BA}=1\implies \frac{CX}{XD}=\frac{9}{5}.</cmath>Then, using Menelaus's Theorem on the points <math>A, F, X</math> on the sides of triangle <math>ECD</math>, we see that <cmath>\frac{EA}{AC}\cdot\frac{CX}{XD}\cdot\frac{DF}{FE}=1\implies \frac{DF}{FE}=\frac{2}{3}.</cmath> Thus, <math>\frac{[CFE]}{[CFD]}=\frac{3}{2},</math> so that <cmath>\frac{[AEF]}{[CFD]}=\frac{[AEF]}{[CFE]}\cdot\frac{[CFE]}{[CFD]}=5\cdot\frac{3}{2}=\frac{15}{2}=\frac{m}{n}\implies m+n=\boxed{17}.</cmath>
 +
 +
----
 +
 
*[[Mock AIME 1 2006-2007 Problems/Problem 6 | Previous Problem]]
 
*[[Mock AIME 1 2006-2007 Problems/Problem 6 | Previous Problem]]
  

Latest revision as of 16:54, 25 November 2015

Problem

Let $\triangle ABC$ have $AC=6$ and $BC=3$. Point $E$ is such that $CE=1$ and $AE=5$. Construct point $F$ on segment $BC$ such that $\angle AEB=\angle AFB$. $EF$ and $AB$ are extended to meet at $D$. If $\frac{[AEF]}{[CFD]}=\frac{m}{n}$ where $m$ and $n$ are positive integers, find $m+n$ (note: $[ABC]$ denotes the area of $\triangle ABC$).

Solution

We can immediately see that quadrilateral $AEFB$ is cyclic, since $\angle AEB=\angle AFB$. We then have, from Power of a Point, that $CE\cdot CA=CF\cdot CB$. In other words, $1\cdot 6 = CF\cdot 3$. $CF$ is then 2, and $BF$ is 1. We can now use Menelaus on line $DF$ with respect to triangle $ABC$:

\[\frac{AE}{EC}\cdot \frac{CF}{FB}\cdot \frac{BD}{DA}=1\]

\[\frac{5}{1}\cdot \frac{2}{1}\cdot \frac{BD}{DA}=1\]

\[\frac{BD}{DA}=\frac{1}{10}\]

This shows that $\frac{BA}{BD}=9$.

Now let $[ABC]=x$, for some real $x$. Therefore $[CFA]=\frac{CF}{CB}\cdot [ABC]=\frac{2x}{3}$, and $[AEF]=\frac{AE}{AC}\cdot [CFA]=\frac{5}{6}\cdot \frac{2x}{3}=\frac{5x}{9}$. Similarly, $[CBD]=\frac{DB}{AB}\cdot [ABC]=\frac{x}{9}$ and $[CFD]=\frac{CF}{CB}\cdot [CBD]=\frac{2}{3}\cdot \frac{x}{9}=\frac{2x}{27}$. The desired ratio is then

\[\frac{[AEF]}{[CFD]}=\frac{\frac{5x}{9}}{\frac{2x}{27}}=\frac{5\cdot 27}{9\cdot 2}=\frac{15}{2}\]

Therefore $m+n=\boxed{017}$.


Alternate Solution

As above, use Power of a Point to compute $CF=2$ and $FB=1$. Since triangles $AFE$ and $CEF$ share the same height, $\frac{[AEF]}{[CFE]}=\frac{CE}{EA}=5$. Similarly, $\frac{[CFE]}{[CFD]}=\frac{EF}{FD}$. Using Menelaus's Theorem on points $E, F, D$ on the sides of triangle $ABC$, we see that \[\frac{AE}{EC}\cdot\frac{CF}{FB}\cdot\frac{BD}{DA}=1\implies \frac{AB}{BD}=9.\]Let $X=AF\cap CD$. Using Ceva's Theorem on points $X, E, B$ lying on the sides of triangle $ACD$, we find that \[\frac{AE}{EC}\cdot\frac{CX}{XD}\cdot\frac{DB}{BA}=1\implies \frac{CX}{XD}=\frac{9}{5}.\]Then, using Menelaus's Theorem on the points $A, F, X$ on the sides of triangle $ECD$, we see that \[\frac{EA}{AC}\cdot\frac{CX}{XD}\cdot\frac{DF}{FE}=1\implies \frac{DF}{FE}=\frac{2}{3}.\] Thus, $\frac{[CFE]}{[CFD]}=\frac{3}{2},$ so that \[\frac{[AEF]}{[CFD]}=\frac{[AEF]}{[CFE]}\cdot\frac{[CFE]}{[CFD]}=5\cdot\frac{3}{2}=\frac{15}{2}=\frac{m}{n}\implies m+n=\boxed{17}.\]


Invalid username
Login to AoPS