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

 
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7. Let <math>\triangle ABC</math> have <math>AC=6</math> and <math>BC=3</math>. Point <math>E</math> is such that <math>CE=1</math> and <math>AE=5</math>. Construct point <math>F</math> on segment <math>BC</math> such that <math>\angle AEB=\angle AFB</math>. <math>EF</math> and <math>AB</math> are extended to meet at <math>D</math>. If <math>\frac{[AEF]}{[CFD]}=\frac{m}{n}</math> where <math>m</math> and <math>n</math> are positive integers, find <math>m+n</math> (note: <math>[ABC]</math> denotes the area of <math>\triangle ABC</math>).
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==Problem==
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Let <math>\triangle ABC</math> have <math>AC=6</math> and <math>BC=3</math>. Point <math>E</math> is such that <math>CE=1</math> and <math>AE=5</math>. Construct point <math>F</math> on segment <math>BC</math> such that <math>\angle AEB=\angle AFB</math>. <math>EF</math> and <math>AB</math> are extended to meet at <math>D</math>. If <math>\frac{[AEF]}{[CFD]}=\frac{m}{n}</math> where <math>m</math> and <math>n</math> are positive integers, find <math>m+n</math> (note: <math>[ABC]</math> denotes the area of <math>\triangle ABC</math>).
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==Solution==
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We can immediately see that quadrilateral <math>AEFB</math> is cyclic, since <math>\angle AEB=\angle AFB</math>. We then have, from Power of a Point, that <math>CE\cdot CA=CF\cdot CB</math>. In other words, <math>1\cdot 6 = CF\cdot 3</math>. <math>CF</math> is then 2, and <math>BF</math> is 1. We can now use Menelaus on line <math>DF</math> with respect to triangle <math>ABC</math>:
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<cmath>\frac{AE}{EC}\cdot \frac{CF}{FB}\cdot \frac{BD}{DA}=1</cmath>
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<cmath>\frac{5}{1}\cdot \frac{2}{1}\cdot \frac{BD}{DA}=1</cmath>
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<cmath>\frac{BD}{DA}=\frac{1}{10}</cmath>
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This shows that <math>\frac{BA}{BD}=9</math>.
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Now let <math>[ABC]=x</math>, for some real <math>x</math>. Therefore <math>[CFA]=\frac{CF}{CB}\cdot [ABC]=\frac{2x}{3}</math>, and <math>[AEF]=\frac{AE}{AC}\cdot [CFA]=\frac{5}{6}\cdot \frac{2x}{3}=\frac{5x}{9}</math>. Similarly, <math>[CBD]=\frac{DB}{AB}\cdot [ABC]=\frac{x}{9}</math> and <math>[CFD]=\frac{CF}{CB}\cdot [CBD]=\frac{2}{3}\cdot \frac{x}{9}=\frac{2x}{27}</math>. The desired ratio is then
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<cmath>\frac{[AEF]}{[CFD]}=\frac{\frac{5x}{9}}{\frac{2x}{27}}=\frac{5\cdot 27}{9\cdot 2}=\frac{15}{2}</cmath>
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Therefore <math>m+n=\boxed{017}</math>.
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----
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==Alternate Solution==
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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>
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*[[Mock AIME 1 2006-2007 Problems/Problem 6 | Previous Problem]]
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*[[Mock AIME 1 2006-2007 Problems/Problem 8 | Next Problem]]
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*[[Mock AIME 1 2006-2007]]

Latest revision as of 15:54, 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 \[\frac{[AEF]}{[CFD]}=\frac{[AEF]}{[CFE]}\cdot\frac{[CFE]}{[CFD]}=5\cdot\frac{3}{2}=\frac{15}{2}=\frac{m}{n}\implies m+n=\boxed{17}.\]

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