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2003 AMC 10A Problems/Problem 22: Difference between revisions

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Proof: <math>\angle CGH=\angle EGA</math>, obviously.
Proof: <math>\angle CGH=\angle EGA</math>, obviously.


<math>\begin{eqnarray}
<cmath>\begin{eqnarray*}
\angle HCE=180^{\circ}-\angle CHG\\
\angle HCE&=&180^{\circ}-\angle CHG\\
\angle DCE=\angle CHG-90^{\circ}\\
\angle DCE&=&\angle CHG-90^{\circ}\\
\angle CEED=180-\angle CHG\\
\angle CEED&=&180-\angle CHG\\
\angle GEA=\angle GCH
\angle GEA&=&\angle GCH
\end{eqnarray}</math>
\end{eqnarray*}</cmath>


Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.
Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.
Line 86: Line 86:
Let <math>GC=x</math>.
Let <math>GC=x</math>.


<cmath>\begin{eqnarray}
<cmath>\begin{eqnarray*}
\dfrac{x}{3}=\dfrac{x+4\sqrt{5}}{5}\\
\dfrac{x}{3}&=&\dfrac{x+4\sqrt{5}}{5}\\
5x=3x+12\sqrt{5}\\
5x&=&3x+12\sqrt{5}\\
2x=12\sqrt{5}\\
2x&=&12\sqrt{5}\\
x=6\sqrt{5}
x&=&6\sqrt{5}
\end{eqnarray}</cmath>
\end{eqnarray*}</cmath>


Also, <math>\triangle GFE\approx \triangle CDE</math>, therefore
Also, <math>\triangle GFE\approx \triangle CDE</math>, therefore

Revision as of 17:55, 10 March 2015

Problem

In rectangle $ABCD$, we have $AB=8$, $BC=9$, $H$ is on $BC$ with $BH=6$, $E$ is on $AD$ with $DE=4$, line $EC$ intersects line $AH$ at $G$, and $F$ is on line $AD$ with $GF \perp AF$. Find the length of $GF$.

$\mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30$

Solution

Solution 1

$\angle GHC = \angle AHB$ (Opposite angles are equal).

$\angle F = \angle B$ (Both are 90 degrees).

$\angle BHA = \angle HAD$ (Alt. Interior Angles are congruent).

Therefore $\triangle GFA$ and $\triangle ABH$ are similar. $\triangle GCH$ and $\triangle GEA$ are also similar.

$DA$ is 9, therefore $EA$ must equal 5. Similarly, $CH$ must equal 3.

Because $GCH$ and $GEA$ are similar, the ratio of $CH\; =\; 3$ and $EA\; =\; 5$, must also hold true for $GH$ and $HA$. $\frac{GH}{GA} = \frac{3}{5}$, so $HA$ is $\frac{2}{5}$ of $GA$. By Pythagorean theorem, $(HA)^2\; =\; (HB)^2\; +\; (BA)^2\;...\;HA=10$.

$HA\: =\: 10 =\: \frac{2}{5}*(GA)$.

$GA\: =\: 25.$

So $\frac{GA}{HA}\: =\: \frac{GF}{BA}$.

$\frac{25}{10}\: =\: \frac{GF}{8}$.

Therefore $GF= \boxed{\mathrm{(B)}\ 20}$.

Solution 2

Since $ABCD$ is a rectangle, $CD=AB=8$.

Since $ABCD$ is a rectangle and $GF \perp AF$, $\angle GFE = \angle CDE = \angle ABC = 90^\circ$.

Since $ABCD$ is a rectangle, $AD || BC$.

So, $AH$ is a transversal, and $\angle GAF = \angle AHB$.

This is sufficient to prove that $GFE \approx CDE$ and $GFA \approx ABH$.

Using ratios:

$\frac{GF}{FE}=\frac{CD}{DE}$

$\frac{GF}{FD+4}=\frac{8}{4}=2$

$GF=2 \cdot (FD+4)=2 \cdot FD+8$

$\frac{GF}{FA}=\frac{AB}{BH}$

$\frac{GF}{FD+9}=\frac{8}{6}=\frac{4}{3}$

$GF=\frac{4}{3} \cdot (FD+9)=\frac{4}{3} \cdot FD+12$

Since $GF$ can't have 2 different lengths, both expressions for $GF$ must be equal.

$2 \cdot FD+8=\frac{4}{3} \cdot FD+12$

$\frac{2}{3} \cdot FD=4$

$FD=6$

$GF=2 \cdot FD+8=2\cdot6+8=\boxed{\mathrm{(B)}\ 20}$

Solution 3

Since $ABCD$ is a rectangle, $CH=3$, $EA=5$, and $CD=8$. From the Pythagorean Theorem, $CE^2=CD^2+DE^2=80\Rightarrow CE=4\sqrt{5}$.

Lemma

Statement: $GCH \approx GEA$

Proof: $\angle CGH=\angle EGA$, obviously.

\begin{eqnarray*} \angle HCE&=&180^{\circ}-\angle CHG\\ \angle DCE&=&\angle CHG-90^{\circ}\\ \angle CEED&=&180-\angle CHG\\ \angle GEA&=&\angle GCH \end{eqnarray*}

Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.


Let $GC=x$.

\begin{eqnarray*} \dfrac{x}{3}&=&\dfrac{x+4\sqrt{5}}{5}\\ 5x&=&3x+12\sqrt{5}\\ 2x&=&12\sqrt{5}\\ x&=&6\sqrt{5} \end{eqnarray*}

Also, $\triangle GFE\approx \triangle CDE$, therefore

\[\dfrac{8}{4\sqrt{5}}=\dfrac{GF}{10\sqrt{5}}\]

We can multiply both sides by $\sqrt{5}$ to get that $GF$ is twice of 10, or $\boxed{\mathrm{(B)}\ 20}$

Solution 4

We extend $BC$ such that it intersects $GF$ at $X$. Since $ABCD$ is a rectangle, it follows that $CD=8$, therefore, $XF=8$. Let $GX=y$. From the similarity of triangles $GCH$ and $GEA$, we have the ratio $3:5$ (as $CH=9-6=3$, and $EA=9-4=5$). $GX$ and $GF$ are the altitudes of $GCH$ and $GEA$, respectively. Thus, $y:y+8 = 3:5$, from which we have $y=12$, thus $GF=y+8=12+8=\boxed{\mathrm{(B)}\ 20}$

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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