Difference between revisions of "2003 AMC 10A Problems/Problem 22"

Revision as of 09:58, 6 March 2008

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

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.

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

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

$FD=6$

$GF=2 \cdot FD+8=2\cdot6+8=20 \Rightarrow B$

Solution 2

Since $ABCD$ is a rectangle, $CD=3$ , $EA=5$ , 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.

$\begin{eqnarray} \angle HCE=180^{\circ}-\angle CHG\\ \angle DCE=\angle CHG-90^{\circ}\\ \angle CEED=180-\angle CHG\\ \angle GEA=\angle GCH \end{eqnarray}$ (Error compiling LaTeX. Unknown error_msg)

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 $GF is twice of 10, or$ 20\Rightarrow \mathrm{(B)}$

@@ Line 7: / Line 7: @@
 == Solution ==
+=== Solution 1 ===
 Since <math>ABCD</math> is a rectangle, <math>CD=AB=8</math>.
@@ Line 40: / Line 41: @@
 <math>GF=2 \cdot FD+8=2\cdot6+8=20 \Rightarrow B</math>
+=== Solution 2 ===
+Since <math>ABCD</math> is a rectangle, <math>CD=3</math>, <math>EA=5</math>, and <math>CD=8</math>. From the [[Pythagorean Theorem]], <math>CE^2=CD^2+DE^2=80\Rightarrow CE=4\sqrt{5}</math>.
+==== Lemma ====
+Statement: <math>GCH \approx GEA</math>
+Proof: <math>\angle CGH=\angle EGA</math>, obviously.
+<math>\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}</math>
+Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.
+Let <math>GC=x</math>.
+<cmath>\begin{eqnarray}
+\dfrac{x}{3}=\dfrac{x+4\sqrt{5}}{5}\\
+x=3x+12\sqrt{5}\\
+x=12\sqrt{5}\\
+x=6\sqrt{5}
+\end{eqnarray}</cmath>
+Also, <math>\triangle GFE\approx \triangle CDE</math>, therefore
+<cmath>\dfrac{8}{4\sqrt{5}}=\dfrac{GF}{10\sqrt{5}}</cmath>
+We can multiply both sides by <math>\sqrt{5}</math> to get that <math>GF is twice of 10, or </math>20\Rightarrow \mathrm{(B)}$
 == 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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