2003 AMC 10A Problems/Problem 22

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

$\angle GCH = \angle ABH$ (Opposite angles are equal).

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

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

Therefore $Triangles\: GFA$ and $ABH$ are similar. $GCH$ and $GEA$ 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 $GF\: = \boxed{20} = \boxed{\mathrm{(B)}}$ .

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.

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

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

$FD=6$

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

Solution 3

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 $20\Rightarrow \mathrm{(B)}$

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=20. B.

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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