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 $GF$ .

$[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0); draw(D--A--B--C--D--F--G--Ep); draw(A--G); label("$F$",F,W); label("$G$",G,W); label("$C$",C,WSW); label("$H$",H,NNE); label("$6$",(6,8),N); label("$B$",B,NE); label("$A$",A,SW); label("$E$",Ep,S); label("$4$",(2,0),S); label("$D$",D,S);[/asy]$

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

Solution

Solution 1

$\angle GHC = \angle AHB$ (Vertical 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

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 $GF=y+8=12+8=\boxed{\mathrm{(B)}\ 20}$

Solution 6

Since $GF\perp AF$ and $AF\perp CD,$ we have $GF\parallel CD\parallel AB.$ Thus, $\triangle CDE\sim GFE.$ Suppose $GF=x$ and $FD=y.$ Thus, we have $\dfrac{x}{8}=\dfrac{y+4}{4}.$ Additionally, now note that $\triangle GAF\sim AHB,$ which is pretty obvious from insight, but can be proven by AA with extending $BH$ to meet $GF.$ From this new pair of similar triangles, we have $\dfrac{x}{8}=\dfrac{y+9}{6}.$ Therefore, we have by combining those two equations, $\dfrac{y+9}{6}=\dfrac{y+4}{4}.$ Solving, we have $y=6,$ and therefore $x=\boxed{\mathrm{(B)}\ 20}$

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

Page

Toolbox

Search