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

(Lemma)
(Solution)
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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
 +
<math>\angle GCH = \angle ABH</math> (Opposite angles are equal).
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<math>\angle F = \angle B</math> (Both are 90 degrees).
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<math>\angle BHA = \angle HAD</math> (Alt. Interior Angles are congruent).
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 +
Therefore <math>Triangles\: GFA</math> and <math>ABH</math> are similar.
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<math>GCH</math> and <math>GEA</math> are also similar.
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 +
<math>DA</math> is 9, therefore <math>EA</math> must equal 5. Similarly, <math>CH</math> must equal 3.
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 +
Because <math>GCH</math> and <math>GEA</math> are similar, the ratio of <math>CH\; =\; 3</math> and <math>EA\; =\; 5</math>, must also hold true for <math>GH</math> and <math>HA</math>. <math>\frac{GH}{GA} = \frac{3}{5}</math>, so <math>HA</math> is <math>\frac{2}{5}</math> of <math>GA</math>. By Pythagorean theorem, <math>(HA)^2\;  =\; (HB)^2\; +\; (BA)^2\;...\;HA=10</math>.
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<math>HA\: =\: 10 =\: \frac{2}{5}*(GA)</math>.
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<math>GA\: =\: 25.</math>
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So <math>\frac{GA}{HA}\: =\: \frac{GF}{BA}</math>.
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<math>\frac{25}{10}\: =\: \frac{GF}{8}</math>.
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Therefore <math>GF\: = \boxed{20} = \boxed{\mathrm{(B)}}</math>.
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=== Solution 2 ===
 
Since <math>ABCD</math> is a rectangle, <math>CD=AB=8</math>.  
 
Since <math>ABCD</math> is a rectangle, <math>CD=AB=8</math>.  
  
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<math>GF=2 \cdot FD+8=2\cdot6+8=20 \Rightarrow B</math>
 
<math>GF=2 \cdot FD+8=2\cdot6+8=20 \Rightarrow B</math>
  
=== Solution 2 ===
+
=== Solution 3 ===
 
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>.
 
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 ====
 
==== Lemma ====

Revision as of 20:40, 15 February 2009

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 $GF$.

2003amc10a22.gif

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

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. ! Missing \endgroup inserted.)

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

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AMC 10 Problems and Solutions
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