Difference between revisions of "2003 AMC 10A Problems/Problem 22"
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
+ | <math>\angle GCH = \angle ABH</math> (Opposite angles are equal). | ||
+ | |||
+ | <math>\angle F = \angle B</math> (Both are 90 degrees). | ||
+ | |||
+ | <math>\angle BHA = \angle HAD</math> (Alt. Interior Angles are congruent). | ||
+ | |||
+ | Therefore <math>Triangles\: GFA</math> and <math>ABH</math> are similar. | ||
+ | <math>GCH</math> and <math>GEA</math> are also similar. | ||
+ | |||
+ | <math>DA</math> is 9, therefore <math>EA</math> must equal 5. Similarly, <math>CH</math> must equal 3. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | <math>HA\: =\: 10 =\: \frac{2}{5}*(GA)</math>. | ||
+ | |||
+ | <math>GA\: =\: 25.</math> | ||
+ | |||
+ | So <math>\frac{GA}{HA}\: =\: \frac{GF}{BA}</math>. | ||
+ | |||
+ | <math>\frac{25}{10}\: =\: \frac{GF}{8}</math>. | ||
+ | |||
+ | Therefore <math>GF\: = \boxed{20} = \boxed{\mathrm{(B)}}</math>. | ||
+ | |||
+ | === 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 | + | === 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 , we have , , is on with , is on with , line intersects line at , and is on line with . Find the length of .
Solution
Solution 1
(Opposite angles are equal).
(Both are 90 degrees).
(Alt. Interior Angles are congruent).
Therefore and are similar. and are also similar.
is 9, therefore must equal 5. Similarly, must equal 3.
Because and are similar, the ratio of and , must also hold true for and . , so is of . By Pythagorean theorem, .
.
So .
.
Therefore .
Solution 2
Since is a rectangle, .
Since is a rectangle and , .
Since is a rectangle, .
So, is a transversal, and .
This is sufficient to prove that and .
Using ratios:
Since can't have 2 different lengths, both expressions for must be equal.
Solution 3
Since is a rectangle, , , and . From the Pythagorean Theorem, .
Lemma
Statement:
Proof: , 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 .
Also, , therefore
We can multiply both sides by to get that is twice of 10, or
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 |