Difference between revisions of "2003 AMC 10A Problems/Problem 22"
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== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
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 === | ||
+ | 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}\ | ||
+ | 5x=3x+12\sqrt{5}\ | ||
+ | 2x=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 == | == See Also == |
Revision as of 09:58, 6 March 2008
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
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 2
Since is a rectangle,
,
, and
. From the Pythagorean Theorem,
.
Lemma
Statement:
Proof: , obviously.
$
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
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 |