Difference between revisions of "2003 AMC 10A Problems/Problem 22"
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=== Solution 3 === | === Solution 3 === | ||
− | Since <math>ABCD</math> is a rectangle, <math> | + | Since <math>ABCD</math> is a rectangle, <math>CH=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 ==== | ||
Statement: <math>GCH \approx GEA</math> | Statement: <math>GCH \approx GEA</math> |
Revision as of 23:45, 4 January 2014
Contents
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 making remote request. No response to HTTP request)
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
Solution 4
We extend such that it intersects at . Since is a rectangle, it follows that , therefore, . Let . From the similarity of triangles and , we have the ratio (as , and ). and are the altitudes of and , respectively. Thus, , from which we have , thus
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 |
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