Difference between revisions of "2003 AMC 10A Problems/Problem 22"
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Proof: <math>\angle CGH=\angle EGA</math>, obviously. | Proof: <math>\angle CGH=\angle EGA</math>, obviously. | ||
− | < | + | <cmath>\begin{eqnarray*} |
− | \angle HCE=180^{\circ}-\angle CHG\\ | + | \angle HCE&=&180^{\circ}-\angle CHG\\ |
− | \angle DCE=\angle CHG-90^{\circ}\\ | + | \angle DCE&=&\angle CHG-90^{\circ}\\ |
− | \angle CEED=180-\angle CHG\\ | + | \angle CEED&=&180-\angle CHG\\ |
− | \angle GEA=\angle GCH | + | \angle GEA&=&\angle GCH |
− | \end{eqnarray}</ | + | \end{eqnarray*}</cmath> |
Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar. | Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar. | ||
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Let <math>GC=x</math>. | Let <math>GC=x</math>. | ||
− | <cmath>\begin{eqnarray} | + | <cmath>\begin{eqnarray*} |
− | \dfrac{x}{3}=\dfrac{x+4\sqrt{5}}{5}\\ | + | \dfrac{x}{3}&=&\dfrac{x+4\sqrt{5}}{5}\\ |
− | 5x=3x+12\sqrt{5}\\ | + | 5x&=&3x+12\sqrt{5}\\ |
− | 2x=12\sqrt{5}\\ | + | 2x&=&12\sqrt{5}\\ |
− | x=6\sqrt{5} | + | x&=&6\sqrt{5} |
− | \end{eqnarray}</cmath> | + | \end{eqnarray*}</cmath> |
Also, <math>\triangle GFE\approx \triangle CDE</math>, therefore | Also, <math>\triangle GFE\approx \triangle CDE</math>, therefore |
Revision as of 18:55, 10 March 2015
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.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.