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 17:55, 10 March 2015
Contents
[hide]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.