2003 AMC 10A Problems/Problem 22
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
Solution 5
Since and
we have
Thus,
Suppose
and
Thus, we have
Additionally, now note that
which is pretty obvious from insight, but can be proven by AA with extending
to meet
From this new pair of similar triangles, we have
Therefore, we have by combining those two equations,
Solving, we have
and therefore
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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