Difference between revisions of "2003 AIME II Problems/Problem 11"
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== Solution == | == Solution == | ||
+ | '''Solution 1''' | ||
+ | |||
We use the [[Pythagorean Theorem]] on <math>ABC</math> to determine that <math>AB=25.</math> | We use the [[Pythagorean Theorem]] on <math>ABC</math> to determine that <math>AB=25.</math> | ||
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Thus, | Thus, | ||
− | <math>[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}=\frac{527\sqrt{11}} {40}.</math> | + | <math>[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}= \frac{527\sqrt{11}} {40}.</math> |
Hence, the answer is <math>527+11+40=\boxed{578}.</math> | Hence, the answer is <math>527+11+40=\boxed{578}.</math> | ||
+ | |||
+ | '''Solution 2''' | ||
+ | |||
+ | By the [[Pythagorean Theorem]] in <math>\Delta AMD</math>, we get <math>DM=\frac{5\sqrt{11}} {2}</math>. Since <math>ABC</math> is a right triangle, <math>M</math> is the circumcenter and thus, <math>CM=\frac{25} {2}</math>. We let <math>\angle CMD=\theta</math>. By the [[Law of Cosines]], | ||
+ | |||
+ | <math>2 \cdot (12.5)^2-2 \cdot (12.5)^2 * \cos (90+\theta).</math> | ||
+ | |||
+ | It follows that <math>\sin \theta = \frac{527} {625}</math>. Thus, | ||
+ | <math>[CMD]=\frac{1} {2} (12.5) (\frac{5\sqrt{11}} {2})(\frac{527} {625})=\frac{527\sqrt{11}} {40}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2003|n=II|num-b=10|num-a=12}} | {{AIME box|year=2003|n=II|num-b=10|num-a=12}} |
Revision as of 13:28, 27 August 2008
Problem
Triangle is a right triangle with
and right angle at
Point
is the midpoint of
and
is on the same side of line
as
so that
Given that the area of triangle
may be expressed as
where
and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime, find
Solution
Solution 1
We use the Pythagorean Theorem on to determine that
Let be the orthogonal projection from
to
Thus,
,
, and
From the third equation, we get
By the Pythagorean Theorem in we have
Thus,
In , we use the Pythagorean Theorem to get
Thus,
Hence, the answer is
Solution 2
By the Pythagorean Theorem in , we get
. Since
is a right triangle,
is the circumcenter and thus,
. We let
. By the Law of Cosines,
It follows that . Thus,
.
See also
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |