Difference between revisions of "2003 AIME II Problems/Problem 11"
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Draw a vertical line through <math>M</math> and a horizontal line through <math>D</math>. Suppose these two lines meet at <math>X</math>. then <math>MX=\frac{24}{7}DX</math> so <math>MD=\frac{25}{7}DX=\frac{25}{24}MD</math> by the pythagorean theorem. | Draw a vertical line through <math>M</math> and a horizontal line through <math>D</math>. Suppose these two lines meet at <math>X</math>. then <math>MX=\frac{24}{7}DX</math> so <math>MD=\frac{25}{7}DX=\frac{25}{24}MD</math> by the pythagorean theorem. | ||
So <math>MX=2.4\sqrt{11}</math> and <math>DX=.7\sqrt{11}</math> so the coordinates of D are <math>(12-.7\sqrt{11},2.5-2.4\sqrt{11})</math>. | So <math>MX=2.4\sqrt{11}</math> and <math>DX=.7\sqrt{11}</math> so the coordinates of D are <math>(12-.7\sqrt{11},2.5-2.4\sqrt{11})</math>. | ||
− | Since we know the coordinates of each of the vertices of <math>\ | + | Since we know the coordinates of each of the vertices of <math>\Delta CMD</math>, we can apply the [[Shoelace Theorem]] to find the area of <math>\Delta CMD, \frac{527 \sqrt{11}}{40}</math>. |
− | [[ | ||
Revision as of 16:01, 21 September 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,
.
Solution 3
Suppose is plotted on the cartesian plane with
at
,
at
, and
at
.
Then
is at
. Since
is isosceles,
is perpendicular to
, and since
and
.
The slope of
is
so the slope of
is
.
Draw a vertical line through
and a horizontal line through
. Suppose these two lines meet at
. then
so
by the pythagorean theorem.
So
and
so the coordinates of D are
.
Since we know the coordinates of each of the vertices of
, we can apply the Shoelace Theorem to find the area of
.
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 |