Difference between revisions of "2015 AIME II Problems/Problem 7"
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After finding the area, solve for the altitude to <math>BC</math>. Let <math>E</math> be the intersection of the altitude from <math>A</math> and side <math>BC</math>. Then <math>AE = \frac{36}{5}</math>. | After finding the area, solve for the altitude to <math>BC</math>. Let <math>E</math> be the intersection of the altitude from <math>A</math> and side <math>BC</math>. Then <math>AE = \frac{36}{5}</math>. | ||
Solving for <math>BE</math> using the Pythagorean Formula, we get <math>BE = \frac{48}{5}</math>. We then know that <math>CE = \frac{77}{5}</math>. | Solving for <math>BE</math> using the Pythagorean Formula, we get <math>BE = \frac{48}{5}</math>. We then know that <math>CE = \frac{77}{5}</math>. | ||
− | Now consider the rectangle <math>PQRS</math>. Since <math>SR</math> is collinear with <math>BC</math> and parallel to <math>PQ</math>, <math>PQ</math> is parallel to <math>BC</math> meaning <math>\ | + | Now consider the rectangle <math>PQRS</math>. Since <math>SR</math> is collinear with <math>BC</math> and parallel to <math>PQ</math>, <math>PQ</math> is parallel to <math>BC</math> meaning <math>\Delta APQ ~ \Delta ABC</math>. |
==See also== | ==See also== | ||
{{AIME box|year=2015|n=II|num-b=6|num-a=8}} | {{AIME box|year=2015|n=II|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:22, 29 March 2015
Contents
[hide]Problem
Triangle has side lengths
,
, and
. Rectangle
has vertex
on
, vertex
on
, and vertices
and
on
. In terms of the side length
, the area of
can be expressed as the quadratic polynomial
Area() =
.
Then the coefficient , where
and
are relatively prime positive integers. Find
.
Solution
If , the area of rectangle
is
, so
and . If
, we can reflect
over PQ,
over
, and
over
to completely cover rectangle
, so the area of
is half the area of the triangle. Using Heron's formula, since
,
so
and
so the answer is .
Solution #2
Diagram: (Diagram goes here)
Similar triangles can also solve the problem.
First, solve for the area of the triangle. . This can be done by Heron's Formula or placing an
right triangle on
and solving. (The
side would be collinear with line
)
After finding the area, solve for the altitude to
. Let
be the intersection of the altitude from
and side
. Then
.
Solving for
using the Pythagorean Formula, we get
. We then know that
.
Now consider the rectangle
. Since
is collinear with
and parallel to
,
is parallel to
meaning
.
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.