Difference between revisions of "2015 AIME II Problems/Problem 7"
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Let <math>F</math> be the intersection between <math>AE</math> and <math>PQ</math>. By the similar triangles, we know that <math>\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}</math>. Since <math>PF+FQ=PQ=\omega</math>. We can solve for <math>PF</math> and <math>FQ</math> in terms of <math>\omega</math>. We get that <math>PF=\frac{48}{125} \omega</math> and <math>FQ=\frac{77}{125} \omega</math>. | Let <math>F</math> be the intersection between <math>AE</math> and <math>PQ</math>. By the similar triangles, we know that <math>\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}</math>. Since <math>PF+FQ=PQ=\omega</math>. We can solve for <math>PF</math> and <math>FQ</math> in terms of <math>\omega</math>. We get that <math>PF=\frac{48}{125} \omega</math> and <math>FQ=\frac{77}{125} \omega</math>. | ||
− | Let's work with <math>PF</math>. We know that <math>PQ</math> is parallel to <math>BC</math> so <math>\Delta APF</math> is similar to <math>\Delta ABE</math>. We can set up the proportion: | + | Let's work with <math>PF</math>. We know that <math>PQ</math> is parallel to <math>BC</math> so <math>\Delta APF</math> is similar to <math>\Delta ABE</math>. We can set up the proportion: |
+ | |||
<math>\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}</math>. Solving for <math>AF</math>, <math>AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega</math>. | <math>\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}</math>. Solving for <math>AF</math>, <math>AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega</math>. | ||
Revision as of 12:40, 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
is similar to
.
Let be the intersection between
and
. By the similar triangles, we know that
. Since
. We can solve for
and
in terms of
. We get that
and
.
Let's work with . We know that
is parallel to
so
is similar to
. We can set up the proportion:
. Solving for
,
.
We can solve for then since we know that
and
.
Therefore, .
This means that .
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.