Difference between revisions of "2015 AIME II Problems/Problem 7"
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==Solution #2== | ==Solution #2== | ||
− | Diagram: | + | Diagram:\ |
(Diagram goes here) | (Diagram goes here) | ||
− | + | \ | |
− | Similar triangles can also solve the problem. | + | Similar triangles can also solve the problem.\ |
− | First, solve for the area of the triangle. <math>[ABC] = 90</math>. This can be done by Heron's Formula or placing an <math>8-15-17</math> right triangle on <math>BC</math> and solving. (The <math>8</math> side would be collinear with line <math>AB</math>) | + | First, solve for the area of the triangle. <math>[ABC] = 90</math>. This can be done by Heron's Formula or placing an <math>8-15-17</math> right triangle on <math>BC</math> and solving. (The <math>8</math> side would be collinear with line <math>AB</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>. | 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>\Delta APQ</math> is similar to <math>\Delta ABC</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</math> is similar to <math>\Delta ABC</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 <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>. | ||
− | We can solve for <math>PS</math> then since we know that <math>PS=FE</math> and <math>FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega</math>. | + | We can solve for <math>PS</math> then since we know that <math>PS=FE</math> and <math>FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega</math>.\ |
− | Therefore, <math>[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2</math>. | + | Therefore, <math>[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2</math>.\ |
This means that <math>\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}</math>. | This means that <math>\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}</math>. | ||
Revision as of 12:38, 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.