Difference between revisions of "2015 AIME II Problems/Problem 7"
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- solution by abvenkgoo | - solution by abvenkgoo | ||
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+ | ==Solution 3== | ||
+ | Heron's Formula gives <math>[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90,</math> so the altitude from <math>A</math> to <math>BC</math> has length <math>\dfrac{2[ABC]}{BC} = \dfrac{36}{5}.</math> | ||
+ | |||
+ | Now, draw a parallel to <math>AB</math> from <math>Q</math>, intersecting <math>BC</math> at <math>T</math>. Then <math>BT = w</math> in parallelogram <math>QPBT</math>, and so <math>CT = 25 - w</math>. Clearly, <math>CQT</math> and <math>CAB</math> are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so | ||
+ | <math></math>\frac{QR}{\frac{36}{5}} = \frac{25 - w}{25}.<math> | ||
+ | Solving gives </math>[PQRS] = \dfrac{36}{5} \cdot \dfrac{25 - w}{25} = \dfrac{36w}{5} - \dfrac{36w^2}{125}<math>, so the answer is </math>36 + 125 = 161$. | ||
+ | |||
+ | - solution by suli | ||
==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 18:48, 29 March 2015
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 1
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
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 .
- solution by abvenkgoo
Solution 3
Heron's Formula gives so the altitude from
to
has length
Now, draw a parallel to from
, intersecting
at
. Then
in parallelogram
, and so
. Clearly,
and
are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so
$$ (Error compiling LaTeX. Unknown error_msg)\frac{QR}{\frac{36}{5}} = \frac{25 - w}{25}.
[PQRS] = \dfrac{36}{5} \cdot \dfrac{25 - w}{25} = \dfrac{36w}{5} - \dfrac{36w^2}{125}
36 + 125 = 161$.
- solution by suli
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.