Difference between revisions of "2015 AIME II Problems/Problem 7"
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==Problem== | ==Problem== | ||
− | Triangle <math>ABC</math> has side lengths <math>AB = 12</math>, <math>BC = 25</math>, and <math>CA = 17</math>. Rectangle <math>PQRS</math> has vertex <math>P</math> on <math>\overline{AB}</math>, vertex <math>Q</math> on <math>\overline{AC}</math>, and vertices <math>R</math> and <math>S</math> on <math>\overline{BC}</math>. In terms of the side length <math>PQ = \omega</math>, the area of <math>PQRS</math> can be expressed as the quadratic polynomial <cmath>Area(PQRS) = \alpha \omega - \beta \omega^2.</cmath> | + | Triangle <math>ABC</math> has side lengths <math>AB = 12</math>, <math>BC = 25</math>, and <math>CA = 17</math>. Rectangle <math>PQRS</math> has vertex <math>P</math> on <math>\overline{AB}</math>, vertex <math>Q</math> on <math>\overline{AC}</math>, and vertices <math>R</math> and <math>S</math> on <math>\overline{BC}</math>. In terms of the side length <math>PQ = \omega</math>, the area of <math>PQRS</math> can be expressed as the quadratic polynomial <cmath>\text\Area(PQRS) = \alpha \omega - \beta \omega^2.</cmath> |
Then the coefficient <math>\beta = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | Then the coefficient <math>\beta = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. |
Revision as of 17:33, 24 September 2020
Contents
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
\[\text\Area(PQRS) = \alpha \omega - \beta \omega^2.\] (Error compiling LaTeX. Unknown error_msg)
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
,
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 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
Solving gives
, so the answer is
.
Solution 4
Using the diagram from Solution 2 above, label to be
. Through Heron's formula, the area of
turns out to be
, so using
as the height and
as the base yields
. Now, through the use of similarity between
and
, you find
. Thus,
. To find the height of the rectangle, subtract
from
to get
, and multiply this by the other given side
to get
for the area of the rectangle. Finally,
.
Solution 5
Using the diagram as shown in Solution 2, let and
Now, by Heron's formula, we find that the
. Hence,
Now, we see that
We easily find that
Hence,
Now, we see that
Now, it is obvious that we want to find in terms of
.
Looking at the diagram, we see that because is a rectangle,
Hence.. we can now set up similar triangles.
We have that .
Plugging back in..
Simplifying, we get
Hence,
Solution 6
Proceed as in solution 1. When is equal to zero,
is equal to the altitude. This means that
is equal to
, so
, yielding
.
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.