Difference between revisions of "2015 AIME II Problems/Problem 7"
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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 | 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 | ||
<cmath>\frac{QR}{\frac{36}{5}} = \frac{25 - w}{25}.</cmath> | <cmath>\frac{QR}{\frac{36}{5}} = \frac{25 - w}{25}.</cmath> | ||
− | 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</math>. | + | Solving gives <math>[PQRS] = \dfrac{36}{5} \cdot \dfrac{25 - w}{25} \cdot w = \dfrac{36w}{5} - \dfrac{36w^2}{125}</math>, so the answer is <math>36 + 125 = 161</math>. |
− | |||
==Solution 4== | ==Solution 4== |
Revision as of 03:16, 5 August 2021
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
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.