Difference between revisions of "2015 AIME II Problems/Problem 7"
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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> | ||
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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 | ||
+ | <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$. | ||
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+ | - 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 17:48, 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 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 |
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