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 16:33, 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. ! Undefined control sequence.)
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.