Difference between revisions of "2015 AIME II Problems/Problem 7"
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Using the diagram from solution 2 above, label AF to be h. Through Heron's formula, the area of ABC turns out to be 90, so using AE as the height and BC as the base yields AE=36/
Using the diagram from solution 2 above, label AFto be h. Through Heron's formula, the area of ABCturns out to be 90, so using AEas the height and BCas the base yields AE=36/. Now, through the use of similarity between triangle APQand ABC, you find w25=h36/5. Thus, h=36w/. To find the height of the rectangle, subtract hfrom 36/to get (365-36w125), and multiply this by the other given side wto get 36w5-36w^2/for the area of the rectangle. 36+125>
Revision as of 12:12, 17 July 2015
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 .
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 the answer is .
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 .
This means that .
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 .
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, .
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