2004 AIME II Problems/Problem 7
Problem
is a rectangular sheet of paper that has been folded so that corner is matched with point on edge The crease is where is on and is on The dimensions and are given. The perimeter of rectangle is where and are relatively prime positive integers. Find
Contents
Solution
Solution 1 (synthetic)
Since is the perpendicular bisector of , it follows that (by SAS). By the Pythagorean Theorem, we have . Similarly, from , we have Thus the perimeter of is , and the answer is .
Solution 2 (analytic)
Let , so and , and let be the length of the rectangle. The slope of is and so the equation of is . We know that is perpendicular to and bisects . The slope of is thus , and so the equation of is . Let the point of intersection of be . Then the y-coordinate of is , so
\frac{14}{l}x &= y-8 = \frac{9}{2}\\ \frac{-l}{14}x &= y-25 = -\frac{25}{2}\\
\end{align*}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)Dividing the two equations yields
The answer is as above.
Solution 3 (Coordinate Bashing)
Firstly, observe that if we are given that and , the length of the triangle is given and the height depends solely on the length of . Let Point . Since , point E is at (8,0). Next, point is at since and point is at since by symmetry. Draw line segment . Notice that this is perpendicular to by symmetry. Next, find the slope of EB, which is . Then, the slope of is -.
Line EF can be written as y=. Plug in the point , and we get the equation of EF to be y=. Since the length of =25, a point on line lies on when . Plug in into our equation to get . . Therefore, our answer is .
See also
2004 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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