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
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
BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \\ BC &= \frac{70}{3}
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)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. Unknown error_msg)Dividing the two equations yields
The answer is as above.
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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