2000 AIME II Problems/Problem 8

Problem

In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$ .

Solution

Call the height of the trapezoid a. Triangles BAC and CBD are similar since AC is perpendicular to BD. Using similar triangles, $CD=a^2/\sqrt{11}$ . Call the foot of the altitude from A to CD H. We know AH=a, but AHD is a right triangle, so by the Pythagorean Theorem, $a=\sqrt{1001-(a^2/\sqrt{11}-\sqrt{11})^2}$ . Letting $a^2=n$ (which is what we're trying to find) and simplifying yields the equation $n^2-11n-10890=0$ , and the positive solution is 110.

2000 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2000 AIME II Problems/Problem 8

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