2000 AIME II Problems/Problem 8

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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, $CD=a^2/11$. 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 $a^2=p$ (which is what we're trying to find) and simplifying yields the equation $n^2-11n-10890=0$, and the positive solution is 110.

See also

2000 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions