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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

Let $x = BC$ be the height of the trapezoid, and let $y = CD$ . Since $AC \perp BD$ , it follows that $\triangle BAC \sim \triangle CBD$ , so $\frac{x}{\sqrt{11}} = \frac{y}{x} \Longrightarrow x^2 = y\sqrt{11}$ .

Let $E$ be the foot of the altitude from $A$ to $\overline{CD}$ . Then $AE = x$ , and $ADE$ is a right triangle. By the Pythagorean Theorem,

$x^2 + \left(y-\sqrt{11}\right)^2 = 1001 \Longrightarrow x^4 - 11x^2 - 11^2 \cdot 9 \cdot 10 = 0$

The positive solution to this quadratic equation is $x^2 = \boxed{110}$ .

size(200); pathpen = linewidth(0.7);pair C=(0,0),B=(0,110^.5),A=(11^.5,B.y),D=(10*11^.5,0),E=foot(A,C,D);D(MP("A",A...

See also

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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2000_AIME_II_Problems/Problem_8"

Category: Intermediate Geometry Problems

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