Question

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 $BC^2.$

Answer

Answer: 110

Concepts:trapezoid coordinate geometry vector quadratic

Difficulty rating: 2450

Solution:

Place $B = (0, 0),$ $A = (\sqrt{11}, 0),$ $C = (0, h),$ and $D = (d, h),$ so that $\overline{BC}$ is vertical and $BC^2 = h^2.$ The diagonals give vectors $\overrightarrow{AC} = (-\sqrt{11}, h)$ and $\overrightarrow{BD} = (d, h),$ and perpendicularity means $-\sqrt{11}\,d + h^2 = 0,$ so $d = \frac{h^2}{\sqrt{11}}.$

Then $AD^2 = (d - \sqrt{11})^2 + h^2 = 1001.$ Setting $u = h^2,$ this becomes $\frac{(u - 11)^2}{11} + u = 1001,$ that is, $u^2 - 11u - 10890 = 0.$ The positive root is $u = \frac{11 + \sqrt{121 + 43560}}{2}$ $= \frac{11 + 209}{2} = 110,$ so $BC^2 = 110.$

2000 AIME II Problem 8

Answer: 110

Difficulty rating: 2450

Solution:

Problem 8 in Other Years