2000 AIME II Problem 8

Attempt Problem 8 of the 2000 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AIME II solutions, or check the answer key.

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8.

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

Small hint:

Place

B

at the origin with

\overline{BC}

vertical:

A = (\sqrt{11}, 0),

C = (0, h),

D = (d, h)

Big hint:

Perpendicular diagonals force

d\sqrt{11} = h^2;

substitute into

AD^2 = 1001

to get a quadratic in

h^2

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.