2021 AMC 12A Problem 17

Problem 17 of 25IntermediateGeometry

Trapezoid

ABCD

has

AB \parallel CD,

BC = CD = 43,

and

AD \perp BD.

Let

O

be the intersection of the diagonals

AC

and

BD,

and let

P

be the midpoint of

BD.

Given that

OP = 11,

the length

AD

can be written in the form

m\sqrt n,

where

m

and

n

are positive integers and

n

is not divisible by the square of any prime. What is

m + n?

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Solution

Place

D = (0,0)

with

B = (b, 0)

on one axis and

A = (0, a)

on the other, so that

AD \perp BD.

Since

CD \parallel AB,

write

C = t(b, -a)

for some

t.

Then

CD = t\sqrt{a^2+b^2}

and

BC^2 = b^2(1-t)^2 + t^2a^2.

Setting

BC = CD

gives

t^2 = (1-t)^2,

t = \tfrac12.

Thus

C = \left(\tfrac{b}{2}, -\tfrac{a}{2}\right),

and

CD = 43

gives

a^2 + b^2 = 4\cdot 43^2 = 7396.

The diagonal

AC

meets

BD

(the

x

-axis) at

O = \left(\tfrac{b}{3}, 0\right),

while

P = \left(\tfrac{b}{2}, 0\right).

Hence

OP = \tfrac{b}{6} = 11,

b = 66.

Then

a^2 = 7396 - 66^2 = 3040,

AD = a = \sqrt{3040} = 4\sqrt{190}.

With

m = 4

and

n = 190,

we get

m + n = 194.

Thus, the correct answer is D.