2021 AMC 12B Problem 14

Problem 14 of 25IntermediateGeometry

Let

ABCD

be a rectangle and let

\overline{DM}

be a segment perpendicular to the plane of

ABCD.

Suppose that

\overline{DM}

has integer length, and the lengths of

\overline{MA},

\overline{MC},

and

\overline{MB}

are consecutive odd positive integers (in this order). What is the volume of pyramid

MABCD?

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Solution

Place

D

at the origin with

A,

C

along the rectangle’s edges and

M

directly above

D.

Then

MA^2=AD^2+DM^2,

MC^2=CD^2+DM^2,

and

MB^2=AD^2+CD^2+DM^2.

Thus

MB^2=MA^2+MC^2-DM^2.

Writing

MA,MC,MB=k,k+2,k+4,

we get

DM^2=k^2+(k+2)^2

-(k+4)^2

=k^2-4k-12.

DM=t

is a positive integer, then

(k-2)^2-t^2=16,

(k-2-t)(k-2+t)=16.

The only positive same-parity factor pair giving

t>0

(2,8),

which yields

k=7

and

t=3.

Thus

AD^2=49-9=40

and

CD^2=81-9=72.

The base area is

AD\cdot CD=\sqrt{40}\cdot\sqrt{72}

=\sqrt{2880}=24\sqrt5,

and the volume is

\tfrac13\cdot 24\sqrt5\cdot 3=24\sqrt5.

Thus, the correct answer is A.