Question

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 $MABCD?$

$24\sqrt5$

$60$

$28\sqrt5$

$66$

$8\sqrt{70}$

Answer: A

Concepts:3D geometry Pythagorean Theorem pyramid

Difficulty rating: 1790

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 $DM^2=k^2+(k+2)^2$ $-(k+4)^2$ $=k^2-4k-12.$

If $DM=t$ is a positive integer, then $(k-2)^2-t^2=16,$ so $(k-2-t)(k-2+t)=16.$ The only positive same-parity factor pair giving $t>0$ is $(2,8),$ 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.

2021 AMC 12B Spring Problem 14

Answer: A

Difficulty rating: 1790

Solution:

Problem 14 in Other Years