# 2021 AMC 12B Problems/Problem 14

## Problem

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 $MABCD?$ $\textbf{(A) }24\sqrt5 \qquad \textbf{(B) }60 \qquad \textbf{(C) }28\sqrt5\qquad \textbf{(D) }66 \qquad \textbf{(E) }8\sqrt{70}$

## Solution 1

This question is just about pythagorean theorem $$a^2+(a+2)^2-b^2 = (a+4)^2$$ $$2a^2+4a+4-b^2 = a^2+8a+16$$ $$a^2-4a+4-b^2 = 16$$ $$(a-2+b)(a-2-b) = 16$$ $$a=3, b=7$$ With these calculation, we find out answer to be $\boxed{\textbf{(A) }24\sqrt5}$ ~Lopkiloinm

## Solution 2

Let $\overline{AD}$ be $b$, $\overline{CD}$ be $a$, $\overline{MD}$ be $x$, $\overline{MC}$, $\overline{MA}$, $\overline{MB}$ be $t$, $t-2$, $t+2$ respectively.

We have three equations: $$a^2 + x^2 = t^2$$ $$a^2 + b^2 + x^2 = t^2 + 4t + 4$$ $$b^2 + x^2 = t^2 - 4t + 4$$

Subbing in the first and third equation into the second equation, we get: $$t^2 - 8t - x^2 = 0$$ $$(t-4)^2 - x^2 = 16$$ $$(t-4-x)(t-4+x) = 16$$ Therefore, $$t = 9$$, $$x = 3$$ Solving for other values, we get $b = 2\sqrt{10}$, $a = 6\sqrt{2}$. The volume is then $$\frac{1}{3} abx = \boxed{\textbf{(A)}24\sqrt{5}}$$ ~jamess2022(burntTacos)

## See Also

 2021 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS