Difference between revisions of "2021 AMC 12B Problems/Problem 14"

Revision as of 16:30, 19 June 2021

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 $\begin{align*} a^2+(a+2)^2-b^2 &= (a+4)^2 \\ 2a^2+4a+4-b^2 &= a^2+8a+16 \\ a^2-4a+4-b^2 &= 16 \\ (a-2+b)(a-2-b) &= 16, \end{align*}$ from which $(a,b)=(3,7).$ 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: $\begin{align*} a^2 + x^2 &= t^2, \\ a^2 + b^2 + x^2 &= t^2 + 4t + 4, \\ b^2 + x^2 &= t^2 - 4t + 4. \end{align*}$ Subbing in the first and third equation into the second equation, we get: $\begin{align*} t^2 - 8t - x^2 &= 0, \\ (t-4)^2 - x^2 &= 16, \\ (t-4-x)(t-4+x) &= 16. \end{align*}$ Therefore, we have $t = 9$ and $x = 3$ .

Solving for other values, we get $b = 2\sqrt{10}$ , $a = 6\sqrt{2}$ . The volume is then $\frac{1}{3} abx = \boxed{\textbf{(A) }24\sqrt5}.$

~jamess2022 (burntTacos)

Video Solution by Hawk Math

https://www.youtube.com/watch?v=p4iCAZRUESs

Video Solution by OmegaLearn (Pythagorean Theorem and Volume of Pyramid)

https://youtu.be/4_Oqp_ECLRw

@@ Line 37: / Line 37: @@
 The volume is then <cmath>\frac{1}{3} abx = \boxed{\textbf{(A) }24\sqrt5}.</cmath>
-~jamess2022(burntTacos)
+~jamess2022 (burntTacos)
 ==Video Solution by Hawk Math==

2021 AMC 12B (Problems • Answer Key • Resources)
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