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

Revision as of 16:13, 20 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

Let $MD=a$ and $MA=b.$ 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 $AD=b$ , $CD=a$ , $MD=x$ , $MC=t$ . It follows that $MA=t-2$ and $MB=t+2$ .

We have three equations: $\begin{align*} a^2 + x^2 &= t^2, \\ a^2 + b^2 + x^2 &= t^2 + 4t + 4, \\ b^2 + x^2 &= t^2 - 4t + 4. \end{align*}$ Substituting in the first and third equations 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)

Solution 3 (Six Variables, Five Equations)

We are given that $\begin{align*} MC&=MA+2, &\hspace{27mm}(1) \\ MB&=MA+4. &\hspace{27mm}(2) \end{align*}$ Applying the Pythagorean Theorem to right triangles $\triangle MDA,\triangle MDC,$ and $\triangle MDB,$ we have $\begin{align*} MA^2&=MD^2+DA^2, &\hspace{5mm}(3) \\ MC^2&=MD^2+DC^2, &\hspace{5mm}(4) \\ MB^2&=MD^2+DB^2 \\ &=MD^2+CA^2 \\ &=MD^2+DA^2+DC^2. &\hspace{5mm}(5) \end{align*}$ Subtracting $(4)$ from $(5)$ and applying $(1)$ and $(2),$ we express $DA^2$ in terms of $MA:$ $\begin{align*} DA^2&=MB^2-MC^2 \\ &=(MA+4)^2-(MA+2)^2 \\ &=4MA+12. \hspace{37mm}(\bigstar) \end{align*}$ We substitute $(\bigstar)$ into $(3),$ then rearrange: $\begin{align*} MA^2&=MD^2+(4MA+12) \\ MA^2-4MA-MD^2&=12 \\ (MA-2)^2-MD^2&=16 \\ (MA+MD-2)(MA-MD-2)&=16. \end{align*}$ As $MA+MD-2$ and $MA-MD-2$ always have the same parity, both factors must be even. Since $MA>MD,$ the only possibility is $\begin{align*} MA+MD-2&=8, \\ MA-MD-2&=2, \end{align*}$ from which $MD=3$ and $MA=7.$

Substituting the current results into $(1),(3),(4),$ we get $MC=9,DA=2\sqrt{10},DC=6\sqrt{2},$ respectively.

Let the brackets denote areas. Finally, the volume of pyramid $MABCD$ is $\begin{align*} V_{MABCD}&=\frac13\cdot[ABCD]\cdot MD \\ &=\frac13\cdot(DA\cdot DC)\cdot MD \\ &=\boxed{\textbf{(A) }24\sqrt5}. \end{align*}$ ~MRENTHUSIASM

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 65: / Line 65: @@
 (MA+MD-2)(MA-MD-2)&=16.
 \end{align*}</cmath>
-Note that <math>MA</math> and <math>MD</math> must have the same parity. Since <math>MA>MD,</math> the only possibility is
+As <math>MA+MD-2</math> and <math>MA-MD-2</math> always have the same parity, both factors must be even. Since <math>MA>MD,</math> the only possibility is
 <cmath>\begin{align*}
 MA+MD-2&=8, \\

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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