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

Let $MA=a$ and $MD=d.$ It follows that $MC=a+2$ and $MB=a+4.$

As shown below, note that $\triangle MAD$ and $\triangle MBC$ are both right triangles. [asy] size(300); import graph3; import solids;  currentprojection=orthographic((0.5,-0.25,-0.5)); triple A, B, C, D, M; A = (-2sqrt(10),0,0); B = (-2sqrt(10),-6sqrt(2),0); C = (0,-6sqrt(2),0); D = (0,0,0); M = (0,0,3); draw(surface(M--A--D--cycle),yellow); draw(surface(M--B--C--cycle),yellow); draw(D--A--B--C^^A--M^^B--M^^C--M^^D--M); draw(C--D,dashed); dot(A^^B^^C^^D^^M,linewidth(4.5)); label("$A$",A,2*dir(A-B)); label("$B$",B,2*dir(B-A)); label("$C$",C,2*dir(C-D)); label("$D$",D,2*dir(D-C)); label("$M$",M,2*dir((1,1,0))); label("$a$",midpoint(M--A),2*dir((-1,-1,0)),red); label("$d$",midpoint(M--D),2*dir((1,1,0)),red); label("$a+2$",midpoint(M--C),2*dir((1,-1,0)),red); label("$a+4$",midpoint(M--B),2*dir((-1,1,0)),red); [/asy] By the Pythagorean Theorem, we have \begin{alignat*}{6} AD^2 &= MA^2 - MD^2 &&= a^2 - d^2, \\ BC^2 &= MB^2 - MC^2 &&= (a+4)^2 - (a+2)^2. \end{alignat*} Since $AD=BC$ in rectangle $ABCD,$ we equate the expressions for $AD^2$ and $BC^2,$ then rearrange and factor: \begin{align*} a^2 - d^2 &= (a+4)^2 - (a+2)^2 \\ a^2 - d^2 &= 4a + 12 \\ a^2 - 4a - d^2 &= 12 \\ (a-2)^2 - d^2 &= 16 \\ (a+d-2)(a-d-2) &= 16. \end{align*} As $a+d-2$ and $a-d-2$ have the same parity, we get $a+d-2=8$ and $a-d-2=2,$ from which $(a,d)=(7,3).$

Applying the Pythagorean Theorem to right $\triangle MAD$ and right $\triangle MCD,$ we obtain $AD=2\sqrt{10}$ and $CD=6\sqrt2,$ respectively.

Let the brackets denote areas. Together, the volume of pyramid $MABCD$ is \[\frac13\cdot [ABCD]\cdot MD = \frac13\cdot (AD\cdot CD)\cdot MD = \boxed{\textbf{(A) }24\sqrt5}.\] ~Lopkiloinm ~MRENTHUSIASM

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 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)

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

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 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

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