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

Revision as of 23:36, 4 November 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 $MD=b$ and $MA=a.$ 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 $(a,b)=(7,3).$ 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 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

@@ Line 37: / Line 37: @@
 ~jamess2022 (burntTacos)
-==Solution 3 (Six Variables, Five Equations)==
-We are given that
-<cmath>\begin{align*}
-MC&=MA+2, &\hspace{27mm}(1) \\
-MB&=MA+4. &\hspace{27mm}(2)
-\end{align*}</cmath>
-We apply the Pythagorean Theorem to right triangles <math>\triangle MDA,\triangle MDC,</math> and <math>\triangle MDB,</math> respectively:
-<cmath>\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*}</cmath>
-Subtracting <math>(4)</math> from <math>(5)</math> and applying <math>(1)</math> and <math>(2),</math> we express <math>DA^2</math> in terms of <math>MA:</math>
-<cmath>\begin{align*}
-DA^2&=MB^2-MC^2 \\
-&=(MA+4)^2-(MA+2)^2 \\
-&=4MA+12. \hspace{37mm}(\bigstar)
-\end{align*}</cmath>
-We substitute <math>(\bigstar)</math> into <math>(3),</math> then rearrange:
-<cmath>\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*}</cmath>
-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, \\
-MA-MD-2&=2,
-\end{align*}</cmath>
-from which <math>MA=7</math> and <math>MD=3.</math>
-Substituting the current results into <math>(1),(3),(4),</math> we get <math>MC=9,DA=2\sqrt{10},DC=6\sqrt{2},</math> respectively.
-Let the brackets denote areas. Finally, the volume of pyramid <math>MABCD</math> is
-<cmath>\begin{align*}
-V_{MABCD}&=\frac13\cdot[ABCD]\cdot MD \\
-&=\frac13\cdot(DA\cdot DC)\cdot MD \\
-&=\boxed{\textbf{(A) }24\sqrt5}.
-\end{align*}</cmath>
-~MRENTHUSIASM
 ==Video Solution by Hawk Math==

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