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

Revision as of 02:00, 28 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 $MA=a$ 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.

DIAGRAM WILL BE READY SOON.

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 $AD=BC,$ 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 $(a,d)=(7,3).$

By the Pythagorean Theorem on right $\triangle MAD,$ we have $AD=2\sqrt{10}.$

By the Pythagorean Theorem on right $\triangle MCD,$ we have $CD=6\sqrt2.$

Together, the volume of pyramid $MABCD$ is $\frac13\cdot AD\cdot CD\cdot MD = \boxed{\textbf{(A) }24\sqrt5}.$ ~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

@@ Line 5: / Line 5: @@
 ==Solution 1==
-Let <math>MA=a</math> and <math>MD=d.</math> It follows that <math>MC=a+2</math> and <math>MB=a+4,</math> as shown below:
+Let <math>MA=a</math> and <math>MD=d.</math> It follows that <math>MC=a+2</math> and <math>MB=a+4.</math>
+As shown below, note that <math>\triangle MAD</math> and <math>\triangle MBC</math> are both right triangles.
 <b>DIAGRAM WILL BE READY SOON.</b>
-Note that <math>\triangle MAD</math> and <math>\triangle MBC</math> are both right triangles. By the Pythagorean Theorem, we have
+By the Pythagorean Theorem, we have
 <cmath>\begin{alignat*}{6}
 AD^2 &= MA^2 - MD^2 &&= a^2 - d^2, \\
@@ Line 24: / Line 26: @@
 As <math>a+d-2</math> and <math>a-d-2</math> have the same parity, we get <math>a+d-2=8</math> and <math>a-d-2=2,</math> from which <math>(a,d)=(7,3).</math>
-By the Pythagorean Theorem on right <math>\triangle MAD,</math> we have <math>AD=2\sqrt{10}.</math> By the Pythagorean Theorem on right <math>\triangle MCD,</math> we have <math>CD=6\sqrt2.</math>
+By the Pythagorean Theorem on right <math>\triangle MAD,</math> we have <math>AD=2\sqrt{10}.</math>
+By the Pythagorean Theorem on right <math>\triangle MCD,</math> we have <math>CD=6\sqrt2.</math>
 Together, the volume of pyramid <math>MABCD</math> is <cmath>\frac13\cdot AD\cdot CD\cdot MD = \boxed{\textbf{(A) }24\sqrt5}.</cmath>

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