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

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draw(surface(M--A--D--cycle),yellow);
 
draw(surface(M--A--D--cycle),yellow);
 
draw(surface(M--B--C--cycle),yellow);
 
draw(surface(M--B--C--cycle),yellow);
−
draw(A--B--C--D--cycle^^A--M^^B--M^^C--M^^D--M);
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draw(D--A--B--C^^A--M^^B--M^^C--M^^D--M);
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draw(C--D,dashed);
 
dot(A^^B^^C^^D^^M,linewidth(4.5));
 
dot(A^^B^^C^^D^^M,linewidth(4.5));
 
label("$A$",A,2*dir(A-B));
 
label("$A$",A,2*dir(A-B));
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BC^2 &= MB^2 - MC^2 &&= (a+4)^2 - (a+2)^2.
 
BC^2 &= MB^2 - MC^2 &&= (a+4)^2 - (a+2)^2.
 
\end{alignat*}</cmath>
 
\end{alignat*}</cmath>
−
Since <math>AD=BC,</math> we equate the expressions for <math>AD^2</math> and <math>BC^2,</math> then rearrange and factor:
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Since <math>AD=BC</math> in rectangle <math>ABCD,</math> we equate the expressions for <math>AD^2</math> and <math>BC^2,</math> then rearrange and factor:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
a^2 - d^2 &= (a+4)^2 - (a+2)^2 \\
 
a^2 - d^2 &= (a+4)^2 - (a+2)^2 \\
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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>
 
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>
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Applying the Pythagorean Theorem to right <math>\triangle MAD</math> and right <math>\triangle MCD,</math> we obtain <math>AD=2\sqrt{10}</math> and <math>CD=6\sqrt2,</math> respectively.
  
−
By the Pythagorean Theorem on right <math>\triangle MCD,</math> we have <math>CD=6\sqrt2.</math>
+
Let the brackets denote areas. Together, the volume of pyramid <math>MABCD</math> is <cmath>\frac13\cdot [ABCD]\cdot MD = \frac13\cdot (AD\cdot CD)\cdot MD = \boxed{\textbf{(A) }24\sqrt5}.</cmath>
−
 
 
−
Together, the volume of pyramid <math>MABCD</math> is <cmath>\frac13\cdot (AD\cdot CD)\cdot MD = \boxed{\textbf{(A) }24\sqrt5}.</cmath>
 
 
~Lopkiloinm ~MRENTHUSIASM
 
~Lopkiloinm ~MRENTHUSIASM
  
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~jamess2022 (burntTacos)
 
~jamess2022 (burntTacos)
 +
 +
 +
==Video Solution (Fast! Just 4 min!)==
 +
https://youtu.be/Bo2EvRZdRnA
 +
 +
~<i>Education, the Study of Everything</i>
  
 
==Video Solution by Hawk Math==
 
==Video Solution by Hawk Math==
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== Video Solution by OmegaLearn (Pythagorean Theorem and Volume of Pyramid) ==
 
== Video Solution by OmegaLearn (Pythagorean Theorem and Volume of Pyramid) ==
 
https://youtu.be/4_Oqp_ECLRw
 
https://youtu.be/4_Oqp_ECLRw
 +
 +
~pi_is_3.14
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021|ab=B|num-b=13|num-a=15}}
 
{{AMC12 box|year=2021|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:38, 23 July 2023

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 (Fast! Just 4 min!)

https://youtu.be/Bo2EvRZdRnA

~Education, the Study of Everything

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

~pi_is_3.14

See Also

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