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

m (→‎Solution 2)
m (→‎Solution 3 (Six Variables, Five Equations))
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==Solution 1==
 
==Solution 1==
−
This question is just about Pythagorean theorem
+
Let <math>MD=a</math> and <math>MA=b.</math> This question is just about Pythagorean theorem
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
a^2+(a+2)^2-b^2 &= (a+4)^2 \\
 
a^2+(a+2)^2-b^2 &= (a+4)^2 \\
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==Solution 2==
 
==Solution 2==
−
Let <math>\overline{AD}</math> be <math>b</math>, <math>\overline{CD}</math> be <math>a</math>, <math>\overline{MD}</math> be <math>x</math>,  
+
Let <math>AD=b</math>, <math>CD=a</math>, <math>MD=x</math>, <math>MC=t</math>. It follows that <math>MA=t-2</math> and <math>MB=t+2</math>.
−
<math>\overline{MC}</math>, <math>\overline{MA}</math>, <math>\overline{MB}</math> be <math>t</math>, <math>t-2</math>, <math>t+2</math> respectively.
 
  
 
We have three equations:
 
We have three equations:
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Subbing in the first and third equation into the second equation, we get:
 
Subbing in the first and third equation into the second equation, we get:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
t^2 - 8t - x^2 &= 0, \\
+
t^2 - 8t - x^2 &= 0 \\
−
(t-4)^2 - x^2 &= 16, \\
+
(t-4)^2 - x^2 &= 16 \\
 
(t-4-x)(t-4+x) &= 16.
 
(t-4-x)(t-4+x) &= 16.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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~jamess2022 (burntTacos)
 
~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>
 +
Applying the Pythagorean Theorem to right triangles <math>\triangle MDA,\triangle MDC,</math> and <math>\triangle MDB,</math> we have
 +
<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 apply <math>(\bigstar)</math> to rewrite <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>
 +
Note that <math>MA</math> and <math>MD</math> must have the same parity. 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>MD=3</math> and <math>MA=7.</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==
 
==Video Solution by Hawk Math==

Revision as of 13:12, 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*} Subbing in the first and third equation 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 apply $(\bigstar)$ to rewrite $(3),$ 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*} Note that $MA$ and $MD$ must have the same parity. 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

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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