Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 3 (Six Variables, Five Equations))
(→‎Solution 2)
 
(24 intermediate revisions by 5 users not shown)
Line 5: Line 5:
  
 
==Solution 1==
 
==Solution 1==
−
Let <math>MD=a</math> and <math>MA=b.</math> This question is just about Pythagorean theorem
+
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.
 +
<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
 +
<cmath>\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*}</cmath>
 +
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+(a+2)^2-b^2 &= (a+4)^2 \\
+
a^2 - d^2 &= (a+4)^2 - (a+2)^2 \\
−
2a^2+4a+4-b^2 &= a^2+8a+16 \\
+
a^2 - d^2 &= 4a + 12 \\
−
a^2-4a+4-b^2 &= 16 \\
+
a^2 - 4a - d^2 &= 12 \\
−
(a-2+b)(a-2-b) &= 16,
+
(a-2)^2 - d^2 &= 16 \\
 +
(a+d-2)(a-d-2) &= 16.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
from which <math>(a,b)=(3,7).</math> With these calculation, we find out answer to be <math>\boxed{\textbf{(A) }24\sqrt5}</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>
 +
 
 +
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.
  
−
~Lopkiloinm
+
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>
 +
~Lopkiloinm ~MRENTHUSIASM
  
 
==Solution 2==
 
==Solution 2==
Line 25: Line 64:
 
b^2 + x^2 &= t^2 - 4t + 4.
 
b^2 + x^2 &= t^2 - 4t + 4.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
Subbing in the first and third equation into the second equation, we get:
+
Substituting the first and third equations into the second equation, we get:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
t^2 - 8t - x^2 &= 0 \\
 
t^2 - 8t - x^2 &= 0 \\
Line 38: Line 77:
 
~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.
+
==Video Solution (Fast! Just 4 min!)==
 +
https://youtu.be/Bo2EvRZdRnA
  
−
Let the brackets denote areas. Finally, the volume of pyramid <math>MABCD</math> is
+
~<i>Education, the Study of Everything</i>
−
<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==
Line 87: Line 88:
 
== 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_14&oldid=196000"