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 1)
Line 5: Line 5:
  
 
==Solution 1==
 
==Solution 1==
−
This question is just about pythagorean theorem
+
This question is just about Pythagorean theorem
−
<cmath>a^2+(a+2)^2-b^2 = (a+4)^2</cmath>
+
<cmath>\begin{align*}
−
<cmath>2a^2+4a+4-b^2 = a^2+8a+16</cmath>
+
a^2+(a+2)^2-b^2 &= (a+4)^2 \\
−
<cmath>a^2-4a+4-b^2 = 16</cmath>
+
2a^2+4a+4-b^2 &= a^2+8a+16 \\
−
<cmath>(a-2+b)(a-2-b) = 16</cmath>
+
a^2-4a+4-b^2 &= 16 \\
−
<cmath>a=3, b=7</cmath>
+
(a-2+b)(a-2-b) &= 16,
−
With these calculation, we find out answer to be <math>\boxed{\textbf{(A) }24\sqrt5}</math> ~Lopkiloinm
+
\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>
 +
 
 +
~Lopkiloinm
 +
 
 
==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>\overline{AD}</math> be <math>b</math>, <math>\overline{CD}</math> be <math>a</math>, <math>\overline{MD}</math> be <math>x</math>,  

Revision as of 06:24, 12 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

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 $\overline{AD}$ be $b$, $\overline{CD}$ be $a$, $\overline{MD}$ be $x$, $\overline{MC}$, $\overline{MA}$, $\overline{MB}$ be $t$, $t-2$, $t+2$ respectively.

We have three equations: \[a^2 + x^2 = t^2\] \[a^2 + b^2 + x^2 = t^2 + 4t + 4\] \[b^2 + x^2 = t^2 - 4t + 4\]

Subbing in the first and third equation into the second equation, we get: \[t^2 - 8t - x^2 = 0\] \[(t-4)^2 - x^2 = 16\] \[(t-4-x)(t-4+x) = 16\] Therefore, \[t = 9\], \[x = 3\] Solving for other values, we get $b = 2\sqrt{10}$, $a = 6\sqrt{2}$. The volume is then \[\frac{1}{3} abx = \boxed{\textbf{(A)}24\sqrt{5}}\] ~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

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=155834"