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

(→‎Solution 1)
(→‎Solution)
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<cmath>a=3, b=7</cmath>
 
<cmath>a=3, b=7</cmath>
 
With these calculation, we find out answer to be <math>\boxed{\textbf{(A) }24\sqrt5}</math> ~Lopkiloinm
 
With these calculation, we find out answer to be <math>\boxed{\textbf{(A) }24\sqrt5}</math> ~Lopkiloinm
 +
===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>,
 +
<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:
 +
<cmath>a^2 + x^2 = t^2</cmath>
 +
<cmath>a^2 + b^2 + x^2 = t^2 + 4t + 4</cmath>
 +
<cmath>b^2 + x^2 = t^2 - 4t + 4</cmath>
 +
 +
Subbing in the first and third equation into the second equation, we get:
 +
<cmath>t^2 - 8t - x^2 = 0</cmath>
 +
<cmath>(t-4)^2 - x^2 = 16</cmath>
 +
<cmath>(t-4-x)(t-4+x) = 16</cmath>
 +
Therefore, <cmath>t = 9</cmath>, <cmath>x = 3</cmath>
 +
Solving for other values, we get <math>b = 2\sqrt{10}</math>, <math>a = 6\sqrt{2}</math>.
 +
The volume is then <cmath>\frac{1}{3} abx = \boxed{\textbf{(A)}24\sqrt{5}}</cmath> ~jamess2022(burntTacos)
  
 
==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}}

Revision as of 04:41, 12 February 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 $MACD?$

$\textbf{(A) }24\sqrt5 \qquad \textbf{(B) }60 \qquad \textbf{(C) }28\sqrt5\qquad \textbf{(D) }66 \qquad \textbf{(E) }8\sqrt{70}$

Solution

Solution 1

This question is just about pythagorean theorem \[a^2+(a+2)^2-b^2 = (a+4)^2\] \[2a^2+4a+4-b^2 = a^2+8a+16\] \[a^2-4a+4-b^2 = 16\] \[(a-2+b)(a-2-b) = 16\] \[a=3, b=7\] 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)

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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