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

(Created page with "==Problem== Let <math>ABCD</math> be a rectangle and let <math>\overline{DM}</math> be a segment perpendicular to the plane of <math>ABCD</math>. Suppose that <math>\overline{...")
 
(→‎Solution)
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==Solution==
 
==Solution==
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{{solution}}
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===Solution 1===
 +
This question is just about pythagorean theorem
 +
<cmath>a^2+(a+2)^2-b^2 = (a+4)^2</cmath>
 +
<cmath>2a^2+4a+4-b^2 = a^2+8a+16</cmath>
 +
<cmath>a^2-4a+4-b^2 = 16</cmath>
 +
<cmath>(a-2+b)(a-2-b) = 16</cmath>
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<cmath>a=3, b=7</cmath>
 +
With these calculation, we find out answer to be
  
 
==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 23:59, 11 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

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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