Difference between revisions of "2021 AMC 12B Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
− | Let <math>MA=a</math> and <math>MD=d.</math> It follows that <math>MC=a+2</math> and <math>MB=a+4 | + | 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. | ||
<b>DIAGRAM WILL BE READY SOON.</b> | <b>DIAGRAM WILL BE READY SOON.</b> | ||
− | + | By the Pythagorean Theorem, we have | |
<cmath>\begin{alignat*}{6} | <cmath>\begin{alignat*}{6} | ||
AD^2 &= MA^2 - MD^2 &&= a^2 - d^2, \\ | AD^2 &= MA^2 - MD^2 &&= a^2 - d^2, \\ | ||
Line 24: | Line 26: | ||
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> | 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> | ||
− | By the Pythagorean Theorem on right <math>\triangle MAD,</math> we have <math>AD=2\sqrt{10}.</math> By the Pythagorean Theorem on right <math>\triangle MCD,</math> we have <math>CD=6\sqrt2.</math> | + | By the Pythagorean Theorem on right <math>\triangle MAD,</math> we have <math>AD=2\sqrt{10}.</math> |
+ | |||
+ | By the Pythagorean Theorem on right <math>\triangle MCD,</math> we have <math>CD=6\sqrt2.</math> | ||
Together, the volume of pyramid <math>MABCD</math> is <cmath>\frac13\cdot AD\cdot CD\cdot MD = \boxed{\textbf{(A) }24\sqrt5}.</cmath> | Together, the volume of pyramid <math>MABCD</math> is <cmath>\frac13\cdot AD\cdot CD\cdot MD = \boxed{\textbf{(A) }24\sqrt5}.</cmath> |
Revision as of 02:00, 28 November 2021
Contents
Problem
Let be a rectangle and let
be a segment perpendicular to the plane of
. Suppose that
has integer length, and the lengths of
and
are consecutive odd positive integers (in this order). What is the volume of pyramid
Solution 1
Let and
It follows that
and
As shown below, note that and
are both right triangles.
DIAGRAM WILL BE READY SOON.
By the Pythagorean Theorem, we have
Since
we equate the expressions for
and
then rearrange and factor:
As
and
have the same parity, we get
and
from which
By the Pythagorean Theorem on right we have
By the Pythagorean Theorem on right we have
Together, the volume of pyramid is
~Lopkiloinm ~MRENTHUSIASM
Solution 2
Let ,
,
,
. It follows that
and
.
We have three equations:
Substituting the first and third equations into the second equation, we get:
Therefore, we have
and
.
Solving for other values, we get ,
.
The volume is then
~jamess2022 (burntTacos)
Video Solution by Hawk Math
https://www.youtube.com/watch?v=p4iCAZRUESs
Video Solution by OmegaLearn (Pythagorean Theorem and Volume of Pyramid)
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.