Difference between revisions of "2021 AMC 12B Problems/Problem 14"
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MRENTHUSIASM (talk | contribs) (Undo revision 180241 by Puffer13 (talk) I did make the diagram and revised the solution to be clearer, so we coauthored.) (Tag: Undo) |
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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> | 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 | + | ~Lopkiloinm ~MRENTHUSIASM |
==Solution 2== | ==Solution 2== |
Revision as of 00:18, 6 November 2022
Contents
[hide]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. By the Pythagorean Theorem, we have Since in rectangle we equate the expressions for and then rearrange and factor: As and have the same parity, we get and from which
Applying the Pythagorean Theorem to right and right we obtain and respectively.
Let the brackets denote areas. 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)
~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.