Difference between revisions of "2021 AMC 12B Problems/Problem 14"
MRENTHUSIASM (talk | contribs) m (→Solution 3 (Six Variables, Five Equations)) |
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(MA+MD-2)(MA-MD-2)&=16. | (MA+MD-2)(MA-MD-2)&=16. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | As <math>MA+MD-2</math> and <math>MA-MD-2</math> always have the same parity, both factors must be even. Since <math>MA>MD,</math> the only possibility is | |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
MA+MD-2&=8, \ | MA+MD-2&=8, \ |
Revision as of 16:13, 20 June 2021
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 This question is just about Pythagorean theorem from which With these calculation, we find out answer to be .
~Lopkiloinm
Solution 2
Let , , , . It follows that and .
We have three equations: Substituting in 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)
Solution 3 (Six Variables, Five Equations)
We are given that Applying the Pythagorean Theorem to right triangles and we have Subtracting from and applying and we express in terms of We substitute into then rearrange: As and always have the same parity, both factors must be even. Since the only possibility is from which and
Substituting the current results into we get respectively.
Let the brackets denote areas. Finally, the volume of pyramid is ~MRENTHUSIASM
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.