Difference between revisions of "2021 AMC 12B Problems/Problem 14"
Lopkiloinm (talk | contribs) (→Solution 1) |
Jamess2022 (talk | contribs) (→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
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
Solution 1
This question is just about pythagorean theorem With these calculation, we find out answer to be ~Lopkiloinm
Solution 2
Let be , be , be , , , be , , respectively.
We have three equations:
Subbing in the first and third equation into the second equation, we get: Therefore, , Solving for other values, we get , . The volume is then ~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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