Difference between revisions of "2021 AMC 12B Problems/Problem 14"
Lopkiloinm (talk | contribs) (→Solution) |
Lopkiloinm (talk | contribs) (→Solution 1) |
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<cmath>(a-2+b)(a-2-b) = 16</cmath> | <cmath>(a-2+b)(a-2-b) = 16</cmath> | ||
<cmath>a=3, b=7</cmath> | <cmath>a=3, b=7</cmath> | ||
| − | With these calculation, we find out answer to be | + | With these calculation, we find out answer to be <math>\boxed{\textbf{(A) }24\sqrt5}</math> |
==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 23:01, 11 February 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
Solution 1
This question is just about pythagorean theorem
![\[a^2+(a+2)^2-b^2 = (a+4)^2\]](http://latex.artofproblemsolving.com/9/2/b/92b312ead36501db86d8c2960c49cf7819bff2c5.png)
![\[2a^2+4a+4-b^2 = a^2+8a+16\]](http://latex.artofproblemsolving.com/b/a/c/bacac51bb0351a1508c0d23ab106b121ece6625e.png)
![\[a^2-4a+4-b^2 = 16\]](http://latex.artofproblemsolving.com/5/b/d/5bdc493669a9a705e70050920f21d33c3fa7168f.png)
![\[(a-2+b)(a-2-b) = 16\]](http://latex.artofproblemsolving.com/5/1/f/51fc21cb0d6793788206a52e0d68554ce6c2e4ad.png)
![\[a=3, b=7\]](http://latex.artofproblemsolving.com/1/5/2/152b590268184f8176ac8c130c45bb6c6db5e539.png)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
