Difference between revisions of "2022 AIME II Problems/Problem 3"
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Solve this equation will give us <math>l = \frac{17}{4}</math>. Therefore, <math>m+n=\boxed{021}</math> | Solve this equation will give us <math>l = \frac{17}{4}</math>. Therefore, <math>m+n=\boxed{021}</math> | ||
| − | ~DSAERF-CALMIT | + | ~DSAERF-CALMIT (https://binaryphi.site) |
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=II|num-b=2|num-a=4}} | {{AIME box|year=2022|n=II|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:28, 18 February 2022
Problem
A right square pyramid with volume 
has a base with side length 
The five vertices of the pyramid all lie on a sphere with radius 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius 
). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: 
. Because of the symmetrical property of the pyramid,
we can imagine that the line of the apex and the (sphere's) center will intersect the square at the (base's) center.
Since the volume is 
, where 
is the height of this pyramid, we have: 
according to pythagorean theorem.
Solve this equation will give us 
. Therefore, 
~DSAERF-CALMIT (https://binaryphi.site)
See Also
| 2022 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
