Problem:
A right square pyramid with volume 54 has a base with side length 6. The five vertices of the pyramid all lie on a sphere with radius nm​, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let ABCD be the base of the pyramid, P be its apex, and h be its height. Furthermore, let O be the center of the sphere that circumscribes the pyramid. Let M be the foot of the altitude from P. Then O,P, and M are collinear. Because the volume of a pyramid is 31​ times the area of its base times its height, 31​⋅62⋅h=54, so MP=h=29​. Because M is the center of square ABCD with side length 6,AM=32​. Setting r=OP=OA gives OM=∣∣∣​29​−r∣∣∣​. Applying the Pythagorean Theorem to △OAM yields
(29​−r)2+(32​)2=r2
from which it follows that r=417​. The requested sum is 17+4=21.
OR
More generally, let the pyramid have height h and base side length s. Let A and C be diagonally opposite vertices of
the point on the sphere such that PQ​ is a diameter of the sphere. Then M is on PQ​, Then AC=s2​, and AM=2​s​.
Let the sphere have radius r. Because △PAQ is a right triangle with altitude AM, triangles △PMA and △AMQ are similar. Thus PMAM​=AMQM​, so
h2​s​​=2​s​2r−h​
Solving for r gives
r=4hs2+2h2​
As in the first solution, h=29​. Thus the required radius is
4hs2+2h2​=4(29​)62+2(29​)2​=417​
as above.
OR
Define A,B,C,D,M,P, and r as in the first solution, where it was shown that PM=29​. Note that the radius of the sphere is also the circumradius of △APC. The isosceles triangle △APC has base AC=AB2​=62​ and height PM=29​, so each of its legs has length
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