Problem:
Let ABCD be a rectangle and let DM be a segment perpendicular to the plane of ABCD. Suppose that DM has integer length, and the lengths of MA,MC, and MB are consecutive odd positive integers (in this order). What is the volume of pyramid MABCD?
Answer Choices:
A. 245
B. 60
C. 285
D. 66
E. 870 Solution:
Answer (A): Let AD=BC=a,AB=CD=b, and DM=h, where h is a positive integer. Then MA=h2+a2,MC=h2+b2, and MB=h2+a2+b2.
Because MA,MC, and MB are consecutive odd positive integers, there exists a nonnegative integer k such that MA=2k+1,MC=2k+3, and MB=2k+5. It follows that
(2k+1)2=h2+a2(2k+3)2=h2+b2, and (2k+5)2=h2+a2+b2.
Adding the first two equations above yields
(2k+1)2+(2k+3)2=2h2+a2+b2.
Subtracting the third equation above from the previous equation gives
(2k+1)2+(2k+3)2−(2k+5)2=h2.
This simplifies to
4k2−4k=15+h2.
Completing the square on the left-hand side gives (2k−1)2=16+h2, so (2k−1)2−h2=16. Factoring the left-hand side of this equation as a difference of squares gives
(2k−1−h)(2k−1+h)=16.
Because 2k−1−h and 2k−1+h have the same parity (so neither factor can equal 1 ) and 2k−1−h<2k−1+h, there is only one possibility, namely
2k−1−h=2 and 2k−1+h=8.
This system has solution (h,k)=(3,3). So MA=7 and MC=9. Then a2=72−32=40 and b2=92−32=72, so (ab)2=40⋅72=8⋅5⋅8⋅9, and hence ab=245. The volume of the pyramid is therefore 31⋅h⋅ab=245.