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:
Let MA=a and MD=d. It follows that MC=a+2 and MB=a+4.
As shown below, note that △MAD and △MBC are both right triangles.
By the Pythagorean Theorem, we have
AD2BC2=MA2−MD2=a2−d2=MB2−MC2=(a+4)2−(a+2)2
Since AD=BC in rectangle ABCD, we equate the expressions for AD2 and BC2, then rearrange and factor:
As a+d−2 and a−d−2 have the same parity, we get a+d−2=8 and a−d−2=2, from which (a,d)=(7,3). Applying the Pythagorean Theorem to right △MAD and right △MCD, we obtain AD=210 and CD=62, respectively.
Together, the volume of pyramid MABCD is
31⋅[ABCD]⋅MD=31⋅(AD⋅CD)⋅MD=(A) 245
OR
Let AD=b,CD=a,MD=x,MC=t. It follows that MA=t−2 and MB=t+2.
We have three equations:
a2+x2a2+b2+x2b2+x2=t2=t2+4t+4=t2−4t+4
Substituting the first and third equations into the second equation, we get:
t2−8t−x2(t−4)2−x2(t−4−x)(t−4+x)=0=16=16
Therefore, we have t=9 and x=3.
Solving for other values, we get b=210,a=62. The volume is then
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