Problem:
A square pyramid with base ABCD and vertex E has eight edges of length 4. A plane passes through the midpoints of AE,BC, and CD. The plane's intersection with the pyramid has an area that can be expressed as p. Find p.
Solution:
Place the pyramid on a coordinate system with A at (0,0,0),B at (4,0,0),C at (4,4,0),D at (0,4,0) and with E at (2,2,22). Let R,S, and T be the midpoints of AE,BC, and CD respectively. The coordinates of R,S, and T are respectively (1,1,2),(4,2,0) and (2,4,0). The equation of the plane containing R,S, and T is x+y+22z=6. Points on BE have coordinates of the form (4−t,t,t2), and points on DE have coordinates of the form (t,4−t,t2). Let U and V be the points of intersection of the plane with BE and DE respectively. Substituting into the equation of the plane yields t=21 and (27,21,22) for U, and t=21 and (21,27,22) for V. Then RU=RV=7,US=VT=3 and ST=22. Note also that UV=32. Thus the pentagon formed by the intersection of the plane and the pyramid can be partitioned into isosceles triangle RUV and isosceles trapezoid USTV with areas of 35/2 and 55/2 respectively. Therefore the total area is 45 or 80, and p=80.
