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+22​z=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​.
