Problem:
Polyhedron ABCDEFG has six faces. Face ABCD is a square with AB=12; face ABFG is a trapezoid with AB parallel to GF,BF=AG=8, and GF=6; and face CDE has CE=DE=14. The other three faces are ADEG,BCEF, and EFG. The distance from E to face ABCD is 12. Given that EG2=p−qr​, where p,q, and r are positive integers and r is not divisible by the square of any prime, find p+q+r.
Solution:
Place a coordinate system on the figure so that square ABCD is in the xy-plane, as shown in the diagram. Let E=(x1​,y1​,12). Because DE=CE, it follows that x1​=6. Because DE=14, it follows that 142=62+(y1​)2+122, so that y1​=4. Let GK be an altitude of isosceles trapezoid ABFG, and notice that the x-coordinates of both G and K are equal to 21​(AB−GF)=3. To find the y-coordinate of G, let ax+by+cz=d be an equation of the plane determined by A,D, and E. Substitute the coordinates of these three points to find that 12b=d, 0=d, and 6a+4b+12c=d, respectively, from which it follows that b=d=0 and a+2c=0. Thus G=(3,y2​,z2​) lies on the plane z=2x, so z2​=6. Because GA=8, it follows that 82=32+(y2​−12)2+62, so y2​=12±19​. Thus
