Spring Mathematical Tournament

Question

The edge DA of a pyramid ABCD is perpendicular to its base ABC, (ABD) (BCD), BDC = 45^ and DB = 2. Find ADB if the sum of the squares of lateral faces of the pyramid equals 8.

Accepted Answer

Since (ABD) (ABC), (BCD), then (ABC) BC and hence ABC = DBC = 90^. Since BDC = 45^, then BC = BD = 2. Let ADB = . Then AB = 2 , AD = 2 and AC = 2^2 + 1. So S_ABD = AB AD2 = 2 , S_ACD = AC AD2 = 2 1 + ^2 , S_BCD = BC BD2 = 2. It follows by the condition of the problem that 8 = 4(^2 ^2 + ^2 (1 + ^2 ) + 1) = 4(^2 - 2 ^4 + 2), i.e. = 22. Thus ADB = 45^.