Romanian Mathematical Olympiad

Question

The diagonals of the faces of the rectangular parallelepiped ABCD'A'B'C'D' satisfy the relation: CA^2 = 2 B'A^2 D'A^2B'A^2 + D'A^2. Prove that AC' 3 AA'.

Accepted Answer

Denote AB = a, AD = b, AA' = c. By Pythagorean Theorem, we have a^2 + b^2 = 2(b^2 + c^2)(c^2 + a^2)b^2 + c^2 + c^2 + a^2, therefore a^4 + b^4 = 2c^4. We also have 4c^4 = 2(a^4 + b^4) (a^2 + b^2)^2 thus 2c^2 a^2 + b^2. In the end, C'A = a^2 + b^2 + c^2 3c^2 = c3.