## English Issue, December 2002 | ||||

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## Solutions of problems B

**B. 3416.** *A convex polyhedron is bounded by quadrilateral faces, its surface area is A**, and the sum of the squares of its edges is Q**. Prove that Q*\(\displaystyle \ge\)2*A**.*

Proposed by *Á. Besenyei,* Tatabánya

**Solution.** First we prove that the sum of the squares of the sides of any quadrilateral is at least four times its area. Using the notations of the *figure,* the area of the quadrilateral is

\(\displaystyle A_{ABCD}=A_{ABD}\pm A_{BCD}\le A_{ABD}+A_{BCD}=\frac{a\cdot d\cdot\sin\alpha}{2}+\frac{b\cdot c\cdot\sin\gamma}{2}\le\frac{ad+bc}{2}. \)

From the inequality between the arithmetic and geometric means, \(\displaystyle ad\le\frac{a^2+d^2}{2}\) and \(\displaystyle bc\le\frac{b^2+c^2}{2}\), and thus 4*A*_{ABCD}\(\displaystyle \le\)*a*^{2}+*b*^{2}+*c*^{2}+*d*^{2}, as stated above.

If the corresponding inequality is set up for each face of the polyhedron, and the inequalities are summed, the left-hand side will be 4*A*, and the right-hand side will be 2*Q*, as each edge belongs to two faces of the convex polyhedron. Therefore, 4*A*\(\displaystyle \le\)2*Q*, and hence 2*A*\(\displaystyle \le\)*Q*, and we are done.

Based on the solution by *A. Babos,* Budapest