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Mathematical Olympiad in Ukraine, Fourth Round (March 24, ), 2010

Geometry Difficulty 4.6 AIME Prove it Ukraine

In a convex quadrilateral $ABCD$ angles $\angle ABC$ and $\angle BCD$ are not less than $120^\circ$ . Prove, that $AC + BD > AB + BC + CD$ .

Solution

Let $AB = a$ , $BC = b$ , $CD = c$ (Fig.18).

Then, $AC^2 = a^2 + b^2 - 2ab \cos \angle B \ge a^2 + ab + b^2$ .

By analogy, $BD^2 \ge b^2 + bc + c^2 \Rightarrow AC + BD \ge \sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2}$ .

Since $\sqrt{a^2 + ab + b^2} > a + \frac{1}{2}b$ and $\sqrt{b^2 + bc + c^2} > c + \frac{1}{2}b$ , then $AC + BD > a + b + c = AB + BC + CD$ .

Fig.18

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