2004 USAMO Problems/Problem 1

From AoPSWiki

Problem

Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that

$\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.$

When does equality hold?

Solution

By a well-known property of tangential quadrilaterals, the sum of the two pairs of opposite sides are equal; hence $a + c = b + d \Rightarrow a - b = d - c \Rightarrow |a - b| = |d - c|$ Now we factor the desired expression into $\frac {|d - c|(c^2 + d^2 + cd)}{3} \le|a - b|(a^2 + b^2 + ab)\le 3|d - c|(c^2 + d^2 + cd)$ . Temporarily discarding the case where $a = b$ and $c = d$ , we can divide through by the $|a - b| = |d - c|$ to get the simplified expression $(c^2 + d^2 + cd)/3\le a^2 + b^2 + ab\le 3(c^2 + d^2 + cd)$ .

Now, draw diagonal $BD$ . By the law of cosines, $c^2 + d^2 + 2cd\cos A = BD$ . Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that $A\in [60^{\circ},120^{\circ}]$ . Cosine is monotonously decreasing on this interval, so by setting $A$ at the extreme values, we see that $c^2 + d^2 - cd\le BD^2 \le c^2 + d^2 + cd$ . Applying the law of cosines analogously to $a$ and $b$ , we see that $a^2 + b^2 - ab\le BD^2 \le a^2 + b^2 + ab$ ; we hence have $c^2 + d^2 - cd\le BD^2 \le a^2 + b^2 + ab$ and $a^2 + b^2 - ab\le BD^2 \le c^2 + d^2 + cd$ .

We wrap up first by considering the second inequality. Because $c^2 + d^2 - cd\le BD^2 \le a^2 + b^2 + ab$ , $\text{RHS}\ge 3(a^2 + b^2 - ab)$ . This latter expression is of course greater than or equal to $a^2 + b^2 + ab$ because the inequality can be rearranged to $2(a - b)^2\ge 0$ , which is always true. Multiply the first inequality by $3$ and we see that it is simply the second inequality with the variables swapped; hence by symmetry it is true as well.