Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 3.6 AMC 10/12 Prove it North Macedonia

Let the quadrangle $ABCD$ be inscribed in a circle of radius $1$ . Prove that the difference between its perimeter and the sum of the lengths of its diagonals is positive and less than $4$ .

Solution

From the triangle inequality we have:
$2L = \overline{AB} + \overline{BC} + \overline{CD} + \overline{DA} + \overline{AB} + \overline{CD} + \overline{DA} > \overline{AC} + \overline{BD} + \overline{AC} + \overline{BD}$
from which we get one of the inequalities. Let us denote the point of intersection of the diagonals by $R$ , and the length of the diameter of the circle by $d$ . Then we have
$\begin{aligned} L &= \overline{AB} + \overline{BC} + \overline{CD} + \overline{DA} < \overline{AR} + \overline{BR} + \overline{BR} + \overline{CR} + \overline{CR} + \overline{DR} + \overline{DR} + \overline{AR} = \\ &= \overline{AC} + \overline{BD} + \overline{AC} + \overline{BD} \le \overline{AC} + \overline{BD} + 2d = \overline{AC} + \overline{BD} + 4 \end{aligned}$
from which we get the other inequality.

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