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Geometry Difficulty 6.0 National Olympiad Prove it Ireland

Suppose $ABCD$ is a simple quadrilateral, with side lengths
$a = |BC|,\ b = |CD|,\ c = |DA|,\ d = |AB|.$
Let $p = |AC|$ . Prove that
$\max(|d-a|,\ |b-c|) < p < \min(d+a,\ b+c).$
Conversely, if five positive numbers $a, b, c, d, p$ satisfy this condition, prove that $a, b, c, d$ are the side lengths of a simple quadrilateral, and that $p$ is the length of a diagonal.

Solution

Suppose $ABCD$ is a quadrilateral. The numbers $d, a, p$ are then the side lengths of the triangle $ABC$ . Hence
$||AB| - |BC|| < |AC| < |AB| + |BC| \iff |d - a| < p < d + a.$
Similarly, $b, c, p$ are the side lengths of the triangle $CDA$ , and so $|b - c| < p < b + c$ , whence $p < \min(d + a, b + c)$ and $\max(|b - c|, |d - a|) < p$ . This gives the required result.

The converse also holds. In the first place, if the stated inequality holds, the numbers $a, d, p$ obey the triangle inequalities, and so they are the lengths of the sides of a triangle $ABC$ , say with $a = |BC|, p = |AC|, d = |AB|$ . With $A$ as centre draw a circle with radius $c$ , and with $C$ as centre draw a circle of radius $b$ . Because of the given condition that $|b-c| < p < b+c$ , these circles intersect at two points. Label one of these $D$ . Then $ABCD$ is a quadrilateral and the length of $AC$ is $p$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.