Isoperimetric inequality revisited (for quadrilaterals)

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Isoperimetric inequality revisited (for quadrilaterals)

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Source: Austrian-Polish 1985, Problem 3

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#1 h Jul 5, 2015, 9:18 AM • 2 Y

Y by Adventure10, Mango247

In a convex quadrilateral of area $1$ , the sum of the lengths of all sides and diagonals is not less than $4+\sqrt 8$ . Prove this.

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#2 h Jul 5, 2015, 12:28 PM • 4 Y

Y by viperstrike, kiyoras_2001, Adventure10, Mango247

Hello!

Let $A,B,C,D$ be the vertices of the quadrilateral,in that order.

We know that $(ABCD)=\frac{1}{2}AC\cdot BD\cdot \sin \omega$ where $\omega$ is the angle formed by the diagonals.Thus $AC\cdot BD\geq 2$ ,

which gives $AC+BD\geq 2\sqrt{AC\cdot BD}=2\sqrt{2}=\sqrt{8} \ \left(\bigstar \right)$ .

Also $S=\frac{1}{2}\left(AB\cdot BC\cdot \sin \hat{ABC}+CD\cdot DA\cdot \sin \hat{CDA}\right)\leq \frac{1}{2}\left(AB\cdot BC+CD\cdot DA\right)$ .

Hence, $AB\cdot BC+CD\cdot DA\geq 2 \ (1)$ .Also,Ptolemy's theorem gives $AB\cdot CD+BC\cdot DA\geq AC\cdot BD\geq 2 \ (2)$ .

Adding up $(1),(2)$ gives

$AB\cdot BC+CD\cdot DA+AB\cdot CD+BC\cdot DA\geq 4\Leftrightarrow 4(AB+DA)(BC+CD)\geq 16\Rightarrow (AB+BC+CD+DA)^2\geq 16$ .

Thus, $AB+BC+CD+DA\geq 4\overset{\left(\bigstar \right)}\Leftrightarrow AB+BC+CD+DA+AC+BD\geq 4+\sqrt{8}$ .

Equality takes place only if the quadrilateral is a square.

This post has been edited 1 time. Last edited by gavrilos, Jul 5, 2015, 12:29 PM

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