the minimum of f(P) in a quadrilateral

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the minimum of f(P) in a quadrilateral

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#1 h Mar 3, 2012, 5:57 AM • 2 Y

Y by Adventure10, Mango247

Given a convex quadrilateral with $\angle B+\angle D<180$ .Let $P$ be an arbitrary point on the plane,define
$f(P)=PA*BC+PD*CA+PC*AB$ .
(1)Prove that $P,A,B,C$ are concyclic when $f(P)$ attains its minimum.
(2)Suppose that $E$ is a point on the minor arc $AB$ of the circumcircle $O$ of $ABC$ ,such that $AE=\frac{\sqrt 3}{2}AB,BC=(\sqrt 3-1)EC,\angle ECA=2\angle ECB$ .Knowing that $DA,DC$ are tangent to circle $O$ , $AC=\sqrt 2$ ,find the minimum of $f(P)$ .

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#2 h Mar 3, 2012, 6:10 PM • 2 Y

Y by Adventure10, Mango247

1: Extended Fermat point, $P= \odot ABC \cup BD$

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#3 h Aug 28, 2022, 4:46 PM

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By Ptolemy's Inequality, $PA \cdot BC + PD \cdot CA + PC \cdot AB \geq PB \cdot AC + PD \cdot AC \geq AC\cdot BD,$ which is constant. Equality occurs when $P, A, B, C$ are concyclic and $P$ lies on $\overline{BD}$ .

The main claim is the following:

Claim. $O$ lies on $\overline{AB}$ .

Let $\theta = \angle ACB$ . I will show that $\theta=90^\circ$ . Notice that by the condition, $\angle EBA = \angle ECA = \frac 23 \theta,$ so by the Law of Sines on $\triangle ABE$ we derive $\frac{\sqrt 3}2 = \frac{AE}{AB} = \frac{\sin \angle EBA}{\sin \angle AEB} = \frac{\sin \frac 23 \theta}{\sin \theta} = \frac{2 \cos \frac 13 \theta}{3 - 4 \sin^2 \frac 13 \theta}.$ This solves easily to get $\cos \frac 13 \theta = \frac{\sqrt 3}2$ , so $\theta = 90^\circ$ , as required. $\blacksquare$

Now let $AO=BO=x$ , so $BC = \sqrt{4x^2-2}$ . By Ptolemy's on $ACBE$ , $2x \cdot EC = \sqrt 3 x \cdot \sqrt{4x^2-2} + \sqrt 2 x.$ The condition $\frac {BC}{EC} = \sqrt 3 - 1 \iff \sqrt{2x^2-1} - 1 = 0,$ so $x=1$ . Thus $AC=\sqrt 2$ and $BD=\sqrt 5$ , from where it follows that $f(P) \geq AC \cdot BD = \sqrt 2 \cdot \sqrt 5 = \boxed{\sqrt{10}}.$

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