Minimum of PA + PB + PC

Question

Problem: A circle $\Gamma$ with center $O$ and an interior point $P$ are given. For three points $(A, B, C)$ on the circle $\Gamma$, angles between three lines $(PA, PB, PC)$ are all $2\pi/3$. Prove that $PA + PB + PC$ is minimum when one of the three lines $(PA, PB, PC)$ is perpendicular to $OP$.

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[My Approach]

Denote the radius of $\Gamma$ as $R$, and let $\angle APO=\theta$, $OP=d$.

Using the cosine law, $f(\theta):=AP=d\cos\theta + \sqrt{R^2-d^2\sin^2\theta}$.

Then $PA + PB + PC$ becomes $g(\theta)=f(\theta)+f(\theta+2\pi/3)+f(\theta+4\pi/3)\\ =\sqrt{R^2-d^2\sin^2\theta}+\sqrt{R^2-d^2\sin^2(\theta+2\pi/3)}+\sqrt{R^2-d^2\sin^2(\theta+4\pi/3)}$.

It is not complicated to show that $g(\theta)$ has extremum at $\theta=\pi/2$ ($0<\theta<2\pi/3)$, but how can we prove that it is the minimum point?

• Since $P$ is a Fermat point of $\triangle ABC$, I've tried to apply the properties of Fermat point, but I couldn't get any conclusion from this approach.

Answer 1

This is, unfortunately, too good to be true. See the geogebra image:

Answer 2

I don't fully understand the problem. If $A,B,C$ are fixed, there is a unique interior point $P$ that forms angles of $\frac{2}{3}\pi$. Therefore, $PA+PB+PC$ is also fixed... What is certain is that, with $A,B,C$ fixed, the point $P$ that minimizes this sum is the Fermat point. But Fermat's point does not necessarily hold that $OP$ is perpendicular to one of the sides.