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#1 h May 29, 2010, 8:50 AM • 1 Y

Y by Adventure10

Supose $a, b , c$ are the vertices of closed equilateral triangle in the plane,
find the maximum value of $|z-a||z-b||z-c|$ where $z$ range over this triangle.

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#2 h Apr 23, 2021, 7:29 AM

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Let $\Omega$ be the interior of the closed triangle $\Delta$ . Clearly $\Omega$ is a bounded region. Now $f(z) := |z-a||z-b||z-c|$ is a continuous, nonconstant function on $\bar{\Omega}=\Delta$ so by the Maximum Modulus Principle we have $|f(z)|<\left\|{f}\right\|_{\partial\Omega}=\left\|{f}\right\|_{\partial\Delta}$ for all $z \in \Omega$ . Thus, $\max_{z \in \Delta}f(z)=\max_{z \in \partial\Delta}f(z)$ That is to say, if $z \in \Delta$ is a point that maximizes $f(z)=|z-a||z-b||z-c|$ , then $z$ lies on the boundary of the triangle.

It is now a matter of elementary geometry to maximize $f$ . Let $z \in \{z \in \partial\Delta| f(z)=\max_{z \in \Delta}f(z)\}$ (the latter set is nonempty). WLOG suppose $z$ lies on $[a,b]$ . To simplify notation, let $s=|a-b|=|b-c|=|c-a$ . Then, let $x=|z-a|$ . We then have $|b-z|=s-x$ . We then get by the Pythagorean Theorem that $|z-c|=\sqrt{\left(x-\frac{s}{2}\right)^2+\left(\frac{s\sqrt{3}}{2}\right)^2}=\sqrt{x^2-sx+s^2}=x(s-x)\sqrt{x^2-s(x-s)}$ Thus, $|z-a||z-b||z-c|=x(s-x)\sqrt{x^2-sx+s^2}$ We now maximize this quantity as $x$ ranges from $[0,s]$ . Taking the derivative with respect to $s$ , we have by the first derivative test that
$(s-2x)\sqrt{x^2-sx+s^2}+(xs-x^2)\frac{1}{2}(2x-s)\frac{1}{\sqrt{x^2-sx+s^2}}=0$ so rearranging, we get $(s-2x)(\frac{3}{2}x^2-\frac{3}{2}xs+s^2)=0$ Remark that $\frac{3}{2}x^2-\frac{3}{2}xs+s^2=\frac{3}{2}(x-\frac{s}{2})^2+\frac{5s^2}{8}>0$ for all $x \in [0,s]$ so we must have $s-2x=0$ , that is $x=\frac{s}{2}$ . In this case, we have $|z-a||z-b||z-c|=\frac{s}{2}(\frac{s}{2})\sqrt{(\frac{s}{2})^2-s(\frac{s}{2})+s^2}=\frac{\sqrt{3}s^3}{8}$ Thus, $\max(|z-a||z-b||z-c|)=\boxed{\frac{\sqrt{3}s^3}{8}}$ where $s=|a-b|=|b-c|=|c-a|$ .

Addendum: There is also an elementary proof using the Erdos-Mordell Inequality. Lets see if someone here is smart enough to think of it.

This post has been edited 2 times. Last edited by 277546, Apr 23, 2021, 7:33 AM

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