## Problem of the Day #68: Placing Two Points in a Triangle
*May 26, 2011*

*Posted by Billy in : potd , trackback*

Albert has an isosceles triangle $ABC$ with sides $AB=AC=3$ and $BC=2$. He then places two points $\mathcal{P}$ and $\mathcal{Q}$ somewhere in the interior of $ABC$. For all points $\mathcal{X}$ that lie within $ABC$ or are on the perimeter of $ABC$, Albert defines $f(\mathcal{X}) :=\min(\mathcal{PX}, \mathcal{QX})$. He also lets $G$ be the maximum value of $f(\mathcal{X})$ for all possible $\mathcal{X}$. If Albert places $\mathcal{P}$ and $\mathcal{Q}$ to minimize $G$, find $G$.

## Comments»

no comments yet - be the first?