Areas dominated by two points on a surface: Equal? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:21:28Z http://mathoverflow.net/feeds/question/85942 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85942/areas-dominated-by-two-points-on-a-surface-equal Areas dominated by two points on a surface: Equal? Joseph O'Rourke 2012-01-18T01:09:22Z 2012-01-19T17:36:37Z <p>Let $S$ be a smooth compact surface in $\mathbb{R}^3$, with two distinct, distinguished points $a,b \in S$. Let $R(a)$ be all the points of $S$ closer to $a$ than to $b$, and $R(b)$ all the points of $S$ closer to $b$ than to $a$, where distance is measured by shortest paths on the surface. The points equally distant from $a$ and $b$ form a <em>bisector</em>, a closed curve on $S$. This is in a sense the Voronoi diagram of the two points on $S$.</p> <hr /> <blockquote> <p><b>Q1</b>. Is the area of $R(a)$ equal to the area of $R(b)$? Suppose $S$ has genus zero, or is even convex&mdash;Do these restrictions change the answer?</p> </blockquote> <p><b>Answer:</b> Jack Huizenga's examples in his comment show the answer is a strong <code>No</code>, even for a convex surface.</p> <blockquote> <p><b>Q2</b>. Following Vitali Kapovitch's comment suggestion, is the sphere the only closed, bounded, genus-zero surface for which $R(a)=R(b)$ for every pair $a,b$?</p> </blockquote> <p><hr /> <b>Addendum</b>. There is no need for $S$ to be smooth, and indeed I am primarily interested in polyhedra, specifically, convex polyhedra. I am now thinking it is an interesting question to identify all the pairs of points $(x,y)$ for which their bisector $B=B(x,y)$ halves the surface area, e.g.: <br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <img src="http://cs.smith.edu/~orourke/MathOverflow/Box2Points.jpg" alt="Box bisector"> <br /> But I should save that for perhaps a separate question...</p>