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Posted on Mar 6, 2026
Two teams of shovelers plan to remove all the snow from a parking lot shaped like a regular hexagon.
Team Vertex places one shoveler at each of the six vertices. Team Centroid places all shovelers at the center.
Each team is responsible for the snow initially closer to someone on their own team than anyone on the other team.
What fraction of the lot does Team Centroid shovel?
The Team Centroid region is the intersection of six half-planes: for each vertex \(V\), keep points \(P\) with \(|PO| < |PV|\), where \(O\) is the center.
For a fixed vertex direction, the boundary \(|PO| = |PV|\) is the perpendicular bisector of \(OV\), so along that direction it sits at distance \(R/2\) from \(O\).
Those six bisectors form another regular hexagon centered at O (the blue one in the diagram).
Now use scaling with distance to edges. For the large regular hexagon, the center-to-edge distance is \((\sqrt{3}/2)R\). For the inner hexagon, each edge is a bisector halfway to a vertex, so center-to-edge distance is \(R/2\).
So the linear scale (inner to outer) is:
\[ \frac{R/2}{(\sqrt{3}/2)R}=\frac{1}{\sqrt{3}}. \]
Area scales as distance squared, so:
\[ \frac{\text{inner area}}{\text{outer area}}=\left(\frac{1}{\sqrt{3}}\right)^2=\frac13. \]
Team Centroid fraction \(=1/3\).
The same setup works for a regular \(n\)-gon with a center shoveler versus shovelers at all \(n\) vertices.
The center region is again a smaller similar regular \(n\)-gon, so area comes from a linear scale factor.
The resulting center fraction is:
\[ f_n=\frac23 \text{ for } n=3,\qquad f_n=\frac{1}{4\cos^2(\pi/n)} \text{ for } n\ge 4. \]
The triangle is special. For \(n=3\), the three perpendicular-bisector boundaries line up with the three opposite sides, so each boundary actually meets the original triangle boundary instead of forming a strictly interior similar polygon. For \(n\ge 4\), those boundaries sit strictly inside and enclose an interior similar \(n\)-gon, which is why the single scaling formula applies there.
Checks: triangle \(f_3=2/3\), square \(f_4=1/2\), hexagon \(f_6=1/3\), and as \(n\to\infty\) (circle boundary limit), \(f_n\to1/4\).
Case \(n=3\) (equilateral triangle): Let the center be \(O\) and vertices \(V_1,V_2,V_3\). The inequalities \(|PO| < |PV_i|\) become three half-planes bounded by the perpendicular bisectors of \(OV_i\). In an equilateral triangle, each such bisector is parallel to (and exactly halfway to) the opposite side. Their intersection with the original triangle removes three corner triangles, each similar to the whole with side scale \(1/3\), so each has area \(1/9\) of the whole. Remaining center area is \(1-3\cdot(1/9)=2/3\).
Case \(n>3\): Put the regular \(n\)-gon on circumradius \(R\). For each vertex \(V\), the boundary \(|PO| = |PV|\) is the perpendicular bisector of \(OV\), hence at distance \(R/2\) from the center. These \(n\) lines form an interior regular \(n\)-gon (the center region). The outer polygon has center-to-edge distance \(R\cos(\pi/n)\); the inner one has center-to-edge distance \(R/2\). So linear scale is \((R/2)/(R\cos(\pi/n))=1/(2\cos(\pi/n))\), and area ratio is its square:
\[ f_n=\frac{1}{4\cos^2(\pi/n)} \quad \text{for } n>3. \]
Choose \(n\) to view the regular \(n\)-gon. Blue is Team Centroid's region; red is Team Vertex's region.
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Now the lot is a circle. Team Vertex and Team Centroid each choose one random point independently and uniformly in the disk.
The perpendicular bisector of the two chosen points splits the disk into two regions; one team gets the larger region.
What is the expected value of that larger area fraction?
Let \(t\) be the distance from the circle center to the bisector line (\(0 \le t \le 1\) for a unit disk).
To get the larger-side fraction, first compute the smaller circular segment cut off by the line.
In a unit disk, let \(\theta\) be the central angle subtending the smaller segment chord. From a right triangle with adjacent side \(t\) and hypotenuse \(1\),
\[ \theta = 2\arccos(t). \]
Segment area = sector area minus isosceles-triangle area:
\[ A_{\text{small}}=\frac{\theta}{2}-\frac{\sin\theta}{2}. \]
Using \(\theta=2\arccos(t)\) and \(\sin(2\arccos t)=2t\sqrt{1-t^2}\),
\[ A_{\text{small}}=\arccos(t)-t\sqrt{1-t^2}. \]
Divide by total disk area \(\pi\) to get the smaller fraction, then subtract from 1:
\[ L(t)=1-\frac{A_{\text{small}}}{\pi} =1-\frac{\arccos(t)-t\sqrt{1-t^2}}{\pi}. \]
Now connect this to two chosen points. If \(A\) and \(B\) are the random points (as vectors), the perpendicular bisector is:
\[ (B-A)\cdot X = \frac{|B|^2-|A|^2}{2}. \]
So the center-to-line distance is:
\[ t=\frac{\bigl||B|^2-|A|^2\bigr|}{2|B-A|}. \]
(That is just point-to-line distance with normal vector \(B-A\).)
Averaging over two independent uniform random points in the disk gives:
Write \(M=(A+B)/2\) and \(D=(B-A)/2\), so \(t=|M\cdot D|/|D|\). Conditioning on \(s=|D|\), the midpoint \(M\) is uniform on the overlap lens of two unit disks centered at \(\pm D\). Integrating the projection of \(M\) onto the \(D\)-direction gives:
Here \(s\) is half the distance between the two random points (\(|A-B|=2s\)). For fixed \(s\), choose coordinates so \(D\) points along the x-axis; then \(x\) is the x-coordinate of \(M\) (equivalently, the signed projection of \(M\) onto the \(D\)-direction). In these coordinates, \(t=|x|\) and the lens half-width is \(1-s\), so \(x\) runs from \(0\) to \(1-s\) by symmetry.
\[ E[L]=\frac{32}{\pi}\int_{0}^{1}\int_{0}^{1-s} s\,L(x)\sqrt{1-(x+s)^2}\,dx\,ds, \] with \[ L(x)=1-\frac{\arccos(x)-x\sqrt{1-x^2}}{\pi}. \]
Evaluating that integral yields:
\[ E[L]=1-\frac{32}{9\pi^2}. \]
Numerically:
\(E[L]=1-\frac{32}{9\pi^2}\approx 0.6397469\).
So the team with the larger assignment gets about 63.97% of the lot on average.
Drag the points, use click-to-place mode, or generate random points. The circle is colored by which point is closer, and the exact larger/smaller fractions are shown.
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Run a Monte Carlo estimate of the expected larger-area fraction \(E[L]\).
Result will appear here.