Basil has a melon in a shape of a ball, $20$ in diameter. Using a long knife, Basil makes three mutually perpendicular cuts. Each cut carves a circular segment in a plane of the cut, $h$ deep ($h$ is a height of the segment). Does it necessarily follow that the melon breaks into two or more pieces if (a) $h = 17$ ? (6 points) (b) $h = 18$ ? (6 points)
Problem
Source: Tournament of Towns Fall 2015 Senior A - level
Tags: combinatorics
MellowMelon
23.02.2017 23:35
I need to solve this problem for my own safety before WOOT gets hungry again. This solution assumes the cross-section of each cut passes through the center, since otherwise cutting the melon into two pieces is trivial for any $h$.
Suppose the cuts are made on the $xy,yz,zx$ planes. If the circular segment of each cut were the entire melon, the melon would be cut into 8 octants. Label each piece with three signs like $+++$, representing the signs of the $x$-coordinates, $y$-coordinates, and $z$-coordinates respectively of the points in that octant of the melon. We can view the problem as making these full height 20 cuts, then readding segments with height $20 - h$ back in to try to put the 8 octants of the melon back together again. (I wonder how many childhoods I just ruined by revealing that humpty dumpty was actually a melon.)
That is, we have a graph with 8 vertices (octants), and each segment we add back in creates edges between some of these vertices by reattaching those octants. We want to try to get a connected graph.
Note that the arc formed by the uncut part of the segment is a quarter of the circle when $h = 10 + \sqrt{50}$. This is a simple computation. For smaller $h$ the arc is larger. For larger $h$ the arc is smaller.
(a) If $h = 17 < 10 + \sqrt{50}$, then the segments we add back each contain more than a quarter of the circumference of the cross-section. So we can add back a segment that intersects the cross-sections of the other two cuts. Therefore, each segment we add back can add three edges to the graph.
For the cut along the $xy$-plane, readd a segment with these edges:
1. $+++$ to $++-$,
2. $+-+$ to $+--$,
3. $-++$ to $-+-$.
For the cut along the $yz$-plane, readd a segment with these edges:
1. $+--$ to $---$,
2. $++-$ to $-+-$,
3. $+-+$ to $--+$.
At this point, the melon is now exactly two pieces, separated only by the cut on the $zx$-plane. So anywhere we readd that last segment will make the melon whole again.
(b) If $h = 18 > 10 + \sqrt{50}$, then the segments we add back each contain less than a quarter of the circumference of the cross-section. So each piece added gives only 2 edges back. In this way, we form a graph of 8 vertices and 6 edges. But to get a connected graph, we need at least 7 edges, so the melon is certain to be cut into two pieces when $h = 18$.