Arithmetic of Pie Sharing


Arithmetic of Pie Sharing


Below are several illustrations of $15=9+7-1$ cuts that solve the first part of the problem.

Arithmetic of Pie Sharing, proof a), illustration

Thinking of a pie as a rectangle (Morth):

Arithmetic of Pie Sharing, proof a), illustration 2

A different method altogether (Gökhan Karahan): Slice into 9 pieces first (white lines) than slice two pieces into 7 (red lines). Total cuts 15.

Arithmetic of Pie Sharing, proof a), illustration 3


So assume the pie is cut into some pieces which could be combined into $1/p^{th}$ and (somehow else) into $1/q^{th}$ portions. Let's form a bipartite graph the following way: place $p$ nodes in one row and $q$ nodes in another row. These represent two groups of guests. Join two nodes (one in the upper row, the other in the low row) if in a combination of the cut-off pieces there is one piece both receive. Thus we get a graph where nodes represent guests and edges represent pieces of the pie.

We want to show that the number of edges (the pieces of the pie) is not less than $p+q-1.$ To this end note that among all graphs with a given number $N$ of nodes, the (spanning) trees have the smallest number of edges which is $N-1.$ Since in our case $N=p+q,$ $p+q-1$ is the smallest number of pieces with which it is possible to fairly satisfy either $p$ or $q$ guests.


This is problem M1232 (by D. Fomin) from the Russian magazine Kvant (1990, n7). The original solution is by A. Tolpygo and D. Fomin.


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