4k-gons

Probability Level 3

Let n be a positive whole number divisible by 4. George wants to distribute distinct numbers on the vertices of a regular n-gon such that the sum of two random adjacent vertices is equal to the sum of the two adjacent vertices diagonally opposite. Find the sum of all possible values of n. If there are no possible values of n submit your answer as 0.


The answer is 0.

Magnus Seidenfaden by Magnus Seidenfaden , 19, Denmark

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1 solution

Chris Lewis
Mar 8, 2019

Let's call the numbers on the vertices x 1 , x 2 , . . . x n x_1,x_2,...x_n . It will be helpful later to define x n + 1 = x 1 x_{n+1}=x_1 (this is just going round the polygon again)

Say n = 4 k n=4k . The information we're given is that x i + x i + 1 = x i + 2 k + x i + 2 k + 1 x_i+x_{i+1}=x_{i+2k}+x_{i+2k+1} for i = 1 , … , 2 k i=1,\ldots,2k .

Explicitly, these sums are

x 1 + x 2 = x 2 k + 1 + x 2 k + 2 \color{#3D99F6}{x_1+x_2=x_{2k+1}+x_{2k+2}}

x 2 + x 3 = x 2 k + 2 + x 2 k + 3 \color{#D61F06}{x_2+x_3=x_{2k+2}+x_{2k+3}}

x 3 + x 4 = x 2 k + 3 + x 2 k + 4 \color{#3D99F6}{x_3+x_4=x_{2k+3}+x_{2k+4}}

⋮ \vdots

x 2 k − 2 + x 2 k − 1 = x 4 k − 2 + x 4 k − 1 \color{#D61F06}{x_{2k-2}+x_{2k-1}=x_{4k-2}+x_{4k-1}}

x 2 k − 1 + x 2 k = x 4 k − 1 + x 4 k \color{#3D99F6}{x_{2k-1}+x_{2k}=x_{4k-1}+x_{4k}}

x 2 k + x 2 k + 1 = x 4 k + x 4 k + 1 = x 4 k + x 1 \color{#D61F06}{x_{2k}+x_{2k+1}=x_{4k}+x_{4k+1}=x_{4k}+x_1}

(the equations whose first entry is odd are coloured in blue, the others in red).

The idea is to use all of these equations and make as many terms cancel as possible. We will add all of the blue equations, and subtract all of the red ones.

This gives

( x 1 + x 2 ) − ( x 2 + x 3 ) + … + ( x 2 k − 1 + x 2 k ) − ( x 2 k + x 2 k + 1 ) = ( x 2 k + 1 + x 2 k + 2 ) − ( x 2 k + 2 + x 2 k + 3 ) + … + ( x 4 k − 1 + x 4 k ) − ( x 4 k + x 1 ) \color{#3D99F6}{(x_1+x_2)} \color{#333333}-\color{#D61F06}{(x_2+x_3)} \color{#333333}+\ldots \color{#333333}+\color{#3D99F6}{(x_{2k-1}+x_{2k})} \color{#333333}-\color{#D61F06}{(x_{2k}+x_{2k+1})} \color{#333333}= \color{#3D99F6}{(x_{2k+1}+x_{2k+2})} \color{#333333}-\color{#D61F06}{(x_{2k+2}+x_{2k+3})} \color{#333333}+\ldots+\color{#3D99F6}{(x_{4k-1}+x_{4k})} \color{#333333}-\color{#D61F06}{(x_{4k}+x_1)}

Cancel, cancel, cancel, until simply:

x 1 − x 2 k + 1 = x 2 k + 1 − x 1 x_1-x_{2k+1}= x_{2k+1}-x_1

and so x 1 = x 2 k + 1 x_1=x_{2k+1} . But we are told that all the numbers are distinct! This contradiction means that there are NO such configurations.

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