# Le Monde puzzle [#1037]

**R – Xi'an's Og**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

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**A** purely geometric Le Monde mathematical puzzle this (or two independent ones, rather):

Find whether or not there are inscribed and circumscribed circles to a convex polygon with 2018 sides of lengths ranging 1,2,…,2018.

In the first (or rather second) case, the circle of radius R that is tangential to the polygon and going through all nodes (assuming such a circle exists) is such that a side L and its corresponding inner angle θ satisfy

L²=R²2(1-cos(θ))

leading to the following R code

R=3.2e5 step=1e2 anglz=sum(acos(1-(1:2018)^2/(2*R^2))) while (abs(anglz-2*pi)>1e-4){ R=R-step+2*step*(anglz>2*pi)*(R>step) anglz=sum(acos(1-(1:2018)^2/(2*R^2))) step=step/1.01}

and the result is

> R=324221 > sum(acos(1-(1:2018)^2/(2*R^2)))-2*pi [1] 9.754153e-05

(which is very close to the solution of Le Monde when replacing sin(α) by α!). The authors of the quoted paper do not seem to consider the existence an issue.

In the second case, there is a theorem that states that if the equations

x¹+x²=L¹,…,x²⁰¹⁸+x¹=L²⁰¹⁸

have a solution on **R**⁺ then there exists a circle such that the polygon is tangential to this circle. Quite interestingly, if the number *n* of sides is *even* there are an infinitude of tangential polygons if any. Now, and rather obviously, the matrix A associated with the above system of linear equations is singular with a kernel induced by the vector (1,-1,…,1,-1). Hence the collection of the sides must satisfy

L¹-L²…+L²⁰¹⁷-L²⁰¹⁸ =0

which puts a constraint on the sequence of sides, namely to divide them into two groups with equal sum, 2018×2019/4, which is not an integer. Hence, a conclusion of impossibility! *[Thanks to my office neighbours François and Julien for discussing the puzzle with me.]*

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