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(This article was first published on <strong><a href="https://xianblog.wordpress.com/2018/01/24/le-monde-puzzle-1037/"> R – Xi'an's Og</a></strong>, and kindly contributed to <a rel="nofollow" href="https://www.r-bloggers.com/">R-bloggers)</a> </div>
<p><strong><img src="https://encrypted-tbn3.gstatic.com/images?q=tbn:ANd9GcRsQd3v7efgSlmsNPKddzl-oQS9fZOF8a68V6KPmf5dnmI_EJ66rw" alt="" width="273" height="185">A</strong> purely geometric <a rel="nofollow" href="https://xianblog.wordpress.com/2011/09/03/le-monde-puzzle-website/" title="Le Monde puzzle [website]" target="_blank">Le Monde mathematical puzzle</a> this (or two independent ones, rather):</p>
<blockquote>
<p><em>Find whether or not there are inscribed and circumscribed circles to a convex polygon with 2018 sides of lengths ranging 1,2,…,2018. </em></p>
</blockquote>
<p>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) <a rel="nofollow" href="https://arxiv.org/abs/1512.06183" target="_blank">is such that</a> a side L and its corresponding inner angle θ satisfy</p>
<p>L²=R²2(1-cos(θ))</p>
<p>leading to the following R code</p>
<pre>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}
</pre>
<p>and the result is</p>
<pre>> R=324221
> sum(acos(1-(1:2018)^2/(2*R^2)))-2*pi
[1] 9.754153e-05
</pre>
<p>(which is very close to the solution of Le Monde when replacing sin(α) by α!). The authors of <a rel="nofollow" href="https://arxiv.org/abs/1512.06183" target="_blank">the quoted paper</a> do not seem to consider the existence an issue.</p>
<p>In the second case, there is <a rel="nofollow" href="https://en.wikipedia.org/wiki/Tangential_polygon" target="_blank">a theorem</a> that states that if the equations</p>
<p>x¹+x²=L¹,…,x²⁰¹⁸+x¹=L²⁰¹⁸</p>
<p>have a solution on <strong>R</strong>⁺ then there exists a circle such that the polygon is tangential to this circle. Quite interestingly, if the number <i>n</i> of sides is <em>even</em> 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</p>
<p>L¹-L²…+L²⁰¹⁷-L²⁰¹⁸ =0</p>
<p>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! <em>[Thanks to my office neighbours François and Julien for discussing the puzzle with me.]</em></p>
	<p>
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