Three non-overlaping regular plane polygons, at least two of which are congruent, all have sides of length $1$. The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^{\circ}$. Thus the three polygons form a new polygon with $A$ as an interior point. What is the largest possible perimeter that this polygon can have?

How can i solve the problem without knowing the number of sides of polygons or at least 2 of them? Please elaborate the solution for me

  • $\begingroup$ The title misled me. I thought I could consider noncongruent polygons. Please try to match the title with the question content. $\endgroup$ – Oscar Lanzi Oct 18 '17 at 1:09

The internal angle of a $n$-sided regular polygon is $\alpha_n=180° - 360°/n$

Suppose two are congruent and three have sides concurrent in one point $A$ such that the sum of the angles is $360°$

We have this constraints


that is

$$\left(180°-\frac{360°}{m}\right)+\left(180°-\frac{360°}{n}\right)+\left(180°-\frac{360°}{n}\right)=360°$$ Or

$m=\dfrac{2 n}{n-4}$ which leads to the following table

$ \begin{array}{ccc|c} poly_1 & poly2 & poly_3 & perim\\ \hline 5 & 5 & 10 & 14\\ 6 & 6 & 6 & 12\\ 8 & 8 & 4 & 14\\ 12 & 12 & 3 & \color{red}{21}\\ \end{array} $

Hope this can help $$...$$


Thanks to the comment of Oscar Lanzi, I realized to have restricted the answer to the case of two congruent polygons.

Actually there is, at least, one result much larger: a triangle, an octagon and a $24$-sided polygon which give a perimeter of $29$ as shown in the second picture below

Re Edit

Oscar Lanzi just commented about a larger polygon which is formed by a triangle an eptangle ($7$ sides) and a $42$-sided regular polygon which has a perimeter of $46$. I missed it because I did not consider decimal angles.

No picture because a $42$-sided regular polygon looks like a circle at this scale :)

enter image description here

enter image description here

  • $\begingroup$ The maximal case does not use any pair of congruent polygons. $\endgroup$ – Oscar Lanzi Oct 18 '17 at 0:57
  • $\begingroup$ @OscarLanzi Thank you for this comment! Influenced by the text of the question, I restricted to two congruent polygons. $\endgroup$ – Raffaele Oct 18 '17 at 8:40
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    $\begingroup$ Now, try $3,7,42$. You are really identifying three-term Egyptian fraction sums for one half, and the sum with the most highly unequal terms $1/3+1/7+1/42$ gives the biggest perimeter of all. $\endgroup$ – Oscar Lanzi Oct 18 '17 at 9:37
  • $\begingroup$ @OscarLanzi Fool me! I did not think to the decimal angles!!! $\endgroup$ – Raffaele Oct 18 '17 at 9:43
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    $\begingroup$ @OscarLanzi some (many) years ago we had a discussion between math teachers about the (in)utility of egyptian fractions... :) $\endgroup$ – Raffaele Oct 18 '17 at 9:49

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