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
$\endgroup$ 11 Answer
$\begingroup$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
$\alpha_n+\alpha_n+\alpha_m=360°$
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 $$...$$
EDIT
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 :)
