Arithmetic Sequence in Sides

consider number of sides to be $(a-d,a,a+d)$ . then the 3 interior angle of the three polygons should add to $2\pi$ or: $2\pi=\pi-\dfrac{2\pi}{a-d}+\pi-\dfrac{2\pi}{a}+\pi-\dfrac{2\pi}{a+d}\to 1 = \dfrac{2}{a-d}+\dfrac{2}{a}+\dfrac{2}{a+d} \\ \to a^3-a d^2 = 6a^2-2d^2 \to d^2 = \dfrac{a^3-6a}{a-2}\to d=\sqrt{\dfrac{a^3-6a^2}{a-2}}$ notice that $d$ is a positive integer so $a-2|a^3-6a^2=a*a*(a-6) \to a-2| 2*2*(-4)$ this means $a-2=1,2,4,8,16 \to a= 3,4,6,10,18$ . non of this yields a positive integer for d and hence there is no solution.

Let the number of sides be $a$ , $a + d$ , and $a + 2d$ for each regular polygon such that $a$ < $a + d$ < $a + 2d$ to follow an arithmetic progression. Then $a \geq 3$ and $a$ is an integer since it is the number of sides of a polygon, $d$ is an integer since $a$ is an integer and $a + d$ is the number of sides of a polygon, and $a < 6$ since $3$ regular hexagons already cover the space around the point and any more sides would cause an overlap of the polygons.

Since the $3$ interior angles for each regular polygon must add up to $2\pi$ to cover the space around the point, and each interior angle for a regular polygon with $n$ sides is $\frac{(n - 2)\pi}{n}$ , we have: $\frac{(a - 2)\pi}{a} + \frac{(a + d - 2)\pi}{a + d} + \frac{(a + 2d - 2)\pi}{a + 2d} = 2\pi$ This simplifies to: $(4 - 2a)d^2 + (12 - 3a^2)d + (6a^2 - a^3) = 0$ Then according to the quadratic equation: $d = \frac{3a^2 - 12a \pm a\sqrt{a^2 - 8a + 48}}{8 - 4a}$

Since $3 \leq a < 6$ and $a$ is an integer, $a = 3$ , $a = 4$ , or $a = 5$ . These values give $a^2 - 8a + 48$ values of $33$ , $32$ , and $33$ respectively, and none of these numbers are perfect squares, and so none of these values give a required integer value for $d$ , and so the situation described in the problem is not possible.