Hexagonizing

Let $a$ be the shortest side of the hexagon, and $k$ be the common increment for each succeeding side.

That said, we can now express its perimeter in terms of $a$ and $k$ .

$a + (a+k) + (a+2k) + ... + (a+5k) = 6a + 15k = 87$

$2a + 5k = 29$

Let us set that aside.

Now we go to the construction of the hexagon.

An equiangular hexagon whose sides in clockwise order are A, B, C, D, E, and F will exist if an only if the following statements are true:

$A + B = D + E$ $B + C = E + F$

Let us assume $B = a+5k$ . Then $A$ can only be $a$ , $(a+k)$ , or $(a+2k)$ . That is because if $A$ becomes $(a+3k)$ or greater, then there would be no choices for $D$ and $E$ to make the first equation true.

So, let us list the ways which will make the first equation true.

$(a+5k) + a = (a+k) + (a+4k)$ $(a+5k) + a = (a+2k) + (a+3k)$ $(a+5k) + (a+k) = (a+2k) + (a+4k)$ $(a+5k) + (a+2k) = (a+3k) + (a+4k)$

If we then assume that $a$ is adjacent to $(a+5k)$ , then all the other equations will follow and we will have the two hexagons is the following order.

Hexagon A

$a$ , $(a+5k)$ , $(a+k)$ , $(a+3k)$ , $(a+2k)$ , $(a+4k)$

Hexagon B

$a$ , $(a+5k)$ , $(a+2k)$ , $(a+k)$ , $(a+4k)$ , $(a+3k)$

Now, we proceed to finding the area of these hexagons.

For an equiangular hexagon with consecutive sides $A$ , $B$ , $C$ , $D$ , $E$ and $F$ , its area is

$[(A+B+C)^{2} - (A^2 + C^2 + E^2)]u$

where $u$ is the unit triangle area, or $\frac{\sqrt{3}}{4}$ sq units. This is derived from the fact that such equiangular hexagons may be viewed as portions of an equilateral triangle whose vertices are cut off with portions of smaller-sized equilateral triangles.

Following such formula, we get the area of Hexagon A to be

$(6a^2 + 30ak + 31k^2)u$

while Hexagon B has an area of

$(6a^2 + 30ak + 29k^2)u$

and so we get

$2k^2u = 18u$

$k = 3$

and since

$2a + 5k = 29$

we get $a =7$ and so we can now compute for the longest side $(a+5k) = 22$ .