Shyan's hexagon

Extend AB to meet CD at G and EF at H. Extend EF and CD to meet at I. Since the hexagon is equiangular and has 6 angles summing to 720, each of the angles is 120 degrees. This means each of the exterior angles is 60 degrees. (HAF, HFA, IED, IDE, GBC, GCB). In triangle AFH, note that angle H is 60 degrees, since the sum of the angles is 180 degrees, and HAF+HFA=120 degrees. Similarly, angles G and I also have measures 60 degrees. Therefore, triangles AHF, GBC, DEI, and GHI are equilateral.

Let AB be x. Then, by the conditions given, CD=2x, and EF=4x. Also, let AF be y. Therefore, by the conditions, BC=2y. Finally, let DE=z. From the equilateral triangles above, FH=AH=y,BG=GC=2y, and EI=DI=z. Therefore, GH has side length x+3y, GI has side length 2x+2y+z, and HI has side length 4x+y+z. Therefore, we obtain the equation x+3y=4x+y+z=2x+2y+z, since GH=GI=HI.

From the equation 4x + y + z = 2x + 2y + z, we obtain 2x = y. From x + 3y = 4x + y + z, we have x + 6x = 4x + 2x + z, and x = z

Since the perimeter is 98, we have x + 2y + 2x + z + 4x + y = 98, so 7x + 3y + z = 98. Plugging in y = 2x and z = x, we have 6x + 7x + x = 98 ⇒ x = 7. We are looking for 4x, which is 28

The first step in ultimately solving this question is in realising that the exercise actually already gives us four different linear equations with the lengths of the sides of the hexagon as variables. In the following, the sides of the hexagon will be denoted by $AB,\ldots,EF,FA$ . The equations that are given, are: $\left\{\begin{array}{l}EF=4AB,\\ CD=2AB,\\ BC=2FA,\\ AB + BC + CD + DE + EF + FA = 98\end{array}\right.$ .

We notice that there are four independent linear equations in six variables so far and in order to solve these, we need two more equations. These are given by the fact that we need to get a hexagon. If we place a point $A$ in the plane, draw a line $AB$ , then make an angle of 120 degrees and continue drawing lines this fashion (it may help to draw a picture at this point), we need to get a closed figure. That is, the line coming from $F$ at a 120 degree angle from $EF$ has to intersect $AB$ precisely in $A$ .

Two different ways of drawing the hexagon then yield the desired extra two equations. First of all we notice that $\cos 60^\circ = \frac{1}{2}$ . If we now draw the hexagon with $BC$ and $EF$ parallel to the $y$ -axis in the real plane, the condition of the figure being closed turns into the linear equation $\frac{1}{2}AB+BC+\frac{1}{2}CD= \frac{1}{2}FA+EF+\frac{1}{2}DE$ .

If we turn the entire hexagon over 120 degrees (now $CD$ and $FA$ are parallel to the $y$ -axis), we can do the same thing to obtain yet another equation: $\frac{1}{2}BC+CD+\frac{1}{2}DE= \frac{1}{2}AB+FA+\frac{1}{2}EF$ .

Filling in the first three equations in the last one, yields $DE=AB$ and the resulting equations then become $\left\{\begin{array}{l}8AB+3FA=98,\\ 3FA-6AB=0.\end{array}\right.$

Solving this yields $AB=DE=7, CD=FA=14, EF=BC=28$ and so the final answer is $EF=28$ .

Consider the hexagon as a closed path or a vector sum of six displacement which is zero.

Since vector addition is associative, we can add them in any order we want.

Assume AB = 1 and FA = L => CD = 2, EF = 4, BC = 2L

Take A = (0,0) i.e. our starting position.

Then add AB + CD + EF this takes us to point (-2, sqrt(3))

Now adding FA + BC should make the y coordinate zero (because we still have DE to move us any distance in x direction)

For this to happen L cos 60° - 2L cos 60° must be -sqrt(3) giving L = 2

This takes us to a point (1, 0) so DE must be 1

The perimeter we now get is 1 + 4 + 2 + 1 + 4 + 2 = 14

Scale it 7 times to get desired perimeter 98

So L must be 7 times what we got, which comes to 2 x 7 = 14