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Equiangular hexagon ABCDEF has side lengths AB = CD = EF = 1 and BC = DE = FA = r. The area of triangle ACE is 70% the area of the hexagon. What is the sum of all possible values of r?

Using Law of Cosines, I found x^2 = r^2 + r + 1. This can be found by applying the formula and using one of the 120 degrees triangles with sides 1, r, and a side of the triangle ACE (which is clearly equilateral). I'm calling the length of the unknown side x.

...Somehow or the other...I made it to 7r = r^2 + r + 1 (In my review notebook) but I am completely blanking on how in the world I got from the first thing I did to this. I probably skipped steps writing it down and now find myself in this confusedness. Would greatly appreciate someone helping me finish this solution.

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Let $A_T$ be the area of the triangle and $A_R$ the area of the rest of the hexagon. Then $$A_R=\tfrac37A_T\quad\hbox{and}\quad A_T=\tfrac14\sqrt3(1+r+r^2)\ .$$ Since "the rest" consists of three triangles with known sides and included angle, $$A_R=3\times\tfrac12r\sin120^\circ=\tfrac34\sqrt3r\ .$$ The rest is easy.

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You need to find the area of the hexagon in terms of $r$. Draw one main diagonal, then draw perpendiculars to each of the other four points. This divided into two rectangles and two pairs of congruent triangles, the areas of which are easy to find. Add the areas to get the total area of the hexagon:$$\frac{(5r+1)\sqrt3}{4}$$Use this to find length $AC$, then use law of sines on $\triangle ABC$ to find $r$

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