Thread: reverse solving of formula

1. reverse solving of formula

Hi folks. I am trying to draw a regular hexagon given a diameter (being the diameter of a circle drawn around the hexagon).

I basically have this formula where a = the length along the 6 sides of the hexagon:

d = a + 2 * cos60 * a (this is right, right?)

So what I am trying to figure out is how to solve for a. I got this far but my old brain is having a hard time with it:

d - a - 2 * cos60 * a = 0,
d - 2 * cos60 * a = a,

now I think I am supposed to divide the left side by a again but that wouldn't make sense since then I am also dividing the right side by a which would give me 1. Thanks for any pointers.

J

2. like a dork I just realized cos60 = 0.5. I'll give that another go now.

3. could it really be that simple?

d = a + 2 * cos60 * a,
d = a + 2 * 0.5 * a,
d = 2a,
d / 2 = a

No way!!! Is this right? Can someone confirm this because it seems to good to be true. Maybe Occham's Razor is right.

4. Originally Posted by jwopitz
could it really be that simple?

d = a + 2 * cos60 * a,
No, this is NOT right. Where did you get it?

d = a + 2 * 0.5 * a,
d = 2a,
d / 2 = a

No way!!! Is this right? Can someone confirm this because it seems to good to be true. Maybe Occham's Razor is right.
No, that's not right. The correct answer is even simpler: Drawing lines from the vertices of the hexagon to the center divides it into 6 equilateral triangles: d= a. That's a well known property or a regular hexagon.

5. I got the formula d = a + 2 * cos60 * a by saying that a regular hexagon has 6 sides whose lengths are = a. If you bisect the hexagon with a line you are left with a trapezoid where the longer side is = d and the 3 other sides are = a. The two acute angles are 60º each and the two obtuse angles are 120º each. This yields the above formula.

I am a bit confused by your response as I came to the same conclusion regarding the 6 equilateral triangles however you contradict your own statement by say d = a. Drawing straight lines through the opposite vertices of the hexagon would yield two sides of two triangles to be parallel and continuous. Given that 2r = d for circles then 2a = d for a hexagon. Maybe it was a typo.

Doing some simple checking on my original formula, it indeed works along with the d = 2a formula. Just a round about way of getting to the same solution. Thanks for the help.