# Thread: Regular Hexagons and a Circle

1. ## Regular Hexagons and a Circle

A circle is inscribed in a regular hexagon, a smaller regular hexagon is inscribed in the circle. Compute the ratio of the area of the larger hexagon to the area of the smaller hexagon.

I have no idea how to solve this problem. I would be much obliged if anyone were to help me.

2. Draw a picture. Even a rough sketch should suffice.
Connect the center to a vertex of the bigger hexagon.
Connect the center to a nearby vertex of the smaller hexagon (which will be the midpoint of a side of the larger...).

Use some angle and side knowledge.
Feel free to ask for more hints!

3. Thank you for your help. I drew the figures as you suggested, but I don't understand how one would know a line drawn from the center of the circle to a nearby vertex of the smaller hexagon passes through the midpoint of a side of the larger hexagon.

4. My phrasing was a little misleading...
It doesn't necessarily pass through the midpoint of the larger side, but you can rotate it without affecting anything else.

5. Originally Posted by NOX Andrew
A circle is inscribed in a regular hexagon, a smaller regular hexagon is inscribed in the circle. Compute the ratio of the area of the larger hexagon to the area of the smaller hexagon.
I have no idea how to solve this problem. I would be much obliged if anyone were to help me.
Look at the diagram. Can you calculate the area of the part in yellow?

6. Sure wish I knew how to make a diagram! It's very simple, once you see the 30-60-90 triangle(s). Maybe Plato will upload a file suggestive of this method...

7. Or go like this:
1: draw the larger hexagon
2: draw its inscribed circle : notice that the 6 tangent points are located at midpoint of the hexagon's sides
3: draw the smaller hexagon by joining these 6 points
Howz dat?
Then I suggest you assign 1 as length of circle's radius; agree Chaz ?

EDIT:
whoops...didn't see Plato's diagram...
mine will look different, and involve 6 congruent triangles as the difference in areas...

8. I would use "1" for the length of the longer... (can't remember the vocab word) center-vertex segment.
This will be the hypotenuse. 1/2 will be the shorter leg... the longer leg is the center-vertex of the smaller pentagon. So this boils down to the ratio of sides of a 30-60-90 triangle (or a trig function).

Using 1 for the radius will make it look more like the unit circle

9. Originally Posted by TheChaz
I would use "1" for the length of the longer... (can't remember the vocab word) center-vertex segment.
This will be the hypotenuse. 1/2 will be the shorter leg... the longer leg is the center-vertex of the smaller pentagon. So this boils down to the ratio of sides of a 30-60-90 triangle (or a trig function).
Using 1 for the radius will make it look more like the unit circle
Right! That's why they pay you the big bucks

10. Hello NOX Andrew,
Drawone sector of Plato's diagram.There are two overlapping equilateral triangles.If the smaller has a side length of 1 its apothem is 1/2*rad3. The larger has an apothem of one and a side lengthof2/rad3.

bjh