diamond circumscribes a circle

Question

The rhombus with a side length $a$ and acute angle $\angle60^{o}$ circumscribes a circle. Calculate the perimeter of the quadrilateral whose vertices are the points of tangency circle with a diamond.

I found $r=\frac{a\sqrt3}{4}$ but I have problem with next steps

You have to make clear what you wrote in the first line: a rhombus can be inscribed in a triangle iff it is a square, as a quadrilateral is inscribable in a circle iff it has complementary opposite angles, so you must have meant something else. — DonAntonio, Jan 31 '14 at 16:11
So you mean: " A rhombus with side $;a;$ and acute angle $;60^\circ;$ circumscribes a circle: — DonAntonio, Jan 31 '14 at 16:17

score 2 · Accepted Answer · answered Jan 31 '14 at 16:36

Draw the figure on a sheet of paper covered by equilateral triangles, like this:

Then you can read the perimeter directly off that picture:

$$\left(2\frac{\sqrt3}4 + 2\frac34\right)a = \frac{3+\sqrt3}2a$$

If you don't trust this visual approach, you could well verify the coordinates of all the points visible in that drawing. Make sure that the lines connecting the toucing points to the center are actually perpendicular to the edges, and actually have the correct radius as well. Shouldn't be too hard, but I prefer to trust my eyes here.

diamond circumscribes a circle

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