# Thread: Circles insribed in equilateral triangles.

1. ## Circles insribed in equilateral triangles.

This is a problem from a discrete set I'm working on, and I could use a little help.

An equilateral triangle has each side length 1000. In this triangle, a circle is inscribed. Then in the space remaining on each corner, inscribe another circle. The second set of circles touch two sides of the triangle and the edge of the previous circle. Keep proceeding in this manner, making smaller and smaller circles that approach the vertices of the triangle.

What i really want to know is some sort of formula to find the radii of each of the circles.

thanks a ton in advance guys

2. Let the triangle have side length x.

The first circle will touch at a points that defines a radii perpendicular to the triangle edges. We can derive a formula for the radius of the first circle using the following:

$h = \sqrt{x^2 - x^2/4} = \sqrt{3}/2x = 2r + y$ where $y$ is the distance from the vertex of a triangle to the circle, $r$ is the radius of te circle, and $h$ is the height of the triangle.

Also
$(y+r)^2 = r^2+(x/2)^2$

Use these two equations to eliminate the variable $y$ and arrive at:

$r = \frac{x}{2 \sqrt{3}}$.

Also observe that $y$ can be used to define the height of new equilateral triangles.

$y = \sqrt{3}x/2 - x/ \sqrt{3} = \frac{x}{2 \sqrt{3}}$.

Then the side length of these new equilaterial trainges $(x_1)$ in which the next circles will be embedded are:

$\left(\frac{x}{2 \sqrt{3}}\right)^2 + \left(\frac{x_1}{2}\right)^2 = (x_1)^2 \Rightarrow x_1 = \frac{x}{3}.$

Now we have a new side length of $\frac{x}{3}$ and we can repeat the procedure to determine the radius of the second circles. By our formula above they will be:

$r_1 = \frac{x_1}{2 \sqrt{3}} = \frac{1}{3}\frac{x}{2 \sqrt{3}}$

And so on.

3. so you used the formula for the altitude of a triangle and basically determined that the side length (and therefore the radius, circumference and area of the subsequent circles) of the triangle will be 1/3 for the second triangle, then 1/3 of that for the third, etc.

thanks a lot.