Okay, I can't get every detail down before class starts, but I'll put down a few basics. I'll insert intermediate steps later if needed.

Given: Three tangential, co-planar circles where no circle is inside another circle with radii $ a,b,c $

Find: Area of the triangle formed by connecting the centers of the circles with line segments.

Many of you will recognize this problem from ENGR 195 CHIPS Homework 5.2 Problem 3. Only, then you were given the actual lengths of the radii.

According to my calculations, if A is the area of the triangle, then:

$ A=\sqrt{abc(a+b+c)} $

I'm not going to show the proof right now so you all can have a chance to mess with it as well. I'll just say that I used the law of cosines to find the cosine of one angle, then the cosine function to find the length of the base of a right triangle with the same angle along the base, and then Pythagoreans theorem to find the height of the triangle, and finally the formula for the area of a triangle to find the area.

Jhunsber

• See discussion for the easy way of showing this (by way of Hero's formula). I went the hard way, and now I finally understand why Hero's Formula works. I never used to get it until I worked through this problem. So in that way this extra, unnecessary work paid off.

The Abbreviated Version of the Proof That Goes About It in a Long Way

Warning, not all the steps are outlined. If you believe my math to be incorrect at any point, call me out.

Let the angle formed by the two radii of length c in circle c be $ \theta $. Then, by law of cosines,

$ (a+b)^2=(b+c)^2+(a+c)^2-2(b+c)(a+c)\cos{\theta} $

Solve for $ \cos{\theta} $

$ \cos{\theta}=\frac{(a+b)^2-(b+c)^2-(a+c)^2}{-2(b+c)(a+c)} $

This is a part where I'm leaving out the messing details of my calculation. But when you simplify the right side you're left with:

$ \cos{\theta}=1-2\frac{ab}{(c+a)(c+b)} $

Now, drop an altitude from circle b to the line between circles c and a (so we now we have a base and a height). Using cosine function, we can find the distance (let it be x) from the center of circle c to the altitude's intersection with the line between circles c and a.

$ \cos{\theta}=\frac{x}{c+b} $

Therefore:

$ \frac{x}{c+b}=1-2\frac{ab}{(c+a)(c+b)} $

Solve for x

$ x=(c+b)\bigg(1-2\frac{ab}{(c+a)(c+b)}\bigg) $

Distribute

$ x=(c+b)-2\frac{ab}{(c+a)} $

Now, to find the height we can use Pythagorean's Theorem

$ (c+b)^2=x^2+h^2 $

Solve for h:

$ h=\sqrt{(c+b)^2-x^2}=\sqrt{(c+b)^2-\bigg((c+b)-2\frac{ab}{(c+a)}\bigg)^2} $

$ h=\sqrt{(c+b)^2-(c+b)^2+4\frac{ab(c+b)}{(c+a)}-4\frac{(ab)^2}{(c+a)^2}} $

$ h=\sqrt{4\frac{ab(c+b)}{(c+a)}-4\frac{(ab)^2}{(c+a)^2}} $

$ h=\sqrt{4ab\bigg(\frac{(c+b)}{(c+a)}-\frac{ab}{(c+a)^2)}\bigg)} $

$ h=\sqrt{4ab\bigg(\frac{(c+b)(c+a)-ab}{(c+a)^2}\bigg)} $

$ h=2\sqrt{ab\bigg(\frac{c^2+ca+cb+ab-ab}{(c+a)^2}\bigg)} $

In this step, if it's not clear, I canceled the ab, then undistributed the c and multiplied by the ab waiting on the outside of the fraction

$ h=2\sqrt{\frac{abc(a+b+c)}{(c+a)^2}} $

$ h=\frac{2}{c+a}\sqrt{abc(a+b+c)} $

Now area is defined as A = 1/2*B*h, so...

$ A=\frac{Bh}{2} $ and $ B=c+a $

Substituting:

$ A=\frac{(c+a)\frac{2}{c+a}\sqrt{abc(a+b+c)}}{2} $

Canceling:

$ A=\sqrt{abc(a+b+c)} $

And there you have it, a special case of Hero's (or Heron's) Formula.