# Homework Help: Geometry - calculating a distance

1. Aug 6, 2013

### Chetlin

Geometry -- calculating a distance

1. The problem statement, all variables and given/known data
I'm trying to calculate the distance from the center of one of these three tangent circles to the center of the entire shape (that is, the center of the "curvy triangular" region between the circles). Each circle has radius R. Here is a picture of the circles:

The distance is apparently $R \sec 30^\circ$ (equal to $\frac{R}{\cos 30^\circ}$), which is $\frac{2R}{\sqrt{3}}$.

2. Relevant equations

I guess the only one would be that $\cos 30^\circ = \frac{\sqrt{3}}{2}$.

3. The attempt at a solution

I was able to solve it just fine by drawing a triangle, using the centers of the three circles as the angles, and drawing some more lines bisecting the angles (also splitting the opposing side into two equal halves), and then using properties of "30-60-90" triangles (which I guess is technically using sines and cosines) to figure out the same answer:

I don't understand, though, how you could see that the answer is $R \sec 30^\circ$ right away without doing all this. Am I able to realize this from the picture I made? I don't see how I could come up with that answer even after drawing the picture.

2. Aug 6, 2013

### symbolipoint

We can call the distance you want, h, and it is the hypotenuse of the 30-60-90 triangle you found. You have take R to be a constant, assume as if known, but h is unknown. You can say
h*cos(30)=R
h=R/cos(30)
h=R/(sqrt(3)/2)
h=2R/sqrt(3)
h=(2/3)*R*sqrt(3)

Your use of making a picture was good. Making a picture like that allows to see and label how several useful parts are related and so this is easier to analyze and solve. Most people can not handle everything, including the picture and labeling, in their heads.

3. Aug 6, 2013

### lurflurf

At a glance consider a 30-60-90 right triangle formed by the center, the center of a circle, and a point of tangency. R is a side and we want to find the hypotenuse so we have R/cos 30 degrees

4. Aug 6, 2013

### Chetlin

Haha, wow, that was pretty simple. I guess I didn't see it because it's not that intuitive when you don't have all the extra lines drawn on (especially the fact that the triangle is 30-60-90).

Thank you to both of you!