# find area of infinite circles

• Feb 15th 2009, 07:30 AM
galactus
find area of infinite circles
Here is a neat little problem involving infinite series I thought was cool.

Please excuse my haphazard 'paint' drawing.

If the circles go off into the vertices to infinity, find the area of the infinite circles inscribed in the equilateral triangle with sides length s.
• Feb 19th 2009, 01:45 AM
.........
(Hi)Hi.

I have used : The inradius of an equilateral triangle has radius r = a/3
where a is the altitude

After making a new tangent to the incircle parralel to the AB,
we get a new triangle similar to the earlier triangle
( ie; an equilateral triangle with side S'= S/3)

the incircle of this triangle is our next required circle
and thus we get its radius(r) as a'/3

$\displaystyle a=\frac{\sqrt{3}S}{2}$

$\displaystyle Area_1 = \pi r^2$

$\displaystyle =\pi (\frac{S}{2\sqrt{3}})^2$

$\displaystyle a' = \frac{\sqrt{3}S}{2\times 3}$

$\displaystyle r' = \frac{ S}{6 sqrt{3}}$

$\displaystyle Area_2 = \pi r'^2$

$\displaystyle =\pi (\frac{ S}{3\times 2 sqrt{3}})^2$

Similarly
$\displaystyle Area_3 = \pi ( \frac{S}{3 \times 3 \times 2 \sqrt{3} } )^2$

Thus the sum now becomes
$\displaystyle =\pi (\frac{S}{2\sqrt{3}})^2 +\pi (\frac{ S}{3\times 2 \sqrt{3}})^2+\pi (\frac{ S}{3\times 3\times 2 \sqrt{3}})^2...........\infty$

This in an infinite GP with first term

$\displaystyle \pi (\frac{S}{2\sqrt{3}})^2$

And common difference
$\displaystyle \frac{1}{9}$

Thus the sum of terms is given by

$\displaystyle \frac{\pi (\frac{S}{2\sqrt{3}})^2}{1-\frac{1}{9}}$

We needed to get it for three sides so sum is

$\displaystyle =\frac{\pi (\frac{S}{2\sqrt{3}})^2}{1-\frac{1}{9}}- 2 \pi (\frac{S}{2\sqrt{3}})^2$
• Feb 20th 2009, 03:35 PM
galactus
Cool, Adarsh. I got the same thing. I was wondering if I was right or not.

Since we agree, I suppose I was. (Clapping)

These types of problems are always fun.