Area of a equilateral triangle with sides of 1 is $\frac { \sqrt3}{2}$

You can find the radius of the largest circle by using pythagorean theorem.

Radius of the greatest circle is $\frac { 1 }{2\sqrt3}$ Making the area of the greater circle $\frac {\pi}{12}$

The radius of the next circle is $\frac { 1 }{4\sqrt3}$

The area is $\frac {\pi}{48}$ There are now three circle so the area of all three circles can simplify to $\frac {\pi}{16}$

The area of the next circle is $\frac {\pi}{192}$ or simplified to $\frac {\pi}{64}$ when adding the area of all three circles. The area of the circles are decreasing at $\pi \sum _{ 2 }^{ \infty }{ \frac { 1 }{ 4 } } ^{ n } = \pi\frac {1}{12}$. So the sum of the total area of the circles is $\frac { \pi }{ 12 } + \frac { \pi }{ 12 } = \frac {\pi}{6}$

I have googled this problem and others have received different results. I don't have a key to check my work and I'm curious what others have received and where is my mistake.



First note that the area of the equilateral triangle is $\frac{\sqrt{3}}4$ not $\frac{\sqrt{3}}2$. Also note that the radius of the inner circle in equilateral triangle is one-third of its altitude. Meaning that $\frac 23$ of the altitude $h$ is reserved by the big circle and the remaining $\frac h3$ is assigned to the smaller circles in a similar way. Let $S_1$ be the area of the big circle, $S_2$ the sum of the areas of the three smaller circles, and so on. We can write: $$\begin{align} S_1&=\pi\left({\frac h3}\right)^2\\ S_2&=3\times\pi\left({\frac 13\times\frac h3}\right)^2=\frac 13 S_1\\ S_3&=3\times\pi\left({\frac 13\times\frac 13\times\frac h3}\right)^2=\frac 1{27}S_1\\&\cdots \end{align}$$ In conclusion: $$\begin{align} S=\sum_{n=1}^\infty S_n&=S_1+\frac 13 S_1\left(1+\frac 19+\frac 1{81}+\cdots\right)\\ &=S_1+\frac 13 S_1\left(\frac 98\right)=\frac{11}8 S_1\\ &=\frac{11}8\pi\left({\frac {\sqrt{3}}6}\right)^2=\frac{11}{96}\pi \end{align}$$

  • $\begingroup$ Thank you. That makes sense and where I made my mistake. $\endgroup$
    – Mrbowtie
    Dec 30 '16 at 1:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.