Endless Circles and Triangles

Relevant wiki: Length and Area Problem Solving

Let the radius of the largest circle be $r_{1}=1$ . Let the side length of the first triangle be $a_{1}$ . Then, let the area between the n:th and the (n+1):th circle be $A_{n}$ . The radius of the n:th circle is $r_{n}$ and the side length of the n:th triangle is $a_{n}$ .

$A_{1}=(\pi r_{1}^{2}-\pi r_{2}^{2} )$

The first triangle can be divided in three triangles congruent to each other, with two sides being 1 unit long and their angle 120 degrees. The area of one triangle is $\frac{1}{2} \times 1 \times 1 \times \sin 120 = \frac{\sqrt 3}{4}$ .

This means the area of the whole triangle is $B_{1}=3 \times \frac{\sqrt 3}{4}$ .

That area is also $\frac{1}{2}a_{1}^{2}\sin 60= \frac{\sqrt{3}}{4}a_{1}^{2}$ . We get an equation:

$\frac{\sqrt{3}}{4}a_{1}^{2}=3 \times \frac{\sqrt 3}{4}$

$a_{1}^{2}=3$

$a_{1}=\sqrt{3}$

Next we will find out the value of $r_{2}$ . We can separate a quadrilateral from the triangle. This quadrilateral has two straight angles, two sides with length $r_{2}$ and two sides with length $\frac{a_{1}}{2}$ . This quadrilateral can be cut in two halves, triangles with a straight angle, a 30 degrees angle and a 60 degrees angle. The side towards the 60 degrees angle is $\frac{a_{1}}{2}$ units long and the one towards the 30 degrees angle is $r_{2}$ units long. We get an equation:

$\tan 30=\frac{ r_{2}}{(\frac{\sqrt{3}}{2})}$

$r_{2}=\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}}$

$r_{2}=\frac{1}{2}$

So $A_{1}=\pi - \pi \times (\frac{1}{2})^{2}= \frac{3 \pi}{4}$

Let the area of the coloured regions between the n:th and the (n+1):th circle be $C_{n}$ . In this case

$C_{1}=\pi \times 1^{2} - B_{1}= \pi - \frac{3 \sqrt{3}}{4}$

Let $A=A_{1}+A_{2}+A_{3}+...$ be the whole area of the largest circle, while $C=C_{1}+C_{2}+C_{3}+...$ is the area of all coloured regions. Due to similarity of the regions between circles,

$\frac{C_{1}}{A_{1}}=\frac{C_{2}}{A_{2}}=\frac{C_{3}}{A_{3}}=...=k$ , where $k$ is constant. We get a result

$\frac{C}{A}=\frac{C_{1}+C_{2}+C_{3}+...}{A_{1}+A_{2}+A_{3}+...}=\frac{kA_{1}+kA_{2}+kA_{3}+...}{A_{1}+A_{2}+A_{3}+...}=k$

$\frac{C}{A}=\frac{C_{1}}{A_{1}}$

This means, we get the percentage of coloured regions inside the largest circle by calculating

$\frac{C_{1}}{A_{1}}= \frac{\pi - \frac{3 \sqrt{3}}{4}}{\frac{3 \pi}{4}}= 0.782004...$

so the approximate percentage with the accuracy of one decimal is $\boxed{78.2}$ %.