Division of Circle into 4 Equal Parts by 3 Equal Length Lines

Problem

Divide a circle into $4$ equal parts by $3$ lines of equal length.

By equal here, we mean of equal area.

Solution

Each of the curved lines is a semicircle whose radius is $\dfrac {n r} 4$, where:

$n$ goes from $1$ to $3$
$r$ is the radius of the main circle.

Proof

Let the main circle have radius $4 a$, to make the numbers convenient.

Let the area of the main circle be $\AA$.

From Area of Circle:

$\AA = \pi \paren {4 a}^2 = 16 \pi a^2$

The area $\AA_1$ of one of the outside shapes is:

half the area of the main circle
less half the area of the circle whose radius is $3 a$
plus half the area of the circle whose radius is $a$

That is:

 $\ds \AA_1$ $=$ $\ds \dfrac {16 \pi a^2} 2 - \dfrac {\pi \paren {3 a}^2} 2 + \dfrac {\pi a^2} 2$ $\ds$ $=$ $\ds \dfrac {\pi a^2} 2 \paren {16 - 9 + 1}$ $\ds$ $=$ $\ds 4 \pi a^2$

The area $\AA_2$ of one of the inside shapes is:

half the area of the circle whose radius is $3 a$
less half the area of the circle whose radius is $2 a$
plus half the area of the circle whose radius is $2 a$
less half the area of the circle whose radius is $a$

That is:

 $\ds \AA_2$ $=$ $\ds \dfrac {\pi \paren {3 a}^2} 2 - \dfrac {\pi \paren {2 a}^2} 2 + \dfrac {\pi \paren {2 a}^2} 2 - \dfrac {\pi \paren a^2} 2$ $\ds$ $=$ $\ds \dfrac {\pi a^2} 2 \paren {9 - 4 + 4 - 1}$ $\ds$ $=$ $\ds 4 \pi a^2$

$\blacksquare$