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

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Problem

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

By equal here, we mean of equal area.


Solution

Circle-into-4-by-3-equal-lines.png

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:

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

$\blacksquare$


Sources

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