Where Should Two Parallel Lines Cut a Circle into Thirds?

Two parallel lines divide a circle into three regions: two outer circular segments and the strip between the lines. For the three areas to be equal, each outer segment must have one-third of the circle’s area. This forces the lines to be symmetric about the center.

The required distance is not found by simply dividing the diameter into three equal widths: the circle narrows toward its edges, so equal areas do not correspond to equal distances. Instead, use the area formula for a circular segment and solve a one-variable equation.

Why the lines are symmetric

Choose coordinates so the parallel lines are vertical, and let their signed distances from the center be $x_1$ and $x_2$, with $x_1<x_2$. The region to the left of the first line and the region to the right of the second line are the two outer segments. If all three regions have equal area, those two caps have equal area.

For a fixed circle, the area of a cap decreases strictly as its line moves from the center toward the circumference. Equal cap areas therefore mean equal distances from the center. Since the lines are distinct and on opposite sides, they must be at $-a$ and $+a$ for some $a>0$. The middle strip then has one-third of the disk’s area as well, because the two caps together account for the other two-thirds.

Set up the circular-segment equation

Let the circle have radius $R$, and place one cutting line at distance $a$ from the center. The area of the cap beyond that line is

$$A(a)=R^2\arccos\left(\frac{a}{R}\right)-a\sqrt{R^2-a^2}.$$

To see where this comes from, take the line to be $x=a$. At each position $x$ between $a$ and $R$, the circle has vertical height $2\sqrt{R^2-x^2}$. Integrating that height gives the cap area. The equal-area condition is $A(a)=\pi R^2/3$.

It is convenient to divide by $R$ and define $t=a/R$. Then solve the dimensionless equation

$$\arccos(t)-t\sqrt{1-t^2}=\frac{\pi}{3},\qquad 0<t<1.$$

The radius cancels, so the same proportion works for every circle. This equation is typically solved numerically rather than by an elementary formula.

Worked numerical solution

Define

$$f(t)=\arccos(t)-t\sqrt{1-t^2}-\frac{\pi}{3}.$$

At $t=0$, $f(t)=\pi/6>0$; at $t=1$, $f(t)=-\pi/3<0$. The function is continuous, so a root lies between 0 and 1. It is also strictly decreasing there, since

f'(t)=-2\sqrt{1-t^2}<0\quad(0\le t<1),$$ so there is only one root. Bisection is a reliable way to find it. Newton’s method also works if you start near $t=0.265$; the update is

t_{n+1}=t_n+\frac{f(t_n)}{2\sqrt{1-t_n^2}}.

The numerical root is $t\approx0.26493$. Thus each line is about $0.26493R$ from the center, on opposite sides. Their separation is

2a\approx0.52986R.

For a circle of diameter $D=2R$, the lines are each about $0.13247D$ from the center, and their separation is about $0.26493D$. For example, if $R=10$ cm, place the lines approximately $2.6493$ cm to either side of the center; they will be about $5.2986$ cm apart. The two outer caps and the intervening strip then each have area $\pi R^2/3$. ## Common mistakes - **Spacing the cuts by thirds of the diameter.** Equal widths do not mean equal areas in a circle, because the cross-section changes with position. - **Putting one line through the center.** That would make one side’s area half the disk, not one-third. - **Using the cap formula without normalizing.** The equation is simplest in $t=a/R$; the answer is a proportion of the radius, not a fixed length. - **Expecting a simple elementary formula.** The defining equation is exact, but its root is normally reported as a numerical approximation. The result assumes two straight, parallel lines intersect the disk and divide its area into three equal regions. Their orientation does not matter; only their distances from the center determine the areas.