Cut a circle into thirds

Required: to divide a circle into three equal-area pieces, with two straight lines. The result will look like this:


Start by isolating the upper part (segment): Label the chord AB, the radius r, and the center O. The angle AOB is θ. C at the top is so we can identify the arc ACB.


The goal is to find the angle θ that makes the area of the segment ACBA equal to 1/3 the area of the circle: (π/3)r2. The angles are in radians.

The area of the sector OACBO is: ½ r2 θ.

The area of the isoceles triangle OAB is: ½ r2 sin θ.

The area of the segment is the area of the sector minus the area of the triangle:

area = ½ r2 θ – ½ r2 sin θ

area = ½ r2 (θ – sin θ)

Set this area to 1/3 the circle area:

(π/3) r2 = ½ r2 (θ – sin θ)

Multiply through by 2/ r2:

(2/3) π = θ – sin θ

This equation doesn’t have a simple analytical solution, like θ = f(x), but it can be approximated as closely as desired. I used a spreadsheet calculation and found

θ = 2.0944

In degrees, the angle is 149.27421 degrees.

Now that we’ve computed the angle, the rest is engineering. Let’s use this result to cut a circular cake (or pie) into three equal-area pieces.

Unless you’re good at measuring angles to within 0.0001 degree, you can use a really close approximation: 150 degrees (a difference of 0.72579 degree (about ¾ degree, or about 43 minutes of arc)). This will give a segment area about 1% too big, and a middle part about 2% too small. This is a lot smaller error than you’d get by using the calculated value and cutting the cake by hand.

Notice that 150 is a multiple of 30, and it’s easy to construct a 30 degree angle. Even simpler, take a clock face: the hour marks are 30 degrees apart. Draw a clock face larger than the cake, put the cake on the clock face diagram. Connect 12 to 5, and 11 to 6. Those are the cut lines.


It’s not hard to see how we can cut the cake into 6 equal-area pieces: just make a cut along a diameter perpendicular to the two lines.


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