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Piece o' Cake (Posted on 2003-12-04) Difficulty: 3 of 5
I was sitting down with Stefanie one day to share a round cake (our birthdays are only two weeks apart). "This is easy enough," I said, "one cut right through the middle divides the cake into two equal pieces."

Then, two more people showed up, but I was undaunted. Two straight cuts will divide the cake into four equal parts, I thought.

Then, I saw another car pulling up. I remembered that three straight lines can divide a circle into at most seven parts, but I was unsure if that could be done so that all the pieces are equal (in volume, not necessarily in shape).

How can I use three straight cuts to divide our cake into all equal parts and accomodate the greatest number of people?

Note: since Stefanie spent so much time decorating the cake, I don't want to rearrange the pieces when I cut them.

See The Solution Submitted by DJ    
Rating: 3.6667 (9 votes)

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Solution solution | Comment 5 of 22 |
Proof that three straight cuts cannot cut a circle into seven pieces with equal area:

In fact, each of the 3 straight line (chords) would have to cut the circle in a 3:4 ratio. If we take the radius of the cake as being our unit of measure, the smaller area cut off would have to be π*3/7.

The area of the smaller piece is the area of the sector that would be produced by connecting the endpoints of the chord and the center, minus the area of the triangle formed by the endpoints and the center of the circle. If the angle at the center subtending the chord is called α, the area of the smaller cut off piece is A = α/2 - sin(α/2)*cos(α/2) when measured in radians. This is the same as (α-sin(α))/2, which is to be set equal to π*3/7, so α-sin(α) = π*6/7.

Using Excel's solver, this happens when α = 2.916221 radians, or 167.0871582 degrees, and the chord is cos(α/2) = 0.112447 units from the center.

As each of the outer "sort of normal cake-slice-shaped" pieces must be the same size, by symmetry the chords must be placed at 120-degree intervals. That leaves the center piece an equilateral triangle with the center 0.112447 units from each base. That triangle then has a base of .389529 and a height of .337342 for an area of .065702, which, out of a total circle area of π, is only 0.020913725 or 1/47.81548878 of the circle's total area.

So seven equal pieces is impossible. There are various ways of getting six.

  Posted by Charlie on 2003-12-04 10:57:38
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