You have to determine a way to cut a circular cake into exactly three portions of equal size. The only marking on the cake is a candle in the very center.

All you have to work with is a knife that is at least as long as the diameter of the cake, and your hands. No other equipment is available..


Hint #1

Though this puzzle requires some understanding of geometry, it is also a practical problem. Use the knife as a way of making measurements and referencing dimensions

Hint #2

The knife doesn't need to be as long as the diameter of the cake, it could be as long as the radius and the same method would work!

  • 1
    $\begingroup$ Is marking a length on the knife (using the icing or keeping a finger somewhere on the knife to measure) allowed? $\endgroup$
    – Zoir
    Commented Sep 19, 2019 at 7:21
  • 1
    $\begingroup$ @Zoir it sure is! $\endgroup$
    – Rorxor
    Commented Sep 19, 2019 at 7:23
  • 8
    $\begingroup$ You've marked this question as a duplicate, but I don't think it is. The answer to the previous (closed) question relies on folding the pizza, which wouldn't work with cake. $\endgroup$ Commented Sep 19, 2019 at 7:55
  • 4
    $\begingroup$ In addition to @Randal'Thor 's comment, the alleged duplicate was closed as 'too broad'; I don't believe that 'too broad' can be applied to this question as formulated. VTR. $\endgroup$ Commented Sep 19, 2019 at 11:21
  • $\begingroup$ Related Numberphile and another $\endgroup$
    – BruceWayne
    Commented Sep 22, 2019 at 14:22

6 Answers 6


Another way to do this

without turning the cake sideways (let's assume we're cutting a 2D circle)


1. Let O be the location of the candle. Cut the cake in the direction from O to any point on its boundary (let's designate this point with A).
2. Measure the radius of the cake with a knife (by placing it inside the OA cut and marking the location of A by e.g. your fingers).
3. Build a chord of this length (the radius) from A to some point B on the boundary. Now AOB is equilateral triangle, so the angle AOB is 60 degrees.
4. Repeat the same operation from B (in the same direction) to get the new point C.
5. Cut from C to O. Since the AOC angle is 120 degrees, so the AOC sector of a cake is exactly 1/3 of it.
6. Repeat steps 3 and 4 to get the location of a new cut and get the other 1/3.
7. Repeat step 5 and cut the second 1/3.
8. The remaining part should be also 1/3. Profit.


Of course, it is not necessary to cut the cake in B, only to mark its location with the knife (in practice, it can be easily done), so this method can help to get exactly 3 slices.

  • 4
    $\begingroup$ Exactly the answer I had in mind! :) $\endgroup$
    – Rorxor
    Commented Sep 19, 2019 at 7:36
  • $\begingroup$ @Rorxor don't forget to accept the answer by clicking the green checkmark. $\endgroup$
    – Brandon_J
    Commented Sep 20, 2019 at 5:18

An answer using the knife as a measure has already been given, but you can also use the cake itself:

I'm going to assume the cake cannot exist in two places at once, so we have to assume it has whipped cream or some other messy stuff on the sides (the candle suggests it's a birthday cake, so that should be a given), so that it leaves that annoying circle of creamy mess wherever it is placed down. (If you have two identical cakes, this is not necessary.)

Using the candle as a guide, mark two pairs of points opposite of each other. This way we can easily find the centre point of the whipped cream ring again after removing the cake:

enter image description here


remove the cake, reconstruct the centre point, and place the cake so that its perimeter just touches the centre point.

enter image description here

Now the creamy ring (or the other cake) will intersect the cake's perimeter at two points, exactly 120° apart from each other. Cut from these points to the candle to get the first slice, then rotate the (now pacman-shaped) cake in place so that one of the cut edges lines up with the other cut's mark, and cut along the same line as before.

  • $\begingroup$ Hopefully this isn't a large cake, or one that'll fall apart easily. Otherwise it'll never survive being moved, ruining the idea of each person getting exactly 1/3rd of the cake. ;-) $\endgroup$ Commented Sep 19, 2019 at 17:29
  • 2
    $\begingroup$ @computercarguy Good news, we all have equal portions. Bad news, 0 = 0 = 0. $\endgroup$
    – corsiKa
    Commented Sep 20, 2019 at 17:46
  • $\begingroup$ @corsiKa, well I was thinking the half not being moved would be fine, so equal portions of X = X = X, but X + X + X = 1/2, meaning each would get only 1/6th of the cake. lol. $\endgroup$ Commented Sep 20, 2019 at 17:52

Divide the cake into 6 pieces.

With the knife,

cut from the candle to an edge. Cut the cake completely in half. Using the knife, hold the point at the centre, and use the length from centre to edge to mark another edge of the cake. Cut from this point to the centre. Continue until you have 6 even slices.

Each person would get

two slices each.

  • $\begingroup$ How do you make sure the cylinders are of equal size? $\endgroup$
    – Rorxor
    Commented Sep 19, 2019 at 7:13
  • $\begingroup$ math.stackexchange.com/a/2465047 $\endgroup$ Commented Sep 19, 2019 at 7:18
  • $\begingroup$ What would you take as the parallel line? (a parallel line is given in that question) $\endgroup$
    – Zoir
    Commented Sep 19, 2019 at 7:24
  • 1
    $\begingroup$ I think you are correct! Except if exactly 3 pieces are required, you can use your same method to rot13(znxr phgf nygreangviryl gb trg rknpgyl 3 fyvprf) :) $\endgroup$
    – Zoir
    Commented Sep 19, 2019 at 7:34
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    $\begingroup$ @Zoir I've given an example in my answer. $\endgroup$
    – trolley813
    Commented Sep 19, 2019 at 7:35

enter image description here

Consider a hexagon filling up the area of the cake. A hexagon is made up of 6 equilateral triangles.

So, each edge of the triangle would be the same as the radius of the cake (candle to edge).

Mark the knife to indicate the radius of the cake going from one point to the candle.

Now, make one cut from the candle to the edge. Then, use the marking on the knife to measure from that cut (on the edge) to the next point on the edge of the cake. Then, do this a second time since we want 3 slices instead of 6. Make the next slice there. Then, repeat for the last slice.

  • 1
    $\begingroup$ Isn't this the same as @trolley813's answer? $\endgroup$
    – Stiv
    Commented Sep 19, 2019 at 20:11
  • $\begingroup$ Doh! My bad. I didn't read through that answer. Sorry. $\endgroup$
    – MER
    Commented Sep 19, 2019 at 20:21
  • 1
    $\begingroup$ No worries - the diagram adds something new! $\endgroup$
    – Stiv
    Commented Sep 19, 2019 at 20:41

I have a different solution, not based on precise measurement but relying on people's love of cake!


First we must assume that each person wants to maximise their portion of cake.

Having your cake

Person 1 makes the first cut from the candle to the edge.
Person 2 makes the second cut from the candle to the edge elsewhere on the cake. Person 3 makes the third cut from candle to edge elsewhere on the cake.

Eating it

Person 1 takes the first portion of cake
Person 3 takes the second portion of cake
Person 2 takes the remaining slice of cake


Person 1 cuts first but takes the first slice. They are in no position to affect the distribution of slice sizes but will penalise 2 and 3 if they cut uneven slices. Person 2 could cut a very small slice, but they will be left with this slice as they choose last. Similarly, Person 3 cuts last and takes the second slice so they will also be penalised if they cut uneven slices.

  • $\begingroup$ Welcome to Puzzling! I think your answer is missing a cut....two cuts yield only two pieces... $\endgroup$ Commented Sep 20, 2019 at 11:39
  • $\begingroup$ Maybe I'm missing something, but if you have one thing and cut once, you have two; cut a second time and you have three pieces. In general pieces = ncuts + 1. If my answer is unclear I'm happy to edit. $\endgroup$
    – RDavey
    Commented Sep 20, 2019 at 12:27
  • $\begingroup$ What you say is true if the cut starts from one point in the perimeter of the object and ends in another. Indeed, If I cut the cake "in half", I have two pieces with one cut. But, if you cut from the centre(candle) to the outside, you still have one piece, the whole cake... $\endgroup$ Commented Sep 20, 2019 at 12:42
  • $\begingroup$ You make a good point; I retract my formula. But if you make two cuts from the centre to the perimeter in two separate places you have three pieces of cake, no? $\endgroup$
    – RDavey
    Commented Sep 20, 2019 at 14:25
  • 1
    $\begingroup$ @RDavey, just two. The first cut gives you a cake with a single cut in it, but still a whole cake. The second cut will give you that first slice, plus the remainder of the cake, but that's still only 2 pieces $\endgroup$
    – Dancrumb
    Commented Sep 20, 2019 at 16:11

You can do this without marking the knife, if you can use it to draw straight lines on the tabletop with icing. That's because

every ruler and straightedge construction can be done with a straightedge alone if you have a fixed circle with its center. That means you can inscribe an equilateral triangle.

It will take you a while.

Instructions here:

Steiner’s Straightedge Problem

  • $\begingroup$ But the cake isn’t an infinite plane to draw on: is it still possible only with lines drawn inside the circle? $\endgroup$
    – boboquack
    Commented Sep 20, 2019 at 7:04
  • $\begingroup$ @boboquack Probably not, I think you'd need the table top. Just a finite part of it, not all the way to infinity. This might make a good small (undergraduate) research project. $\endgroup$ Commented Sep 20, 2019 at 10:35
  • $\begingroup$ The OP says there is no other equipment to work with, which technically seems to rule out a table.... (I agree that's ridiculous) $\endgroup$ Commented Sep 20, 2019 at 20:38
  • $\begingroup$ @GregMartin Fair comment. There's another solution that calls for moving the cake elsewhere on the table. And I am looking into whether the straightedge calculation can be done entirely on the surface of the cake. $\endgroup$ Commented Sep 20, 2019 at 20:45

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