Consider an undecorated cylindrical cake and a perfect knife. We want to find the center of the cross-sectional circle.

If we can only score the surface of the cake, this reduces to finding the center of a circle with just a straightedge, which is impossible.

But using a knife allows additional constructions. For example, Sarvesh Iyer mentions:

The big difference between straight lines on a circle and knives on a cake is that you can shift any cake pieces you cut, around the shape and match them up with other pieces. The piece that's in a darker shade of orange can be removed and used to replicate that particular angle around the center subtended by it. I don't think one can create duplicate angles with just a straightedge, hence the difference.

Similarly, YNK claims that

Cake cut sevenfold

determines the center in seven cuts, although he has not explained the construction procedure.

Here are some rules modeling our use of the knife.

  1. You can't guarantee any nice properties of the lines like being perpendicular or parallel to another, just chords.
  2. The "canvas" for connections is only the circle (you can't be cutting the table, just the cake).

What is the minimal number of cuts necessary to find the circle? Is it YNK's 7?

  • $\begingroup$ @GerryMyerson because your cannot guarantee lines are perpendicular $\endgroup$
    – atul ganju
    Commented Mar 11, 2022 at 5:27
  • 1
    $\begingroup$ @justaguy The accepted answer given there should apply directly to this problem. $\endgroup$
    – Arthur
    Commented Mar 11, 2022 at 5:28
  • 2
    $\begingroup$ @YNK, you can't ping someone who has not taken part in the discussion of the question. There's no reason to expect Xander will see your comment. Also, Xander had no rôle in closing this question. What you can do is you can edit the question to make it crystal clear why it's not a duplicate. That will put it in the Review Queue, and users will be able to decide whether to reopen. $\endgroup$ Commented Mar 12, 2022 at 23:35
  • 1
    $\begingroup$ @David, show us how! $\endgroup$ Commented Mar 14, 2022 at 8:17
  • 1
    $\begingroup$ @SarveshRavichandranIyer You have just described the twist I was trying to describe in my comments and edit to the last letter. $\endgroup$
    – YNK
    Commented Mar 14, 2022 at 8:23

2 Answers 2



The seven figures given above show how we can determine the center of a cake by drawing seven line segments with a knife. In a few hours time, I will add text to this answer describing things like why we need to cut a portion segment that subtend an angle greater than $90^o$. So the interested parties can get themselves ready with a cake and knife similar to those shown in the diagrams given below.

Cake and Knife


In the seven-cut answer we are able to use pieces of the cake for such things as measuring an equal chord inside the circle. In this answer I assume we can also do such things as place two pieces so that their arcs are tangent to each other. The construction then uses four cuts.

Given a cake with center $A$, first cut off a segment along the chord $BC.$ This is the first cut.

enter image description here

Move the segment so it lies along the arc $BD$ and cut along the chord $BD.$ This is the second cut.

enter image description here

Place one of the cut-off segments along the chord $BC$ but with the arc facing "inwards". Place the other segment so that one end is at $C$ and the arcs of the two cut-off segments are tangent at $C.$ Let $E$ be the other end of the second segment's arc; cut along the line $BE.$ This cuts the circle along diameter $BF.$ This is the third cut.

enter image description here

Explanation: The angle between the chord $BC$ and the arc $BC$ is half the angle of arc $BC$, which is also half the angle of arc $BD$, which is the angle $\angle BCD,$ so the "inward facing" chord $BC$ is tangent to $CD.$ Triangle $\triangle BCE$ is isosceles with $BC=CE$ and $CD$ bisects $\angle BCE,$ so $CD$ is perpendicular to $BE,$ so $BE$ is the perpendicular bisector of $CD$ in isosceles triangle $\triangle BCD$ inscribed in the original circle.

Move one of the segments so that it lies along the arc $FG$ on the circle. Make a cut from $G$ to $C$. This cuts the circle along diameter $CG.$ This is the fourth cut.

enter image description here

The center of the circle has now been located at the intersection of the last two cuts.

Explanation: as chords $BD$ and $FG$ are congruent, chord $DG$ is parallel to diameter $BF$ and therefore also perpendicular to chord $CD.$ Triangle $\triangle CDG$ is therefore a right triangle with hypotenuse $CG.$

Incidentally, it is not necessary to be able to make the cuts exactly parallel to the cylindrical axis of the cake. We can turn the cut-off segments upside down after cutting so that when we place them on the top of the cake they will occupy the correct regions. The last two cuts would still intersect at the center of the circle on the top of the cake, but not necessarily intersect along the cake's cylindrical axis.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .