»  Solutions to puzzles in my VDARE.com monthly Diary

  May 2022


The Math Corner of my May Diary included this brainteaser:

Brainteaser:  You have a perfectly circular cake, iced on the top but nowhere else. You make a cut from the precise center of the cake to the perimeter (Cut 1). Now you move your knife counterclockwise precisely one radian and make another cut from center to perimeter (Cut 2).

You have cut a sector with two straight sides, angle between them one radian, and one curved side, the same length as a straight side.

Remove this sector, turn it upside down, and replace it in the cake. The top surface of the cake is no longer completely iced. Your sector, roughly one-sixth (actually one-2πth) of the surface, has no icing, only cake.

Proceed another one radian counterclockwise from Cut 2 and perform Cut 3, for a new sector identical in size and shape to the previous one. Remove it, turn it upside down, and replace it. Now almost one-third (actually one-πth) of your cake's top has no icing.

Repeat … repeat … repeat …

It is a true fact — although a counterintuitive one, I have always thought — that after a finite number N of these cuts, the top of the cake will again be completely iced.

What is the value of N?


•  Solution

Some preparatory stuff before I get into the actual problem: two wologs and a reminder. ("Wolog" is math-speak for "without loss of generality." You declare some simplifying condition that does nothing to weaken your proof.)

Wolog 1: Unit circle.  I'll assume the cake has radius one unit. I don't care what the unit is: one decimeter, one foot, one FDA-approved cake radius, one parsec, … I don't care. It's a unit circle. Note this means that the cake's circumference measures 2π units, i.e. 6.283185307179586476925286766559 and change.
Wolog 2: Rotating cakestand.  I'll assume the cake is on a freely-rotating cakestand. I'll make my first cut along the three-o'clock line, from center of cake to the eastern-most point of the circumference. Then, for ever after, my basic operation will be: (1) advance knife one radian counter-clockwise; (2) cut cake; (3) remove the one-radian sector I just cut, flip it upside-down, replace it in cake; (4) rotate entire cake one radian clockwise. I shall refer to this basic operation as "advance-cut-flip-turn."
Reminder: Concurrent cycles.  At the heart of the solution is the idea of two repeating cyclic processes of different cycle length, clicking away together but occasionally finding themselves in sync.

Consider for example this cycle of length four: ABCDABCDABCDABCDABCD … Got it? Now get it going at the same click rate as this one, length six: abcdefabcdefabcdefabcdef …

Start with Aa, then Bb, then Cc, then Dd, then — whoops! — Ae. Now the cycles are out of sync. Never mind, press on: Bf, Ca, Db, Ac, Bd, Ce, Df, Aa … Hey. they're back in sync!

How many clicks does it take for the two cycles to get back (temporarily, of course) in sync? Answer: It happens after N clicks, where N is the lowest common multiple, the LCM, of the cycle lengths, in this case four and six. N needs to be the smallest integer that both four and six divide into exactly. That is of course twelve. There you go: Aa, Bb, Cc, Dd, Ae, Bf, Ca, Db, Ac, Bd, Ce, Df, Aa.

End of preparatory stuff.

Just to remind you of the basic operation here — the "click" in concurrent-cycle language, the "advance-cut-flip-turn" in mine:

OK, but what makes a repeating cycle of these "clicks"? And how are there going to be two cycles?

You need to notice that six complete segments, which of course you will attain after Cut 7, do not quite fill the circle. By the definition of "radian," the curved side of each cake segment is length one, but the cake's entire circumference is length 2π, which is to say 6.283….

So after you've "clicked" six times from Cut 1, created and flipped six sectors — think of them as Sector 1, Sector 2, …, Sector 6 — and rotated your cake one radian clockwise six times, there is one wee sector left iced, angle 0.283… radians, its bottom radius along the three-o'clock line, its upper radius 0.283… radians above that on the clock face (that's 2.704… in minutes, if you must know, so at the clockface 12.295… minutes mark). The six sectors you've made are not iced. They just show dough.

Let's call this thin sector the Cut 7 sector in honor of the cut you made that caused you to encounter it.

And if you now advance your knife one radian counterclockwise from Cut 7 to make Cut 8 it'll be back in your first sector but 0.283… radians short of Cut 2. A new wee sector! — although this one showing dough, not icing. Call this the Cut 8 sector, because it was Cut 8 that created it.

Now you really have to do some serious visualizing. That one-radian sector you just cut consists of:

When you pull this sector out, flip it over, and replace it, directly counter-clockwise of Sector 6 will be:

Key points: After six advance-cut-flip-turn operations, all the big sectors — five with one-radian angles, the sixth with 0.716… radians — have gone from showing icing to showing dough. After seven operations, there are two of the thin sectors, angles 0.283…, both flipped from icing to dough.


A little more exploratory diddling like this will convince you that a true picture of the cake at any stage (although not necessarily in this precise orientation) is the one in my diagram here. There are thirteen cut lines, six fat sectors (angle 0.716… radians), and seven thin ones (angle 0.283…).

(And if you want to object that the radius just south of the three-o'clock line, if you advance one radian counterclockwise from it, needs a new radius drawn to roughly the clockface 8-minute mark, you're forgetting that we don't just cut, we also flip, so that the issue of which two thin sectors are adjacent is always changing, as happened with Cut 8. You really have to play with this a while …)

After six advance-cut-flip-turn operations the fat sectors all go from icing to dough. After another six they go from dough all back to icing.

After seven operations (it's easy to show) the thin sectors all go from icing to dough. After another seven they go from dough all back to icing.

So there are two concurrent cycles, cycle lengths twelve and fourteen. Least common multiple: 84. The cake goes from all-icing back to all-icing on an 84-click super-cycle.

That makes the solution to my brainteaser as stated 85, since I counted the very first cut as Cut 1. It would have been neater to make it Cut 0, since it isn't strictly a part of the advance-cut-flip-turn cycle … Sorry.