You're about to cut three pieces from a large cake to put in a round box of radius 1. If the pieces must be congruent triangles, and cannot overlap, what shape gives you the maximum amount of cake?

enter image description here

  • $\begingroup$ May the pieces be rotated or flipped? $\endgroup$
    – bobble
    Apr 17, 2022 at 20:49
  • $\begingroup$ @bobble Of course. $\endgroup$
    – Eric
    Apr 17, 2022 at 23:48
  • $\begingroup$ Are we cutting three wedges out of a round cake, or can these triangles be any shape? Wedges from a round cake would have to be isosceles triangles but the third side is an arc. $\endgroup$
    – will
    Apr 18, 2022 at 14:46
  • $\begingroup$ @will Any shape. $\endgroup$
    – Eric
    Apr 18, 2022 at 14:47

2 Answers 2


An "improved" version of AxiomaticSystem's solution:

enter image description here

The idea is to group all the most acute angles together at the bottom in a fan-like pattern. If $\alpha$ is the most acute angle, the points of the right part are at $(0,-1)$, $(\sin(\alpha), \cos(\alpha))$ and $(\sin(3\alpha), \cos(3\alpha))$.

The area of the right slice can be found using the cross product of 2 sides of the triangle and simplifying the expression. The area is $(\sin(3\alpha) + \sin(2\alpha) - \sin(\alpha))/2$. A numeric maximum can be found by zeroing the derivative.

The largest area is found at $\alpha = 0.53478228$ in radians or $30.64076777$ degrees. The total area is then $3 * 0.68338743 = 2.050162291$.

PS: I realize the layout is actually the same as AxiomaticSystem, an optimal $\theta$ will put the more acute angle at the bottom as I propose. In the end, the only difference in my solution is in the area calculation that uses another approach.


The coordinates of the points can be easily computed using the Inscribed Angle Theorem.

![![enter image description here

It says that an arc seen from the center of the circle covers exactly twice the angle as the same arc seen from any point of the circle. In the picture, the blue angles $\alpha$ at the bottom translate into angles $2\alpha$ measured at the center.

Using the symmetry of the layout you can find that the blue lines cross the circle at angles $\alpha$ and $3\alpha$ from the vertical, seen from the center.

  • 1
    $\begingroup$ When looking at AxiomaticSystem's picture it looks like the central slice doesn't have its acute angle at the bottom. An "obvious improvement" was to put the acute angle at the bottom for more space. Then the 2 side slices can be flipped upside down without effect on the solution. So it looked to me that my solution is better. But you are right, it is actually the same layout, just far from the optimal theta. The difference shows one of us must have made a mistake in his calculations. $\endgroup$
    – Florian F
    Apr 18, 2022 at 14:11
  • 1
    $\begingroup$ Can you explain the $(\sin(\alpha),\cos(\alpha))$ and $(\sin(3\alpha),\cos(3\alpha))$ coordinates? $\endgroup$
    – Dr Xorile
    Apr 19, 2022 at 15:27
  • 3
    $\begingroup$ sympy actually finds the following analytical solution -sqrt(22 - 4*sqrt(10))/12 - sqrt(11 - 2*sqrt(10))*(-sqrt(5)/6 + sqrt(2)/6)/(7 - 2*sqrt(10)) - sin(3*atan(sqrt(11 - 2*sqrt(10))*(-sqrt(5) + sqrt(2))/3))/2 which evaluated numerically and multiplied by 3 equals the value given in this answer. $\endgroup$
    – loopy walt
    Apr 20, 2022 at 13:54
  • 1
    $\begingroup$ Do we know if this is optimal? I am still working on my computer solution... $\endgroup$ Apr 22, 2022 at 10:14
  • 1
    $\begingroup$ It looks optimal to me, but without a real proof. (My "proof" of optimality so far is that I can't think of a better way...). $\endgroup$
    – Florian F
    Apr 22, 2022 at 22:19

Answering the general question looks really tricky, so let's consider a promising special case to get a lower bound.
It's clear that sticking the triangles together is usually ideal, so let's do that as much as we can. Specifically, consider a

"fan" shape like this: enter image description here
The long sides of the triangles (and the outer two middle-sides) are reflections over the y-axis, and the central triangle is the reflection of one of the outer triangles along its long side.
Let x be the common measure of the three bottom angles. Upscaling the polar formula for a circle ($r = \sin 2\theta$) from a diameter of 1 to a radius of 1 (with $\frac{x}{2}$ for $\theta$), we get that the long sides have length $2 \sin x$. Similarly, the middle sides have length $2 \sin 3x$.
This makes the area of each triangle $\text{A}=\frac{(2\sin x)(2\sin 3x)}{2}\sin x = 2\sin^2x\sin 3x$. Maximizing this is easy with logarithmic derivatives: $\frac{A'}{A} = \frac{\sin^2x}{\sin^2x}+\frac{\sin 3x}{\sin 3x} = 2 \cot x + 3 \cot 3x$, with a solution of approximately $0.7353...$ radians.
This gives our three triangles a total area of $2.17345...$, or just over $2\frac{1}{6}$.

  • 1
    $\begingroup$ Improvement: rotate IBH so that I coincides with F, IB collinear with FJ. This allows you to enlarge the triangles. $\endgroup$
    – Eric
    Apr 18, 2022 at 4:07
  • $\begingroup$ A square inscribed inside the circle has an area of 2. Looking at the figure, it seems this scheme can't produce more area than that. Are you sure about the calculation? $\endgroup$
    – Eric
    Apr 18, 2022 at 6:11
  • 2
    $\begingroup$ I think the long sides have length $2 \sin \theta$, not $2 \sin x$ $\endgroup$
    – Florian F
    Apr 18, 2022 at 14:28
  • $\begingroup$ @Eric I think that's the same solution, if you rotate F to be at the bottom, and flip horizontally. $\endgroup$
    – justhalf
    Apr 18, 2022 at 14:43
  • $\begingroup$ @justhalf I see. Then he made a mistake in calculation. $\endgroup$
    – Eric
    Apr 18, 2022 at 14:48

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