What is the maximal area covered by two triangles in a unit circle? There are no restrictions other than that. They can overlap, touch the circle, not touch the circle etc.

So far I have shown

  1. In an optimal solution all six vertices have to lie on the unit circle.
  2. If the two triangles are not overlapping, the optimal solution has area $2$.
  3. Using two equilateral triangles, the maximal area is $\sqrt 3\approx 1.7320508$.


AD 1: This is true since any triangle with not all vertices touching the unit circle will be entirely contained in a larger triangle where all three vertices lie on the unit circle.

AD 2: The optimal non-overlapping solution is a square formed by two triangles meeting in a diameter of the circle. This can be derived from a few simple principles:

(a) Given a triangle with all vertices on the unit circle we can always increase the area by fixing two vertices using their distance as base and then move the third vertex to the middle of the circular segment between the other two to maximize the height with out changing the base. This makes optimal non-overlapping solutions consist of isosceles triangles.

(b) Any non-overlapping triangle contained within one half of the circle can be enlarged by a parallel displacement of one side until it either meets the diameter of the circle or a side of the other triangle. This effectively makes the two triangles share one side so they form a 4-gon.

(c) The optimal 4-gon with vertices on a circle is a square which can be obtained by applying principle (a) four times namely to each of the triangles formed by dividing the 4-gon by a diagonal.

So the optimal non-overlapping solution will be an inscribed square of area $2$.

AD 3: I wrote a "dirty" equation for the area as a function of the radian distance between a vertex from each triangle and found that the area was maximized when the six vertices where evenly spread out forming a hexagram (star of David).


I suspect the optimal solution is actually $2$ so that the triangles do not need to overlap. In other words, we gain nothing from allowing the triangles to overlap, if I am right. I hope someone can provide a nice and clear way to either prove or disprove this conjecture.

  • $\begingroup$ Assuming no conditions other than "two triangles in unit circle", you might as well solve "one triangle in unit circle", and then make two copies of that one triangle. For if in the ideal solution the two triangles had different areas, you'd replace the smaller with a copy of the larger, right? $\endgroup$ – John Hughes May 28 '14 at 12:23
  • $\begingroup$ @John: I do not understand how solving "one triangle in unit circle" helps. The optimal solution to that is an equilateral triangle and two such triangles can only cover an area of $\sqrt 3$ because $2/3$ of their areas overlap. This is less than what is covered by two right isosceles (and non-overlapping) triangles with a diameter as common side which form a square of area $2$. Each of those right isosceles triangles are far from optimal solutions to the "one triangle in unit circle" problem. $\endgroup$ – String May 28 '14 at 14:17
  • $\begingroup$ There's no reason for the vertices to be in any orientation other than evenly spaced, so the only real candidate shapes are the square, the pseudo-pentagram created when the triangles share one vertex and the hexagram. The pseudo-pentagram may have a larger area than the square but I sort of doubt it. $\endgroup$ – Mr. G May 28 '14 at 15:56
  • $\begingroup$ I misunderstood the problem as "find the pair of triangles in the unit circle for which the sum of the areas of the interiors is maximal", rather than "for which the area of union of the two interiors is maximal." The latter is, of course, a much more interesting problem, but one on which I have essentially no insight. :( $\endgroup$ – John Hughes May 29 '14 at 0:17
  • 1
    $\begingroup$ @String Draw $5$ vertices along the outside of the circle. The lines $1-4, 4-3, 3-1$ make the first triangle and $5-4, 4-2, 2-5$ make the second. What I was saying about the evenly spaced vertices is true for the case in which the triangles share two vertices (square) and zero vertices (hexagram,) but for the case where they share one vertex (pseudo-pentagram) I realized that it may not be. However, there is one minor simplification which is that the shape must be symmetric about a line passing through vertex $4$ and the center of the circle so you only need to consider one half. $\endgroup$ – Mr. G May 29 '14 at 14:12

Here's an approach to a proof.

We know the six vertices lie on the circle. Now divide into three cases.

  1. The two triangles are disjoint.

  2. The two triangles have overlapping interiors

  3. The two triangles meet along their boundaries, either by sharing one vertex or sharing two.

CASE 1: In case 1, consider a chord of the circle that separates the two triangles but doesn't meet either of them. This chord will be longer than at least one of the two nearby edges. Replacing that edge with the chord will increase the area, proving that the original configuration was not optimal.

Case 3: In case 3, consider the case where the triangles share just one vertex. WLOG, suppose this vertex, $S$, is at the bottom of the circle. Reading clockwise from it, we have vertices $W$, $N_1$, $N_2$, and $E$ forming two triangles $SWN_1$ and $SN_2 E$, where the letters are informally meant to suggest "west,", "north", and "east".

subcase 1: If the two edges $SN_1$ and $SN_2$ are equal length, then one lies west of north and one lies east of north; replacing both $N_1$ and $N_2$ by $N = (0, 1)$ will increase the area, showing this case is not optimal.

Subcase 2: If the x-coordinates of $N_1$ and $N_2$ have opposite signs, we can increase the areas of both triangles by replacing N_1$ and $N_2$ by $N$, again showing this situation is not optimal.

Subcase 3: WLOG, suppose the $x$-coordinates $x_1$ and $x_2$ of $N_1$ and $N_2$ are both non-negative, with $N_2$ being strictly to the right of $N_1$. Then the perpendicular distance from $N_2$ to the edge $SE$ is less than the perpendicular distance from $N_1$ to $SE$. [Proof: some trig/geometry stuff I don't feel like working out :) ] Replacing $N_1$ with $N_2$ therefore would increase the area of the right-hand triangle, showing this situation is also not optimal.

Conclusion, so far: the optimal solution either involves two interior-disjoint triangles that share an edge, or two overlapping triangles. In the first of these cases, I'm pretty sure a gradient argument will show that if you fix the shared edge, the area is optimized by having each triangle be isoceles. You then compute the exact areas as a function of the location of the shared edge's second vertex (assume the first is at the south pole) and find it's optimal when the other vertex is at the north pole, and you're done.

What about the overlapping triangles case? In this case, the total area is a smooth function of the five vertex locations (I'm still fixing the sixth at the south pole). That means you can compute the derivative with respect to each of the six angles and show that the gradient's always nonzero. So following the flow of this gradient field must lead you to a point NOT in the "overlapping interiors" domain, i.e., one of the other cases. And the only one that can conceivably be the optimum is (by prior argument) the "two triangles erected on a chord that turns out to be a diameter" situation.

I haven't computed the gradients; I haven't done the geom/trig proof in the middle of one of the cases...but it least it's a sketch of how one might get to the end result.

  • $\begingroup$ I have to read this answer a bit more carefully, but it seems quite interesting! BTW I already proved that if the triangles do not overlap then a square is optimal. So it all comes down to excluding the overlapping cases. $\endgroup$ – String May 30 '14 at 22:23

I whipped up a quick Matlab program that computes area stochastically (generate 20000 points; count how many are inside either triangle), and then ran it on 10,000 randomly generated triangle pairs (i.e., 6 random points on the unit circle; the first 3 define one triangle; the next three define the other. WLOG, I made the first angle be 0). Each time a new "largest area" was found, I printed out the triangles. It instantly found an area of ~1.25; then steadily improved this to 1.87 over the next couple of minutes. A previous instance of the program reached 1.97 or so...and this one just jumped to ~1.92, with a solution that's very nearly the two 45-90-45 triangles erected on a diameter.

That suggests to me that the correct answer is indeed this pair of triangles, with area 2 as the optimum.

>> testCircles
area = 1.2587; angles are [0 4.13265878886387 1.69578766384379 2.78030964099446 4.4988452659303 2.81878425776269]
area = 1.4833; angles are [0 1.03997694447012 4.97128628805005 5.53558413650189 4.31034051247746 2.03741390656017]
area = 1.5065; angles are [0 2.583714405482 3.95028117360417 3.41185810372066 4.81492678793123 1.47312279640955]
area = 1.8173; angles are [0 4.99055083947817 1.93427156493409 1.86260140259647 3.04043013930809 5.0276959720494]
area = 1.8517; angles are [0 3.01875721392096 1.30210531664363 4.12564514448203 2.38462146306761 5.8391459560257]
area = 1.87; angles are [0 1.78060248058588 4.66075033844476 2.08372315689924 5.4697587833186 3.76689571284346]
area = 1.9184; angles are [0 4.43108946197308 1.4092884007527 4.31113371053254 2.84429076671212 1.32017294630991]

Here's a picture of the most recent success: a near-optimal pair of triangles

The associated angles/area are

area = 1.9428; angles are [0 1.51926494165392 3.23093144636552 0.0773804130776986 3.34779844425289 4.92681116635565]
  • $\begingroup$ Very interesting! $\endgroup$ – String May 29 '14 at 14:28

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