Maximal area covered by two triangles in unit circle
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:
The associated angles/area are
area = 1.9428; angles are [0 1.51926494165392 3.23093144636552 0.0773804130776986 3.34779844425289 4.92681116635565]
Here's an approach to a proof.
We know the six vertices lie on the circle. Now divide into three cases.
The two triangles are disjoint.
The two triangles have overlapping interiors
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.