Among all triangles inscribed in the unit circle, how can the one with the largest area be found?
[Math] Finding the largest triangle inscribed in the unit circle
areacirclesgeometryoptimizationtriangles
Related Solutions
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]
Let a triangle be inscribed in a unit circle, and let $A$ and $B$ mark two vertices. Let $\theta$ be half the length of the arc connecting $A$ and $B$, and let $\ell$ be the length of the chord, you have that from elementary trigonometry
$$ \ell / 2 = \sin (\theta) $$
Now let $\theta_1, \theta_2, \theta_3$ be half the lengths of the three arcs demarcated by the vertices of the triangle, from the above we have that the perimeter is equal to
$$ 2 \left[ \sin (\theta_1) + \sin(\theta_2) + \sin(\theta_3) \right] $$
and we know that
$$ \theta_1 + \theta_2 + \theta_3 = \pi $$
while
$$ \theta_1, \theta_2, \theta_3 \in [0,\pi] $$
Now from the fact that $\sin$ is a concave function on $[0,\pi]$, we have that the perimeter is equal to
$$ 6 * \frac13 \left[ \sin(\theta_1) + \sin(\theta_2) + \sin(\theta_3) \right] \leq 6 * \sin (\pi / 3) $$
by Jensen's inequality, with the optimum (being the case of the equality) holding when $\theta_1 = \theta_2 = \theta_3$.
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Best Answer
Take an arbitrary triangle inscribed in the circle and let one of the sides subtend the central angle $\alpha$.
Keeping this side fixed and moving the opposite vertex to form an isoceles triangle, we get a larger triangle, and the two other sides will both subtend the central angle $\pi-\dfrac\alpha2$.
Repeating with one of the other sides, we establish the recurrence $\alpha_{k+1}=\pi-\dfrac{\alpha_k}2$. This sequence always converges to $\alpha=\dfrac{2\pi}3$, which yields the largest area.
(Actually it suffices to say that a non-equilateral triangle can always be enlarged.)