Calculate the area of an triangle that is inscribed into an ellipse such that the elliptical sectors are of equal size.

Question

Let us inscribe a triangle $ABC$ into a ellipse such that the sectors $S_1$, $S_2$ and $S_3$ have an equal sized area. This situation is depicted by the figure below. How we can calculate the triangle's area by the semimajor axis $a$ and semiminor axis $b$ of this ellipse?

My ideas / what I know so far:

The are of the ellipse is $\pi\cdot a\cdot b$ and we can express it as the sum of the triangle's area $T_{\triangle ABC}$ and the three equally sized areas of the elliptical sectors $A_{S_1}+A_{S_2}+A_{S_3}=3\cdot A_{S_1}$. Althought this problem seems to be an elementary geometrical one, it does not seem to be so trivial, and I would appreciate any help.

Answer 1

Thanks to the hint of markvs, we can simplify the problem as follows:

• We set $a=b$ and have a circle (with radius $b=a$) instead of the ellipse.
• The elliptical sectors remain equally sized.
• This leads to the (new/projected) triangle being equilateral.
• The area of an equilateral triangle inscribed in a circle with radius $b$ is $\frac{3}{4}\sqrt{3}b^2$.

Stretching the area back to the ellipse (along the x-axis) leads to the area:

$$T_{\triangle ABC}=\frac{a}{b}\cdot\frac{3}{4}\sqrt{3}b^2$$