Inscribed Sector

There are few approaches for this problem.

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Method 1: Large Equilateral Triangle Method

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Consider a semicircle of radius $r$ and maximum area enclosed by the equilateral triangle.

Figure 1. Large equilateral triangle of the small triangle

Since the semicircle bounded by the equilateral triangle has the maximum area, we can slice $\Delta AEI$ with $\overline{FJ}$ to obtain the smaller equilateral triangle and color triangle $\Delta FGK$ to remove the unneeded area. The side length to decrease for $\Delta AEI$ is determined by the side length of $\Delta BCK$ , which is an equilateral triangle. This guarantees that the right-angle circular sector has the maximum area.

Observe that the quadrilateral $CFGK$ is a rectangle, consisting of two $30$ - $60$ - $90$ triangles. Since $|GK| = |CF| = r$ , then $|CK| = |FG| = \dfrac{r}{\sqrt{3}}$ . Since $\Delta BCK$ is equilateral, $|CK| = |BK| = |CB| = \dfrac{r}{\sqrt{3}}$ , which shows that to obtain the smaller triangle we decrease the side length by $\dfrac{r}{\sqrt{3}}$ .

Notice that $\Delta CEF$ and $\Delta FGK$ since $\overline{CF} \parallel \overline{AG}$ , $\overline{CK} \parallel \overline{EI}$ and $\overline{FK} \parallel \overline{AE}$ . Therefore, $|EG| = \dfrac{2r}{\sqrt{3}}$ . Since $\Delta AEI$ is symmetric about $\overline{AG}$ , $|EG| = |GI| = \dfrac{2r}{\sqrt{3}}$ , which implies that $|EI| = \dfrac{4r}{\sqrt{3}}$ . So by decreasing each side length by $\dfrac{r}{\sqrt{3}}$ , we obtain $\dfrac{3r}{\sqrt{3}}$ .

Thus, for the problem, $\text{radius of circle} = r = \dfrac{\sqrt{3}}{3}$ which gives $\text{Area of sector} = \dfrac{\pi}{4} r^2 = \boxed{\dfrac{\pi}{12}}$

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Method 2: Direct Geometric Approach

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Figure 2. Starting equilateral triangle

This problem can also be solved without considering the large equilateral triangle. For the sector to have maximum area, bound its side and vertices on the triangle's segments as illustrated above.

Because sector $BDE$ is cyclic and tangent to $\overline{AF}$ , this forces $\angle DGF = 90^{\circ}$ . But since $\overline{BD} \perp \overline{CF}$ at point $D$ , $\angle CDB = \angle DGF = 90^{\circ}$ . Noticing that $\Delta CDB$ and $\Delta DFG$ share common angles of $60^{\circ}$ as $\Delta ACF$ , and that $|BD| = |DG| = r$ , by Angle-Angle-Side Postulate , this shows that $\Delta DFG$ and $\Delta CDB$ are the same triangles of same lengths and angles.

In that case, $|CD| = |GF| = \dfrac{r}{\sqrt{3}}$ and $|BC| = |DF| = \dfrac{2r}{\sqrt{3}}$ , which implies that $|CF| = \dfrac{3r}{\sqrt{3}}$ . We achieve the same result as we did by Method 1 .