I've attempted various solutions for the problem below, to no avail:
Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?
Methods attempted:
- First, I tried drawing the problem "bare-bones". I tried finding the area of the segment using: $\frac { \theta }{ 360 } *\pi { r }^{ 2 }-[\triangle BOC]\quad =\quad area\quad of\quad segment$. I would divide this by the area of an equilateral triangle ($\frac { { s }^{ 2 }\sqrt { 3 } }{ 4 } $).
- I then tried extending the tangent lines to form another equilateral triangle, as such:
- Lastly, (very similar to previous), I tried making similar triangles ($\triangle ADE\sim \triangle ABC$), as shown below, but quickly realized it wouldn't help.
I take your second figure as reference, for the naming of points. But this naming doesn't coincide with the naming in the statement of the initial question (where $B$ and $C$ are for the points of tangency).
Let $a$ be the side of the equailateral triangle, and $r$ the radius of the circle, with:
$$\tag{0}a=r\sqrt{3}$$
The area of equilateral triangle ADE is
$$\tag{1}\sqrt{3}a^2/4.$$
The value of the area of triangle ADE minus the circular segment is
$$\tag{2}\underbrace{ar}_{\text{area of quadrilateral ODEA}}-\underbrace{(2 \pi/3)r^2}_{\text{area of circ. sector with angle 120°}}$$
Note the use of radians, without reference to degrees.
It suffices now to take the ratio between (2) and (1), and take (0) into account, yielding a unit-free result (a constant, independent of $r$ or $a$).