Find distance to vertex of equilateral triangle enclosing circle

Question

If I have a circle with radius $r$, how do I find the distance to a vertex of the minimal equilateral triangle that encloses it? Image source

Answer 1

Since EBD is $60^0$, and by property of circle EB=BD, thus triangle EBO is an equilateral triangle.

Also EOD is an isosceles triangle & EOD is $120^0$. So, EOB is $60^0$. Thus $b=r\cos 60=\frac{r}{2}$.

Finding the length of BF now, We know that $a=r\sin 60=\frac{r\sqrt3}{2}$. ED=$r\sqrt3$. ED=EB=BD.

So, BF=EB$\sin 60=\frac{3r}{2}$.

Hence, OB=OF(i.e. b)+BF=$\frac{r}{2}+\frac{3r}{2}=2r$

Thus, the distance from the vertex to the circle is...

2r-r=r

Answer 2

You can clearly see that the angle $EOD$ is $120$ degrees because it is carving out a third of the circle. This is due to the symmetry of an equilateral triangle. Knowing this, because the segment $BO$ bisects the angle $EOD$, we have that $EOB$ is a $60$ degree angle and therefore the triangle $EOB$ is a $30-60-90$ triangle. That makes the length of the segment $BO$ equal to $2r$ and consequently the distance from the circle to the vertex $2r-r =r$ because we subtract out the radius of the circle.