I am currently trying to prove that the smallest bounding circle of an acute circle sector is the circumscircle of the corresponding isosceles triangle, and I have run into this step. I have figured out how to prove that a circle that bounds an acute isosceles triangle and is tangent to the triangle at exactly 0 or 2 vertices can not have a radius less than that of the circumcirle. However, I can not rigorously prove that a bounding circle tangent to the given triangle at only one point can not be the minimum enclosing circle.
My knowledge of math is limited to the very beginnings of calculus (I have only recently started working with derivatives). I found an explanation that seems to rely on matrices, but as of yet I do not know enough to understand it.
My question, restated in picture form:
- Prove that the enclosing circle of the iscosolese acute triangle is never the smallest enclosing circle if it is tangent to the triangle at only one point.
- Prove that the enclosing circle of the iscosolese acute triangle is never the smallest enclosing circle if it is tangent to the triangle at only one point.
I will leave the details to you, but the idea is as follows:
Assume there exists a smallest bounding circle for a triangle with just one point $A$ on its circumference. Given such a circle, show that you can construct a smaller circle which still contains the triangle by moving the centre of the circle in the direction of $A$, and shrinking the circle by an amount (which will depend on how much you moved the centre). Since this is a smaller circle, your assumption that there existed such a smallest circle was false, and so no smallest bounding circle can have just one point on its circumference.