Circumscribed Circle vs. Minimum Enclosing Circle III

Let me begin by apologizing for the recent infrequency of site updates and additions, in particular on the subject of the minimum enclosing circle of finitely many points in the plane. This post is long overdue.

I first want to comment some more on computing the radius of the circumscribed circle of a triangle \triangle ABC, where A, B, and C are the vertices. It turns out that the diameter 2r of the circumscribed triangle is length of any side divided by the sine of the angle opposite that side. Note that by the law of sines, we only need to prove

\displaystyle2r=\frac{a}{\sin A},

where we abuse notation and label the angle (in radians) formed by sides \overline{AB}, \overline{AC} as A.

We break down the proof into three cases based on whether the triangle \triangle ABC is acute, right, or obtuse. The proof for a right triangle is fairly straightforward, and I leave it to the reader.

Suppose that \triangle ABC is acute. Denote the circumcenter by O. Consider the isosceles triangle \triangle BOC. If we let E denote the midpoint of side \overline{BC}, then we have \angle BOE=\angle COE. I claim that the common value of these angles is A. Indeed, define real numbers \alpha,\beta,\gamma by

\displaystyle \alpha:=\angle OAC,\indent\beta:=\angle OAB,\indent\gamma:=\angle OBC

Since the triangles \triangle AOB and \triangle AOC are also isosceles, we have

A=\alpha+\beta,\indent B=\beta+\gamma,\indent C=\alpha+\gamma

Recalling that the sum of the angles of a triangle is \pi, we see that \alpha+\beta+\gamma=\frac{\pi}{2}, and therefore

\displaystyle\angle BOE=\frac{\pi}{2}-\gamma=\alpha+\beta=A

The length of the sides \overline{BO},\overline{CO} is r, therefore

\displaystyle\frac{a}{2}=r\sin A\Longleftrightarrow 2r=\frac{a}{\sin A}

Now suppose that \triangle ABC is obtuse. Relabeling if necessary, we may assume that angle A is obtuse. I claim that angle \angle BOC=2(\pi-A). Consider the isosceles triangles \triangle BOA and \triangle AOC. Observe that A=\angle ABO+\angle ACO and

\displaystyle\angle ABO=B+\theta and \displaystyle\angle ACO=C+\theta,

Since A+B+C=\pi, we conclude that \theta=\frac{\pi}{2}-B-C. Thus,

\displaystyle\frac{a}{2}=r\sin(\pi-A)=r\sin A\Longleftrightarrow 2r=\frac{a}{\sin A}

I now want to turn to the question of when the circumscribed circle coincides with the smallest enclosing circle of \triangle ABC. Not surprisingly, the answer hinges on whether \triangle ABC is acute, right, or obtuse.

Theorem. If \triangle ABC is acute or right, then the circumscribed circle is the smallest enclosing circle. If \triangle ABC is obtuse, then the circumscribed circle is strictly larger than the smallest enclosing circle. Moreover, the smallest enclosing circle is centered at the midpoint of the longest side with radius half the length of this side.

Proof. In all cases, let O and r denote the center and radius, respectively, of the circumscribed circle. A week or so ago, we proved the following classification:

  • O is inside the triangle if and only if \triangle ABC is acute;
  • O is the midpoint of the hypotenuse if and only if \triangle ABC is right;
  • O is outside the triangle if and only if \triangle ABC is obtuse.

Also recall that the minimum enclosing disk centered at O' with radius r' must contain at least two points of \left\{A,B,C\right\} on its boundary. Without loss of generality, assume that A,B lie on the boundary. Since O' is equidistant from A and B, O' lies on the perpendicular bisector of side \overline{AB}.

Suppose \triangle ABC is acute. If O'\neq O, then since r'\leq r, we have that \left\|\frac{A+B}{2}-O'\right\|<\left\|\frac{A+B}{2}-O\right\|. Looking at the orthogonal projection of C onto the perpendicular bisector of \overline{AB}, we see that \left\|O'-C\right\|>r\geq r'. But then the point C is not contained in the minimum enclosing disk, which is a contradiction. 

Suppose \triangle ABC is right. If O'\neq O, then it is evident from the Pythagorean theorem that \left\|A-O'\right\|>r or \left\|B-O'\right\|>r.

Now suppose that \triangle ABC is obtuse. Since O lies outside the triangle, it is clear that O\neq O' and r>r'. Without loss of generality, suppose that C is the obtuse angle, so that \overline{AB} is the longest side. If O' does not lie on the perpendicular bisector of \overline{AB}, then either \left\|A-O'\right\|>r or \left\|B-O'\right\|>r. If O' lies on the perpendicular bisector to \overline{AB} but O'\neq\frac{A+B}{2}, then

\displaystyle\left\|A-O'\right\|=\left\|B-O'\right\|>r

by the Pythagorean theorem. \Box

Advertisements
This entry was posted in cs.DS, math.OC and tagged , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s