- 6.1 Amusement 1:
- 6.2 Amusement 2:
- 6.3 Circles and circles
- 6.4 Orthogonal circles
- 6.5 Tangent circles

- Start with arbitrary triangle
.
- Construct the lines , and , parallel to , and , respectively.
- These three lines form a new triangle,
; label these so that is parallel to , is parallel to and is parallel to .
- Observe that the sides of
divide
into four triangles.
- Since the sides are parallel, these four triangles are all similar to the big triangle; in particular,
.
- Since they have some sides in common, the four smaller triangles are all congruent.
- It follows that is the midpoint of ; is the midpoint of ; and is the midpoint of .
- Construct the perpendicular bisectors of , and ; we know these all meet at a point ; call it . ( is the center of the circumscribed circle for
; this center is called the
*orthocenter*). - The lines , and , when extended, are altitudes of
.
- We have shown that the three altitudes of an arbitrary triangle meet at a point.

- Let
be an arbitrary triangle.
- Let and be medians. Let be the point of intersection of these two lines.
- Draw the line .
- Observe that is parallel to . (This was part of a homework assignment.)
- Then
.
- We know that
. Hence,
and
.
- Repeat the above argument, using the medians from and .
- Conclude that the three medians of an arbitrary triangle meet at a point. (This point is called the
*centroid*of the triangle; it is at the center of gravity.) - We also have shown that the centroid divides each median into two segments; the segment between the centroid and the vertex is twice as long as the segment between the centroid and the opposite side.

It is essentially immediate that two circles with the same center but different radii are disjoint.

Since a line is the shortest distance between two points, if we have two circles where the distance between the centers is greater than the sum of their radii, then the circles are necessarily disjoint.

If we have two circles with the property that the distance between their centers is exactly equal to the sum of their radii, then the line between their centers contains a point on both circles.

The three points cannot be collinear, for a line intersects a circle in at most two points. Since the three points are not collinear they form a triangle. Then both circles are circumscribed about . Since the circumscribed circle about a triangle is unique, the two circles are the same.

**Remark:** We could have chosen the angle between the circles to be
. The reason for our choice is that, if two circles are tangent, and each lies outside the other, then, by continuity, the angle between them is 0, while if one lies inside the other, then the angle between them is .

Observe first that, by , and . It follows that and that . Hence .

Since the angle of intersection at and the angle of intersection at have the same measure, we can simply call it the *angle of intersection* of the two circles.

We remark that, since the tangent to at is orthogonal to , and the tangent to at is orthogonal to , then the angle between the lines and has the same measure as one of the angles between these tangents.

Since , the point lies outside the circle centered at , and since , the point lies outside the circle centered at . Hence neither circle lies inside the other, and so the circles intersect. Let be one of the points of intersection, and observe that and .

To prove existence, find the point on , on the same side of as , and at distance

To show that and intersect, it suffices to show that there is a point on both and , or equivalently, that there is a triangle with side lengths, , and . This follows from the above proposition, once we observe that , , and .

We remark without proof that, given two orthogonal circles and , there is a 1-parameter family of circles orthogonal to both and . However, given three mutually orthogonal circles, there is no fourth circle orthogonal to all three.

There are three possible orientations for the two tangent circle. We can have that lies outside and lies outside , or we can have that lies inside , or we can have that lies inside .

**Exercise:** Suppose , and are three given points on a line. How many distinct triples of mutually tangent circles are there, where one of the circles is centered at , one is centered at , and the third is centered at .

It is an exercise in linear algebra to show that these equations have a unique solution.

We close with the remark that, given three mutually tangent circles, there exist exactly two disjoint circles that are tangent to all three. If one has four mutually tangent circles, then there can be no fifth.