Subsections

# 6 Some Amusements

## 6.1 Amusement 1:

Fill in the steps in the construction below, and observe the result.

2. Construct the lines , and , parallel to , and , respectively.

3. These three lines form a new triangle, ; label these so that is parallel to , is parallel to and is parallel to .

4. Observe that the sides of divide into four triangles.

5. Since the sides are parallel, these four triangles are all similar to the big triangle; in particular, .

6. Since they have some sides in common, the four smaller triangles are all congruent.

7. It follows that is the midpoint of ; is the midpoint of ; and is the midpoint of .

8. 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).

9. The lines , and , when extended, are altitudes of .

10. We have shown that the three altitudes of an arbitrary triangle meet at a point.

## 6.2 Amusement 2:

1. Let be an arbitrary triangle.

2. Let and be medians. Let be the point of intersection of these two lines.

3. Draw the line .

4. Observe that is parallel to . (This was part of a homework assignment.)

5. Then .

6. We know that . Hence, and .

7. Repeat the above argument, using the medians from and .

8. 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.)

9. 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.

## 6.3 Circles and circles

Two circles and either are disjoint, or they meet at a point, in which case they are said to be tangent, or they meet at two points, in which case, they intersect.

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.

Proposition 6.1   If two circles have three points in common, then they are identical.

Proof. Label the three points as , and , and draw the lines , and .

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.

Proposition 6.2   Suppose the circles and intersect at the points and . Let be the center of and let be the center of . Then the line is the perpendicular bisector of the line segment .

Proof. Since is a chord of (), the perpendicular bisector of the chord passes through the center (). Hence the perpendicular bisector of the line segment is the line determined by and .

Proposition 6.3   Suppose the circles and are tangent at . Then the line connecting the centers of these circles, passes through .

Proof. Let be the center of and let be the center of . Suppose the line does not pass through . Construct the perpendicular from to the line , and let be the point where this perpendicular bisector meets . Now construct the point on the line so that lies between and and so that . Then, by , . Hence , from which it follows that lies on . Using the same argument, , from which it follows that lies on . We have constructed a second point of intersection of and ; since we assumed these circles had only one point in common, we have reached a contradiction.

Corollary 6.4   If the circles and are tangent at , and is the line tangent to at , then is tangent to .

Proof. Since and are tangent at the line connecting their centers passes through . Hence the radius of at lies on the line , and so is orthogonal to . The same argument shows that the radius of lies on the line . Since the tangent to at is the line orthogonal to the radius at , it is .

We now return to the case of intersecting circles. Suppose the circles and intersect at the points and . Then the line joining the centers and is the perpendicular bisector of the line segment . We draw the radii, , , and . We define the angle of intersection of these two circles at to be . Likewise, the angle of intersection at is .

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 .

Proposition 6.5   If the circles and intersect at and , then the angle of intersection at has the same measure as the angle of intersection at .

Proof. We draw the line , which is a chord for both circles. We know that is the perpendicular bisector of this chord; let be the point of intersection of the lines and .

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.

Proposition 6.6   Suppose we are given three positive real number, , and , where , and . Then there is a triangle with sides , and .

Proof. Consider the line segment , where . Draw the circle of radius centered at , and draw the circle of radius centered at . Since there are points on that lie inside both circles. Hence either one circle lies inside the other, or the circles intersect.

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 .

## 6.4 Orthogonal circles

Two circles are orthogonal if the angle between them is .

Proposition 6.7   Let be a circle with center and radius . Let be some point on , and let be any real number. Then there is a unique circle of radius , orthogonal to , where the center of lies on the line , and lies on the same side of as does .

Proof. We first prove uniqueness. Suppose we have such a circle . Let be one of the two points of intersection of and . Then the triangle is a right triangle with right angle at . Hence, by the Pythagorean theorem, . This shows that the distance from to is determined; hence the circle is determined.

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

from . Then construct the circle of radius at that point.

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.

## 6.5 Tangent circles

Proposition 6.8   Let be a given circle of radius and center . Let be any point, where and does not lie on . Then there is a circle , centered at , where and are tangent.

Proof. Draw the line . This line intersects in two points; let be one of them. Draw the circle of radius about . This circle certainly meets at . Since lies on the line connecting the centers of the circles, the circles are tangent at .

We remark that we have in fact shown that there are exactly two circles centered at that are tangent to .

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 .

Proposition 6.9   Let be given. Then there are three mutually tangent circles, centered at , centered at , and centered at . If we require that each of the three circles lies outside the others, then the radii of these circles are determined by the lengths of the sides of the triangle.

Proof. We need to find the radii; call these , and , where is the radius of , is the radius of and is the radius of . Then we must solve the equations:

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.