CHAPTER 13

INTRODUCTION TO INVERSIVE GEOMETRY

13.1 Inversion in the Euclidean Plane

We introduce the concept of inversion with a simple example, that of constructing the midpoint of a line segment using only a compass.

Example 13.1.1. Given a line through A and B, find the midpoint of the segment AB using only a compass.

Solution. With center A and radius r = AB, draw the circle C(A, r) and locate the point P on the line AB so that B is the midpoint of AP.

With center P, draw the circle C(P, AP) intersecting the first circle at C, as in the figure on the following page.

Finally, draw C(C, r) intersecting the line AB at P′. Then P′ is the midpoint of AB.

To sec that this is the case, note that the triangles AP′C and ACP are similar isosceles triangles since they share a vertex angle at A, so that

which implies that

and this implies that

Note that with

we have AP · AP′ = r². This relationship between P and P′ is called an inversion. More generally, we have the following definition:

Definition. (Inverse of a Point)

Given a circle C(O, r) and a point P other than O, the point P′ on the ray is the inverse of P if and only if OP · OP′ = r².

The circle C(O, r) is called the circle of inversion, the point O is called the center of the inversion, r is called the radius of inversion, and r² is called the power of the inversion.

Remark. Suppose P is a point other than the center of inversion. If P is outside the circle of inversion, then its inverse P′ is interior to the circle of inversion. If P is on the circle, then it is its own inverse. If P is inside the circle, then its inverse P′ is exterior to the circle of inversion.

Compass Method of Finding the Inverse

Note that given the ray and the circle C(O, r), the compass-only construction described above also works to find the inverse of P when P is outside the circle of inversion.

With center P and radius OP, draw an arc intersecting C(O, r) at Q. With center Q and radius OQ, draw an arc intersecting OP at P′. Then P′ is the inverse of P.

To see this, note that the isosceles triangles OQP and OP′Q are similar by the AA similarity condition so that

and therefore

The Tangent Method of Finding the Inverse

Another construction for finding the inverse using a compass and straightedge is as follows (several construction lines have been omitted).

Here we are given the circle C(O, r) and the point P outside the circle.

Draw the segment OP, and construct the tangents PS and PT to the circle with S and T being the points of tangency, as in the figure below.

Let P′ = ST ∩ OP, Then P′ is the inverse of P.

We leave the proof as an exercise.

Note that an easy modification works to find P′ when P is inside the circle: draw the line through P perpendicular to OP intersecting the circle at S and T. Then draw the tangents at S and T meeting at P′.

The Perpendicular Diameter Method of Finding the Inverse

Another method that works when P is inside or outside the circle, as in the figure on the following page, is as follows:

Then P′ is the inverse of P.

The proof is left as an exercise.

The Inversive Plane

Given a circle C, every point P in the plane has an inverse with respect to C except the center of the circle O. The point O has no inverse and is not itself the inverse of any point. As far as inversion is concerned, the point O may as well not exist.

In order to overcome this omission, we append a single point at infinity I to the plane so that the inversion maps O to I and vice-versa. The point I is also called the ideal point, and it is considered to be on every line in the plane. The Euclidean plane together with this single ideal point is called the inversive plane.

Note. There is only one ideal point in the inversive plane, in contrast to the extended Euclidean plane discussed in the earlier chapters, which has infinitely many ideal points that make up the ideal line.

When we want to exclude the ideal point from the discussion of the inversive plane, we refer to the nonideal points as ordinary points.

In the inversive plane, all lines pass through the ideal point. Two lines that meet at a single point in the Euclidean plane meet at two points in the inversive plane. Two lines that are parallel in the Euclidean plane meet only at the ideal point in the inversive plane.

Technically speaking, there are no parallel lines in the inversive plane, although we continue to use the term “parallel lines” to mean that the lines meet only at the ideal point As in the Euclidean plane, lines that coincide are also said to be parallel.

The following facts are immediate consequences of the definition of inversion in C(O, r):

Theorem 13.1.2.

Example 13.1.3. Suppose that P and Q are points on the ray . Let P′ and Q′ be the respective inverses. Show that if OQ = k · OP, then OP′ = k · OQ′.

Solution. Let r be the radius of inversion. Then

Multiplying both sides of the equation

by

we get

which implies that

Thus,

13.2 The Effect of Inversion on Euclidean Properties

A Euclidean property is one that is preserved by the Euclidean transformations (translations, rotations, reflections, and combinations thereof). Euclidean properties include distance, shape, and size. Inversion does not preserve Euclidean properties, but it does affect them in useful ways.

In this section, we denote the inversion in the circle C(O, r) by I(O, r²). Note that given a figure , its inverse ′ is obtained by taking the inverse of each point of . Thus, it follows that ′ is the inverse of if and only if G is the inverse of ′.

Lines and Circles

Before proving the result about how inversion affects lines and circles, we first prove a useful lemma.

Lemma 13.2.1. Under the inversion I(O, r²), suppose that P and Q have inverses P′ and Q′, respectively. Then ΔOPQ ~ ΔOQ′P′.

Proof. We have

which implies that

Since ∠POQ ≡ ∠Q′OP′, then ΔOPQ ~ ΔOQ′P′ by the sAs similarity criterion.

Remark. Note that in the similar triangles above we have

Note also that if we were discussing signed or directed angles, the direction of the angles would be reversed; that is,

where m(∠ABC) denotes the measure of the angle ∠ABC. as defined in Part I, Section 1.1.

And now the theorem describing the effect of inversion on lines and circles.

Theorem 13.2.2. The inversion operator I(O, r²) affects lines and circles as follows:

Proof.

(1) The first assertion is straightforward. However, note that a point of the line inverts to a different point of the line, except for the point where the line intersects the circle of inversion. The points I and O of the line are inverses of each other.

(2) Let Q be the foot of the perpendicular from O to the line, and let Q′ be the inverse of Q. Let P be any other point on the line other than I, and let P′ be the inverse of P, as in the figure on the following page. It follows from Lemma 13.2.1 that ΔOP′Q′ ~ ΔOQP and, therefore,

Thus, P′ is on the circle ω with diameter OQ′ by the converse to Thales′ Theorem. In a similar way, every point X on ω is the inverse of some point X′ on the line.

(3) This follows from the fact that since a circle through O is the inverse of a line not through O, then the line not through O is the inverse of a circle through O.

(4) Referring to the figure below, where α is the circle with diameter PQ to be inverted, Lemma 13.2.1 tells us that

Thus,

Now,

so that

From the External Angle Theorem, we have

while from Thales’ Theorem, we have

Thus, ∠QR′P′ = 90°, and from the converse to Thales’ Theorem we can conclude that R′ is on β, the circle with diameter Q′P′. Similarly, any point on the circle β is the inverse of some point on the circle α.

Inversion and Distances

Theorem 13.2.3. Let P and P′ be inverse points with P outside the circle of inversion, and let B be the point where PP′ meets the circle of inversion, as in the figure below. Then,

Proof. We have

That is,

The proof that is similar.

Theorem 13.2.4. If P′ and Q′ are inverse points for P and Q, respectively, under the inversion I(O, r²), then

Proof. We consider three distinct cases.

Case (i). P, Q, and O are not collinear.

By Lemma 13.2.1, ΔOP′Q′ ~ ΔOQP, so that

which implies that

That is,

Case (ii). P, Q, and O are collinear, with O not between P and Q.

We may assume that OP < OQ, as in the figure below.

This implies that OQ′ < OP′, since OP · OP′ = OQ · OQ′, so that

That is,

Case (iii). P, Q, and O are collinear, with O between P and Q.

The proof is almost the same as Case (ii) and is left as an exercise.

Theorem 13.2.5. (Ptolemy’s Theorem)

If ABCD is a convex cyclic quadrilateral, then

Proof. Consider the effect of I (A, r²) on the circumcircle of the cyclic quadrilateral. The circle inverts into a straight line, and the inverse points B′, C′, and D′ are on this line, as shown in the figure below.

The convexity of ABCD guarantees that C′ is between B′ and D′, so that

and thus,

from which the theorem follows.

Inversion and Angles

The angle between two curves at a point of intersection P is defined as the angle between the tangents to these curves at P.

If one or both of the curves fail to have a tangent at P, then the angle is not defined. We will be dealing only with curves that do have tangents.

Given two curves ω and ξ, there are two different magnitudes that are associated with the angle from ω to ξ: one measured counterclockwise from ω to ξ, shown as angle α in the figure below, and the other measured clockwise, as shown by β. In general, when we refer to the angle from ω to ξ without mentioning the direction, we mean the counterclockwise one.

The main result concerning inversion and angles is:

Theorem 13.2.6. Inversion preserves the magnitude of the angle between intersecting curves but reverses the direction.

We omit the proof.

For us, the main consequence of Theorem 13.2.6 is that inversion preserves tangencies and orthogonality. For example, if two circles are tangent to each other, then their inverses are also tangent to each other. If two lines meet at 90°, then their inverses, which may be circles, also meet at 90°.

13.3 Orthogonal Circles

Theorem 13.3.1. If two circles meet at P and Q, then the magnitude of the angles between the circles is the same at P and Q.

Proof. Referring to the figure below, we have ΔAPB ≡ ΔAQB by the SSS congruency condition, so ∠APB ≡ ∠AQB. Since the tangents to the circles at P are perpendicular to the radii AP and BP, it follows that the angle between the tangents at P is equal in measure to m(∠APB). Likewise, the angle between the tangents at Q is equal in measure to m(∠AQB).

Remark. The previous theorem means that to determine the magnitude of the angle between two circles intersecting at P and Q, we only need to check one of the angles. Note, however, that the directions of the angles at P and Q are opposite.

Definition. Two intersecting circles α and β are said to be orthogonal if the angle between them is 90°. We sometimes write α ⊥ β to indicate orthogonality.

Theorem 13.3.2. If two circles α and β are orthogonal, then:

Proof.

(1) This follows because a line through the point of tangency perpendicular to the tangent must pass through the center of the circle.

(2) Let P be a point on the circle β. Join P to O, the center of α, and let r be the radius of α. Let Q be the other point where the ray meets β. Let T be the point of intersection of the two circles so that OT is the tangent to β by (1) above By the power of the point O with respect to α, we have

showing that the inverse of any point on β is another point on β.

The preceding theorem has the following converse:

Theorem 13.3.3. Two intersecting circles α and β are orthogonal if any one of the following statements is true.

(a)

(b)

Proof.

The Arbelos Theorem

Theorem 13.3.4. (The Arbelos Theorem, a.k.a. Pappus’ Ancient Theorem)

Suppose that P, Q, and R are three collinear points with C, D, and K₀ being semicircles on PQ, PR, and QR, respectively. Let K₁, K₂, …, be circles touching C and D, with K₁ touching K₀, K₂ touching K₁, and so on.

Let h_n be the distance of the center of K_n from PR and let r_n be the radius of K_n. Then h_n = 2nr_n.

Proof. In the figure below, let t be the tangential distance from P to the circle K_n and apply I(P, t²).

K_n is orthogonal to the circle of inversion, so it is its own inverse.

C inverts into a line ℓ.

D inverts into a line m parallel to ℓ.

K₀ inverts into a semicircle K′₀ tangent to ℓ and m, since inversion preserves tangencies.

K_i inverts into a circle K′_i tangent to ℓ and m.

Thus, all of the K_i’s have the same radius, namely r_n, and the theorem follows.

Steiner’s Porism

Given a point P outside a circle α, a point X of α is said to be visible from P if the segment PX meets α only at X.

Figure (b) above shows the set of points that are visible from P, namely, the two tangent points and the points on the arc between the tangent points. In other words, a point X of α is visible from P if and only if

• either PX is tangent to α
• or the tangent to α at X has α and P on opposite sides.

Note also that if a line m is tangent to α at X, and if P is on the same side of m as α but not on the line m, then X is not visible from P.

Lemma 13.3.5. Suppose the line PQ misses the circle α. Then

The figure below illustrates how to find points X and Y, Let m be a line parallel to PQ and tangent to α. X is the point of tangency of m with α. There are two lines from P tangent to α. Let ℓ be the tangent line such that α and Q are both on the same side of ℓ. Then Y is the point where ℓ is tangent to α.

Lemma 13.3.6. Given two circles ω and ξ and given a point P not on either circle, there is a circle through P orthogonal to both ω and ξ.

Proof. Let Q be the inverse of P with respect to ω, and let S be the inverse of P with respect to ξ. Then there is a unique circle through P, Q, and S, and this circle must be orthogonal to both ω and ξ by Theorem 13.3.3.

Note. If O, U, and P are collinear, then the orthogonal “circle” is a line. If O, U, and P are not collinear, then the orthogonal circle is a true circle.

Lemma 13.3.7. Let ω and ξ be two nonintersecting circles with centers O and U, O ≠ U. Then we can find points X and Y that are inverses to each other with respect to both ω and ξ.

Proof. Let α be any circle other than a line that is orthogonal to both ω and ξ. We claim that the line OU intersects α in two points. In this case, the two points are X and Y and they are inverses to each other with respect to both ω and ξ.

There are two cases to consider: (i) when the circles are exterior to each other and (ii) when one circle is inside the other.

Definition. Let ω and ξ be two nonintersecting circles.

Let α₁ be a circle tangent to both ω and ξ.

Let α₂ be a circle tangent to α₁, ω, and ξ.

Let α₃ be a circle tangent to α₂, ω, and ξ.

Continuing in this fashion, if at some point α_k is tangent to α₁, then we say that

is a Steiner chain of k circles.

Remark. Given two circles ω and ξ, there is no guarantee that a Steiner chain exists for ω and ξ.

In order to prove the next theorem, we need the following lemma:

Lemma 13.3.8. Given two nonintersecting circles ω and ξ that are not concentric, there is an inversion that transforms them into concentric circles.

Proof. Using Lemma 13.3.6, we can find two circles α and β simultaneously orthogonal to both ω and ξ. These two circles intersect at the points X and Y referred to in Lemma 13.3.7.

Perform the inversion I(X, r²) for some radius r. Then:

• α transforms to α′, a straight line through Y′ and not through X.
• β transforms to β′, a straight line through Y′ and not through X.
• ω transforms to a circle ω′ and

since orthogonality is preserved.

• ξ transforms to a circle ξ′ and

Since the circle ω′ is orthogonal to the line α′, then ω′ must be centered at some point of α′. Similarly, ω′ must be centered at some point of β′. Thus, ω′ is centered at Y′. By the same argument, ξ′ must also be centered at Y′.

Now we are able to prove the following:

Theorem 13.3.9. (Steiner’s Porism)

Suppose that two nonintersecting circles ω and ξ have a Steiner chain of k circles. Then any circle tangent to ω and ξ is a member of some Steiner chain of k circles.

Proof. In the figure below, we invert α and β into concentric circles. The inversion preserves the Steiner chain of k circles.

Using the same inversion, we transform ξ into a circle ξ′, as in the figure below.

The circle ξ′ is obviously part of a Steiner chain of k circles, so by the reverse inversion, ξ must also be part of a Steiner chain of k circles.

13.4 Compass-Only Constructions

We will use some special notation for this section only:

• A(P)   the circle with center A passing through the point P.
• A(r)   the circle with center A and radius r.
• A(BC)   the circle with center A and radius BC.

Note that when we say “draw A(B),” we will often draw only an arc of the circle rather than the entire circle. Also, in this section only, “construct” means “construct using only a compass.”

The objective of this section is to show that, insofar as constructing points is concerned, a compass alone is just as powerful as the combination of compass and straightedge. To put it another way, even if a construction involves drawing straight lines, we can carry out the construction in such a manner that we postpone drawing the straight lines until the very end. Furthermore, at the end we only use the straightedge to draw lines between existing points—we never need to use the straightedge to perform any new construction. This does not mean, however, that using a compass alone will be as efficient as a compass and straightedge together.

We begin with some examples that have been discussed previously.

Example 13.4.1. Given points A and B, construct the point C such that B is the midpoint of AC.

Solution. We perform the following constructions, as shown in the figure on the following page:

The justification is as follows. As in the figure, ΔABD, ΔDBE, and ΔEBC are all equilateral triangles, so that ∠ABD = ∠DBE = ∠EBC = 60°. This means that ABC is a straight line, and so AC is a diameter of B(A). Therefore, B is the midpoint of AC.

Example 13.4.2. Given a point P outside a circle ω with center A, construct the inverse of P with respect to ω.

Solution. We perform the following constructions, as in the figure below:

Then Q is the inverse of P.

The justification is as follows. As in the figure, P′ is on B(A) and the line AP. Also, P′ is on C(A) and the line AP, so P′ is the point B(A) ∩ C(A) other than A; that is, P′ = Q.

Example 13.4.3. Given points A and B, find the midpoint C of AB.

Solution. Use Example 13.4.1 to construct the point D such that B is the midpoint of AD.

Now draw A(B), and use Example 13.4.2 to find the inverse C of D with respect to A(B). Then C is the midpoint of AB.

Example 13.4.4. Given a circle ω with center A and radius r, and given a point P inside ω, construct the inverse of P with respect to ω.

Solution. Repeatedly use Example 13.4.1 to construct points P₁, P₂, …, P_k so that P_k is outside ω. with

For example, with k = 3, we would have the following figure.

Use Example 13,4.2 to find the inverse S of P_k. Then AS · AP_k = r².

Now use Example 13.4.1 to find points S₁, S₂, …, S_k so that

Then S_k is the inverse of P, since

The following example is a famous problem due to Mohr.

Example 13.4.5. Given a circle α with unknown center A, construct its center.

Solution. We perform the following constructions, as shown in the figure below.

The justification is as follows. Let A be the center of α, and referring to the figure above, note that if we knew where A was, then Q would be the inverse of A by Example 13.4.2, so A must be Q′.

Example 13.4.6. Given an arc BC with center A, construct the midpoint of the arc.

Solution. It suffices to construct a point X of arc BC that is on the right bisector of chord BC, as in the analysis figure below.

Construction:

Justification:

The Mohr-Mascheroni Theorem

A straightedge and compass construction allows only the following:

• Drawing a straight line through two different points.
• Drawing a circle centered at one point passing through another point.
• Drawing a circle centered at one point with a particular radius (typically specified by two other points).

All constructions are just a sequence of these basic operations. By using these operations we can construct new points and then use the new points to carry out more of the basic operations.

In the Euclidean plane, there are only three ways to construct new points:

The purpose of this section is to show that we can carry out all of these constructions with a compass alone. In a sense, this means that any construction we can perform with a straightedge and compass we can also perform with a compass alone. Thus, the objective here is to accomplish (2) and (3) using only a compass. We restate (2) as an example:

Example 13.4.7. Given points A, B, C, and D, construct the intersection of C (D) with the line AB.

Solution, There are two cases to consider:

Case (i). A, B, C are not collinear. The analysis figure is shown below.

Construction:

Justification:

Case (ii). A, B, C are collinear. The analysis figure is shown below.

Construction:

The justification is left as an exercise.

Accomplishing (3), that is, constructing the intersection of two lines, requires the construction of 12 different circles. We restate (3) as an example:

Example 13.4.8. Given points A, B, C, and D, construct the intersection of the lines AB and CD.

Solution. As in the figure below, we make the following constructions.

1. Draw A(C), B(C), yielding H. Then AB is the right bisector of CH because ACBH is a kite.

2. Draw B(D) and AD, yielding K. Then AB is the right bisector of KD because AKBD is a kite. Note that this means that HK = CD because of trapezoid KCHD.

Next we perform the following constructions, as shown in the figure on the following page.

Now the justification. We need to show that X = AB ∩ CD.

Since ACXH is a kite with XC = XH, then X is on the right bisector of CH, but AB is the right bisector of CH by step 1.

To show that X is also on CD, we will show that ∠HCX = ∠HCD.

Consider the triangles HCX and HGK. We have

while from steps 2 and 4 we have

Thus,

However, since ΔHCF and ΔHGE are isosceles with a common vertex angle at H, they are similar, and

Therefore,

which shows that triangles HCX and HGK are similar by the sss similarity condition, so that ∠HCX = ∠HGK.

Hence ∠HGK = ∠HCD, and since CD || GK, then ∠HCX = ∠HCD; that is, the points C. X, and D are collinear, which completes the proof.

The solutions to Example 13.4.7 and Example 13.4.8 prove:

Theorem 13.4.9. (The Mohr-Mascheroni Construction Theorem)

Any Euclidean construction, insofar as the given and required elements are points, may be completed with the compass alone.

13.5 Problems

1. Suppose that P and Q are inverse points with respect to a circle with center S, that SP = m, and that the radius of the circle of inversion is n. Find SQ.

2. Given that A and A′ are two inverse points (A ≠ A′) with respect to some circle ω, find:

(a) the radius of ω and

(b) the center O of ω.

3. Given points P and Q with PQ = 8, draw all circles ω of radius 3 such that P and Q are inverses with respect to ω.

4. In the figure below, the tangents from P to the circle C(O, r) meet the circle at S and T. The point Q is the intersection of OP and ST. Prove that P and Q are inverses of each other with respect to the circle.

5. In the figure below, O is the center of the circle. The diameter ST is perpendicular to OP. PT intersects the circle at R, and SR intersects OP at Q. Prove that P and Q are inverses of each other with respect to the circle.

6. If the circle ξ passes through the center O of the circle ω, and if a diameter of ω meets the common chord of ω and ξ at P and meets the circle ξ at Q, show that P and Q are inverse points with respect to ω.

7. Draw the figure obtained by inverting a square with respect to its circumcircle.

8. What is the image under inversion I(O, r²) of the set of lines passing through P, where P is different from O and I? Include a sketch.

9. What is the inverse of a set of parallel lines?

10. Let P and P′ be inverses under I(O, r²) with P outside the circle of inversion. Let B be the point where PP′ meets the circle of inversion. Show that

11. Let P and Q have inverses P′ and Q′, respectively, under I(O, r²), with O between P and Q. Show that

This is called the distortion teorem.

12. If A and B are two distinct points inside some circle α, use inversion to show that there are exactly two circles through both A and B that are tangent to α.

13. A circle and an intersecting line (nontangential) can be inverses to each other in two different ways. Illustrate this by showing how to find two circles of inversion α and β such that the line and the given circle are inverses of each other.

14. If ABCD is a convex noncyclic quadrilateral, show that

15. Given a triangle ABC with circumcenter O, let A′, B′, and C′ be the images of the points A, B, and C under the inversion I(O, r²). Prove that

16. If a quadrilateral with sides of length a, b, c, and x is inscribed in a semicircle of diameter x, as shown, prove that

17. If PQ and RS are common tangents to two circles PAR and QAS, respectively, prove that the circles PAQ and RAS are tangent to each other.

18. Given a circle ω with center O and a point A outside ω, construct the circle with center A orthogonal to ω.

19. Given a circle ω and two noninverse points P and Q inside ω, construct the circle through P and Q orthogonal to ω.

20. Two circles α and β intersect orthogonally at P. O is any point on a circle γ tangent to both former circles at Q and R. Prove that the circles ω and ξ through OPQ and OPR, respectively, intersect at an angle of 45°.

21. Construct (using a straightedge and compass) a circle orthogonal to a given circle having within it one-third of the circumference of the given circle.

22. Construct (using a straightedge and compass) a circle orthogonal to a given circle so that one-third of the circumference of the constructed circle lies within the given circle.

23. Let AC be a diameter of a given circle and chords AB and CD intersect (produced if necessary) in a point O. Prove that the circle OBD is orthogonal to the given circle.

24. Let ω and ξ be orthogonal circles intersecting at P and Q, Let AB be a straight line tangent to both circles at A and B. Show that one of ∠APB and ∠AQB is 45° and the other is 135°.

25. In the Arbelos Theorem, show that the points of contact of K_i and K_i+1, i = 0, 1, …, lie on a circle.