CHAPTER 14

RECIPROCATION AND THE EXTENDED PLANE

14.1 Harmonic Conjugates

If A and B are two points on a line, any pair of points C and D on the line for which

is said to divide AB harmonically. The points C and D are then said to be harmonic conjugates with respect to A and B.

Lemma 14.1.1. Given ordinary points A and B, and given a positive integer k where k ≠ 1, there are two ordinary points C and D such that

One of the points C and D is between A and B, while the other is exterior to the segment AB.

Proof. Choose a point C on the line AB such that

Since k > 0, then CB < AB, and we may assume that C lies between A and B.

Now, we have

so that

that is,

Now we find the point D, which will be exterior to the segment AB—beyond B if k > 1 and beyond A if 0 < k < 1.

Assuming that k > 1, we set

and solve for AD to get

Therefore, if k is a positive number such that k > 1 and C satisfies

then there always exists a point D ≠ C such that

We simply choose D such that

A similar argument will find the point D when 0 < k < 1.

Note. The midpoint C of AB satisfies

and we will adopt the convention that

where I is the ideal point in the inversive plane.

Using this convention, given two ordinary points A and B, for every positive number k there are harmonic conjugates C and D with respect to A and B for which

Recall that three positive numbers a, b, and c form a harmonic progression if and only if

form an arithmetic progression. A similar definition holds for an infinite sequence of positive numbers.

For example, the sequence

forms a harmonic progression, since the sequence

forms an arithmetic progression.

Theorem 14.1.2. Given four ordinary points A, B, C, and D, if AB is divided harmonically by C and D, then CD is divided harmonically by A and B.

This terminology is explained by the following:

Theorem 14.1.3. Suppose that P, Q, R, and S are consecutive ordinary points on a line and that Q and S divide PR harmonically. Then the sequence of distances PQ, PR, and PS forms a harmonic progression.

Proof. The hypothesis says that

We want to show that

are in an arithmetic progression; that is, that

From the first equation, we have

which implies that

which in turn implies that

which is what we wanted to show.

The Circle of Apollonius

If we are given points A and B and a positive number k ≠ 1, we can find precisely two ordinary points X on the line AB such that

However, there are also points X not on the line AB for which

and in fact, as we show in the following theorem, the set of all such points X lie on a circle.

Theorem 14.1.4. (Circle of Apollonius)

Given two ordinary points A and B, and a positive number k ≠ 1, the set of all points X in the plane for which

forms a circle called the Circle of Apollonius for A, B, and k.

Proof. Let C and D be the two points on AB for which

and let ξ be the circle with diameter CD.

We show first that every point X for which AX/XB = k is on ξ.

Let X be a point such that AX/XB = k, and since AC/CB = k and AD/DB = k, we know from the Angle Bisector Theorem that XC and XD are, respectively, internal and external bisectors of angle AXB. Referring to the figure below, we see that α + β = 90°; that is, ∠CXD is a right angle. Therefore, by the converse to Thales’ Theorem, this means that X is on the circle ξ.

We show next that every point X on the circle ξ satisfies AX/XB = k.

Let X be a point on the circle ξ and draw BP || DX and BQ || CX, as shown below.

Since X is on the circle, then ∠CXD = 90°, and it follows that ∠PBQ = 90°.

Also, since

we have the following:

Since

it follows that

from which we get XP = XQ.

Now, ∠PBQ is a right angle, and so the point B is on the circle centered at X with radius XP, by Thales’ Theorem. Thus, XB = XP, so that

Now we give some facts concerning the Circle of Apollonius.

Theorem 14.1.5. Let O be the center and r the radius of the Circle of Apollonius for A, B, and k. Then:

Proof. Statements (1) and (4) follow directly from Theorem 14.1.4.

Harmonic Conjugates and Inverses

Theorem 14.1.6. A and B are harmonic conjugates with respect to C and D if and only if A and B are inverses with respect to the circle with diameter CD.

Proof. Suppose that A and B are harmonic conjugates for CD. Then C and D are harmonic conjugates for AB; that is,

Letting r be the radius of the circle ω with diameter CD, as in the figure below.

We want to show that OA · OB = r², and the proof proceeds as in the proof of statement (3) of Theorem 14.1.5. We have

that is,

which simplifies to OA · OB = r².

Conversely, suppose that A and B are inverses with respect to the circle ω with diameter CD. Assuming that A is outside ω, as shown above, then to prove that A and B are harmonic conjugates for CD, it suffices to show that

Referring to the figure above, we have

and since OA · OB = r², we get

which completes the proof.

The relationship between harmonic conjugates and inverses allows us to show how a straightedge alone can be used to find the inverse of a point P that is outside the circle of inversion.

Example 14.1.7. Given a point P outside a circle ω with center O, construct the inverse of P using only a straightedge.

Solution. The analysis figure is shown below.

Construction:

Then Q is the inverse of P.

Justification:

By Theorem 14.1,6, this means that P and Q are inverses with respect to ω.

Inversion and the Circle of Apollonius

We state here several theorems that are easy consequences of the results in the preceding sections.

Theorem 14.1.8. If ω is the Circle of Apollonius for A, B, and k, then A and B are inverses with respect to ω.

Theorem 14.1.9. The Apollonian circle for A, B, and k is the same as the Apollonian circle for B, A, and 1/k.

Remark. Note the change in the order of the points A and B in the previous theorem.

Theorem 14.1.10. If A and B are inverse points for a circle ω, them ω is the Circle of Apollonius for A, B, and some positive number k.

Theorem 14.1.11. If α and β are orthogonal circles, then whenever either circle intersects a diameter of the other, it divides that diameter harmonically.

Proof. Referring to the figure, we know that if β cuts the diameter AB of α at P and Q, then P and Q are inverse points for α, since β is its own inverse by Theorem 13.3.2.

Thus, P and Q are harmonic conjugates with respect to A and B by Theorem 14.1.6.

The following is the converse of the previous theorem.

Theorem 14.1.12. If α and β are two circles and β divides a diameter of α harmonically, then the two circles are orthogonal.

14.2 The Projective Plane and Reciprocation

We augment the Euclidean plane with infinitely many ideal points, or points at infinity, in such a way that:

Collectively, the ideal points form the ideal line or line at infinity.

The resulting structure is called the extended plane or the projective plane.

Nonideal points and nonideal lines are called ordinary points and ordinary lines, respectively. The words “point” and “line” by themselves may refer to either an ordinary or ideal point and line.

Unlike the situation in inversive geometry, we can illustrate ideal points by using arrows or vectors to indicate the ideal point. Parallel vectors indicate the same ideal point. In the figure below, all three arrows indicate the same ideal point, and any line parallel to these vectors passes through that ideal point.

Immediate consequences of the definitions in the projective plane are as follows:

Reciprocation

Definition. Given a circle ω centered at an ordinary point O and given an ordinary point P ≠ O, the polar of P is the line p that is perpendicular to OP passing through the inverse P′ of P.

The circle ω is called the circle of reciprocation. The center O is called the center of reciprocation.

Note. We use the convention that uppercase letters denote points and corresponding lowercase letters denote the polars of those points.

• The polar of O is defined to be the ideal line.
• If P is an ideal point, then its polar is a line through O perpendicular to OP, that is, perpendicular to the arrow that points to P, as in the figure below.

Definition. If m is a line, then the pole of m is the point M such that m is the polar of M, as in the figure below.

• The pole of the ideal line is the center O of the circle ω.
• If m is a line through the center O of the circle ω, then the pole of m is the ideal point M on any line perpendicular to m, as in the figure on the following page.

The most useful theorem about poles and polars follows:

Theorem 14.2.1. (Reciprocation Theorem)

P is on q if and only if Q is on p.

Proof. We will show that if P is on q, then Q is on p, and the converse can then be obtained by interchanging P and Q. We consider three cases.

Case (i). P is an ordinary point and q is an ordinary line.

Suppose that P is on q and let Q′ be the inverse of Q.

On OP, let X be the foot of the perpendicular from Q, as in the figure below. Then from the AA similarity condition, we have

Therefore,

which implies that

Thus, X is the inverse of P.

The definition of p now shows that p is the line QX, which means that if P is on q, then Q is on p.

Case (ii). P is an ideal point and q is the ideal line.

Since q is the ideal line, Q is the center of the reciprocating circle; that is, Q = O. Therefore, p is the line through O perpendicular to any line pointing to P, as in the figure below. Thus, Q is on p.

Case (iii). P is an ideal point and q is an ordinary line passing through P.

Let Q be the pole of q. Since q is an ordinary line, we can draw the line OQ so that OQ is perpendicular to q, as in the figure on the following page.

Thus, the line OQ is the polar of P; that is, p = and Q is on p.

We leave as an exercise the situation where P = O, the center of the reciprocating circle.

 

Theorem 14.2.1 can be stated in different ways:

• P is on the polar of Q if and only if Q is on the polar of P.
• p is on the pole of q if and only if q is on the pole of p.

Definition. A range of points is a set of collinear points, and a pencil of lines is a set of concurrent lines.

An immediate consequence of Theorem 14.2.1 is the following:

Corollary 14.2.2. The polars of a range of points on a line m are a pencil of lines concurrent at M, and vice versa; that is, the poles of a pencil of lines are a range of points.

Example 14.2.3. Given a circle ω with center O, suppose that P, Q, and R are a range of points that lie on the line m, and let M be the pole of m. Show that the polars p, q, and r are concurrent at M.

Solution. We have the following:

Therefore, the polars p, q, and r of P, Q, and R, respectively, are concurrent at the point M; that is, polars p, q, and r form a pencil of lines through M, as in the figure below.

In the extreme situation where all the points are ideal points, we have a range of points on the ideal line, and their polars form a pencil of lines through the center of the circle of reciprocation, as in the figure below, where we are given ideal points P, Q, R, and S.

In the following, denotes the line through A and B.

Theorem 14.2.4. Let ω be the circle of reciprocation. Then:

Proof. We will prove statements (4) and (5) and leave the others as exercises.

Duality

Given a figure that consists of lines (entire lines, not just segments) and points (which may or may not be on the lines), then the polar or dual of is the figure that is obtained by taking the poles of the lines of and the polars of the points of .

Example 14.2.5. Draw the dual of the figure on the following page.

Solution. The dual is shown below.

The following is a translation table for obtaining the dual of a figure or the dual of a statement.

• To obtain the dual, any word or phrase that appears in one column must be replaced by the corresponding word or phrase in the other column.
• The symbol ω is the circle of reciprocation.

The translation table is a direct consequence of Theorem 14.2.1.

Theorem 14.2.6. (Principle of Duality)

If a statement that involves only points, lines, and their incidence properties is true, then the dual statement is automatically true.

We will illustrate the use of the translation table by using it to obtain the dual of Pascal’s Mystic Hexagon Theorem. The dual theorem is called Brianchon’s Theorem, and it was discovered by Brianchon by taking the dual, as illustrated on the following page.

In the left column, we state Pascal’s Theorem using only the incidence properties of points and lines. The right column is the translation obtained by using the table above.

The points

are the vertices of a hexagon, so in more familiar language, Brianchon’s Theorem says:

Theorem 14.2.7. (Brianchon’s Theorem)

If a hexagon is circumscribed about a circle, then the lines joining the opposite vertices are concurrent.

It is worth mentioning that in Brianchon’s Theorem the hexagon is considered to circumscribe the circle if each edge, possibly extended, is tangent to the circle. In the figure above, the “hexagons” are shown as bold and the lines joining the opposite vertices are lighter.

If you try the same exercise to dualize Desargues’ Theorem, you will find that you get nothing new—Desargues’ Theorem is self-dual.

14.3 Conjugate Points and Lines

Let ω be a fixed circle with center O.

Two points A and B are said to be conjugate points with respect to ω if each lies on the polar of the other (that is, if A lies on b and B lies on a).

Two lines a and b are conjugate lines with respect to ω if each passes through the pole of the other.

A point or line which is conjugate to itself is said to be self-conjugate.

Examples

The following properties concerning conjugate points and conjugate lines with respect to a circle ω are illustrated in the figure below and are easily proven.

Example 14.3.1. Each point on a line has a conjugate point on that line.

Solution. Let A be on ℓ and let B = a ∩ ℓ. Note that if ℓ and a are parallel, then B is an ideal point.

If ℓ and a are not parallel, then B is on a and, by the basic reciprocation theorem, A is on b.

Example 14.3.2. Each line through a point A has a conjugate line through A.

Solution. This is the dual of the previous example.

A direct proof is as follows:

Let b be a line through A. Then is a line conjugate to b.

Recall that the pencil of lines at B is the set of all lines conjugate to b.

Example 14.3.3. Of two distinct conjugate points on a line that cuts the circle of reciprocation, one point is inside or on the circle and the other point is outside the circle.

Solution. Let ω be the circle of reciprocation and suppose that A and B are conjugate points on the line ℓ that cuts ω.

If A is inside ω, then a misses ω, and since B is on a, B must be outside ω.

If A is on ω, then a is tangent to ω at A, and since B is on a and is different from A, then B must be outside ω.

If A is outside ω, then a cuts ω. Suppose for a contradiction that B is also outside ω, and let L be the pole of ℓ. Since we are given that ℓ cuts ω, L is outside ω. The situation is as shown in either of the two diagrams below.

Now, since A is on l, L is on a, and since B is conjugate to a, B is on a. Together these imply that a = .

Let X and Y be the points of tangency from L to ω. Then B is outside the segment XY and LB misses ω. However, this contradicts the fact that a cuts ω, Thus, we must conclude that B is on or inside ω.

The following is the dual of the previous example and we leave the direct proof as an exercise.

Example 14.3.4. Of two distinct conjugate lines that intersect outside a circle ω, one cuts the circle or is tangent to it, and the other misses the circle.

Self-Polar Triangles

A triangle is self-polar if each vertex is the pole of the opposite side. Here, the sides are considered as lines.

Remark. For a self-polar triangle:

• Any two vertices are conjugate points.
• Any two sides are conjugate lines.
• Given any two conjugate points, A ≠ B, they are the vertices of some self-polar triangle, and the third vertex is C = a ∩ b; that is, c = .

Theorem 14.3.5. Every nondegenerate self-polar triangle is obtuse, with the obtuse angle inside the circle of reciprocation ω.

Proof. Let ABC be self-polar. Then exactly one vertex must be inside ω.

Here are the reasons:

This proves that exactly one vertex is inside ω. Supposing that A is inside ω, it remains to show that ∠BAC is obtuse.

In the figure below, A is on b. Join OB. Then b is the line through A perpendicular to OB. Note that C is on b and B is on c, since C and B are conjugates.

A is on c. Join OC. Then c is the line through A perpendicular to OC.

Referring to the figure, for angles α, β, and γ, we have

showing that β is obtuse.

Theorem 14.3.6. Every obtuse triangle ABC is self-polar with respect to a unique circle ω, which is called the polar circle for the triangle.

Proof. In the figure on the following page, let O be the orthocenter of ΔABC.

Referring to the figure above,

Therefore, AECD is cyclic, and by the power of the point O with respect to the circumcircle of AECD, we have

Similarly, ADBF is cyclic, and by the power of the point O with respect to the circumcircle of ADBF, we have

Let OA · OD = k². Then

Now, let ω be the circle with center O and radius k. Then ω is the polar circle for triangle ABC.

Note that this works because:

• OA · OD = k², which means that D = A′, so BC = a (with respect to ω),
• OC · OE = k², which means that E = C′, so AB = c, and
• OF · OB = k², which means that F = B′ so AC = b.

Thus, each vertex is the pole of the opposite side, and the obtuse triangle ABC is self-polar with respect to ω.

Example 14.3.7. Show that given an obtuse triangle, the circumcircle and the 9-point circle invert into each other with respect to the polar circle.

Solution. Let ABC be an obtuse triangle with the obtuse angle at vertex B. From the previous theorem, ΔABC is self-polar with respect to the polar circle, and the vertices A, B, and C invert into the points A′, B′, and C, respectively, as in the figure below.

The circle through A, B, and C is the circumcircle of ΔABC. Since A′, B′, and C′ are the feet of the altitudes, the inverse of ΔABC with respect to the polar circle is just the 9-point circle of ΔABC.

14.4 Conies

Duality Revisited

A complete quadrangle consists of four points, A, B, C, and D, no three of which are collinear, and the six joins or lines determined by these points.

A complete quadrilateral consists of four lines, a, b, c, and d, no three of which are concurrent, and the six points determined by these lines.

Exercise 14.4.1. Show that the reciprocal of a complete quadrangle is a complete quadrilateral, and vice-versa.

We can think of an n-gon as being composed of n points and the successive lines between these points, or, in the opposite aspect, as n lines and the successive points of intersections of these lines. In general, the dual or reciprocal of an n-gon is another n-gon of the opposite aspect.

Reciprocals of Circles

The problem we are concerned with here is this: what is the image of a circle under reciprocation?

We can view a circle as a locus of points or as an envelope of lines, as in the figures on the following page, and in fact, every smooth curve can be viewed in these two aspects.

To find the reciprocal of the circle, view the circle in one aspect and see what curve is generated by the reciprocal aspect. We will use ω throughout to denote the reciprocating circle.

Special Cases

The reciprocal of the reciprocating circle ω is ω itself.

If the circle α is concentric with ω, and if the radius of α is s and the radius of ω is r, the reciprocal of α is another circle β concentric with ω with radius r²/s.

Theorem 14.4.2. Let α and ω be two nonconcentric circles with centers A and O, respectively. The reciprocal of a with respect to ω is

In each case, the focus of the conic section is O and the directrix is the polar of A. If the radius of α is s, the eccentricity of the conic is given by

Proof. We will prove case (2). The proofs of (1) and (3) are similar.

Recall that a parabola with focus O and directrix a is the set of all points P such that dist(P, a) = PO. (See the next two sections for the connection between the focus-directrix definition and the more common Cartesian definition.)

As in the figure below, let p be tangent to α, let P be its pole, and let P′ be the inverse of P.

Let M = p ∩ , let M′ be the inverse of M, and let m be the polar of M. Note that P is on m since M is on p.

We will show that PK/PO = 1.

We have

That is,

and since ΔMP′O ~ ΔMTA, we have

Focus-Directrix Definition of a Conic

Let d be a fixed line, let F be a fixed point not on the line, and let dist(X, d) denote the perpendicular distance from the point X to the line d, If is a fixed positive constant, then the set of all points X for which

is a conic section. The point F is the focus of the conic and the line d is the directrix of the conic. The positive constant is called the eccentricity of the conic, and:

• If = 1, the conic is a parabola.
• If 0 < < 1, the conic is an ellipse.
• If > 1, the conic is a hyperbola.

In the next section, we will show how to recover the Cartesian definitions of the conic sections from the focus-directrix definitions.

Cartesian Definitions from Focus-Directrix Definitions

Let d be the vertical line through (−r, 0) and let F be the point (0,0), as in the figure on the following page.

We have

which implies that

which in turn implies that

After rearranging, we get

If = 1, the x² term disappears and the equation takes the form

which we recognize as the Cartesian equation for a parabola.

If ≠ 1, then after some additional rearrangement we get

This is of the form

which is the Cartesian equation for an ellipse if < 1 or a hyperbola if > 1.

Pascal’s Mystic Hexagon Theorem

Two facts that we will not prove in this text:

Theorem 14.4.3. Pascal’s Mystic Hexagon Theorem holds for conies.

Proof. Let the vertices of the hexagon inscribed in the conic be A, B, C, D, E, and F, and let α and ω be circles such that the conic is the reciprocal of α with respect to ω. Then α is the reciprocal of the conic.

The points A, B, C, D, E, and F on the conic have polars a, b, c, d, e, and f that are tangent to the circle α. Brianchon’s Theorem, which holds for the circle α, tells us that the joins of a ∩ b and d ∩ e, b ∩ c and e ∩ f, c ∩ d and a ∩ f are concurrent. Thus, taking reciprocals, the points AB ∩ DE, BC ∩ EF, and CD ∩ AF are collinear.

Using the same idea, the following can be seen to be true.

Theorem 14.4.4. Brianchon’s Theorem holds for conic sections.

Another approach to showing that Pascal’s Mystic Hexagon Theorem and Brianchon’s Theorem are true for conies is to use the fact that all proper conies can be obtained via a central perspectivity of a circle. (See Theorem 16.6.1.)

14.5 Problems