Compass Geometry

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Introduction

With compasses alone, all the points that can be constructed with straightedges and compasses can be constructed. That means that straightedges are only necessary for the actual drawing of lines. One would not want to dispense with straightedges, however, since the constructions with compasses alone are much more complicated.

The geometry of compasses was developed independently by G. Mohr in Denmark in 1672, and by L. Mascheroni in Italy in 1797. The easiest way, however, to show that compasses are sufficient depends on circle inversion which wasn't invented until 1828 by Jacob Steiner.

References

1. R. Courant and H.E. Robbins, What is Mathematics? Oxford Univ. Pr., New York, 1953.
2. H.S.M. Coxeter, Introduction to Geometry, Wiley, New York, 1961.
3. Euclid, Elements, aleph0.clarku.edu/~djoyce/java/elements/elements.html.
4. D. Pedoe, Circles, Dover, New York, 1957.

April, 1998; March, 2002.
David E. Joyce
Department of Mathematics and Computer Science
Clark University
Worcester, MA 01610

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1. Euclidean constructions

In Book I of Euclid's Elements there are four postulates for constructions in plane geometry. Postulates 1 and 2 construct straight lines given two points. Postulate 3 constructs a circle given a center and radius Postulate 5 constructs points of intersections for nonparallel lines. Although Euclid did not state postulates for constructing intersection points for two circles, and intersection points for a circle and a line, such unstated postulates are used. For instance, an intersection point for two circles is needed in Proposition I.1, and an intersection point for a circle and a line is used in I.2.

The rest of the constructions in Euclidean geometry are built out of the constructions of the postulates. Such constructions include angle bisection (I.9), line bisection (I.10), perpendicular lines (I.11 and I.12), triangles given the three sides (I.22), parallel lines (I.31), and many others.

Our construction of a compass is precisely that of Euclid's Postulate 3. This doesn't correspond precisely to what a physical compass can do, since a physical compass can also transfer distances; given three points A, B, and C, a physical compass can be set to an opening of size AB, then moved to draw a circle with center C and radius equal to size AB. Nonetheless, Postulate 3 is enough, as Euclid shows in I.3 how to transfer distances.

We are interested in constructing the points of intersections of circles and lines, rather than the circles and lines themselves. Although we cannot construct a line with a compass that only draws circles, we will be able to find the intersection points of that line with other lines and with circles.

Essentially, there are three methods to construct points in Euclidean geometry, CC, LC, and LL.

CC: construct the intersections of two circles given their centers and a point on each circumference.

For example, in the diagram at the right, the circle with center A and radius AB intersects the circle with center C and radius CD at the two points E and F.



LC: construct the intersections of a line given by two points and a circle given its center and a point on its circumference.

For example, the intersections of the line GH with the circle with center C and radius CD are the two points K and L.



LL: construct the intersection of two lines, each given by two points.

For example, the line GH intersects the line MN at the point P.

Frequently, the circles and lines don't intersect, but it is only when they do intersect that the points are constructed, so the possibility of nonintersection is not relevant here.

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2. Reduce Euclidean constructions to circular constructions

It is evident that the first construction is possible by compasses alone; just draw the two circles. We will first show that the other two constructions are possible by means of two other constructions, PC and CN, which we'll show later can be performed by compasses alone.

PC: construct the inverse P' of a point P in a circle given its center and point on its circumference

CN: construct the center of a circle given three points on its circumference

It takes a few steps to use the three circular constructions, CC, PC, and CN, to perform the two constructions involving straight lines, LC and LL.

Step 1. Select an arbitrary circle outside the figure, that is, a circle containing no parts of the figure. Call it the bounding circle. Use PC to invert in that bounding circle all the specified points of the original figure. The inverted points will all lie inside the bounding circle. This inversion converts the data for circles and lines of the original figure to data for circles in the inverted figure.

Step 2. We'll need the centers of the circles in the inverted figure in order to draw the circles. The construction CN constructs those centers so long as we have three points on each circle, and we can get those by using PC to invert enough points from the original circles and lines.

Step 3. Each required intersection of the original figure — whether of two circles, of a circle and a line, or of two lines — corresponds to an intersection of two circles in the inverted figure. Use CC to find the intersections in the inverted figure.

Step 4. Use PC again to invert the intersections in the inverted figure back to the intersections of the original figure.

Thus, the constructions CC, PC, and CN are sufficient to perform the constructions LC and LL.

The above diagram shows the original figure of two circles and two lines inverted in a bounding circle RST. The large circle RST on the right is outside the original figure, and the inverted figure is displayed inside it. Note how the blue and orange circles are inverted to blue and orange circles, but the lines GH and MN are inverted to (white) circles. All the straight lines from the original figure invert to circles passing through the center of RST. The inverted points A' and B' are not shown because they are not the centers of the inverted circles and play no role in the construction.

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又见《近代欧氏几何学》§63

3. Summary of inversion

We don't need all the properties of inverion in a circle for this investigation, just a few of them. Of course, we need a definition.

Fix a circle with center O and radius OA = r. For each point P other than the center O, the point inverse to P in the circle O is that point P' on the line OP (extended in the direction of P if needed) so that the proportion

OP / r = r / OP'

is satisfied. An equivalent condition is that

OP OP' = r².

In order to cover the special case that occurs when P is the center O of the circle, we'll add a point ∞ to the plane, and let O and ∞ be inverses of each other.

We'll use the notation P▷C to indicate the inverse point P'. Read "P▷C" as "P through C." More generally, when B is any plane geometric figure (point, line, circle, etc.) and C is a circle, we'll use the notation B▷C for the result of inverting B in the circle C. We'll also use this notation for reflecting across a straight line. So, if B is any plain geometric figure, and L is a straight line, then B▷L is the figure that results from reflecting B across L.

Note that inversion in a circle is an involution. That is, application of inversion in the same circle a second time undoes the first application. Otherwise said, it is an operation that when done twice yields the identity operation. Algebraically, this observation is the identity (B▷C)▷C = B, where B is any plane figure.

Also note that a point P is its own inverse, that is, P▷C = P, if and only if it lies on the circumference of the circle. In particular, C▷C = C.

A standard construction for inversion

The standard construction to invert a point in a circle uses both compass and straightedge, so we won't be able to use it. Our construction using just a compass will come later. Still, the standard construction is interesting. We only need to consider points P which either lie outside the circle, or lie inside the circle but are not the center of the circle; points on the circle are their own inverses, and the center has ∞ as its inverse.

Let there be a circle with center O and radius OA = r, and let P be a point. To invert P in the circle, first draw the line OP. Now, if P happens to be a point outside the circle, then draw the tangents PB and PC to the circle, then draw the line BC. Where these two lines meet will be the inverse point P'. Since the two right triangles OBP and BP'P are similar, therefore the proportion OP / r = r / OP' holds.

But if P happens to be a point inside the circle (except the center), then draw the line BC perpendicular to OP, and let B and C be the points of intersection of that line with the circle. Then draw tangents to the circle at B and C. In this case, the inverse point P' is the intersection of the two tangent lines. Note again the two similar right triangles, from which it follows that the identical proportion holds.

A third case occurs when P lies on the circle, and in that case let the inverse point P' be the same as P itself.

The fourth case occurs when P is the center O of the circle, and as mentioned above, O and ∞ are inverses.

Inverting lines in a circle.

We will need to know some properties of inversion. The first is that when you invert a straight line in a circle, the result is a circle. Also, when you invert one circle in another, then the result is a circle (or in a special case, a straight line).

We will invert lines and circles in our circle with center O and radius OA = r.

First, consider a line BC which doesn't pass through the center O. We'll show that its inverse is a circle passing through the center of the circle. Let D be the foot of the perpendicular drawn from O to the line BC. Let D' be the point inverse to D. Draw the circle with diameter OD'. Let E be an arbitrary point on the line BC, and let E' be where the ray OE meets the circle with diameter OD'. Then triangle OE'D' is a right triangle, and it's similar to the right triangle ODE. Hence, we have the proportion

OE / OD = OD' / OE'. Therefore, OE OE' = OD OD' = r², and we conclude E' is inverse to E. As E was an arbitrary point on the line, we may conclude that the circle with diameter OD' is inverse to the line BC.

Since inversion is an involution, it is clear that any circle passing through the center of our circle will invert to a straight line.

Should a line BC pass through the center O, then it is clear that its inverse will be itself.

Inverting circles in a circle.

Next, we'll invert a circle BCD in our circle with center O and radius OA = r. There are two cases to consider—when O lies outside the circle BCD, and when it lies inside that circle. (The third case is when O lies on the circle, but that's taken care of above, and the inverse of BCD is a straight line.)

We'll start with the case where O lies outside the circle BCD.

Let E be any point on the circle BCD, and F the other point where the line OE intersects the circle BCD. Let OT be a line from O tangent to the circle BCD at the point T, and let t denote the length OT. Then by Euclid's Proposition III.35, OE OF = t².

We're looking for a point E' on the line OE such that OE OE' = r². Dividing this equation by the equation OE OF = t², we recognize that what we need is a point E' so that

OE' / OF = t² / r².

Therefore, OE' is a constant times OF. This means that when we apply a scaling to the plane by a factor of t² / r² with the fixed point of the plane being O, then points on the circle BCD will be inverted to points in the scaled circle. Note that both the original and scaled circles have the same tangent lines from the point O.

The other case occurs when O lies inside the circle BCD. It may seem that because there are no tangents from O to the circle BCD we've run into serious problems. But we didn't actually use the tangent lines in the first case; all we needed to know was that the product OE OF was a constant. In this case, it's also a constant.

As before, let E be any point on the circle BCD, and F the other point where the line OE intersects the circle BCD. (This time, F is on the other side of O.) Then by Euclid's Proposition III.36, the product OE OF is a constant, call it k.

We're still looking for a point E' on the line OE such that OE OE' = r². Dividing this equation by the equation OE OF = k, we recognize that what we need is a point E' so that

OE' / OF = k / r².

Therefore, OE' is a constant times OF. This means that when we apply a scaling to the plane by a factor of k / r² with the fixed point of the plane being O, then points on the circle BCD will be inverted to points in the scaled circle.

We conclude that when one circle is inverted in another, and the center of the other is not on the circumference of the first, then the result is a circle. But if the center of the other is on the circumfere nce of the first, then the result is a straight line.

Reflection in a line as a limit of inversions

Kepler's principle of continuity can be used to relate reflection in a line to inversion in circles. Kepler observed that some figures are limits of others. For instance, a parabola is a limit of ellipses in the sense that if one focus of a ellipse is left fixed, and the other moves to infinity along a line, then the ellipse becomes closer and closer to a parabola. A more basic example of continuity is the one we can use here, namely that a line is the limit of circles. Let ST be a straight line, and consider a circle AT tangent to the line at T. Let the center of the circle be O, and let the radius OT grow by moving O further away from T along a line perpendicular to ST. As the circle grows, the circumference gets closer and closer to the line ST.

Now examine what happens to the inverted point D = B▷OT of a point B in the circle OT as the circle enlarges. In the diagram, as you drag point A toward S, the circle OT approaces the line ST and the inverse point D approaches C = B▷ST, the reflection of C in the line ST. Thus, the operation of reflection in a line is a limit of inversions in circles.

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又见《近代欧氏几何学》§144-149

4. Stereographic projection

So that the point at infinity, ∞, is not treated separately, the ordinary Euclidean plane is replaced by the sphere. This is most easily seen by means of the well-known stereographic projection. A nice feature of using the sphere is that straight lines in the Euclidean plane correspond to circles on the sphere, so that straight lines and circles are put on the same footing. Furthermore, reflecting across a straight line becomes inversion in a circle on the sphere.

Here's how it works. Take a sphere and a plane in three-space. Typically, the plane is taken to cut through the center of the sphere, or it's taken tangent to the sphere, but it doesn't matter. What's important is the selection of the projection point N. It must be on the sphere and on a line through the center of the sphere perpendicular to the given plane. (It also can't be a point in the plane.) In this diagram, the sphere is tangent to the plane at the point S (south pole), O is the center of the sphere, and N (north pole) is the projection point.

A point A on the sphere corresponds to a point B on the plane when the three points N, A, and B are collinear. The only point on the sphere that doesn't correspond to a point in the plane is the projection point N, and it corresponds to the point at infinity, ∞. Thus, we have a correspondence between the plane with a point of infinity with the sphere.

It's fairly easy to see why a straight line in the plane, like BD, corresponds to a circle on the sphere passing through the projection point N. Just consider the plane that includes the line BD and the point N. It will cut the sphere in a circle ACN. Conversely, given a circle ACN passing through the projection point N, the plane of the circle will intersect the base plane in a line BD.

It is more difficult to see why a circle in the plane corresponds to a circle on the sphere not passing through the projection point N. In the diagram below, we'll show that the circle ACE on the sphere is projected to the circle BDF on the plane.

The easiest way to see that is to recognize stereographic projection as an inversion in a sphere. Inversions in a sphere are three-dimensional analogues of inversions in a circle. The defining proportion is the same.

Inversion in a sphere has properties similar to inversion in a circle. For instance, inverting a sphere A in a sphere B results in a sphere A▷B, unless the center of the B sphere lies on A, in which case the result is a plane.

Take the first case where the center P of B doesn't lie on the the sphere A. The sphere A is generated by rotating a circle around an axis, so take the axis to be line OP connecting the centers P and O of the spheres. Then the sphere A intersects that axis at two points C and D, and A is the sphere on the diameter CD. Let a plane containing this axis rotate around the axis. That plane intersects A in a circle with diameter CD, and the inverse of that circle is the circle in that plane on the diameter C'D', where C' and D' are the points inverse to C and D, respectively. As the plane rotates around the axis, the circles on the diameter CD are inverted to the circles on the diameter C'D'. Thus, a sphere A▷B is generated by the rotating inverse circles.

The other case where the center of B does lie on A is relevant to stereographic projection. The argument that A▷B is a plane is similar to the previous argument, but it depends on the fact that a plane is generated by a rotating line perpendicular to an axis at a point on that axis.

We'll show now that the stereographic projection of a sphere with diameter NS to the plane tangent to S is actually inversion in a sphere with center N and radius NS. The diagram to the left shows a cross section, that is, an intersection with a plane containing the axis NS. The point A on the sphere is projected to the point B. Note that we have three similar right triangles: NSB, NAS, and SAB. Therefore, NA : NS = NS : AB. Thus, when A is inverted in the circle with radius NS, the resulting point is B. We conclude that stereographic projection is inversion in a sphere.

Now we can answer why stereographic projection preserves circles. Consider any circle on the sphere that doesn't pass through N. It is the intersection of the sphere with a plane. Since stereographic projection is an inversion in a circle, that plane will be inverted to a sphere, and that sphere will intersect the tangent plane in a circle. Thus the image of a circle not passing through N is a circle.

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又见《近代欧氏几何学》§66

5. Construction PC to invert a point in a circle

Let there be given a circle C with center O and point A on its circumference, and a point P either outside the circle C or inside but more than half the radius away from O. Here is a construction to invert P in the circle C to get P' = P▷C. Draw the circle with center P and radius PO. Then this new circle meets the circle C at two points, Q and Q'. Draw the two circles with centers Q and Q' and radii QO and QO', respectively. Besides O, they will meet at another point P'.

This point P' will be the image of the point P inverted in the circle C. The usual definition of the inverted point P' is that it is situated on the line OP from the center of the circle to P so that OP:OA = OA:OP'. From the symmetry of the construction, P' is clearly on that line. To see that the proportion is valid, note that the two triangles QOP and P'OQ are each isosceles, and they have the same base angles, so they're similar. The proportion immediately follows.

Now, this construction doesn't work if P is within half the radius of O since the two circles won't intersect to produce the points Q and Q', so some other construction is needed to invert such P. But if P is at least a quarter of the radius away from O, then we will find the point R twice as far from O as P is, then invert R in the circle C to get R' = R▷C by the method described above, and then P' will be the point twice as far from O as R' is. This requires a doubling construction 2OP.

2OP: given two points O and P, construct the point R on the line OP twice as far from O as P is.

The construction 2OP can be effected as follows. Draw the circles with centers O and P and radius OP. Let them intersect at Q and Q'. Draw the circle with center Q and radius QQ'. It will intersect the circle with center P at the points Q' and R.

Now we can invert any point P in a circle so long as it is at least a quarter of a radius away from the center of the circle.
When P is very close to O, all we have to do is repeatedly double the distance from O until we've got a point at least half the radius from O, invert that point, and double the resulting distance the same number of times to get P'.

Bisecting and multisecting lines

Before going on, we can use inversion in circles to effect a construction to bisect a line segment, or more generally, cut off one n-th of a line segment. To bisect a line segment OP, use construction 2OP to double it to OR, then construction PC to invert R in the circle with center O and radius OP to get the midpoint M of OP.

We can trisect OP by doubling OP to OR then doubling PR to PS, so that OS is triple OP, then inverting S in the circle with center O and radius OP to get S'. Then OS' will be one-third of OP

More generally, we can cut off an n-th of OP by first extending OP by a factor of n (which is easily seen as a variant of the doubling construction), then inverting in that same circle.

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6. Construction C to find the center of a circle given three points on its circumference

We're not done yet in showing that compasses alone are sufficient to construct all the constructable points in Euclidean geometry. The problem is that all the circle constructions depend on knowing the centers of the circles, and when we invert one circle C in another D to get a circle E = C▷D, then the center of E is not the inverse of the center of C inverted in D. The center of E is determined, however, by three of the points on E, and three points on E can by found by taking any three points on C and inverting them in D. Thus, we'll need the construction C to find the center of a circle given three points on its circumference.

Let P,Q, and R be three given points. We are to construct the center O of the circle passing through them. We'll find O by inverting everything in the circle C with center P and radius PQ.

First, invert the point R in C to get the point R' = R▷C. Next draw two circles, one with center Q and radius QP, the other with center R' and radius R'P. Besides at P, these two circles will intersect at another point O'. It is evident that this point O' is the reflection of P in the line QR', that is, O' = P▷QR'. Finally, invert O' in C to get O.

Next, we'll show O is the center of the circle PQR. Now, when the circle PQR is inverted in the circle C, then the result is the straight line QR', since P will be sent to the point ∞ at infinity, Q is fixed, and R is sent to R'. Algebraically, that says PQR▷C = QR'. To show that O is the center of the circle PQR, we only need to show that O = ∞▷PQR. That will follow from a property of ▷, namely, ▷ distributes over itself on the right, as shown in the next section.

∞▷PQR = (P▷C) ▷ (QR'▷C) = (P▷QR')▷C = O'▷C = O.

Thus, we have constructed O, the center of the circle passing through the three points P, Q, and R.

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7. Inversive geometry and involutory quandles.

Consider now the algebraic operation of inversion, ▷, and its properties on circles. First, circles remain fixed upon inversion in themselves. Second, inverting twice in the same circle is the identity operation. Third, inversion preserves inversion in the sense that if B inverts A and A', then B▷C inverts A▷C and A'▷C.

In the diagram above, the initial three circles are A, B, and C. Then D = A▷B, E = A▷C, F = B▷C, and G is both (A▷B)▷C, and (A▷C)▷(B▷C). In other words, ▷ distributes over itself on the right.

In order to verify this statement, it suffices to show that

(P▷B)▷C = (P▷C)▷(B▷C)

when P is a point. For if the identity holds for all points P of A, then it holds for A itself.

In the diagram to the right, P is a given point, and B and C are given circles. The point Q = P▷B, R = P▷C, the circle F = B▷C as before, and the point S will equal both Q▷C and R▷F.

To see that S does equal both of these, it is enough to note that Q is characterized as the point that all the circles pass through which are orthogonal to B and pass through P, that R▷F is analogously characterized as the point that all the circles pass through which are orthogonal to F and pass through R, and that inversion through C preserves circles and orthogonality of circles.

Involutive quandles

Algebraically, the three properties of inversion in circles are the following three identities.

Q1. A▷A = A.

Q2. (A▷B)▷B = A.

Q3. (A▷B)▷C = (A▷C)▷(B▷C).

A set equipped with a binary operation ▷ satisfying these three axioms is called a involutory quandle. We'll follow the notational convention for ▷ that when parentheses are left out of an expression, operations are to be performed from left to right.

It may appear at first sight that these properties are somewhat mysterious, but they're not. Any time you consider the involutions in a group (elements of order 2 in a group), or a conjugacy class of involutions, then conjugation as a binary operation forms a involutory quandle. That is, when ▷ is defined by x▷y = yxy^-1, which is the same as yxy when y is an involution, then ▷ satisfies Q1 through Q3. Each of the three axioms is easily verified.

That's just the situation we have when we consider inversions in circles in inversive geometry. The group is the Möbius group of inversive transformations, transformations that preserve circles. The inversions form a conjugacy class of involutions. (They're not all the involutions; for instance, half-turns are also involutions.)

Incidentally, when any conjugacy class is considered, a quandle results, but it's only an involutory quandle conjugacy classes of involutions are considered. For a quandle, axiom Q2 is weakened.

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In 1943, Mituhisa Takasaki (高崎光久) introduced an algebraic structure which he called a Kei (圭), which would later come to be known as an involutive quandle.

en.wikipedia.org/wiki/Racks_and_quandles

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hbghlyj 发表于 2022-7-17 14:33
7. Inversive geometry and involutory quandles.

Consider ...

又见《近代欧氏几何学》§80


§80 定理 设两个已知点 $P, Q$ 关于圆 $c$ 互为反演, $P, Q$ 及圆 $c$ 关于另 一个圆 $b$ 的反形为 $P^{\prime}, Q^{\prime}$ 及圆 $c^{\prime}$, 则 $P^{\prime}, Q^{\prime}$ 关于圆 $c^{\prime}$ 互为反演.
过 $P, Q$ 任作两个圆 $j, k$, 则它们都与圆 $c$ 正交, 它们关于 $b$ 的反形 $j^{\prime}, k^{\prime}$ 与 $c^{\prime}$ 正交. 因此 $j^{\prime}$ 与 $k^{\prime}$ 的交点 $P^{\prime}, Q^{\prime}$ 关于 $c^{\prime}$ 互为反演.
这个定理可以解释成 “反演性经反演后不变”. 即两个互为反形的图形, 连同它们的反演圆, 受到一个反演的作用, 所得的图形仍互为反形.