How can we draw a straight line? One way would be to use a "straight edge" something that we accept as straight. Notice that this is different from the way that we would draw a circle. When using a compass to draw a circle, we are not starting with a figure that we accept as circular; instead, we are using a fundamental property of circles that the points on a circle are a fixed distance from the center. Can we use the symmetry properties of a straight line to construct a straight line? Remember the examples earlier in this chapter. Is there a tool (serving the role of a compass) that will draw a straight line? For an interesting discussion of this question see How to Draw a Straight Line: A Lecture on Linkages by A.B. Kempe [Z: Kempe], which shows the apparatus depicted in Figure 0. For a photograph of a working model click here.
The links labeled with the same letter must have the same length. The fact that this apparatus draws a straight line is the subject of the problems below. See [SE: Hilbert, pp. 27273] for another discussion of this topic. The discovery of this linkage about 1870 is variously attributed to the French army officer, Charles-Nicolas Peaucellier (1832-1913), and to Lippman Lipkin, who lived in Lithuania and studied in Saint Petersburg. (See also Phillip Davis' delightful little book The Thread [Z: Davis], Chapter IV.) For a more detailed history (that goes back to the 13th Century) see the article by Daina Taimina here.
Figure 0 Apparatus for drawing a straight line
Before we can study the mathematics behind the Peaucellier-Lipkin linkage we need to explore some geometry of the circle:
Circles in the Plane
Q: How does a geometer capture a lion in the desert?
A: Build a circular cage in the desert, enter it, and lock it. Now perform an inversion with respect to the cage. Then you are outside and the lion is locked in the cage.
a mathematical joke from before 1938
In our study of circles we will need two criteria for similar triangles:
AAA criterion: If two triangles are similar, then the corresponding sides of the triangles are in the same proportion to one another.
SAS criterion: If two triangles have an angle in common and if the corresponding sides of the angle are in the same proportion to each other, then the triangles are similar.
Problem 1 Angles and Power Points of Plane Circles
a. If an arc of a circle subtends an angle 2a from the center of the circle, then the same arc subtends an angle a from any point on the circumference.
Use Figures 1 and 2. Draw a segment from the center of the circle to the point A and use Isosceles Triangle Theorem. Note the four different locations for A.
Figure 1 Angles subtended from outside the arc
Figure 2 Angles subtended from on the arc
b. If two lines through a point P intersect a circle at points A, A¢ (possibly coincident) and B, B¢ (possibly coincident), then
This product is called the power of the point P with respect to the circle.
Use Figures 3 and 4 and draw the segment joining A to B¢ and the segment joining A¢ to B. Then apply Part a and look for similar triangles.
Figure 3 Power of a point outside with respect to a circle
Figure 4 Power of a point inside with respect to a circle
Problem 2 Inversions in Circles
Figure 5 Inversion with respect to a circle
Note that an inversion takes the inside of the circle to the outside and vice versa and that the inversion takes any line through the center to itself. Because of this, inversion can be thought of as a reflection in the circle.
We strongly suggest that the reader play with inversions by using dynamic geometry software such as Geometers Sketchpad®, Cabri®, or Cinderella®. Using any of these you may construct the image, P¢, of P under the inversion through the circle G as follows:
Sample constructions can be found here .
a. Prove that these constructions do construct inversive pairs.
The purpose of this part is to explore and better understand inversion, but it will not be directly used in the other parts of this problem.
b. Show that an inversion takes each circle orthogonal to the circle of inversion to itself. See Figure 6.
Figure 6 A circle orthogonal to G inverts to itself
Two circles are orthogonal if, at each point of intersection, the angle between the tangent lines is 90°. (Note that, at these points, the radius of one circle is tangent to the other circle.)
c. Show that an inversion takes a circle through the center of inversion to a line not through the center, and vice versa. What happens in the special cases when either the circle or the straight line intersects the circle of inversion? Note that the line is parallel to the line tangent to the circle at C.
Look at Figure 7 and prove that DCPQ and DCQ¢P¢ are similar triangles.
Figure 7 Circles through the center invert to lines
d. An inversion takes circles not through the center of inversion to circles not through the center. Note: The circumference of a circle inverts to another circle but the centers of these circles are on the same ray from C though not an inversive pair.
Look at Figure 8. If P, Q, X invert to P¢, Q¢, X¢, then show that
by looking for similar triangles. Thus, argue that as X varies around the circle with diameter PQ then X¢ varies around the circle with diameter Q¢P¢.
Figure 8 Circles invert to circles
Parts e and f are not needed to understand the Peaucellier-Lipkin linkage, but these parts are included here because that express important properties of inversions.
e. Inversions are conformal. A transformation is called conformal if it preserves the measure of every angle.
Look at two lines that intersect and form an angle at P. Look at the images of these lines.
The following result demonstrates the close connection between inversion through a circle in the plane and reflections through great circles on a sphere.
f. Let S be a sphere tangent at its South Pole to the plane P and let f: S ® P be a stereographic projection from the North Pole. If G is the circle that is the image under f of the equator and if g is the intrinsic (or extrinsic) reflection of the sphere through its equator (or equatorial plane), then show that the transformation
is the inversion of the plane with respect to the circle G.
Imagine a sphere tangent to a plane at its South Pole, S. Stereographic projection is the transformation that maps each point, Q, (other than the North Pole, N) on the sphere to the point on the plane that is on the ray from N to Q. Stereographic projection was known already to Hipparchus (Greek, Second Century b.c.). Show that the triangle DSNP is similar to DRNS, which is congruent to DQSN, which is similar to DSP¢N.
Figure 9 Stereographic projection and inversion
Part e provides another route to prove that inversions are conformal. It can be shown that f, stereographic projection, is conformal. In addition, g (being an isometry) is conformal. Thus, the inversion
is conformal.
Problem 3 Applications of Inversions to the Peaucellier-Lipkin Linkage
a. Show that for the linkage in Figure 10 the points P and Q are the inversions of each other through the circle of inversion with center at C and radius
.
Draw the circle with center R and radius d and note that C, P, Q, are collinear.
Figure 10 A linkage for constructing an inversion
b. Show that the point Q in the linkage in Figure 11 always traces a straight line.
Figure 11 Peaucellier-Lipkin linkage for drawing a straight line
If we modify the Peaucellier-Lipkin linkage by changing the distance between the anchor points then:
c. The point Q in the linkage in Figure 12 always traces the arc of a circle. Why? Show that the radius of the circle is expressed by r2f / (g2 - f 2).
Figure 12 Peaucellier-Lipkin linkage modified to draw the arc of a circle