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Mathematics LibreTexts

15.6: Moving points to infinity

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Theorem 15.5.1 makes it possible to take any line in the projective plane and declare it to be ideal. In other words, we can choose a preferred affine plane by removing one line from the projective plane. This construction provides a method for solving problems in projective geometry which will be illustrated by the following classical example:

Theorem 15.6.1 Desargues' theorem

Desargues’ theorem[thm:desargues] Consider three concurrent lines (AA), (BB), and (CC) in the real projective plane. Set

X=(BC)(BC),Y=(CA)(CA),Z=(AB)(AB).

Then the points X, Y, and Z are collinear.

截屏2021-02-26 上午8.53.46.png

Proof

Without loss of generality, we may assume that the line (XY) is ideal. If not, apply a perspective projection that sends the line (XY) to the ideal line.

截屏2021-02-26 上午8.54.51.png

That is, we can assume that

(BC)(BC)    and    (CA)(CA)

and we need to show that

(AB)(AB).

Assume that the lines (AA), (BB), and (CC) intersect at point O. Since (BC)(BC), the transversal property (Theorem 7.3.1) implies that OBC=OBC and OCB=OCB. By the AA similarity condition, OBC\zOBC. In particular,

OBOB=OCOC.

The same way we get that OACOAC and

Therefore,

By the SAS similarity condition, we get that OABOAB; in particular, OAB=±OAB.

Note that AOB=AOB. Therefore,

By the transversal property (Theorem 7.3.1), we have (AB)(AB).

The case (AA)(BB)(CC) is done similarly. In this case the quadrangles BBCC and AACC are parallelograms. Therefore,

BB=CC=AA.

Hence BBAA is a parallelogram and (AB)(AB).

Here is another classical theorem of projective geometry.

Theorem 15.6.2 Pappus' theorem

Assume that two triples of points A, B, C, and A, B, C are collinear. Suppose that points X, Y, Z are uniquely defined by

X=(BC)(BC), Y=(CA)(CA), Z=(AB)(AB).

Then the points X, Y, Z are collinear.

Pappus’ theorem can be proved the same way as Desargues’ theorem.

Idea of the Proof

Applying a perspective projection, we can assume that Y and Z lie on the ideal line. It remains to show that X lies on the ideal line.

In other words, assuming that (AB)(AB) and (AC)(AC), we need to show that (BC)(BC).

Exercise 15.6.1

Finish the proof of Pappus’ theorem using the idea described above.

Hint

Assume that (AB) meets (AB) at O. Since (AB)(BA), we get that OABOBA and OAOB=OBOA.

Similarly, since (AC)(CA), we get that OAOC=OCOA.

Therefore OBOC=OCOB. Applying the SAS similarity condition, we get that OBCOCB. Therefore, (BC)(CB).

The case (AB)(AB) is similar.

The following exercise gives a partial converse to Pappus’ theorem.

Exercise 15.6.2

Given two triples of points A, B, C, and A, B, C, suppose distinct points X, Y, Z are uniquely defined by

X=(BC)(BC),Y=(CA)(CA),Z=(AB)(AB).

Assume that the triples A, B, C, and X, Y, Z are collinear. Show that the triple A, B, C is collinear.

截屏2021-02-26 上午9.07.55.png

Hint

Observe that the statement is equivalent to Pappus’ theorem.

Exercise 15.6.3

Solve the following construction problem

  1. using Desargues’ theorem;
  2. using Pappus’ theorem.
Hint

To do (a), suppose that the parallelogram is formed by the two pairs of parallel lines (AB)(AB) and (BC)(BC) and =(AC) in the notation of Desargues' theorem (Theorem 15.6.1)

To do (b), suppose that the parallelogram is formed by the two pairs of parallelogram is formed by the two pairs of parallel lines (AB)(AB) and (BC)(BC) and =(AC) in the notation of Pappus' theorem (Theorem 15.6.2).

Problem 15.6.1

Suppose a parallelogram and a line are given. Assume the line crosses all sides (or their extensions) of the parallelogram at different points are given. Construct another line parallel to with a ruler only.


This page titled 15.6: Moving points to infinity is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Anton Petrunin via source content that was edited to the style and standards of the LibreTexts platform.

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