扩展：射影变换

The Cross Ratio

Given four collinear points $A$, $B$, $X$, $Y$ we define their cross ratio to be

$$(A,B;X,Y) = \frac{XA}{XB} : \frac{YA}{YB}.$$

Here the ratios are directed, meaning $\frac{XA}{XB}$ is positive if $X$ lies outside segment $AB$ and negative otherwise.

For example:

Next, we define the cross ratio of four concurrent lines.

The cross ratio of four concurrent lines $a$, $b$, $x$, $y$ is defined as

$$(a,b;x,y) = \frac{\sin\angle(x,a)}{\sin\angle(y,a)} : \frac{\sin\angle(x,b)}{\sin\angle(y,b)}$$

where the signs are chosen so that the cross ratio is negative if lines $x$ and $y$ are in different regions of $\angle(a,b)$ and positive otherwise.

This definition is set up so that the following theorem holds.

Theorem: In the figure shown, let $a$, $b$, $x$, $y$ denote the lines $OA$, $OB$, $OX$, $OY$. Then $$(A,B;X,Y) = (a,b;x,y).$$ Proof: Law of Sines

In particular, one may take any other blue line as displayed, and one obtains the relation $$(A,B;X,Y) = (a,b;x,y) = (A_1, B_1; X_1, Y_1).$$

Finally, we define the cross ratio for four points of a circle as follows. We pick a point $O$ on the circle and take the bundle $OA$, $OX$, $OB$, $OY$, and define the ratio accordingly.

Because the angles on a circle don't change, this doesn't depend on the choice of $O$. In fact:

Thus, these theorems let us "chase" cross ratios across a diagram by projecting from points onto lines and circles. Moreover,

Projective Transformations

A projective transformation is the most general type of transformation. It sends lines to lines and conics to conics. Moreover, it preserves the cross ratio.

It turns out

Theorem: There is always a unique projective transformation for each of the following.

(a) Sending any four points in general position to four other points in general position.

(b) Fixing any circle while sending any point in it to any other point.

(c) Fixing any circle while sending a line outside to it the point at infinity.

Here is an example application.

Theorem: Let $P$ be outside a circle $\omega$ and let $PX$, $PY$ be tangents. A line through $P$ intersects $\omega$ again at $A$ and $B$. Then the cross ratio of the four points $A$, $X$, $B$, $Y$ is $-1$. (Such quadrilaterals are called harmonic quadrilaterals).

Proof: We'd like to take a projective transformation.

Of course, a square is clearly harmonic (since $\frac11:\frac11=1$), which completes the proof.

We define a pencil as the set of all lines through a point, and we define a range as the set of all points on a line. When we draw these two geometric objects together, we see a bijection between the range and the pencil.

In the diagram shown above, three lines of the pencil and three points of the range are shown.

Now suppose we were to combine these bijections. Then we get a transformation called a perspectivity.

Combining multiple perspectivities gives us what is called a projectivity (or also called a projective transformation). An interesting fact about projectivities is that when you are given three points and the three corresponding points mapped by the projectivity, the projectivity is unique. This is called the fundamental theorem of projective geometry.

In the figure below, let's prove using projective transformations that $(T,D;B,C) = -1$. (It's also possible to produce a proof using Ceva and Menelaus.)

First, we need the following lemma.$\\$ Lemma: Consider two points $X$, $Y$, the midpoint $M$ of $XY$, and the point at infinity $P$ along line $AB$.

By taking a projective transformation, we may assume $ABC$ is equilateral with center $P$.

Thus $T$ is now the point infinitely far away on line $BC$. However, $D$ is the midpoint of $BC$. So by the lemma, $(T,D;B,C)$ is harmonic as desired.

In particular, the position of $T$ does not depend on $P$ once $D$ has been chosen.