Cross-Ratios

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.

The cross ratio is a number that is assigned to any set of four collinear points in a projective plane. In this post, we will be working in the real projective plane, also denoted as $\mathbb{R}P^2$.

In the real projective plane, we define a cross ratio as follows:

Definition 1. Let $A$, $B$, $C$, $D$ be four collinear points in $\mathbb R P^2$. Then the cross ratio $(A,C;B,D)$ is $$\frac{\overrightarrow{AB}}{\overrightarrow{AD}} : \frac{\overrightarrow{CB}}{\overrightarrow{CD}}$$where each length are directed.

Now here's a seemingly unrelated definition.

Definition 2. Suppose a line goes through two diagonal points $A$ and $C$ of a complete quadrangle and intersects the two remaining sides at $B$ and $D$, such that $A$, $B$, $C$, $D$ lies on the line in that order. Then we say that $A$, $B$, $C$, $D$ is a harmonic set of points.

Now, how are these two things related?

Theorem 1. $A$, $B$, $C$, $D$ are a harmonic set of points if and only if $(A,C;B,D) = -1$.

Try to prove it. (hint: what theorems have we seen that use directed lengths?)

Proof.

(NOTE: This proof is unfinished.)

Also, here is another important fact that involves cross ratios.

Theorem 2. Suppose a projectivity maps $ABCD$ (a collienar set of points) to $A'B'C'D'$ (which are another collinear set of points). Then the cross ratio is preserved.

(NOTE: This proof is unfinished.)

This is part 1 of the cross ratio series. The others are: Part 2

Angle-chasing is one of the most important technique in Olympiad geometry. However, it may not always work, and here projective methods might come to rescue. So, let me demonstate the power projective geometry!

Problem 1 (source: unknown): Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$. Let lines $AB$ and $CD$ meet at $E$ and $AD$ and $BC$ meet at $F$. Let $M$ be the projection of $O$ onto line $EF$. Prove that $MO$ bisect $\angle BMD$.

Solution: At first sight, it seems like there are no angles to start chasing with. Power of point theorem is also unapproachable. So, why not try using harmonic division?

Let us start with a nice and very useful lemma:

Lemma 1: Let points $A, C, B, D$ lie on a line in this order. $P$ is a point not on on this line. Then any two of the following conditions imply the third:

1. $(A, B; C, D)$ is harmonic.

2. $PB$ is the angle bisector of $\angle CPD$.

3. $AP \perp PB$.

Proof of lemma: I will left it for the readers to ponder with. (Hint: Sine Law)

Back to the main proof. We quickly realize that the cyclic condition doesnt really helps. So..... why not generalize it? Let us prove a more general statement:

''Let $ABCD$ be a convex quadrilateral. Let lines $AB$ and $CD$ meet at $E$ and $AD$ and $BC$ meet at $F$ and diagonals $AC$ and $BD$ meet at $N$. Let $M$ be the projection of $N$ onto line $EF$. Prove that $MN$ bisect $\angle BND$.''

Solution: Complete the harmonic bundle!! Let $P$ be the intersection of $BD$ and $EF$ and $Q$ be the intersection of $AC$ and $EF$. Then $(E,F;P,Q)$ is harmonic. Perspectivity at A implies $(P,B;M,D)$ is also harmonic. Then since $MN\perp NP$, from the lemma $MN$ is the angle bisector of $\angle BND$.

To solve the original problem, simply notice that by Brokard's theorem, $O$ and the intersection of diagonals $AC$ and $BD$ is perpendicular to $EF$. Hence we also solved the original problem. Q.E.D

So why not trying projective geometry in your future problem solving?

This is part 2 of the cross ratio series.

The only way to improve math is by examples and lots of practice, so let me continue further with more lemmas and problems.

We continue journey with a problem from the IMO 1995 Shortlist.

Problem 2: Let $ABC$ be a triangle, and let $D$, $E$, $F$ be the points of tangency of the incircle of triangle $ABC$ with the sides $BC$, $CA$ and $AB$ respectively. Let $X$ be in the interior of $ABC$ such that the incircle of $XBC$ touches $XB$, $XC$ and $BC$ in $Z$, $Y$ and $D$ respectively. Prove that $EFZY$ is cyclic.

Dont be afraid with the fact that this is from the shortlist. In fact, we just need a lemma, and one extra point to solve the problem.

Here I present another useful lemma:

Lemma 2: In a triangle $ABC$ consider three points $X, Y , Z$ on the sides $BC$, $CA$, respective $AB$. If $W$ is the point of intersection of $YZ$ with the extended side $BC$, then the four points $(B,X;C,W)$ are harmonic if and only if the cevians $AX$, $BY$ and $CZ$ are concurrent.

Hint: Try to prove that $\frac{BX}{XC}=\frac{BW}{WC}$. Ceva and Menelaus may help.

Back to the main proof.

Solution: Denote $T = BC \cap EF$. Because of the lines $AD, BE, CF$ concur at the Gergonne point of triangle $ABC$, we deduce that $(T,B;D,C)$ is harmonic. Similarly, the lines $XD$, $BY$ and $CZ$ are concurrent in the Gergonne point of triangle $XBC$, so by the lemma 2, $T \in YZ$. Now expressing the power of point $T$ with respect to the incircle of triangle $ABC$ and the incircle of triangle $XBC$ we have that $TD^2 = TE·TF$ and $TD^2 = TZ ·TY$. So $TE · TF = TZ · TY$ , therefore the quadrilateral $EFZY$ is cyclic.

Great! You have just solved an IMO shortlist! Now you are going to apply it again to the following exercise:

Let $ABC$ be a right triangle with $\angle A = 90^{\circ}$ and let $D$ be a point on side $AC$. Denote by $E$ the reflection of $A$ across the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular to $BC$ at $D$ . Prove that $AF$, $DE$ and $BC$ are concurrent.

Hint: Try power of point. Use Lemma 2.

There you see how useful harmonic conjugates can be. When you are desperate for a solution, think of this technique. Try it and it may works sometimes in unexpected ways.

Let's start with the definition of the ratio.

Definition: If $A,B,C$ are three points on a line, not all equal, then we define their ratio to be $$(A,B;C) = \frac{AC}{BC},$$ where the ratio is taken to be positive if the rays $AC$ and $BC$ point in the same direction, and negative otherwise.

If $l_1,l_2,l_3$ are three directed lines passing through a point, not all equal, then their ratio is defined by $$(l_1,l_2;l_3) = \frac{\sin\angle l_1l_3}{\sin \angle l_2l_3},$$ where the angles are oriented in the counterclockwise sense.

Exercise: (a) Show that if $A\ne B$ then there is a bijection between points $C$ on the line $AB$ and ratios $(A,B;C)$. Thus we can use the ratio as a coordinate on the line $AB$.

(b) Show that the ratio $(l_1,l_2;l_3)$ does not depend on the orientation of line $l_3$. Show that if $l_1 \ne l_2$ we can use the ratio $(l_1,l_2;l_3)$ as a coordinate on the set of lines through the point $l_1\cap l_2$.

Important exercise: Suppose that points $A,B,C$, not all equal, are on a line, and that point $P$ is not on that line. Show that $$\frac{(A,B;C)}{(PA,PB;PC)} = \frac{|PA|}{|PB|}.$$

Definition: If $A,B,C,D$ are four points on a line, no three of them equal, then we define their cross ratio to be $$(A,B;C,D) = \frac{(A,B;C)}{(A,B;D)} = \frac{AC}{CB}\bigg/\frac{AD}{DB}.$$ If $l_1,l_2,l_3,l_4$ are four lines passing through a point, no three of them equal, then their cross ratio is defined by picking an orientation for each line, and then setting $$(l_1,l_2;l_3,l_4) = \frac{(l_1,l_2;l_3)}{(l_1,l_2;l_4)} = \frac{\sin\angle l_1l_3}{\sin \angle l_3l_2}\bigg/\frac{\sin\angle l_1l_4}{\sin\angle l_4l_2}.$$

Main theorem of cross ratios: If $A,B,C,D$ are on a line, no three of them equal, and if $E$ is a point not on that line, then $$(EA,EB;EC,ED) = (A,B;C,D).$$ Easy exercise: Prove the main theorem of cross ratios.

Exercise: Let $ABC$ be a triangle, let $M$ be the midpoint of $AC$, and let $N$ be a point on line $BM$ such that $AN$ is parallel to $BC$. Let $P$ be any point on line $AC$, and let $Q$ be the intersection of line $BP$ with line $AN$. Use cross ratios to prove that $$\frac{AQ}{QN} = \frac{1}{2}\frac{AP}{PM}.$$

Exercise: (a) Show that if $A\ne B$ and $(A,B;C,X) = (A,B;C,Y)$ then $X=Y$.

(b) Show that if $(A,B;C,D) = 1$ then either $A=B$ or $C=D$.

(c) Show that if $A\ne B$, $C\ne D$, and $(A,B;C,D) = (A,B;D,C)$ then $(A,B;C,D) = -1$.

Definition: If $(A,B;C,D) = -1$, then the four points $A,B,C,D$ are called harmonic. We also say that $D$ is the harmonic conjugate of $C$ with respect to $A,B$. Sometimes we say that $A,B,C,D$ are harmonic when three of them are equal.

Examples: (i) If $M$ is the midpoint of $AB$ and if $\infty$ is the infinite point along line $AB$, then $(A,B;M,\infty) = -1$.

(ii) If $ABC$ is a triangle, and if $X,Y$ are the feet of the internal and external angle bisectors through $C$, then $(A,B;X,Y) = -1$ by the angle bisector theorem.

(iii) We have $(1,-1;x,\frac{1}{x}) = -1$ and $(0,\infty;x,-x) = -1$ for any $x$.

Quadrilateral Theorem: Let $ABCD$ be any quadrilateral. Let $E$ be the intersection of sides $AB$ and $CD$, and let $F$ be the intersection of sides $BC$ and $DA$. Let $X$ be the intersection of diagonal $AC$ with the line $EF$, and let $Y$ be the intersection of diagonal $BD$ with line $EF$. Then $$(E,F;X,Y) = -1.$$

Proof 1, using Ceva and Menelaus: By Ceva applied to triangle $AEF$ and point $C$, we have $$\frac{AB}{BE}\frac{EX}{XF}\frac{FD}{DA} = 1.$$ By Menelaus applied to triangle $AEF$ and line $BD$, we have $$\frac{AB}{BE}\frac{EY}{YF}\frac{FD}{DA} = -1.$$ Dividing these two equations, we get $(E,F;X,Y) = -1$.

Proof 2, using cross ratios: Let $P$ be the intersection of the diagonals $AC$ and $BD$. We have $$(E,F;X,Y) = (AE,AF;AX,AY) = (B,D;P,Y) = (CB,CD;CP,CY) = (F,E;X,Y).$$ Since $E\ne F$ and $X\ne Y$, we conclude that $(E,F;X,Y) = -1$.

If $EA,EB,EC,ED$ intersect a line $l$ at points $A',B',C',D'$, it often saves space to abbreviate the inference $$(A,B;C,D) = (EA,EB;EC,ED) = (A',B';C',D')$$ by just writing $$(A,B;C,D) \stackrel{E}{=} (A',B';C',D').$$ Now let's use this notation to give a compact proof of Desargues' Theorem:

Desargues' Theorem: Suppose that triangles $ABC$ and $XYZ$ are perspective from a point, that is, suppose that the lines $AX,BY,CZ$ all meet at a point $P$. Then the triangles $ABC$ and $XYZ$ are perspective from a line, that is, the intersections $AB\cap XY$, $BC\cap YZ$, $CA\cap ZX$ all lie on a line.

Proof: Let $U = BC\cap YZ$, $V = CA\cap ZX$, $W = AB\cap XY$. We want to show that $U,V,W$ lie on a line, so we may as well suppose that $V\ne W$. Let $Q,M,N$ be the intersections of line $BY$ with the lines $WV$, $AC$, $XZ$, respectively. Then we have $$(W,V;Q,BC\cap VW) \stackrel{B}{=} (A,V;M,C) \stackrel{P}{=} (X,V;N,Z) \stackrel{Y}{=} (W,V;Q,YZ\cap VW).$$ Thus $BC\cap VW = YZ\cap VW$, so the three lines $BC,YZ,VW$ meet at the point $U$.