Moving points tutorial

by yayups, Jan 5, 2019, 9:50 AM

I think a lot of people will really like this. (original link here https://artofproblemsolving.com/community/c6h1763036)

Hi All,

I've been getting many requests lately to explain this technique, so here is a short writeup which explains the method of projective maps from scratch. Any suggestions are welcome, and I hope it helps!

Basic familarity with projective geometry is assumed. Mainly just good intuition with cross ratios is required.

Let $\mathcal{C}$ be a conic, a line, or a pencil of lines through a fixed point (note that this last one is often ommited from the definition, but I find it to be quite important). These objects have a structure of cross ratio built into them, specifically, if $A,B,C,D\in\mathcal{C}$ , then we can talk about $(AB;CD)$ (note that if $\mathcal{C}$ is a pencil of lines, then $A,B,C,D$ are lines). The definition of a projective map is now very simple.

Definition A projective map $f$ from $\mathcal{C}_1$ to $\mathcal{C}_2$ where these are either conics, lines, or a pencil of lines is a function that preserves cross ratio. In particular, if $A,B,C,D\in\mathcal{C}_1$ , then
$(AB;CD) = (f(A)f(B);f(C)f(D)).$ And that's all!

The reason this is useful is the following theorem.

Theorem If $f,g:\mathcal{C}_1\to\mathcal{C}_2$ are projective, then $f\equiv g$ if $f$ and $g$ coincide on three different input values.

Proof: Suppose $f(A)=g(A)$ , $f(B)=g(B)$ , $f(C)=g(C)$ . Then,
$(AB;CD)=(f(A)f(B);f(C)f(D))=(g(A)g(B);g(C)g(D))=(f(A)g(B);f(C)g(D)).$ The important equation is $(f(A)f(B);f(C)f(D))=(f(A)g(B);f(C)g(D))$ . Since the cross ratio is bijective, we have $f(D)=g(D)$ , as desired. $\blacksquare$

When solving problems, if we can phrase the problem as in the theorem, then actually we only have to check the problem for 3 cases! Before doing examples, let's cover some basic transformations that are projective maps. Your intuition with cross ratios is going to be really helpful here. Firstly, note that the composition of two projective maps is projective, and the inverse of a projective map is projective (Exercise: Prove this from the definition!). Here is a list I could come up with for projective maps (not exhaustive by any means).

Given a line $\ell$ and a point $P$ , the map from $\ell$ to $\mathcal{C}_P$ (the pencil of lines through $P$ ) given by $X\mapsto PX$
Given a conic $\gamma$ and a point $P$ on the conic, the map from $\ell$ to $\mathcal{C}_P$ given by $X\mapsto PX$
Given a conic $\gamma$ and any point $P$ , the map from $\gamma$ to $\gamma$ by $X\mapsto PX\cap\gamma\not=X$
This one's a little off beat, but still useful. Given two clines $\gamma_1$ and $\gamma_2$ , any inversion (or Mobius transform for that matter) that sends $\gamma_1$ to $\gamma_2$ is projective. This is because of the famous fact that inversion preserves cross ratio.

You might be wondering where projecting from a line to a line fits in. It turns out this is actually a composition of two of the maps we listed above. In particular, if $P$ is a point and $\ell_1,\ell_2$ are lines, then we first map $\ell_1\to\mathcal{C}_P$ using the first map, then map $\mathcal{C}_P\to\ell_2$ using the inverse of the first map. Their composition is the classic perspectivity map. It's time to do some examples!

USA Winter TST for IMO 2019 Problem 1 wrote:

Let $ABC$ be a triangle and let $M$ and $N$ denote the midpoints of $\overline{AB}$ and $\overline{AC}$ , respectively. Let $X$ be a point such that $\overline{AX}$ is tangent to the circumcircle of triangle $ABC$ . Denote by $\omega_B$ the circle through $M$ and $B$ tangent to $\overline{MX}$ , and by $\omega_C$ the circle through $N$ and $C$ tangent to $\overline{NX}$ . Show that $\omega_B$ and $\omega_C$ intersect on line $BC$ .

Merlijn Staps

$[asy] size(8cm); pair A = dir(129); pair B = dir(220); pair C = dir(320); filldraw(A--B--C--cycle, invisible, black); pair M = 0.5*A + 0.5*B; pair N = 0.5*A + 0.5*C; pair O = circumcenter(A,M,N); pair Q = rotate(-90,A)*O; pair X = 4.2*Q - 3.2*A; pair Y = intersectionpoints(circle(X,abs(X-A)),circumcircle(A,M,N))[0]; pair Z = 0.5*A + 0.5*Y; filldraw(circumcircle(C,Y,N), invisible, blue+opacity(0.5)); draw(A--X,red); pair S = circumcenter(C, N, Y); pair T = foot(S, B, C); pair D = 2*T - C; pair P = extension(S, midpoint(N--C), X, N); draw(N--X); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(0)); dot("$M$", M, dir(135)); dot("$N$", N, dir(80)); dot("$X$", X, dir(X)); dot("$D$", D, dir(225)); [/asy]$

Diagram made by v_Enhance. Let $\ell_1$ be the tangent at $A$ to $(ABC)$ and let $\ell_2=BC$ . Let $f$ be the map from $\ell_1$ to $\ell_2$ given by $f(X)=D=\omega_C\cap BC$ . Note that this means $\angle XND=\angle C$ (directed angles).

So consider the map from $\mathcal{C}_N$ to $\mathcal{C}_N$ given by rotation by angle $\angle C$ . This is clearly projective. We have the map from $\ell_1\to\mathcal{C}_N$ by $X\mapsto NX$ , and we have $\mathcal{C}_N\to\ell_2$ by $w\mapsto w\cap\ell_2$ . Therefore, composing all these projective maps in the order
$\ell_1\to\mathcal{C}_N\to\mathcal{C}_N\to\ell_2,$ we have that $f$ is a projective map that maps $X\mapsto D$ .

Similarly, we define a map $f':\ell_1\to\ell_2$ that sends $f'(X)=D'=\omega_B\cap BC$ . We have again that its projective. We want to show $D=D'$ , so by the theorem, we just have to check this for three choices of $X$ . This is now a geometry problem that is left to the reader!

Note: This proof is written much shorter like this. Let $D=\omega_C\cap BC$ . Then,
$X\mapsto NX\mapsto ND\mapsto D$ is projective, so $X\mapsto D$ is projective. Similarly if $D'=\omega_B\cap BC$ , then $X\mapsto D'$ is projective. So it suffices to check for three values of $X$ .

The next example will have the proof again written nice and short like the above.

IMO 2010/2 wrote:

Given a triangle $ABC$ , with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$ . Let $E$ be a point on the arc $BDC$ , and $F$ a point on the segment $BC$ , such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$ . If $G$ is the midpoint of $IF$ , prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$ .

$[asy] import olympiad; import cse5; size(8cm); defaultpen(fontsize(9pt)); pair A = dir(110); dot("$A$", A, dir(A)); pair B = dir(210); dot("$B$", B, dir(B)); pair C = dir(330); dot("$C$", C, dir(C)); pair I = incenter(A, B, C); dot("$I$", I, dir(45)); pair O = origin; pair D = extension(A, I, O, B+C); dot("$D$", D, dir(225)); pair E = dir(310); dot("$E$", E, dir(E)); pair F_1 = B*C/E; pair F = extension(B, C, A, F_1); dot("$F$", F, dir(F)); pair G = midpoint(I--F); dot("$G$", G, dir(-15)); pair K = extension(E, I, D, G); dot("$K$", K, dir(K)); pair I_A = 2*D-I; pair P = extension(D, K, A, F); pair Z = extension(I, P, I_A, F); draw(A--B--C--cycle, red); draw(unitcircle, blue); draw(A--E--K--D--cycle, green); draw(I--F); [/asy]$

Diagram from v_Enhance. We will animate $E$ on $\Gamma$ , which means that we have mappings generated by varying $E$ . Let $K_1=EI\cap\Gamma$ and $K_2=DG\cap\Gamma$ . We have that $E\mapsto K_1$ is projective by the third item on our list.

We'll now show that $E\mapsto K_2$ is projective. Note that $E\mapsto AE\mapsto AF\mapsto F$ is projective since $AE\mapsto AF$ is a reflection which clearly preserves cross ratio. Also, $F\mapsto G$ is projective since it is a homothety at $I$ with factor $1/2$ (here the codomain is the scaling of $BC$ by $1/2$ at $I$ ). Thus, $E\mapsto F\mapsto G$ is projective. Projecting through $D\in\Gamma$ , we have that $G\mapsto K_2$ is projective, so $E\mapsto G\mapsto K_2$ is projective, as deisred.

Therefore, to show $K_1=K_2$ , it suffices to check three values of $E$ . It is now a geometry problem (an easy one at that) to check the problem for $E=B,D,C$ (possible hint: fact 5).

This post has been edited 1 time. Last edited by yayups, Mar 27, 2019, 11:40 PM

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