Projective Bary

by yayups, May 11, 2019, 6:06 AM

This solution shows a nontrivial usage of the projective nature of barycentric coordinates.

USAMO Shortlist 2014, 110 Geometery Problems #77 wrote:

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incircle $\gamma$ . Let $A'$ be the mixtilinear excircle touch point, and let $A'B'$ and $A'C'$ be tangent to $\gamma$ with $B',C'\in\Gamma$ . Let $X$ be the tangency point of $B'C'$ with $\gamma$ which exists by Poncelet's Porism. Show that $(XBC)$ is tangent to $\gamma$ .

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ unitsize(0.2inches); import graph; size(0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10, xmax = 14, ymin = -22.62, ymax = 13.46; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((0.26,7.92)--(-1.52,-2.18)--(9.54,-2.2)--cycle, linewidth(2) + zzttqq); /* draw figures */ draw((0.26,7.92)--(-1.52,-2.18), linewidth(2) + zzttqq); draw((-1.52,-2.18)--(9.54,-2.2), linewidth(2) + zzttqq); draw((9.54,-2.2)--(0.26,7.92), linewidth(2) + zzttqq); draw(circle((4.017668907550992,2.050905875698934), 6.968955550057753), linewidth(2)); draw(circle((2.278232171040842,1.001527516890443), 3.1883907164579206), linewidth(2)); draw((5.8094995867931365,8.78556876819826)--(0.07556713698788581,3.306761151157378), linewidth(2)); draw((0.07556713698788581,3.306761151157378)--(-2.791521751662705,0.5672401366735647), linewidth(2)); draw((-2.791521751662705,0.5672401366735647)--(-5.658856324457237,-2.1725156305163527), linewidth(2)); draw((-5.658856324457237,-2.1725156305163527)--(-1.52,-2.18), linewidth(2)); draw((-2.791521751662705,0.5672401366735647)--(5.215750162966643,-4.8142920649363115), linewidth(2)); draw((5.215750162966643,-4.8142920649363115)--(5.8094995867931365,8.78556876819826), linewidth(2)); /* dots and labels */ dot((0.26,7.92),dotstyle); label("$A$", (0.34,8.12), N * labelscalefactor); dot((-1.52,-2.18),dotstyle); label("$B$", (-1.44,-1.98), NE * labelscalefactor); dot((9.54,-2.2),dotstyle); label("$C$", (9.62,-2), E *3* labelscalefactor); dot((2.2724665553022745,-2.1868579865376176),linewidth(4pt) + dotstyle); label("$D$", (2.36,-2.02), NE * labelscalefactor); dot((5.215750162966643,-4.8142920649363115),linewidth(4pt) + dotstyle); label("$A'$", (5.3,-4.66), S * 4*labelscalefactor); dot((-2.791521751662705,0.5672401366735647),linewidth(4pt) + dotstyle); label("$B'$", (-3.48,0.62), NW*0.01 * labelscalefactor); dot((5.8094995867931365,8.78556876819826),linewidth(4pt) + dotstyle); label("$C'$", (5.88,8.94), NE * labelscalefactor); dot((-5.658856324457237,-2.1725156305163527),linewidth(4pt) + dotstyle); label("$T$", (-6.02,-1.92), NE * labelscalefactor); dot((0.07556713698788581,3.306761151157378),linewidth(4pt) + dotstyle); label("$X$", (0.02,3.66), N * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Firstly, the problem is clearly just asking us to show that $B'C'$ is tangent to $(XBC)$ . Letting $T=B'C'\cap BC$ , we see that this is equivalent to $TB\cdot TC=TX^2$ , but $TD=TX$ , so all we need to show is that
$TB\cdot TC=TD^2.$ We'll actually just compute the barycentric coordinates of $T$ and verify this explicitly. However, finding $B'$ and $C'$ is really messy, and this doesn't seem tractable immediately. However, the beauty of this solution is getting a simple way to find the coordinates of $T$ .

Let $P$ be a variable point on $BC$ , and define the composite map $\phi$ given by
$P\mapsto P'\mapsto P_1P_2\cap BC$ where $P'$ is the $\sqrt{bc}$ inverse of $P$ and $P_1P'$ and $P_2P'$ are tangent to $\gamma$ with $P_1,P_2\in\Gamma$ . It's not easy to see that this map $\phi$ is projective but its true.

Why?

Now, we check that
$B\mapsto C\mapsto AB\mapsto B,$ so $\phi(B)=B$ , and similarly $\phi(C)=C$ . Also,
$\infty_{BC}\mapsto A\mapsto BC,$ but uh-oh, what's $BC\cap BC$ ? The fix is that we actually have the tangent to $\gamma$ at some point infinitesimally close to $D$ , which then intersects $BC$ really close to $D$ , so we have $\phi(\infty_{BC})=D$ .

Now, use projective coordinates for $BC$ given by the last two coordinates of the barycentric coordinates of a point on $BC$ . The map $\phi$ looks like some matrix
$\begin{pmatrix}w & x\\ y& z\end{pmatrix}$ that is considered the same under scaling. We see that $B=(1:0)$ and $C=(0:1)$ map to themselves, so $x=y=0$ . The point $\infty_{BC}$ is given by $(1:-1)$ , and it maps to $D=(s-c:s-b)$ , so the matrix is
$\phi\equiv \begin{pmatrix}s-c & 0\\ 0 & -(s-b)\end{pmatrix}.$ Now, we see that $T=\phi(D)$ since $D$ and $A'$ are $\sqrt{bc}$ inverses (the incircle gets sent to the mixtilinear excircle), so
$T=\begin{pmatrix}s-c & 0\\ 0 & -(s-b)\end{pmatrix}\begin{pmatrix}s-c \\ s-b\end{pmatrix}=\begin{pmatrix}(s-c)^2 \\ -(s-b)^2\end{pmatrix},$ so $T=((s-c)^2:-(s-b)^2)$ . The rest is a boring bash which we do below.

Homogenizing, we have $D=\left(\frac{s-c}{a},\frac{s-b}{a}\right)$ , $T=\left(\frac{(s-c)^2}{a(b-c)},-\frac{(s-b)^2}{a(b-c)}\right)$ , $B=(1,0)$ , and $C=(0,1)$ . By taking simply the first coordinate, we get a coordinate system on $BC$ that preserves the lengths in the original setup, so
$TB=\frac{(s-c)^2}{(b-c)a}-1=\frac{(s-b)^2}{(b-c)a}$ and $TC=\frac{(s-c)^2}{(b-c)a}$ , and
$TD=\frac{(s-c)^2}{(b-c)a}-\frac{s-c}{a}=\frac{s-c}{a}\cdot\frac{s-b}{b-c}=\frac{(s-b)(s-c)}{a(b-c)}.$ Evidently, we have $TB\cdot TC=TD^2$ , as desired. As we showed before, this means $TB\cdot TD=TX^2$ , so $(XBC)$ is tangent to $B'C'$ , which is tangent to $\gamma$ at $X$ , so $(XBC)$ is tangent to $\gamma$ . $\blacksquare$

This post has been edited 3 times. Last edited by yayups, May 11, 2019, 10:35 PM

geometry

projective geometry

Olympiad

Combinatorial Games

by yayups, Feb 14, 2019, 2:58 AM

Here are 2 nice combinatorial games problems that I solved recently. They are very similar as explained in the remark below.

USAMO 1999/5 wrote:

The Y2K Game is played on a $1 \times 2000$ grid as follows. Two players in turn write either an S or an O in an empty square. The first player who produces three consecutive boxes that spell SOS wins. If all boxes are filled without producing SOS then the game is a draw. Prove that the second player has a winning strategy

Solution

ISL 2015 C4 wrote:

Let $n$ be a positive integer. Two players $A$ and $B$ play a game in which they take turns choosing positive integers $k \le n$ . The rules of the game are:

(i) A player cannot choose a number that has been chosen by either player on any previous turn.
(ii) A player cannot choose a number consecutive to any of those the player has already chosen on any previous turn.
(iii) The game is a draw if all numbers have been chosen; otherwise the player who cannot choose a number anymore loses the game.

The player $A$ takes the first turn. Determine the outcome of the game, assuming that both players use optimal strategies.

Proposed by Finland

Solution

Remark, also a spoiler

combinatorics

Olympiad

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

projective geometry

geometry

Olympiad

Submit

i searched up moving points and found this
what the actual orz
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by 554183, Oct 18, 2021, 3:32 PM
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by lneis1, Jul 26, 2021, 2:17 PM
Are u surprised by him making IMO
looking at his posts, it was very likely any way
by 554183, Jul 8, 2021, 6:05 AM
WAIT WHAT HE MADE IMO 2020

HOW TO BE SO GOD TIER

ORZORZORZORZ
by OlympusHero, Jun 6, 2021, 3:00 AM
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by yofro, Apr 13, 2021, 1:44 AM
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by Gaussian_cyber, Sep 20, 2020, 10:37 AM
did you make IMO 2020?
which contestant are you?
by Orestis_Lignos, Sep 18, 2020, 2:27 PM
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by fukano_2, Sep 10, 2020, 6:30 AM
how do u know he made IMO?
by Puffer13, Sep 6, 2020, 12:12 PM
Congrats on USA IMO!
by Imayormaynotknowcalculus, Aug 15, 2020, 4:52 PM
IMO 2020
by cmsgr8er, Aug 7, 2020, 8:16 PM
Nice blog! dang you seem pretty clever
by sunfishho, Sep 6, 2017, 3:45 AM
Nice blog
by niyu, Jul 7, 2017, 10:33 PM
thanks bro
by yayups, Jul 7, 2017, 5:10 AM
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by pinetree1, Jul 2, 2017, 1:20 AM
thanks
by yayups, Jun 28, 2017, 2:58 AM
Nice first posts, keep it up!
by pohoatza, Jun 28, 2017, 2:39 AM
lol
we can make a collaborative blog if you want though
but I kind of want to keep this blog more of a personal blog
by yayups, Jun 28, 2017, 1:17 AM
Nice blog!

@pinetree1 He told me that he's not giving contribs on this blog
by niyu, Jun 28, 2017, 1:12 AM
hi
That problem is very cool
I was considering putting it, but then decided not to since it was a little too difficult IMO
by yayups, Jun 28, 2017, 12:54 AM
hello yayups!
by JasperL, Jun 28, 2017, 12:43 AM
Contrib?
by pinetree1, Jun 27, 2017, 6:14 PM
Thanks
You guys seem pretty good
by yayups, Jun 27, 2017, 6:04 PM
Nice blog!
by pinetree1, Jun 27, 2017, 3:28 PM
First shout, cool blog
by Ankoganit, Jun 27, 2017, 8:00 AM

93 shouts

The blog of yayups

Projective Bary

by yayups, May 11, 2019, 6:06 AM

Combinatorial Games

by yayups, Feb 14, 2019, 2:58 AM

Moving points tutorial

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