IMO 2012 Problem 4

by liberator, Oct 5, 2018, 8:45 PM

This was a pretty tricky functional equation, especially for a P4...

Problem: Find all functions $f:\mathbb Z\rightarrow \mathbb Z$ such that, for all integers $a,b,c$ that satisfy $a+b+c=0$ , the following equality holds:
$f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).$ (Here $\mathbb{Z}$ denotes the set of integers.)

Proposed by Liam Baker, South Africa

My solution

algebra

IMO Shortlist 2017 Geometry

by liberator, Oct 4, 2018, 9:40 PM

Here are my solutions to some of the geometry problems from this year's IMO Shortlist.

G1. Let $ABCDE$ be a convex pentagon such that $AB=BC=CD$ , $\angle{EAB}=\angle{BCD}$ , and $\angle{EDC}=\angle{CBA}$ . Prove that the perpendicular line from $E$ to $BC$ and the line segments $AC$ and $BD$ are concurrent.

My solution

G3. Let $O$ be the circumcenter of an acute triangle $ABC$ . Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$ , respectively. The altitudes meet at $H$ . Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$ .

My solution

G4. In triangle $ABC$ , let $\omega$ be the excircle opposite to $A$ . Let $D, E$ and $F$ be the points where $\omega$ is tangent to $BC, CA$ , and $AB$ , respectively. The circle $AEF$ intersects line $BC$ at $P$ and $Q$ . Let $M$ be the midpoint of $AD$ . Prove that the circle $MPQ$ is tangent to $\omega$ .

My solution

geometry

ISL

The Hales-Jewett theorem

by liberator, Apr 2, 2018, 7:50 PM

This is just some stuff I've been looking at recently.

In this blog post, we will discuss the powerful Hales-Jewett theorem, one of the central results of Ramsey theory. We use it to prove the famous van der Waerden theorem, as well as its $n$ -dimensional generalisation, Gallai's theorem.

1. Definitions and Examples

For any positive integer $n$ , we write $[n]=\{1,2,\dots,n\}$ . Given a finite set $X$ , we denote by $X^n$ the $n$ -dimensional hypercube on the alphabet $X$ ; that is, the set of all possible $n$ -tuples $(x_1,x_2,\dots,x_n)$ , with $x_i\in X$ .

A $k$ -colouring of $X^n$ is a map $X^n\to[k]$ ; that is, we assign each element of $X^n$ one of $k$ colours, labelled by the elements of $[k]$ .

A subset $L$ of $X^n$ is called a combinatorial line if there exists a non-empty set of indices $I=\{i_1,i_2,\dots,i_k\}\subseteq[n]$ and, for each $i\not\in I$ , an $a_i\in X$ such that $L=\{(x_1,\dots,x_n)\in X^n\mid x_{i_1}=x_{i_2}=\dots=x_{i_k} \text{ and }x_i=a_i \text{ for } i\not\in I\}.$ We call $I$ the set of active coordinates of $L$ , since these are the coordinates that vary: the others are "fixed". More informally, a combinatorial line is the set of $n$ -tuples we get when we fix the values of some of the components, and set the other components equal to each other, allowing this common value to vary among the elements of $X$ .

For example, in $[3]^2$ , we have the following combinatorial lines:

$\begin{tabular}{|c|c|} \hline Active coordinates $I$ & Combinatorial line $L$ \\ \hline $~$ & $\{(1,1),(2,1),(3,1)\}$ \\ $\{1\}$ & $\{(1,2),(2,2),(3,2)\}$ \\ $~$ & $\{(1,3),(2,3),(3,3)\}$ \\ \hline $~$ & $\{(1,1),(1,2),(1,3)\}$ \\ $\{2\}$ & $\{(2,1),(2,2),(2,3)\}$ \\ $~$ & $\{(3,1),(3,2),(3,3)\}$ \\ \hline $\{1,2\}$ & $\{(1,1),(2,2),(3,3)\}$ \\ \hline \end{tabular}$ Displayed more pictorially,
$[asy] unitsize(1cm); real e = 0.5; for (int x=0; x<=2; ++x) { for (int y=0; y<=2; ++y) { label("$("+string(3-y)+","+string(x+1)+")$",(x,y)); } } DPA((-e,2)--(2+e,2)^^(-e,1)--(2+e,1)^^(-e,0)--(2+e,0),red); DPA((0,-e)--(0,2+e)^^(1,-e)--(1,2+e)^^(2,-e)--(2,2+e),blue); DPA((-e,2+e)--(2+e,-e),green); [/asy]$
Visual representations of combinatorial lines can be good when $n\leq3$ ; however, for $n\geq4$ , I find them less useful, essentially because I struggle to visualise in $4$ or more dimensions

.

Consider the standard "product order" on $[m]^n$ , given by $(x_1,x_2,\dots,x_n)\preceq(y_1,y_2,\dots,y_n)\iff x_i\leq y_i$ for all $i$ . If we restrict our attention to a combinatorial line, this partial order $\preceq$ is in fact a total order. Thus given a combinatorial line $L$ , we define $\alpha(L)$ to be the last element in our ordering of $L$ and $\beta(L)$ to be the first element.

Given a $k$ -colouring of $[m]^n$ and a positive integer $r\leq k$ , a rainbow $r$ -windmill is a set of $r$ combinatorial lines $L_1,L_2,\dots,L_r$ such that

There exists $f\in[m]^n$ such that $\alpha(L_i)=f$ for all $i$ . We call $f$ the focus of the rainbow $r$ -windmill.
Each individual $L_i\setminus\{\alpha(L_i)\}$ is monochromatic; i.e. all the points of an $L_i$ , ignoring the focus, are the same colour.
No two $L_i\setminus\{\alpha(L_i)\}$ have the same colour.

Finally, we introduce a definition needed for Gallai's theorem. Given a finite set $S\subset\mathbb N^d$ , a homothetic copy of $S$ is a set of the form $a+\lambda S$ , where $a\in\mathbb N^d$ and $\lambda\in\mathbb N$ . Geometrically, we are enlarging the set $S$ by a positive integer scale factor from the origin, then translating the enlarged image by some vector in $\mathbb N^d$ . As an example, when $d=1$ , a homothetic copy of $[m]$ is an arithmetic progression of length $m$ .

2. The Theorem

We are now ready to state the main result of this post.

Theorem. [Hales-Jewett] Given $m,k\in\mathbb N$ , there exists $n\in\mathbb N$ such that for any $k$ -colouring of $[m]^n$ , there is a monochromatic combinatorial line.

The smallest such $n$ with the desired property is denoted by $HJ(m,k)$ .

Proof. We go by induction on $m$ ; the base case $m=1$ is trivial. Now consider the following claim:

Claim. For $r\leq k$ , there exists $n$ such that for any $k$ -colouring of $[m]^n$ , there is either a monochromatic combinatorial line or a rainbow $r$ -windmill.

Note that our theorem follows from this claim: take $r=k$ , and look at the colour of the focus of the rainbow $k$ -windmill.

We prove our claim by induction on $r$ . For the base case $r=1$ , take $n=HJ(m-1,k)$ . Now assume that $n$ works for $r$ ; we show that $n+d$ , where $d=HJ\left(m-1,k^{m^n}\right)$ , works for $r+1$ .

Suppose that we have $k$ -coloured $[m]^{n+d}$ such that there are no monochromatic combinatorial lines. Note that $[m]^{n+d}$ is isomorphic to $[m]^n\times[m]^d$ . There are $k^{m^n}$ possible $k$ -colourings of $[m]^n$ , so by our choice of $d$ , there is a (monochromatic) combinatorial line $L$ in $[m]^d$ , with active coordinate set $I$ , such that for any $a\in[m]^n$ and $b\in L\setminus\{\alpha(L)\}$ , the colour of $(a,b)$ in our $k$ -colouring of $[m]^{n+d}$ is the same as the color of $a$ in the $k$ -colouring of $[m]^n$ .

Now take a rainbow $r$ -windmill $L_1,\dots,L_r$ for our $k$ -colouring of $[m]^n$ . Suppose the respective active coordinate sets are $I_1,\dots,I_r$ , and let $f$ be the focus. For each $1\leq i\leq r$ , let $L_i'$ be the combinatorial line through the point $(\beta(L_i),\beta(L))$ and with active coordinate set $I_i\cup I$ . Then $L_1',\dots,L_r'$ is a rainbow $r$ -windmill with focus $(f,\alpha(L))$ . Combining this with the combinatorial line through the point $(f,\beta(L))$ with active coordinate set $I$ gives us the desired rainbow $(r+1)$ -windmill.

This completes the proof of the claim, and hence of the theorem. $\square$

3. Applications

We now see how the Hales-Jewett theorem can be used to prove van der Waerden's theorem and Gallai's theorem easily.

Theorem. [van der Waerden] Given $m,k\in\mathbb N$ , there exists $n\in\mathbb N$ such that for any $k$ -colouring of $[n]$ , there is a monochromatic arithmetic progression of length $m$ .

Proof. Given a $k$ -colouring $\mathcal C$ of $\mathbb N$ , consider the $k$ -colouring $\mathcal C'$ of $[m]^n$ , with $n$ sufficiently large, given by $\mathcal C'((x_1,\dots,x_n))=\mathcal C\left(\sum_{i=1}^nx_i\right).$ By the Hales-Jewett theorem, there is a monochromatic combinatorial line in $[m]^n$ , which corresponds to an arithmetic progression of length $m$ in $\mathbb N$ . Explicitly, if $I$ is the set of active coordinates of our combinatorial line, then our arithmetic progression is $A+|I|, \quad A+2|I|, \quad \dots\quad,\quad A+m|I|,$ where $A=\sum_{i\not\in I}x_i$ . $\square$

Gallai's theorem is essentially a generalisation of van der Waerden's theorem to $\mathbb N^d$ . Recall that a homothetic copy of $S\subset\mathbb N^d$ is a set of the form $a+\lambda S$ , where $a\in\mathbb N^d$ and $\lambda\in\mathbb N$ .

Theorem. [Gallai] Given a finite set $S\subset\mathbb N^d$ , for any $k$ -colouring of $\mathbb N^d$ , there exists a monochromatic homothetic copy of $S$ .

Proof. Let $S=\{S(1),\dots,S(m)\}$ . Given a $k$ -colouring $\mathcal C$ of $\mathbb N^d$ , consider the $k$ -colouring $\mathcal C'$ of $[m]^n$ , with $n$ sufficiently large, given by $\mathcal C'((x_1,\dots,x_n))=\mathcal C\left(\sum_{i=1}^nS(x_i)\right).$ By the Hales-Jewett theorem, there is a monochromatic combinatorial line in $[m]^n$ , which corresponds to a monochromatic homothetic copy of $S$ in $\mathbb N^d$ . Explicitly, if $I$ is the set of active coordinates of our combinatorial line, then our homothetic copy of $S$ is $\sum_{i\not\in I}S(x_i)+|I|S$ . $\square$

combinatorics

Ramsey Theory

Proving collinearities with the Newton-Gauss line

by liberator, Apr 11, 2017, 10:29 PM

So apparently I'm being peer pressured into posting

Here is the desired blog entry, a whole 364 days after the last! I hope that this satisfies your requests/demands or whatever you want to call it.

This is just to illustrate a nice little trick that I've used to solve several (read: three) problems.

To show that three points are collinear, we can use the converse of the Newton-Gauss line property, by proving that circles on appropriate diameters are coaxal (orthocentres often come in handy here), which can give us the fourth side of a complete quadrilateral. This is perhaps best demonstrated by three examples:

Problem 1: Let $ABC$ be a triangle. Let $\triangle P_aP_bP_c$ and $\triangle Q_aQ_bQ_c$ be the circumcevian triangles of points $P$ and $Q$ respectively. Prove that $\triangle ABC$ and the triangle determined by lines $P_aQ_a,P_bQ_b,P_cQ_c$ are in perspective.

[Jerabek's theorem]

$[asy] unitsize(2cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); path carc(pair A, pair B, pair C, real d=0, bool dir) { pair O=circumcenter(A,B,C); return arc(O,circumradius(A,B,C),degrees(A-O)+d,degrees(C-O)-d,dir); } pair A=dir(110), B=dir(200), C=dir(-20), Q=(-0.1,-0.05), P_a=-A, P_b=-B, P_c=-C, Q_a=IP(unitcircle,L(A,Q,-0.1,10)), Q_b=IP(unitcircle,L(B,Q,-0.1,10)), Q_c=IP(unitcircle,L(C,Q,-0.1,10)), X=extension(P_a,Q_a,B,C), Y=extension(P_b,Q_b,C,A), Z=extension(P_c,Q_c,A,B); D(X--B--A--C); D(unitcircle,heavygreen); D(carc(A,foot(A,B,C),X,CCW),red); D(X--Q_a--A--foot(A,B,C)); D("A",A,NW); D("B",B,SW); D("C",C,SE); D("Q",Q,E); D("Q_a",Q_a); D("P_a",P_a,N); D("X",X,E); D("H",A+B+C,SSE); [/asy]$
Solution. Let $X=P_aQ_a\cap BC$ , and define $Y,Z$ analogously. Take a homography fixing $(ABC)$ and sending $P$ to the centre. It is easy to see that the circles on diameters $\overline{AX},\overline{BY},\overline{CZ}$ are coaxal, with common radical axis $QH$ . Then by the converse of the Newton-Gauss line, it follows that $X,Y,Z$ are collinear. $\square$

[Sorry for the lack of complete diagrams for the next two problems. Some of the points are very far apart, meaning that the diagrams would probably not be very clear anyway]

Problem 2: A circle $\omega$ with centre $P$ intersects the sides $BC,CA,AB$ of triangle $ABC$ at $A_1$ and $A_2$ , $B_1$ and $B_2$ , $C_1$ and $C_2$ respectively. Let $A'$ be the circumcentre of triangle $A_1A_2P$ , $B'$ the circumcentre of triangle $B_1B_2P$ , $C'$ the circumcentre of triangle $C_1C_2P$ . Prove that $AA'$ , $BB'$ and $CC'$ are concurrent.

[UK Training for RMM 2017]

$[asy] unitsize(2cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); pair P=origin, A_1=dir(230), A_2=dir(-50), B_1=dir(-20), B_2=dir(95), C_1=dir(120), C_2=dir(185), A=extension(B_1,B_2,C_1,C_2), B=extension(C_1,C_2,A_1,A_2), C=extension(A_1,A_2,B_1,B_2), Ap=circumcenter(A_1,A_2,P), Bp=circumcenter(B_1,B_2,P), Cp=circumcenter(C_1,C_2,P), X=2/(conj(IP(A--Ap,unitcircle))+conj(OP(A--Ap,unitcircle))); D(A--B--C--cycle); DPA(A--Ap^^B--Bp^^C--Cp); D(unitcircle,heavygreen); D("A",A,NW); D("B",B,SW); D("C",C,SE); D("A'",Ap,E); D("B'",Bp); D("C'",Cp,SW); D("P",P,E); [/asy]$
Solution. Let the polar of $A$ in $\omega$ meet the polar of $A'$ at $X$ , and define $Y, Z$ analogously. It suffices to show that $X, Y, Z$ are collinear. Let $\triangle DEF$ be the triangle formed by the polars of $A', B', C'$ in $\omega$ , which is homothetic to $\triangle ABC$ with centre $P$ . Consider the circles on diameters $\overline{DX},\overline{EY},\overline{FZ}$ . Easily, $P$ and the orthocentre of $\triangle DEF$ lie on a common radical axis of these circles. Thus the midpoints of $\overline{DX},\overline{EY},\overline{FZ}$ are collinear, so the result follows by the converse of the Newton-Gauss line. $\square$

Problem 3: A circle $\omega$ with centre $P$ intersects the sides $BC,CA,AB$ of triangle $ABC$ at $A_1$ and $A_2$ , $B_1$ and $B_2$ , $C_1$ and $C_2$ respectively. Let the lines tangent to $\omega$ at $A_1$ and $A_2$ meet at $A'$ and define similarly $B'$ and $C'$ . Prove that the lines $AA'$ , $BB'$ , and $CC'$ are concurrent.

[Silouanas Brazitikos, UK IMO Training and Selection Camp April 2017]
$[asy] unitsize(2cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); pair P=origin, A_1=dir(230), A_2=dir(-50), B_1=dir(-20), B_2=dir(95), C_1=dir(120), C_2=dir(185), A=extension(B_1,B_2,C_1,C_2), B=extension(C_1,C_2,A_1,A_2), C=extension(A_1,A_2,B_1,B_2), Ap=2/(conj(A_1)+conj(A_2)), Bp=2/(conj(B_1)+conj(B_2)), Cp=2/(conj(C_1)+conj(C_2)), X=extension(B,C,tangent(A,P,1),tangent(A,P,1,2)); D(A--B--C--cycle); DPA(A--Ap^^B--Bp^^C--Cp); D(unitcircle,heavygreen); D("A",A,NW); D("B",B,SW); D("C",C,SE); D("A'",Ap); D("B'",Bp,NE); D("C'",Cp,NW); D("P",P,E); [/asy]$
Solution. Let the polar of $A$ in $\omega$ meet $BC$ at $X$ , and define $Y,Z$ analogously. It suffices to show that $X,Y,Z$ are collinear. Consider the circles on diameters $\overline{AX},\overline{BY},\overline{CZ}$ . Easily, $P$ and the orthocentre of $\triangle ABC$ lie on a common radical axis of these circles. Thus the midpoints of $\overline{AX},\overline{BY},\overline{CZ}$ are collinear, so the result follows by the converse of the Newton-Gauss line. $\square$

geometry

Parallelogram Isogonality Lemma

by liberator, Apr 12, 2016, 1:43 PM

Well I thought that I better make a blog post within the year so here it is.

This nice lemma allows us to solve a variety of olympiad geometry problems. I first encountered it as BrMO2 2013/2. It is a generalisation of A Simple Lemma.

Lemma. The point $P$ lies inside triangle $ABC$ so that $\angle ABP = \angle PCA$ . The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\angle QAB = \angle CAP$ .
$[asy] unitsize(2.8cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10pt); real x=0.54; pair A=dir(110), B=dir(205), C=dir(-25), X=WP(C--A,x), Y=IP(circumcircle(B,C,X),A--B), P=IP(B--X,C--Y), Q=B+C-P, Ap=A+B-P; D(unitcircle,heavygreen); D(A--B--C--cycle,heavygrey+linewidth(1)); D(Q--A--P,purple); DPA(B--P--C--Q--cycle^^Ap--A^^Ap--B^^Ap--Q); /* Angle marks */ DPA(anglemark(P,B,A,5)^^anglemark(A,C,P,5),orange); /* Dots and labels */ D("A",A,dir(A)); D("B",B,dir(B)); D("C",C,dir(C)); D("P",P,dir(130)); D("Q",Q,dir(Q)); D("A'",Ap,dir(Ap)); [/asy]$
Proof. Let $A'$ be such that $APBA',ACQA'$ are parallelograms. Then $\angle A'AB=\angle PBA=\angle PCA=\angle A'QB$ , so $AA'BQ$ is cyclic. But then $\angle BAQ=\angle BA'Q=\angle PAC$ , as required. $\square$

Some example problems:

Problem 1: Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$ . The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$ . Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$ . Prove that $D,H,F,G$ are concyclic.

[ISL 2012 G2]

My solution

Problem 2: Let $P$ be a point inside triangle $ABC$ , and suppose lines $AP$ , $BP$ , $CP$ meet the circumcircle again at $T$ , $S$ , $R$ (here $T \neq A$ , $S \neq B$ , $R \neq C$ ). Let $U$ be any point in the interior of $PT$ . A line through $U$ parallel to $AB$ meets $CR$ at $W$ , and the line through $U$ parallel to $AC$ meets $BS$ again at $V$ . Finally, the line through $B$ parallel to $CP$ and the line through $C$ parallel to $BP$ intersect at point $Q$ . Given that $RS$ and $VW$ are parallel, prove that $\angle CAP = \angle BAQ$ .

[Taiwan TST2 2014 Problem 6]

My solution

Problem 3: Let $ABCD$ be a cyclic quadrilateral, and let diagonals $AC$ and $BD$ intersect at $X$ .Let $C_1,D_1$ and $M$ be the midpoints of segments $CX,DX$ and $CD$ , respecctively. Lines $AD_1$ and $BC_1$ intersect at $Y$ , and line $MY$ intersects diagonals $AC$ and $BD$ at different points $E$ and $F$ , respectively. Prove that line $XY$ is tangent to the circle through $E,F$ and $X$ .

[EGMO 2016 Problem 2]

My solution

This post has been edited 1 time. Last edited by liberator, Apr 15, 2016, 8:21 PM
Reason: added EGMO 2016/2

geometry

isogonality lemma

IMO 2011 Problem 6

by liberator, Jul 21, 2015, 4:57 PM

Problem: Let $ABC$ be an acute triangle with circumcircle $\Gamma$ . Let $\ell$ be a tangent line to $\Gamma$ , and let $\ell_a, \ell_b$ and $\ell_c$ be the lines obtained by reflecting $\ell$ in the lines $BC$ , $CA$ and $AB$ , respectively. Show that the circumcircle of the triangle determined by the lines $\ell_a, \ell_b$ and $\ell_c$ is tangent to the circle $\Gamma$ .

Proposed by Japan

My solution

This post has been edited 2 times. Last edited by liberator, Jul 22, 2015, 1:35 PM

IMO 2015 Problem 3

by liberator, Jul 14, 2015, 2:48 PM

Problem: Let $ABC$ be an acute triangle with $AB > AC$ . Let $\Gamma$ be its cirumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$ . Let $M$ be the midpoint of $BC$ . Let $Q$ be the point on $\Gamma$ such that $\angle HQA = 90^{\circ}$ and let $K$ be the point on $\Gamma$ such that $\angle HKQ = 90^{\circ}$ . Assume that the points $A$ , $B$ , $C$ , $K$ and $Q$ are all different and lie on $\Gamma$ in this order.

Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

Proposed by Ukraine

My solution

This post has been edited 2 times. Last edited by liberator, Jul 14, 2015, 8:58 PM

Disjoint discs tangent to 6 others

by liberator, Jul 5, 2015, 9:19 PM

Problem: We are given a family of discs in the plane, with pairwise disjoint interiors. Each disc is tangent to at least six other discs of the family. Show that the family is infinite.

[Austrian-Polish Mathematical Olympiad 1978 Problem 6]

My solution

combinatorics

graph theory

Austria-Poland MO

Coaxal circles in incenter/excenter configuration

by liberator, Apr 15, 2015, 7:18 PM

Problem: Let $ABC$ be a triangle, whose excircle (opposite $A$ ) touches $BC,CA,AB$ at $P,Q,R$ respectively. Denote $D$ as the intersection of the lines $PQ$ and $AB$ , and $E$ as the intersection of the lines $RP$ and $CA$ . If $I_a$ is the excenter of $\triangle ABC$ , opposite $A$ , prove that the circumcircles of triangles $PQE, PRD$ and $PI_aA$ are coaxal.

Commentary: We may replace "excircle" with "incircle", "excenter" with "incenter", and the result still holds.

See interactive diagram here.

My solution

Attachments:

geometry

Reim s theorem

power of a point

Partition a group of nine people

by liberator, Apr 10, 2015, 6:43 PM

Problem: In a group of nine people it is not possible to choose four people such that every one knows the three others. Prove that this group of nine people can be partitioned into four groups such that nobody knows anyone from his or her group.

[Bulgarian IMO TST 2005 Problem 6]

My solution (long)

combinatorics

Bulgarian TST

graph theory

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1000 VISITS!!!

IN 10 DAYS!!!
by liberator, Aug 23, 2014, 3:47 PM
YOU ARE SO PRO!!!!

Have you been to the IMO? If you have, then on not bragging about it. If you haven't, HOW ARE YOU SO CLEVER??!!?
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Cool blog, great geometry problems!
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Extremely good at Geometry

Could you consider joining an Online Math Open Team with me?
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thanks Cyclicduck!
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Hi, you are very good at geometry!
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Posts? What are those, exactly?
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@kmw2001, have 5 posts first.
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How do i blog?
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thx bcp123, I guess its because I only have 10 posts, not much to brag on about
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why no one is writing here?
keep this going. nice collection of geoish things
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thx!
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nice blog
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