[/quote]
,
has an incenter 
Suppose 
.
be the midpoint of 
.
.
meets 
again at point 
.
and 
and 
.
lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet 
at 
and 
.
.
and 
angle bisectors with the opposite sides. Let 
.
where 
is the major 
and 
again at 
or 
and 
concur on 
be a positive integer. Alex plays on a row of 9 squares as follows. Initially, all squares are empty. In each turn, Alex must perform exactly one of the following moves:
Choose a number of the form 
,
a non-negative integer, and place it in an empty square.
Choose two (not necessarily consecutive) squares containing the same number, say 
and erase the number in the other square.
,
and circumcentre 
.
and 
lie on 
and 
.
meet 
meet 
and 
have an intersection lie on line 
.
and![\[ f(x+y)+f(x^2+f(y))=f(f(x))^2+f(x)+f(y)+y,\forall x,y\in\mathbb{R}\]](http://latex.artofproblemsolving.com/c/8/9/c895ae7fdf8d7f284ac9fc94cc077d6edad6cbf0.png)
be a convex hexagon with 
and 
.
and 
.
and 
are the centers of mass of 
and 
,
.
.
the reflections of 
and 
are collinear if and only if 
coincide, prove that 
given that there exist positive integers 
satisfying
and 
as follows:
where 
are all the positive divisors of 
.
Find all positive integers 
.
Find all positive integers 
.
be an odd prime number. How many 
of 
are there, the sum of whose elements is divisible by 
be real numbers satisfying
a variable point on segment 
is constructed such that 
and 
.
and 
.
board, two opposite sides have been joined, forming a cylinder. Determine whether it is possible to place 
.
.
Y by megarnie, mathematicsy, jhu08, centslordm, Adventure10, Mango247
The inscribed circle of triangle 
touches the sides 
, 
and 
at 
, 
and 
respectively. Let 
denote the other point of intersection of 
and the inscribed circle. Prove that 
extended passes through the midpoint of 
if and only if 
.
touches the sides 
, 
and 
at 
, 
and 
respectively. Let 
denote the other point of intersection of 
and the inscribed circle. Prove that 
extended passes through the midpoint of 
if and only if 
.
,
,
.
.
,![\[ t = \frac{AQ}{AD} = \frac{a(s-a)^2}{c^2(s-c)+b^2(s-b)-a(s-b)(s-c)}. \]](http://latex.artofproblemsolving.com/e/0/b/e0b2ec4aae9f7eed9bc7b28954c4ead44de41209.png)
![\[ Q = \left(1-t, t \cdot \frac{s-c}{a}, t \cdot \frac{s-b}{a} \right). \]](http://latex.artofproblemsolving.com/6/8/b/68bad710fc1e008302367c6827d12c32b67aa8c2.png)
as the midpoint of 
,![\[ \det \left[ \begin{array}{ccc} 1-t & t \cdot \frac{s-c}{a} & t \cdot \frac{s-b}{a} \\ s-c & 0 & s-a \\ c+(s-b) & s-a & 0 \end{array} \right] = 0 \]](http://latex.artofproblemsolving.com/4/d/2/4d2159cff8b9bf1848c3f13d8880cc4a4ab8f66f.png)
![\[ \left[ c(s-a)(s-c) + 2(s-a)(s-b)(s-c) \right] = a(s-a)^2 \left( \frac{1}{t} - 1 \right) \iff (s-a)(s-c)(c+2(s-b)) = a(s-a)^2 \left( \frac{1}{t} - 1 \right)\]](http://latex.artofproblemsolving.com/9/0/7/907dc41896d240bbc687ba6470625d3c53ad589e.png)
,![\[ c^2(s-c) + b^2(s-b) - a(s-b)(s-c) - a(s-a)^2 = (s-a)(s-c)(c+2(s-b)) \]](http://latex.artofproblemsolving.com/2/f/1/2f1f1719219e12a25e1f342d67cdeb659c4456a2.png)
,
,
,
.
,
and 
.
gives
.
.
.
and 
.
is harmonic, so 
.
,
,
,
,
;
,
.
,
,
,
.
such that 
.
.
since 
.
,
.
,
But 
,
and 
.
.
is the point at infinity along line 
.
,
which implies the problem.
Notice ![\[-1=(Q,D;E,F)\stackrel{E}=(P,\overline{DE}\cap\overline{AB};A,F)\]](http://latex.artofproblemsolving.com/1/a/d/1ad45a237a0b0f6b5dcc322c6a5122702fcddd80.png)
so 
if and only if 
But 
so 
is equivalent to 
as desired. 
,
is harmonic. Let 
is parallel to 
so 
,
but since 
.
.
Thus, since 
,
is the point at infinity along line 
iff 
.
and 
be the point on the incircle such that 
.
.
,
passing through the intersection of the tangents through 
is cyclic, by Reim's 
should be as well, hence by power of a point 
,
is harmonic. Let ![\[ (EF;QD) \overset{E}= (A,F;M,\overline{ED} \cap \overline{AB}) = -1. \]](http://latex.artofproblemsolving.com/8/4/d/84df8505c76cb88d3de9bccbea3ffc172b8b64c3.png)
This in turn in only holds if 
which holds when ![\[ \frac{AC}{BC} = \frac{EC}{DC} = 1. \]](http://latex.artofproblemsolving.com/9/8/6/986a3472e684914b7b769a69b91a63bd444701fb.png)
is harmonic. Projecting through
.
is a harmonic pencil. On the other hand 
.
and we are done. 
As its given 
by POP we get
hence 
.
and 
.
are tangents and 
we have that,![\[(QD;EF)=-1\]](http://latex.artofproblemsolving.com/0/b/5/0b57400ce853755d55c9c2ac3348359a72f619cd.png)
Now, if 
.![\[(AF;MP_\infty) \overset{E}{=}(EF;QD)=-1\]](http://latex.artofproblemsolving.com/4/1/e/41eb1b90b1042af2ceb696e730916872202a9e6d.png)
This means that ![\[-1=(EF;QD) \overset{E}{=} (AF;MP)\]](http://latex.artofproblemsolving.com/3/0/3/3036bd81204aeada5eb1b940b285e6990272b771.png)
where 
.
which implies that 
and thus, 
is isoceles with 
![[asy]
import olympiad; import cse5;
defaultpen(fontsize(10pt));
usepackage("amsmath"); usepackage("amssymb"); usepackage("gensymb"); usepackage("textcomp");
size(8cm);
pair MRA (pair B, pair A, pair C, real r, pen q=anglepen) {
r=r/2;
pair Bp=unit(B-A)*r+A;
pair Cp=unit(C-A)*r+A;
pair P=Bp+Cp-A;
D(Bp--P--Cp,q);
return A;
}
pointpen=black+linewidth(2);
pen polyline=linewidth(pathpen)+rgb(0.6,0.2,0);
pen polyfill=polyline+opacity(0.1);
pen angleline=linewidth(pathpen)+rgb(0,0.4,0);
pen anglefill=angleline+opacity(0.4);
markscalefactor=0.01;
size(12cm);
pair A=dir(180+50),B=dir(-50),C=dir(90);
// filldraw(A--B--C--cycle,polyfill,polyline);
D(A--B--C--cycle,polyline);
pair I=incenter(A,B,C);
pair D=foot(I,B,C),E=foot(I,C,A),F=foot(I,A,B);
pair Q=2*foot(I,A,D)-D;
pair M=(A+F)/2;
D(CP(I,D));
D(A--D);
D(E--M,pathpen+dashed);
D("A",D(A),A);
D("B",D(B),B);
D("C",D(C),C);
D("I",D(I),dir(90));
D("D",D(D),unit(D-I));
D("E",D(E),unit(E-I));
D("F",D(F),unit(F-I));
D("Q",D(Q),dir(174));
D("M",D(M),S);
[/asy]](http://latex.artofproblemsolving.com/6/d/4/6d401bcc37e4157a91878d96feae70826e172c7b.png)
![\[-1 = (DQ; EF) \stackrel{E}{=} (DE \cap AB, EQ \cap AF; AF)\text{.}\]](http://latex.artofproblemsolving.com/4/d/0/4d0f84eaa6e3add938b0b232d8df4489ab479516.png)
Now if 
so 
must necessarily be 
then 
and so 
.
is a harmonic quadrilateral, we have 
so 
.
.
,
.
and from 
,
Thus, 
is the midpoint of 
.
we know by similar triangles 
be 
be 
be 
as ![\[-1 = (A, F; \overline{QE} \cap \overline{AF}, P_\infty) \stackrel E= (A, \overline{EF} \cap \overline{AD}; Q, \overline{EP_\infty} \cap \overline{AD}).\]](http://latex.artofproblemsolving.com/f/b/f/fbffa4bac924aa9fcb1f58e8cad4755598f0eb78.png)
But 
