AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a function satisfying the following conditions:
, if 
and 
, then 
;
is surjective.
.
be a triangle with 
and 
. Let 
and 
be the midpoints of 
and 
respectively. The point 
is inside of 
such that 
is equilateral. Let 
and 
. Prove that 
.
. This has to be covered by three types of tiles - red, blue and black. The red tiles are of size 
, the blue tiles are of size and the black tiles are of size 
. Let 
denote the number of ways this can be done. For example, clearly 
because we can have either a red or a blue tile. Also 
since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile.
for all 
.
for all 
.![\[ \binom{m}{r} = \begin{cases}
\dfrac{m!}{r!(m-r)!}, &\text{ if $0 \le r \le m$,} \\
0, &\text{ otherwise}
\end{cases}\]](http://latex.artofproblemsolving.com/b/8/d/b8d653661e421db1fce7d93837e7bce31337bb30.png)
for integers 
.
can be written in the form
where 
for some non-negative 
such that
for every 
.
be real numbers such that ![\[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]](http://latex.artofproblemsolving.com/3/d/0/3d015292c249a69a44d5f923092201d7a271b896.png)
Find the smallest possible value of 
. such that the following statement holds: Suppose real numbers 
, 
, 
, 
, 
, 
, , 
satisfy 
for all 
. Then there exists 
, 
, , 
, each of which is either 
or 
, such that![\[ \left| \sum_{i=1}^n \varepsilon_i a_i \right| + \left| \sum_{i=1}^n \varepsilon_i b_i \right| \le 1. \]](http://latex.artofproblemsolving.com/6/5/8/658ddc5e8626e7d5ef29809c8c3ac641e3a87757.png)
vertices is given. To each vertex, a natural number from the set 
is assigned. Prove that there are 
vertices 
such that if the numbers 
are assigned to them respectively, then 
and 
is a parallelogram.
such that 
. Show that 
. When will equality hold?
.
has 
obtuse. Let , , be the feet of the altitudes from 
, 
, 
respectively. Let 
, 
, 
be arbitrary points on 
, , 
respectively. The circles with diameter 
, 
, 
are drawn. Show that the lengths of the tangents from the orthocentre of to these circles are equal.
be triangle with point 
and on 
and 
are on the same side of 
be midpoint of segment and 
be midpoint of segment 
be intersection of 
with 
and 
be triangle with point and on and are on the same side of be midpoint of segment and be midpoint of segment be intersection of with and 
be triangle with point and on and are on the same side of be midpoint of segment and be midpoint of segment be intersection of with and 
true for D E F symmetrically ?
![[asy]
size(9cm);
defaultpen(fontsize(11pt));
pair A = (-0.25,2);
pair B = (-1,0);
pair C = (2,0);
path C1 = circle(B,1);
path C2 = circle(C,1);
pair E1 = intersectionpoint(C1, A--B);
pair F = intersectionpoint(C2, A--C);
pair M = (E1+F)/2;
pair N1 = (B+C)/2;
pair P = 2N1-A;
pair Q = 2M-A;
pair R = extension(Q,F,B,P);
pair S1 = extension(E1,Q,C,P);
pair T = extension(B,F,C,E1);
pair D = extension(T,P,B,C);
pair U = extension(F,Q,C,E1);
draw(A--B--C--cycle, black);
draw(E1--Q--F, black);
draw(B--P--C, black);
draw(R--Q--S1, black+dashed);
draw(E1--F, black);
draw(B--F,blue);
draw(C--E1,blue);
draw(T--P, blue+dashed);
draw(M--N1, red+dashed);
dot(A);
dot(B);
dot(C);
dot(E1);
dot(F);
dot(P);
dot(Q);
dot(R);
dot(S1);
dot(M,red);
dot(N1,red);
dot(T,blue);
dot(D);
label("$A$", A, N, black);
label("$B$", B, SW, black);
label("$C$", C, E, black);
label("$E$", E1, NW, black);
label("$F$", F, NE, black);
label("$Q$", Q, 1.3W+0.5S, black);
label("$P$", P, S, black);
label("$R$", R, SW, black);
label("$S$", S1, SE, black);
label("$N$", N1, SW, black);
label("$M$", M, N, black);
label("$G$", T, S+0.5W, black);
label("$D$", D, NE, black);
[/asy]](http://latex.artofproblemsolving.com/4/a/5/4a5c8083b91859c80df2379908b173908762b3c8.png)
and 
denote the reflection of across and , respectively.
, and 
, and 
. It implies that 
, and 
are parallelograms., we have 
. This indicates that 
is a rhombus. Furthermore, the line 
must bisect 
.:
so, lies on the angle bisector of ; that is 
. also lies on , and it could be proceed using Menelaus' theorem on 
and the line 
:
The converse of Menelaus' theorem also holds for 
, so 
. is the homothetic center sending 
. Therefore, 
.
as desired.
and . linearly along 
, then moves linearly along and moves linearly. Assume WLOG 
, then let the value of when 
be . Then when 
and we get that lies on the line through and the midpoint of 
. A homothety of scale factor 
from takes this line through parallel to the -angle bisector. A simple bary (or MMP) argument shows that moves on a line (since it's degree 
). When 
we get 
which does lie on the desired line. When 
we get 
which does lie on the desired line. 
, 
, and 
. Then, 
and 
, so 
has direction 
. In addition, 
, so 
also has direction 
. Therefore, and are parallel.
or ) that is parallel to the angle bisector of 
. by the angle bisector property.
or ) that is parallel to the angle bisector of . by the angle bisector property.
is the reflection of wart to then lies on the angle bisector of , for which it suffices to prove that the distances from to and are equal. Since 
it suffices to prove that ![$[G'CF]=[G'EB]$](http://latex.artofproblemsolving.com/5/3/a/53a4c302f9c1b01a0f80107fa69645346f39d588.png)
. Now, observe that![$$[G'CF]=[G'CB]=[G'EB],$$](http://latex.artofproblemsolving.com/8/9/5/8954d64f559da05554039b7c37f44bf721f45902.png)
because 
and 
. Done.
be the arc midpoint of 
and intersect and at 
and 
respectively. will be the center of spiral similarity that sends 
and 
and 
then let 
is the center of spiral similarity that sends 
and a fixed point 
is cyclic
is a fixed point
such that 
is a parallelogram the problem reduces to-
be a point on the angle bisector of and let 
and 
be lines from and such that and is parallel to and . Let 
and 
. Prove .
such that 
.
.
