AoPS Academy
and 
are the incenter and the 
-excenter of triangle 
, respectively.
be the midpoint of arc 
on the circumcircle, and 
be the foot of the to 
. The line 
intersects the circumcircle again at 
. For
on the circumcircle of triangle , let 
intersect at 
. Prove
are concyclic.
squares, we cover the area with several S-Tetrominos (=Z-Tetrominos) along the square so that in every square, there are two or fewer tiles covering that (tiles can be overlap). Find the maximum possible number of squares covered by at least one tile.
satisfies
for any non-negative integers 
and 
. Determine the number of possible sets of integers 
.
be a function from the Euclidean plane to the real numbers such that 
for any acute triangle 
with circumcenter 
, centroid 
and orthocenter 
. Prove that 
is constant.
be the positive real number. Prove that
, such that for all positive reals 
, the following is true:
grid. Alice goes first, and they take turns drawing a black point from the coordinate set![\[\{(i, j) \mid i, j \in \mathbb{N}, 1 \leq i, j \leq n\}\]](http://latex.artofproblemsolving.com/4/3/f/43f7493d1dff43079eb5481ee17bc8f16d7b9553.png)
There is a constraint that the distance between any two black points cannot be an integer. The player who cannot draw a black point loses. Find all integers such that Alice has a winning strategy. grid, where 
. Prove that the remainder can be covered by copies of the figures of 
or 
unit squares depicted in the drawing below. for which there exists a positive integer 
with the property that is divisible by and 
is divisible by 
.
denote the set of real numbers. Find all functions 
such that![\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]](http://latex.artofproblemsolving.com/5/b/e/5be34cb26d276bae3c60fb299ad66d83bc158a01.png)
.
, feet of altitude from to , 
diameter in 
, 
, 
and 
midpoint of 
.
by McLaurin we have 
so 
is tangent to 
in addition as 
we must have tangent to as well, also note that 
sre colinear due to the 
angles so as 
and the given colinearity we have that 
therefore 
are colinear and now we finish with 
therefore 
are colinear and now this means that since 
we in fact have 
as desired thus we are done 
.
then 
hence in the radical axis of 
and 
is also on the radical axis of and 
is the radical center of 
is on the radical axis of 
and which is 
where 
their intersections;
is diameter then 
i.e. on the polar of 
so is on the polar of 
is the polar of which leads to the desired result.
. Let 
, where 
and let 
Since 
, we will be done if we show that 
. Since 
from the Ceva/Menelaus picture, we obtain that 
Hence, 
Therefore, 
, so 
where we used Heron's formula. Thus, 
as desired.
. We wish to show that 
are collinear. and is the midpoint of 
, we obtain that 
.
and since 
, we obtain that 
. is the diameter in , thus 
, so are collinear, as desired.
, feet of altitude from to , diameter in , , and midpoint of . by McLaurin we have so is tangent to in addition as we must have tangent to as well, also note that sre colinear due to the angles so as and the given colinearity we have that therefore are colinear and now we finish with therefore are colinear and now this means that since we in fact have as desired thus we are done .
,obviously 
so well known X is the center of circle appolonius for points 
such that 
.By 2002 G7 
is tangent to 
and since is a chord tangent to incircle, 
is bisector of 
so 
so is on the circle appolonius.
(the line of centers). Since is the same as 
we are done
be a point on incircle such that 
tangent to incircle.Then 
. Since 
and midpoint of , from harmonic properties we get that , from 
we get that 
, so tangent to the 
. From IMO 2002 G7 we get that 
collinear, so 
. Let the 
excircle touches at 
.
and meet the incircle again at 
and ,
.
and 
.
i.e. 
are concyclic. This implies 
are collinear and 
.
, we are done.
at and be the antipode of .
, , and are collinear. Let 
intersect the incircle again at . By construction 
is harmonic so 
is tangent to the incircle. So 
as desired.
and 
are similar so triangles 
and 
are also similar finishing the problem.
, 
, 
, 
, inradius
and -exradius
is midpoint of , we obtain 
.
be a point which -excircle is tangent to 
![$$XD\cdot DA’=\frac{(a+c-b)(a+b-c)}{4}=(s-b)(s-c)=\frac{s(s-a)(s-b)(s-c)}{s(s-a)}=\frac{4[ABC]^2}{(a+b+c)(b+c-a)}=\frac{2[ABC]}{a+b+c}\cdot \frac{2[ABC]}{b+c-a}=rR$$](http://latex.artofproblemsolving.com/c/1/d/c1d252b9b3413deb8340b6ecb1232699d0a181f0.png)
So, 
is a diameter in 
we get 
and 
. Notice that ![\[-1=(P,D;X,\infty_{BC})\overset{I}{=}(Q,V;K,R)\]](http://latex.artofproblemsolving.com/1/1/6/11619d3d0e46d2b4061ec993ae8bf13dbf45eabc.png)
Now, it suffices to prove that , and 
are collinear. If we let meet again at 
, it suffices to then prove 
.![\[-1=(P,D;B,C)\overset{I}{=}(Q,V;B,C)\overset{I_a}{=}(I_aQ\cap BC,\infty_{BC};B,C)\]](http://latex.artofproblemsolving.com/1/3/b/13b308d0a8dc5ec23fe353012eb07e7f0864c6e8.png)
Hence , and are collinear and since 
, by butterfly theorem we get that , and 
are also collinear. Therefore ![\[(Q,V;K',R)\overset{D}{=}(U,I;I_a,S)\overset{R}{=}(M,\infty_{BC};D,T)=-1. \ \ \blacksquare\]](http://latex.artofproblemsolving.com/0/a/9/0a9028a4b2c26900500c99cbf67130ea639eb9b7.png)
.
. Since 
and is the midpoint of , from EGMO Lemma 9.17 we have 
. Therefore has equal power wrt and so 
collinear. But 
so 
is tangent to . Similarly, 
is tangent to , hence 
is the pole of wrt to . By La Hire’s Theorem, lies on the pole of wrt . Since 
is tangent to , lies on the pole of too, so the conclusion follows 
be the feet of the altitude from to , be the midpoint of 
,
be the antipode of . We know that 
are colinear. Because of homothety from , we have 
.
tangent to 
are colinear 
from two of this, 
We are done.
,
at 
.
meets 
and 
is by realicing 
and knowing what 
does) and then you can realice what was everything about, spotting harmonic boundles is extremely importsnt in oly
meet 
,
meet 
meet 
meet 
,
meet 
since 
we get that 
so 
.