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 scalene triangle with point 
and 
on the plane such that 
is an equilateral . Let 
intersect 
and 
at 
and 
respectively and 
intersect 
and 
at 
and 
respectively .
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of square numbers. Prove that: 
is the sum of square numbers
real number on the board such that each number is more than 
.
on the board and write 
.
and 
.
?
be triangle and point 
be 
altitude of 
.
be a point on and such that 
and 
. Point 
is the intersection of 
and 
. Point be intersection of 
and 
. Line 
intersect circumcircle of again at .
.
bisect .
, define a sequence of integers 
by letting 
and 
for 
. Let 
. Determine the largest possible 
such that, for some positive integer , the numbers 
are all prime.
cities in a country, where 
is an integer. Some pairs of cities are connected by direct (two-way) flights. For two cities 
and 
we define:
A 
between and as a sequence of distinct cities 
, 
, such that there are direct flights between 
and 
for every 
;
A 
between and as a path between and such that no other path between and has more cities;
A 
between and as a path between and such that no other path between and has fewer cities. and in the country, there exist a long path and a short path between them that have no cities in common (except and ). Let 
be the total number of pairs of cities in the country that are connected by direct flights. In terms of , find all possible values 
such that 
for all real number 
and 
be a sequence of perfect squares such that 
can be obtained by concatenating a digit to the right of 
. Determine all such sequences that are of maximum length.
, 
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.
a
.
. Prove that
,
at 
.
meets 
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 
,
be the foot from 
meet 
since 
we get that 
so 
.