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 an isosceles triangle with 
Let 
be its circumcircle and let 
be the centre of . let 
meet 
in 
Draw a line parallel to 
thrugh 
Let it intersect 
at 
Suppose 
.Prove that is an equilateral triangle.
with 
has at least two cycles in it.
, where 
and 
are positive integers satisfying:
be the final value obtained after repeatedly summing the digits of until a single-digit number is reached. (For example: 
, because the digit sum of 
is 
and the digit sum of is 
). Evaluate the sum:
be positive real numbers such that 
. Prove that:
is a nonconstant polynomial with integer coefficients with the following property:
and 
are both odd.
be the set of non-negative integer powers of 
and 
, 
. For every 
and 
in satisfying 
Find the minimum value of 
. be a triangle with incentre 
and circumcircle 
. Let 
denote the second point of intersection of line 
and . The line through perpendicular to intersects line 
, segment ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
, and segment ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
at the points 
, 
, and 
, respectively. The circumcircle of triangle 
meets again at 
, and lines 
and intersect at 
. Prove that lines 
and 
intersect on .
such that 
, 
, and are pairwise tangent to each other. Find the radius of the circle that is externally tangent to all three of these circles.
, , or , such that no two consecutive stars have the same letter.
problems in a meeting. For each 
, each leader will talk about the 
-th problem for -th minutes. The chair can assign one leader to talk about a problem of his choice, but he would have to wait for the leader to complete the entire talk of that problem before assigning the next leader and problem. The next leader can be the same leader. The next problem can be a different problem. Each leader’s longest idle time is the longest consecutive time that he is not talking.
so that the chair can ensure that the longest idle time for any leader does not exceed .
-th
then statement for 
-th number is wrong, so there are ![$[\frac{i}{2}]$](http://latex.artofproblemsolving.com/3/4/e/34efc9714888cd72af5c2dd22f380e4430b274d2.png)
wrong statements for 
and 
then statement for -th number is wrong, so there are ![$101-[\frac{i+102}{2}]=50-[\frac{i}{2}]$](http://latex.artofproblemsolving.com/6/6/d/66d226855219417655205994f666b730501cf186.png)
wrong statements for 
statements are false, and so not more than is true.
to prove, that is possible
