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 point inside a triangle 
such that![\[ \angle APB - \angle ACB = \angle APC - \angle ABC.
\]](http://latex.artofproblemsolving.com/a/6/e/a6e3dc27d0457682fa22af4f918f58d9cd8403bc.png)
, 
be the incenters of triangles 
, 
, respectively. Show that the lines 
, 
, 
meet at a point.
cups labeled 
, where the 
-th cup has capacity liters. In total, there are liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.
Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most 
steps.
Prove that in at most 
steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
for which 
divides 
.
and 
, 
cannot be a power of 
.
of natural numbers such that for every pair of natural numbers 
and 
, the following inequality holds:![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
such that 
for all 
such that for all real numbers 
and 
,![\[f(xf(y))+f(y)=f(x+y)+f(xy).\]](http://latex.artofproblemsolving.com/d/2/f/d2ff6ed448cf39e7ec9bce6b944965d5e89b9878.png)
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number 
, divisible by , such that there exists an divisor table, or prove that such does not exist.
, a line 
to tangent , and another circle 
disjoint from such that and lie on opposite sides of . The tangents to from a variable point 
on meet at 
and 
. Prove that, as varies over , the circumcircle of 
is tangent to two fixed circles.
be a triangle and let 
be its incircle. Denote by 
and 
the points where is tangent to sides 
and 
, respectively. Denote by 
and 
the points on sides and , respectively, such that 
and 
, and denote by 
the point of intersection of segments 
and 
. Circle intersects segment at two points, the closer of which to the vertex 
is denoted by 
. Prove that 
.
![\begin{align*}
\frac{a}{c} &= b-\sqrt{b}+c \\
\sqrt{\frac{a}{c}} &= \sqrt{b}+1 \\
\sqrt[4]{\frac{a}{c}} &=\sqrt[3]{b}-1
\end{align*}](http://latex.artofproblemsolving.com/0/c/d/0cd22d36acedbcb10d76d1b884aa35418273ed52.png)
be positive integers. Prove that the expression![\[\frac{\gcd(a+b,ab)}{\gcd(a,b)}\]](http://latex.artofproblemsolving.com/1/b/7/1b78a2027c3b7156fbb566901d2cfd088cc09f00.png)
is always a positive integer, and determine all possible values it can take.
be a convex pentagon such that 
, 
, and 
. Prove that the perpendicular line from 
to and the line segments and 
are concurrent.
students. Let, for 
, the 
student have 
many friends. For 
let us define to be the number of students who have strictly more than friends. Show that 
for all , but 
(and all other 
), so the equation becomes 
. Perhaps means the number of students with strictly more than friends. is counted once for all 
, namely times. Thus the sum of 's counts each student by the number of friends they have, which is exactly the left hand side.
friends is counted in 
.i.e times . so in the RHS too it is counted times . so these are two different ways of counting the sum 
and hence it is proved
chaotic_iak already gave that "simple solution" of yours. Why bother repeating it?
for all , but (and all other ), so the equation becomes . Perhaps means the number of students with strictly more than friends. is counted once for all , namely times. Thus the sum of 's counts each student by the number of friends they have, which is exactly the left hand side.
-gon with vertices 
(
) as persons.We colour the segment 
red if and 
are friends,otherwise we colour it as blue.We shall now show that both the sides number twice the number of red sides..It is counted once when is counted and once when 
is counted.Thus LHS part is proved.
.While we are counting 
,we are only counting 
redlines emanating from 
.This proves our claim for RHS.(Note that thus summing over the 's gives the sum of exact number of redlines originating from each vertex).
means the number of students with strictly more than friends. , then what is 
? 0 ?
be a simple graph with 
vertex. Suppose 
And we connect 
if they are friend 
. Suppose 
.And 
denote number of vertices having degree more than 
.
.
.
.
.
.If it is isolated then 
,
count number of vertex having degree more than in the new graph with 
vertex. we are done. is not isolated. . Suppose it joins with 
,
. Then 
.
,
.
will as same as .
.One must see that number of edges in new graph is 
.
new edges
.
..
for all , but (and all other ), so the equation becomes . Perhaps means the number of students with strictly more than friends. is counted once for all , namely times. Thus the sum of 's counts each student by the number of friends they have, which is exactly the left hand side.
is NOT equal to , it is indeed 
. Observe the question carefully.. means the number of students having MORE than (i.e. 
) friends.. which according to your assumption that everyone hates everyone is indeed . For the same reason, the iteration of stops at 
and not because no one can possibly have 
friends.. hope I could make myself clear..
students. Let, for , the student have many friends. For let us define to be the number of students who have strictly more than friends. Show that 
means the number of students with strictly more than friends. , then what is ? 0 ?
satisfying
Prove
This is unexpected. And sad. Because I just proved a wrong problem. 
array with the columns named as 
and the rows named as 
.
.. Suddenly the problem becomes so much more interesting, isn't it ? 
.
,
boxes, i.e. all the boxes below 30. So to the count we add 30, and to the boxes, we're adding 30 balls.
.
number of students who have greater than or equal to 
)
if 
or,
if 
