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 
, and let 
be its incenter. Incircle touches sides 
at 
, respectively. Foot of altitudes from 
to 
are 
, respectively. Rays 
intersect 
at 
, respectively. Prove that 
touches incircle at 
.
be an acute traingle with 
. Point 
is a foot of altitude through vertex 
. Point 
is a circumcenter of . Prove that areas of quadrilaterals 
and 
are equal.
be a triangle with incenter , incircle 
and circumcircle 
. Let 
be the midpoints of sides 
, 
, 
and let be the tangency points of with and , respectively. Let 
be the intersections of line 
with line 
and line 
, respectively, and let 
be the midpoint of arc 
of . lies on ray 
.
bisects 
.
. 
amount of people stand on coordinates 
where 
. Every person got a water cup and two people are considered to be neighbour if the distance between them is 
. At the first minute, the person standing on coordinates 
got litres of water, and the other 
people's water cup is empty. Every minute, two neighbouring people are chosen that does not have the same amount of water in their water cups, and they equalize the amount of water in their water cups. will not have more than 
litres of water.
be the set of integers, and let 
be a function. Prove that there are infinitely many integers 
such that the function 
defined by 
is not bijective. is bijective if for every integer 
, there exists exactly one integer 
such that 
.
such that ![$1+[\sqrt{2n}]~$](http://latex.artofproblemsolving.com/e/6/a/e6aa1bf5d761758196027add39f2e3cd77db1c1c.png)
divides 
.
, ![$[x]$](http://latex.artofproblemsolving.com/b/c/e/bceb7b14e55d33a8bca29b7863ad3cdae95afce4.png)
denotes the largest integer not exceeding . )
let 
denote the intersection of diagonals 
and 
, and let 
denote the intersection of diagonals 
and 
. Prove that if 
and 
, then 
bisects angle 
.x table such that the unit squares are colored arbitrary in black and white, such that exactly three of the squares placed in the corners of the table are white, and the other one is black. Prove that there exists a 
x square which contains an odd number of unit squares white colored.
be a scalene triangle with circumcenter , incenter , and incircle 
. Let touch the sides , , and at points , 
, and 
respectively. Let 
be the projection of to 
. The line 
intersects the circumcircle of again at point 
. The circumcircles of 
and 
intersect again at points 
and 
respectively. Prove that the lines 
, 
, and 
are concurrent. for which the following statement holds: there exists at least one triple 
of integers such that
and all triples of real numbers, satisfying the equations, are such that 
are integers.
, where both and 
are integers between 
and 
inclusive, such that 
.
can be factored as 
. Thus, we consider three cases: 
, 
, or both.
pairs. The second case yields 
pairs. The third case yields pair. This yields 
pairs.
is a solution then divide by 
, sub 
to get 
, so 
. Casework gives 
.
at the end
and 
can equal 
can equal 
so we have to subtract for overcount.
,assume some constant 
allows us to factor by grouping, thus the expression becomes 
or 
. In order to factor we want 
,we find 
and we can factor and solve as above
We get 
or 
so 
,assume some constant allows us to factor by grouping, thus the expression becomes or . In order to factor we want ,we find and we can factor and solve as above
, where both and are integers between and inclusive, such that .) this equation is equivalent to:
We now have two mutually exclusive cases:
. Since 
, we can constraint 
which rearranges to 
. Likewise, since 
, 
. All such 
with will produce a valid and unique 
satisfying the original equation, so it just remains to count the number of solutions for which is 
.
and 
and we proceed to bound . Since 
, we have 
, and similarly 
. All such 
with will produce a valid and unique solution for except for 
, which violates the constraint. Thus our answer for this case is (subtracting at the end so as not to count ) 
.
solutions.
, and solve over , we have![\[
x=\frac{-y\pm17y}{24}
\]](http://latex.artofproblemsolving.com/a/6/c/a6c330741d1e09ecbd859c435f9b3dbae664ebbc.png)
so either 
, or 
.
if , then we have 
, hence there are 
solutions. Similarly for , there are 
solutions.
as a solution, so the answer is 
.
satisfying
Prove
,
or 
.
,
,
but not 
,
non-zero values of 
ordered pairs.
non-zero values of 
ordered pairs.
non-zero values of 
ordered pairs.
.
.
to 
.
.
,
The pairs of integers that satisfy this equation are 
and 
for 
For the first type of solution, we have 
so the number of solutions of that type are 
Then, for the second type, we get 
so the number of solutions of the second type is 
However, note that the solution 
So when we get that our first case is 3x+2y and 
that then we get 
but we over count (0,0) so we have 
then just find x and y and make sure not to double counted the zero 