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.
and 
connect the points 
and 
These arcs lie in the same half-plane defined by line 
in such a way that arc 
lies between the arcs 
and 
Point 
lies on the segment 
Let 
, and 
be three rays starting at 
lying in the same half-plane, 
being between 
and 
For 
denote by 
the point of intersection of 
and 
(see the Figure below). Denote by 
the curved quadrilateral, whose sides are the segments 
and arcs 
and 
We say that this quadrilateral is 
if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals 
are circumscribed, then the curved quadrilateral 
is circumscribed, too.
Prove that
be integers, let 
be a set with 
elements, and let 
be pairwise distinct non-empty, not necessary disjoint subset of . A function 
is called nice if there exists an index 
such that ![\[\sum_{x \in X_k} f(x)>\sum_{x \in X_i} f(x) \quad \text{for all } i \ne k.\]](http://latex.artofproblemsolving.com/3/d/a/3da5ad1bcade98ee08caa35763de226027bcb2c6.png)
Prove that the number of nice functions is at least 
.
be positive real numbers such that 
Find the minimum value of 
, , and 
are fixed on a line in this order. Let 
be a circle passing through and whose center does not lie on the line . Denote by 
the intersection of the tangents to at and . Suppose meets the segment 
at 
. Prove that the intersection of the bisector of 
and the line does not depend on the choice of .
be a triangle with circumcentre 
. The points 
and 
are interior points of the sides 
and 
respectively. Let 
and 
be the midpoints of the segments 
and 
. respectively, and let 
be the circle passing through and . Suppose that the line is tangent to the circle . Prove that 
. Prove that for any given positive integer , the product
has at most distinct prime divisors.
such that:
for all 
.
upon division by 
be 
. Then consider the numbers 
for 
. Suppose that not all pairwise sums are different then![\[202(a+b)+r_a+r_b=202(c+d)+r_c+r+d\]](http://latex.artofproblemsolving.com/b/b/8/bb883b0f628b01552bf9324c56b682fdd3b9dcf9.png)
for some 
, all distinct. Note that 
. Since 
for all 
, 
and 
is either 
or has absolute value at least 
. The latter is impossible so it is zero, meaning 
. Now, 
and 
implies 
and as a result, 
.
then adding 
gives 
, contradiction. If 
then adding gives 
, also contradiction. Therefpre, all pairwise sums are different, and the largest number is between 
and 
.
be a prime. Let 
be the set of vectors 
for 
. Then the pairwise sums of elements of are distinct.
and 
then 
, and by unique factorization of 
in 
we have 
. 
. All vectors lie in 
, so we can apply the mapping 
to , which will give 
integers in ![$[1, 2p^2+p+1]$](http://latex.artofproblemsolving.com/a/4/d/a4dbcf6cc9cc909010afeefc0f0ee4c94891a487.png)
, and preserve the "pairwise sums are distinct" property. 
Prove that
.![\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]](http://latex.artofproblemsolving.com/b/3/2/b32e0bcba967ca847e878ac186c3168cd6c5fc66.png)
Prove that
Prove that
Where 
satisfying
Prove
and 
Prove that
positive integers not exceeding 
,
![$[n]_p$](http://latex.artofproblemsolving.com/4/c/e/4ce12a27e3995de190e85f6e406874b072388b4c.png)
denote the remainder when ![$f(n) = 2pn + [n^2]_p$](http://latex.artofproblemsolving.com/0/4/b/04b846ff6a7c3b904bf843946cf6d6327d2ec997.png)
.
satisfies the conditions of the problem.
,![$f(i) = 2pi + [i^2]_p < 20200 + 101 < 25000$](http://latex.artofproblemsolving.com/3/b/9/3b9328d0da5f115abc1c08972add54a63e21d78d.png)
.
,
implies ![$|2p(i-j)| = |[i^2]_p - [j^2]_p| < p$](http://latex.artofproblemsolving.com/e/7/9/e797e83b2284885db1ebc3643de8e330d8a3d4f9.png)
,
.
are distinct and less than 100.
for 
,
.![$2p((a+b)-(c+d)) = [c^2]_p + [d^2]_p - ([a^2]_p + [b^2]_p)$](http://latex.artofproblemsolving.com/6/d/3/6d39b1023d75a2febae71cebc184c523772f0f51.png)
.
and has absolute value less than 
.
,
.
as well, 
and 
must be the two roots of the quadratic 
mod 
.
.
,
,
.
numbers 
less than 
with their pairwise sums different (then we just plug in 
to get the result)
is an integer then define 
to be the remainder obtained when 
for some 
and 
between 
and 
and 
,
(
)
are multiples of 
and 
or 
and 
for stewing up a contradiction,hence we have exhausted all possibilities and the result is clear by substituting ![$\sqrt{n}+0.998\sqrt[4]{n}$](http://latex.artofproblemsolving.com/7/6/d/76d052c913d23a956c47e00225e4f9cd945bd4d3.png)