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.
cause there are 10 ways to choose 1 element that will be repeated. Similarly for 2 same elements it would be 
the answer would be ![$(2^{10}-1)-([A_1+A_3+A_5+A_7+A_9]-[A_2+A_4+A_6+A_8+A_{10}].$](http://latex.artofproblemsolving.com/7/3/1/73130cab8af6ffb5c2638548fdf6af3ecc70a6ed.png)
But this number turned out to be 
But I realized that 
So I was really confused of why i got the right answer somehow in my calculations.
written on the board. We apply the following rule: At each step, we will add all the numbers that are the sum of the two numbers on the board so that the sum does not appear on the board. For example, if the two initial numbers are 
; then the numbers on the board after step 1 are 
; after step 2 are 
; 
, prove that the number 2024 cannot appear on the board.
; 
, prove that the number 2024 can appear on the board.
, points 
and 
lie on 
and 
respectively such that 
. Points 
and 
are the reflections of and with respect to the midpoints of sides and , respectively. Point 
lies on segment 
such that the circumcenter of triangle 
coincides with the circumcenter of triangle . passes through the midpoint of segment 
.
that satisfies![\[
a^3+b^3=1957
\]](http://latex.artofproblemsolving.com/5/d/8/5d8d2a40734ec6b0a9149f9a4ae3a5828d5055f3.png)
![\[
(a+b)(a+1)(b+1)=2014
\]](http://latex.artofproblemsolving.com/3/0/0/300f2d16e669a1d8d9c54b357a6eaf7a8eb5d826.png)
Hence, find the value of 
be a positive integer larger than 
. Build an infinite set 
of subsets of 
having the following properties: distinct sets of have exactly one common element;
distinct sets of have void intersection.
that satisfy the equation 
?
and 
Find the value of 
be a triangle with a circle 
with center 
tangent to sides 
at 
respectively. Suppose the circle with diameter 
intersects the circumcircle of again at 
is the reflection of over . Suppose points 
lie on such that 
are parallel to . Prove that: The intersection of 
lie on the circumcircle of 
![\[f(xf(y) + x) = xy + f(x)\]](http://latex.artofproblemsolving.com/9/9/f/99f580ebc50846e6bc2c004667559922749a4dfa.png)
be an acute triangle with 
and let 
be its orthocenter. Let be a point on the perpendicular bisector of 
such that 
and and 
are on different sides of 
, 
a point on the perpendicular bisector of 
such that 
and 
a point on the perpendicular bisector of 
such that 
and 
lie on the opposite side of w.r.t 
. Prove that 
and are collinear.
square is filled with numbers 
.The numbers inside four 
squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?
be a triangle with points 
lie on the perpendicular bisector of such that 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points to the circumcircle of 
intersects at point 
Prove that: 
are parallel.
:
Square again and subtract to get 
Thus, 
Roots are 
Negatives are bad because lhs and rhs would have differing signs, and 
fails. Thus, all are bogus except 
. Now observe that for 
, 
, so we must have 
. So, 
, giving 
.
for can be proven by casework.)
:
.
.
because the LHS is always positive, so the RHS is also always positive. Now we split it into cases:
This gives 
solving gives 
This gives 
so 
again.
