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 
are written on a blackboard, where 
and 
are relatively prime positive integers. At any point, Evan may pick two of the numbers 
and 
written on the board and write either their arithmetic mean 
or their harmonic mean 
on the board as well. Find all pairs 
such that Evan can write 1 on the board in finitely many steps.
such that 
let 
be its circumcircle and let 
be a point on such that 
. Let the projections from 
to 
be 
respectivily. Let 
meet again at 
respecitivily and let 
hit 
at 
. Show that one of the tangents from to bisects the common chord of 
.
hit at 
, then the common chord of 
is segment 
be non-negative real numbers, such that 
Prove that![\[52 + 17(\sqrt a + \sqrt b + \sqrt c + \sqrt d)^2\geqslant 9(ab+ bc + ca + da + db + dc)^2.\]](http://latex.artofproblemsolving.com/6/e/0/6e06719b43ee8df511295af3581b487ddfb17f5c.png)
be the set of positive real numbers. Determine all functions 
such that for all positive real numbers 
and 
![\[f(x+f(xy))+y=f(x)f(y)+1\]](http://latex.artofproblemsolving.com/c/6/a/c6a968a17f6fd5fd637c1ee5d855bdff11e5423a.png)
, denote 
as the foot of the perpendicular from 
to 
and 
to 
.
with respect to 
as 
.
hits at point 
. as 
. Prove that 
.
such that there exists a pair 
of positive integers, such that 
is not divisible by the cube of any prime, and 
, 
, and 
. A circle with center 
and radius 
intersects 
at points 
and 
. Moreover 
and 
have integer lengths. What is 
?
.
from? Sorry, I don't see it!
so we arrive at 
and thus 
.
have to do with anything?
and since BX has to be an integer, 
<this is wrong!! D:. the first one is a contradiction so we conclude that it is 0.5 and so 
. Now this gives a contradiction since it turns out that BX=86 but the options listed for BC=(BX+CX) are much lower!
as they are both radii. Let BY and CY be 
and 
. Take the power of C. We see that:
, will be longer than the diameter, which we can't have.
, it won't work either.
, it'll work. So our answer is 
.
instead of 
. However I believe that's not correct.
![[asy]
//Made by samrocksnature
size(8cm);
pair A,B,C,D,E,X;
A=(0,0);
B=(-53.4,-67.4);
C=(0,-97);
D=(0,-86);
E=(0,86);
X=(-29,-81);
draw(circle(A,86));
draw(E--C--B--A--X);
label("$A$",A,NE);
label("$B$",B,SW);
label("$C$",C,S);
label("$D$",D,NE);
label("$E$",E,NE);
label("$X$",X,dir(250));
dot(A^^B^^C^^D^^E^^X);
[/asy]](http://latex.artofproblemsolving.com/7/7/c/77ce7565827bb55748707258399412ab0383d8b9.png)
intersect at 
and as shown. We apply Power of a Point on point with respect to circle 
This yields the diophantine equation
This generates the four solution pairs for 
: 
we see that 
This implies that we must have 
be . Then,![\[CX\cdot CB = Pow_{\omega}(C) = CO^2-R^2 = 97^2-86^2=11\cdot 183=3\cdot 11\cdot 61\]](http://latex.artofproblemsolving.com/d/5/e/d5e6ae36e5af4d52deef6aaa3fd8ac8016363415.png)
Note that by the triangle inequality, 
, and 
. Thus, the only way to select integers 
is with 
be reals satisfying 
,
and 
prove that![\[\sqrt a + \sqrt b + \sqrt c \geqslant ab+bc+ca.\]](http://latex.artofproblemsolving.com/c/e/9/ce9dc64acc63c6bcce137e963ca8f86ab75018bf.png)
[/quote]
be non-negative real numbers. Prove that![\[4\left(\frac{1}{2x + y} + \frac{1}{2y + x}\right) \ge \frac{3}{x + y} + \frac{21(x + y)}{7x^2 + 22xy + 7y^2}.\]](http://latex.artofproblemsolving.com/7/9/d/79d8d7f94bec30ce0abdd28378e71195b0163bdf.png)
and 
.
,
.
is a divisor of 
.
that are greater than 
.
by PoP. 
clearly and 
and from that it is implied 
are distinct.
.
intersect the circle at points 
you have 
.
and 
.
so 
.
,
and the 
.
,
.
.
,
and 
.
.
,
.
at 
.
,
.
.
.
and 
.
.
,
.
is the only option or else 
which is a contradiction! As a result 
.