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.
with 
and the circle 
of radius 
centered at 
Let 
be the midpoint of 
The line 
intersects a second time at 
Let 
be a point on such that 
Let 
be the intersection of 
and Prove that 
distinct natural numbers not exceeding 
are given. Prove that among these numbers we can find four 
such that 
is divisible by 
be . A point 
lies on segment 
, and lies on segment 
. Let ray 
. A point , which lies on , bisects 
and it is on the other side of 
with respect to . Ray 
, ray 
, and 
. Prove that 
are cyclic.
+1 w
and 
such that
be a real number. Determine all natural numbers 
for which there exist distinct natural numbers 
, 
, 
, such that
denotes the greatest common divisor of natural numbers 
and 
).
with 
, point lies on the circumcircle of such that 
. The line through parallel to 
intersects 
in 
and 
in 
. Prove that the center of the circumcircle of triangle 
lies on the circumcircle of triangle 
.
, no two of which are 
. Prove that:
points and some chords between them is given.
In every step we can take two chords 
with a common point other than 
and erase exactly one of and draw 
let 
be the minimum of chords after some steps. Find the maximum of over all initial positions.
In every step we can take two chords with a common point other than and erase both of and draw let be the minimum of chords after some steps. Find the maximum of over all initial positions.
be a positive integer and let 
and 
be positive integers congruent to 1 modulo 4. Prove that there exists a positive integer 
such that at least one of the numbers 
and 
is divisible by 
be the lengths of the sides of a triangle. Prove that![\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]](http://latex.artofproblemsolving.com/7/4/b/74b7aa5621bc52977d9acd448d74293dc27a8e00.png)
be the intersection of 
and 
![[asy]
pair O=(0,0), C=(1,0), B=(cos(2pi/9), sin(2pi/9)), A=(cos(2pi/3), sin(2pi/3)), E=(cos(13pi/9), sin(13pi/9)), D=(cos(-4pi/9), sin(-4pi/9));
draw(A--B--C--D--E--cycle);
draw(circle(O, 1));
pair F=-A;
draw(A--F);
draw(C--E);
draw(D--B);
label("$C$", C, dir(0));
label("$B$", B, dir(40));
label("$A$", A, dir(120));
label("$E$", E, dir(260));
label("$D$", D, dir(-80));
label("$O$", O, W);
draw(D--O--C--O--B--O--E);
real a=sin(2pi/9), b=cos(2pi/9), c=cos(4pi/9), d=sin(4pi/9);
real x=(b*d+a*c)/(a+d+b*sqrt(3)-c*sqrt(3));
pair G=(x, -x*sqrt(3));
label("$G$", G+(0,0.1), dir(100));
[/asy]](http://latex.artofproblemsolving.com/2/4/8/248e2e816ab6d7616e9a881eff05296214182cf6.png)
and 
Then, 
and 
Thus, 
so 
Now, 
so 
and 
and 
and 
and 
so 
and since 
we have 
Also, by the inscribed angle theorem, 
and 
As such, 
and 
We prove this below.
by sum-to-product and double angle.
so 
as desired. 
be a cyclic pentagon inscribed in a circle of centre 
which has angles 
.
and 
.
.
and 
and let 
be the point on the circumcircle such that 
is a diameter. We'll perform some angle chasing before we continue. Let ![\[a=\overarc{AB},b=\overarc{BC},c=\overarc{CD},d=\overarc{DE},e=\overarc{EA},\]](http://latex.artofproblemsolving.com/e/9/4/e9470b7a6c9283a2313703364d7de9904c1ed0c3.png)
. By pentagon properties 
.
Solving we get 
.
.
and 
so 
.
Now we'll consider 
.
.
and 
.
.
so 
.
so:
In order for 
so 
.
.
Which is true. So