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 the set of all positive integers. Determine all subsets 
of 
such that
and 
intersect at 
and 
. Let 
be the circumcircle of the triangle 
; the circle intersect the circles centered at and at points 
and 
, respectively. The point is inside triangle 
and 
intersects at 
. The point 
on is such that 
. Let 
and 
meet at 
, prove that 
.
. Prove that:
?
be a positive integer. A Japanese triangle consists of 
circles arranged in an equilateral triangular shape such that for each 
, 
, 
, , the 
row contains exactly 
circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with 
, along with a ninja path in that triangle containing two red circles., find the greatest 
such that in each Japanese triangle there is a ninja path containing at least red circles.
, such that 
and 
.
such that 
is prime, 
with 
and ?
then prove \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2}\cdot\frac{abc}{ab+bc+ca}\ge \frac{5}{3}.$$
and B 
were marked, and then the coordinate system was erased. Restore the coordinate system at the two marked points.
![[asy]
size(300);
pair A = (1, 2), B = (3, 1), C = (3, 2), D = (1, 1);
draw(A--C--B--D--cycle);
draw(circle(A, 2));
draw(circle(A, 1));
draw(circle(B, 1));
draw(circle(B, 2));
dot("$A$", A, NW);
dot("$B$", B, SE);
dot("$C$", C, NE);
dot("$D$", D, SW);
[/asy]](http://latex.artofproblemsolving.com/0/a/0/0a0fd6b580c46d1ed575827ad7724cf2e1d4bc55.png)
and . Since we know the side lengths of the rectangle (1 and 2), this is trivialized by considering the intersection points of a circle centered at with radius 2 and a circle centered at with radius 1 and cyclic permutations, as shown above. The construction trivially works.
which is a perfect square root. Constructing lengths of 1 and 2 should be trivial enough.
and by finding the apollonius circle that is the locus of all points whose distances to and are in ratio 1:2 (it's not too bad to construct this circle). Then find this circle's intersection with the circle whose diameter has endpoints on and .
then rotate 
clockwise, then you can get the unit vector by pentisecting the distance from 
to 