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 a positive integer. There are two numbers written on a blackboard, one red and one blue. Initially, both are 0. We define the following procedure: at each step, we choose a nonnegative integer 
(not necessarily distinct from the previously chosen ones), and, if the red and blue numbers are 
and 
respectively, we replace them with 
and 
, which we color blue and red (in this order). We keep doing this procedure until the blue number is at least 
.
are positiv real number, such that 
prove inequality 
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 ?
be the incenter of triangle 
and suppose that the perpendicular lines drawn from on 
intersect a given tangent line of the incircle of triangle at points 
respectively. Prove that 
, 
and 
concur.
be the incenter of triangle and suppose that the perpendicular lines drawn from on intersect a given tangent line of the incircle of triangle at points respectively. Prove that , and concur.
be the tangency points of 
with the incircle. We use complex numbers with 
as the unit circle.
. Also, since 
, we have 
for some 
. Since lies on the tangent to the unit circle at 
, we obtain 
. Solving for 
yields 
, so 
. Analogously, 
.
lies on if and only if 
, which simplifies to ![\[g(x(e+f)+ef-x^2)+\bar g(efx^2+ef(e+f)x-e^2f^2)=4efx.\]](http://latex.artofproblemsolving.com/7/0/d/70da0e60727116340ea36c156ba3e1a4bfc8b2f6.png)
We now compute 
and see that it is symmetric in 
, which will finish the problem. The equation for 
is 
by symmetry. Using cramer's rule on these two equations gives ![\[g=\frac{4efx(dex^2+dex(d+e)-d^2e^2)-4dex(efx^2+efx(e+f)-e^2f^2)}{(x(e+f)+ef-x^2)(dex^2+dex(d+e)-d^2e^2)-(x(d+e)+de-x^2)(efx^2+efx(e+f)-e^2f^2)}.\]](http://latex.artofproblemsolving.com/9/6/e/96e864930475b23b480560e8780cbdaf2b13e701.png)
The numerator factors as 
. Thus, we should divide the denominator by 
. Indeed, doing polynomial long division shows ![\[g=-\frac{4defx}{x^3+x^2(d+e+f)-x(de+ef+fd)-def},\]](http://latex.artofproblemsolving.com/f/8/e/f8e4c0c388782e336dece573f8eb83a68c15759f.png)
which is symmetric in , as desired.
be feet from to 
, respectively. be the tangency point of and line through 
. Apply inversion wrt . Then we get the following problem:
with as its circumcenter. Let be point on . Let be the midpoints of 
, respectively. Let be foot from onto line through perpendicular to 
. Define 
, similarly. Prove that 
are coaxial.
be feet from 
to 
, respectively.
. Therefore, 
lies on 
. Similarly, 
lies on 
and 
lies on 
. Note that are collinear by Simson Line of wrt . Let 
be the foot from to 
. Thus, lies on 
and due to right angles.
and 
.
