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 triangle with 
let 
be the circumcircle of 
and let 
be its radius. Point 
is chosen on 
such taht 
and point 
is the foot of the perpendicular from to 
. Ray 
mets again at 
. Point 
is chosen on line 
such that 
and 
lie on a line in that order. Finally, let 
be a point satisfying 
and 
. Prove that lies on .
, so that the equation 
has two real roots, so that the one is the square of the other.
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both and 
. Integers and are coprime if 
.
such that: ![\[ x = \sqrt{ x - \frac{1}{x} } + \sqrt{ 1 - \frac{1}{x} } \]](http://latex.artofproblemsolving.com/e/c/f/ecf7fb3c2805290f2cb07b7fdd7c73e1c7a55b99.png)
be a convex pentagon such that 
. Suppose that the midpoint of 
is the circumcenter of triangle 
. Let 
be the circumcenter of triangle 
.
passes through the midpoint of segment 
.
such that for all 
the following holds:![\[f(x)+f(y)=f(xy)+f(f(x)+f(y))\]](http://latex.artofproblemsolving.com/a/c/a/aca098172ea7ce5663e9dd0dd10c2f1495630230.png)
be a cyclic pentagon inscribed in a circle of centre which has angles 
. Show that the diagonals 
and 
meet at a point belonging to the diameter .
and 
are the incenter and the 
-excenter of triangle , respectively.
be the midpoint of arc 
on the circumcircle, and be the foot of the to 
. The line 
intersects the circumcircle again at 
. For
on the circumcircle of triangle , let 
intersect at 
. Prove
are concyclic.
be an acute triangle. Points 
, and 
lie on a line in this order and satisfy 
. Let and be the midpoints of 
and 
, respectively. Suppose triangle 
is acute, and let 
be its orthocentre. Points and lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
be the circumcircle of triangle , is the midpoint of and be the second intersection of 
and . Tangent line of at intersects at , let 
be a transversal of and 
be intersections of 
with . Let 
be a tangent line of . Prove that 
is cyclic
be integers such that there are consecutive integers 
satisfy 
.
is a perfect square.
on one side and a on the other. Harry has 
of these coins arranged in a line from left to right. He repeatedly performs the following operation: if there are exactly 
coins showing , then he turns over the 
th coin from the left; otherwise, all coins show and he stops. For example, if 
the process starting with the configuration 
would be 
, which stops after three operations., let 
be the number of operations before Harry stops. For example, 
and 
. Determine the average value of over all 
possible initial configurations .
, 
, 
and 
.
and 
: triangles 
and 
are similars, we put 
and 
.
and 
the midpoints of 
and , respect.
, 
; 
, 
and 
, 
the lemma follows. 
, 
; 
, 
; using the lemma and the fact that 
we get (from the cosine law) : ![\[ \cos \angle EE'M= \frac{b^2+(ak)^2-EM^2}{abk}=\frac{a^2+(bk)^2-MF^2}{abk}=\cos \angle FF'M \]](http://latex.artofproblemsolving.com/c/e/5/ce5c27e74252caf4cd464ebcc796d1997a4cb0df.png)
![\[ \Rightarrow b^2+a^2k^2=a^2+b^2k^2 \]](http://latex.artofproblemsolving.com/9/5/a/95aa130fbb3b5bd5ff7dee111d496157fa866200.png)
![\[ \Rightarrow (k^2-1)(a^2-b^2)=0 \]](http://latex.artofproblemsolving.com/6/0/b/60b5a52a575ecf024e0992546445824a45845852.png)
since 
we get: 
, and triangles and are equilaterals. Since 
and 
we get: 
, with this, clearly the triangles 
and 
are congruents, so 
and since 
, the triangle 
is equilateral, so 
.
paralel with 
, 
in the line 
that 
( bettwen and 
). in the line 
that 
.
and 
are 'paralelogramos' (
, etc).
, 
(because 
is a rotation of 
), then 
and 
.
and the midpoints of 
and 
.
and 
, 
and 
, then 
and 
are homotetic, then 
and 
.
and 
, 
and 
, then 
and 
are homotetic, then 
and 
and the angle is 
and 
. Let 
, then 
(because and are similar isosceles). Then 
and 
.
(contradiction). and , with center and angle 
and 
, with 
in the line 
and in the line . Then 
, because , 
, and 
, 
.
, then 
is a 'trapecio' isosceles, then 
and contradiction.. Finally, 
, , and 
are equilateral triangles. Well, not is difficult to see that $<(RB,SA')=EMF=60º$ (because and 
are similar triangles, with 
).
is between 
and 
.Other cases can be proven analogously.
be the origin.
,
,
,
and 
,
.
.This simplifies to(after not-so-long calculations 
) to 
.Since ,this implies 
since ![$\theta\in (0,\pi]$](http://latex.artofproblemsolving.com/6/7/c/67cabe9e99fb2235bb286af82eeaff187a39653c.png)
.
(multiplying the nominator and the denominator by 
).
.So,
.
a
Prove that
,
be a triangle obtained from 
be the midpoints of the segments 
and 
respectively. If 
,
.
