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 complex numbers such that [center]
for any 
with 
[/center]
and 
for any 
+1 w
, 
, 
we have ![\[ 3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}. \]](http://latex.artofproblemsolving.com/a/9/8/a98b6c29802355c491016bcce58228a020423526.png)
and two lines 
and 
are given in the space. For each nonnegative integer 
we denote by 
the projection of 
on 
and by 
the projection of on 
Prove that there exist two segments ![$[A'A''] \subset d_1$](http://latex.artofproblemsolving.com/8/8/6/88667ee717ddd630efd34a9c4c83ee359c4897ed.png)
and ![$[B'B''] \subset d_2$](http://latex.artofproblemsolving.com/3/7/4/374927492266c8adaed5d34e3871d250baba9c1b.png)
of length 
and a nonnegative integer 
such that ![$A_n \in [A'A'']$](http://latex.artofproblemsolving.com/4/8/5/4851110dd8f0bf0ee7e42239ee051da76f5fbd65.png)
and ![$B_n \in [B'B'']$](http://latex.artofproblemsolving.com/5/e/7/5e7d78dd940f5c81673b779a9b45e83f4dc05b4d.png)
for any 
, 
, 
, 
be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, 
. Prove that 
for all 
.
opposite the vertex 
be tangent to the side 
at the point 
. Define the points 
on 
and 
on 
analogously, using the excircles opposite 
and 
, respectively. Suppose that the circumcentre of triangle 
lies on the circumcircle of triangle . Prove that triangle is right-angled. is great if the following holds: for any point 
on the side , if 
and 
are the feet of the perpendiculars from to the lines and 
, respectively, then the reflection of in the line 
lies on the circumcircle of the triangle . Prove that triangle is great if and only if 
and 
.
. Let be the foot of the altitude of triangle issuing from the vertex , and let 
be the circumcenter of triangle . Assume that 
. Prove that 
.
be a parallelogram such that 
Let 
and 
be the midpoints of and 
respectively. Assuming that 
is a cyclic quadrilateral, find 
be a triangle, and let be an arbitrary point on line 
, where is the circumcenter of . Define 
and 
as the orthocenters of triangles 
and 
. Prove that 
passes through a fixed point which is independent of the choice of .
be a triangle inscribed in a circle, where the tangents to the circle at points and intersect at the point . Let 
be a point on the arc (not containing ) such that 
and 
. Let the lines and 
intersect at point . Let 
be the reflection of with respect to the line . The lines 
and 
intersect at point , and intersects the circumcircle of again at point . lies on the circumcircle of 
.
be circumscribed about circle 
, and let 
be the orthocenter of . The circle touches line at . The tangent to the circle 
at meets at 
. Let 
be the midpoint of 
, and let the line 
meet again at 
. The tangent to parallel to meets the line 
at 
. Prove that 
is tangent to .
, inscribed in the circle 
, satisfies 
and 
, and 
is the intersection point of the diagonals and 
. The circle with center and radius 
intersects in two points 
and 
. Prove that the line 
is tangent to the circles with diameters 
and 
.
be a right triangle in with 
. Let be the midpoint of and a point on such that 
. Let 
.
and the ratio 
.
let 
and 
be the boundary points of the incircle with the sides 
and 
. Show that if 
holds, then the triangle is isoceles.
and let 
.
. So stewarts theorem gives: 
. In the same way 
with equality only when 
.
and 
and see that the expression is convex. Thus max is attained either when 
or 
. (Since ![$ c \in [|a - b|;a + b]$](http://latex.artofproblemsolving.com/0/9/2/092b5b412d9c96d06f6c0f2e4c4d9dc03348eae9.png)
): 
.
(Assume wlog 
) 
with equality only if .
we get that 
. And this is the only case which makes sense, since there is an inscribed circle.
isosceles
holds, then the triangle is isoceles.
