AoPS Academy
are non-negative real numbers such that 
Prove that![\[\sqrt{a^2+6ab}+\sqrt{b^2+6bc}+\sqrt{c^2+6ca} \ge 5\sqrt{1 + \frac{153abc}{50(a+b+c)}}.\]](http://latex.artofproblemsolving.com/4/2/9/4295e20975f0a7c9902a73695bebb665385f01b3.png)
of real numbers such that![\[ a_{m^2 + m + n} = a_{m}^2 + a_m + a_n\]](http://latex.artofproblemsolving.com/b/b/9/bb97053e86e7733fe5c26f7859c87566eb444b52.png)
for all positive integers 
and 
.
be a convex quadrilateral with 
and 
. Let 
and 
be points on segments 
and 
, respectively, such that line 
intersects lines 
and 
at points 
and 
, respectively. It is given that 
.Let the midpoint of 
be 
and the midpoint of be 
.Prove that the points 
and 
lie on a circle.
points are marked on it, dividing it into equal arcs. Petya and Vasya are playing the following game. Petya chooses a positive integer 
and announces this number to Vasya. To win the game, Vasya needs to color all marked points using colors, such that each color is assigned to exactly two points, and for each pair of same-colored points, one of the arcs between them contains exactly 
marked points. Find all for which Petya will be able to prevent Vasya from winning.
be the incenter of , and let be its projection onto . Also let be the midpoint of . must coincide both with the projection of 
onto and with that of 
, so, in particular, these two must coincide. It folows from this that 
(so 
, since 
).
, we also have 
, but 
have lengths equal to one third of the , and, respectively, 
-altitude in 
(
). Moreover, these altitudes cut in the same point , which is the image of through a homothety of center and ratio 
. This means that 
are congruent.
(which is equivalent to 
). If is the intersection between the plane 
and , then, since and lie on different sides of the plane 
and must lie on different sides of , i.e. 
. The areas ![$[ACD],[BCD]$](http://latex.artofproblemsolving.com/4/d/8/4d88030211c600374ffe2415f253670d6fb7ce79.png)
of the triangles 
satisfy ![$\frac{[ACD]}{[BCD]}=\frac{AQ}{QB}$](http://latex.artofproblemsolving.com/9/7/4/974c14abdf2a2e4ba274add866e2264286c889f5.png)
, so ![$[ACD]<[BCD]\ (*)$](http://latex.artofproblemsolving.com/e/5/8/e5817083ffd915daf90b153ca15dfc9b7a48592a.png)
. At the same time, from it follows that ![$[ACD]>[BCD]$](http://latex.artofproblemsolving.com/2/f/2/2f2120bdd1edd155f3a24a957b9a16bffc19816c.png)
, contradicting 
. Something similar happens if we assume that 
, so it must be that 
, i.e. 
.
, implies ![$[BDA]=[BDC]$](http://latex.artofproblemsolving.com/f/8/f/f8ff92f296f05d4f71e0a6d11076dbf357c21614.png)
for example, and since the plane 
divides 
into a ratio equal to the ratio of the areas ![$[BDA],[BDC]$](http://latex.artofproblemsolving.com/b/0/a/b0af7d5194420257807b64cbcaa9e4629c02184c.png)
, it must cut in the middle, i.e. 
are coplanar. Similarly, 
are coplanar, so 
are collinear. In the same way we show that 
are collinear. In the isosceles triangle 
the altitudes 
divide the sides 
respectively in the ratio 
. From here (and from ) we find that 
, exactly what it should be in order to make (and thus all faces of ) equilateral.
or 
