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 midpoint of angle bisector of triangle 
. Circle 
with diameter 
cuts 
at 
, and circle 
with diameter cuts 
at 
.Prove that 
, , and are concyclic.
otherwise. Let the parallel to intersect at . Since 
is an external angle bisector. Therefore, 
so 
. So, 
or 
are concyclic where 
is the incentre of . Indeed, by symmetry, also lies on the circle.
as it is median inside 
, we can get in Stewarts theorem,and 
we can get using Stewarts theorem in triangles 
and 
and Pythagoras theorem in triangle 
. Analgously, we get and 
. Now this yields 
which yields 
which implies , , and are concyclic.
be the midpoint of 
, respectively .
be the intersection of 
with 
, respectively .
are collinear .
and 
,
. i.e. 
under Inversion 
,
are concyclic .
.
by segment 
such that 
.Triangle 
is isosceles so 
be the internal bisector of 
where 
is on 
.We have 
we conclude that 
is the midpoint of 
and so 
.By angle chasing now we have 
which means that 
is tangent to the circumcircle of triangle 
and hence we obtain 
.
etc) we get that 
and so 
so 
is cyclic.We also get that 
which gives that 
is tangent to the circumcircle of 
and so 
which finally gives the desired result.
to (lets call it 
) and then just invert the whole picture through . The result will be very symmetric and it will be suddenly all super-obvious (see the attachment).
parallel to intersect external bisector of 
(the bisector which cuts nearer to ) at point 
. By easy lenght chasing we get that contains (we use Ceva theorem).Now as 
we get that 
is cyclic so 
which implies that is tangent to the circumcircle of triangle 
which implies 
. Analogously we get that 
so 
is cyclic.
cut for a second time at 
, respectively, and let 
be the radii of 
, respectively. Let 
be the midpoints of 
, respectively (i.e. the centers of ). Note that is the midpoint of 
which implies that 
Meanwhile, 
by the Angle-Bisector Theorem. Therefore, is the internal center of homothety that swaps 
Since are the images of 
under this homothety, we find that 
Then note that 
because 
are concyclic. It follows that 
are concyclic as well. 
, 
, 
and 
. Furthermore let be the midpoint of and , and let be the foot of the altitude from on 
. Since 
is the foot of the altitude from to 
, and is the bisector of 
and 
, the homothety centered at with factor 
sends to a point on . Thus lies on the midline 
of 
parallel to . Analogously 
lies on the midline 
parallel to . Note that lies on both and so 
and 
is inscribed in a circle 
. Since is the circumcenter of 
and and pass through the centers of and we have 
hence 
is tangent to at , so the inversion centered at with power 
fixes and and swaps and , so it swaps and , hence it swaps with , and with , so is cyclic.
be the incenter and 
be the excenter of . Define 
as the points on segments and , respectively such that 
Note that 
so we conclude that the six points, 
are concyclic. Finally, we have 
from where we conclude 
. Similarly, 
and the result follows.
be midpoint of angle bisector of triangle . Circle with diameter cuts at , and circle with diameter cuts at .Prove that , , and are concyclic.
as HM-points in 
and 
. Drop perpendiculars 
on line .
and 
are cyclic.
. Hence 
, so 
proving cyclic. Likewise, we see is cyclic too. 
are on 
, respectively. being HM-points. It turns out that we exploit their properties in both solutions. Although here we don't make ad-hoc constructions and appear more naturally.
be midpoint of angle bisector of triangle . Circle with diameter cuts at , and circle with diameter cuts at .Prove that , , and are concyclic.
; 
be a point on extension of side 
such that 
and 
be the median then the circle with diameter 
intersects at 
which is the HM-point with respect to vertex 
. 
is the HM-point of 
with respect to vertex 
. Thus, we have 
and similarly, in 
, we have 
. Thus 
and thus 
is cyclic.
be points on 
such that 
. Also let 
and 
be the midpoints of segments 
and 
respectively. Finally denote the foot of the -altitude by 
. Then we get 
This means that 
. Also, by homothety at , we get that 
, or equivalently that 
are collinear. As 
, so by converse of Reim's Theorem, we have 
. Similarly also lie on this circle, and so 
are concyclic.
, which means that is the circumcenter of 
. Considering the power of wrt 
, we get that 
, which is equivalent to saying that is the inverse of in . Thus, 
, and so 
is tangent to 
. This gives 
be the intersection of and the perpendicular line through on 
we get, 
lies on the circle with diameter 
, where 
lies on the circle with diameter 
be the second intersection of the two circles, which lies on 
is cyclic.
and 
similar.
or 
.
are concyclic with some angle chase. Since, 
is cyclic and thus, 
But since, 
we have: