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.
, note 
., such that 
isn’t divisible by 
but 
is divisible by 
, 
, 
, the inequality![\[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}
\]](http://latex.artofproblemsolving.com/f/a/9/fa964e29de91f0e2f3cb32e335a52d4521bc38fd.png)
is a rational number.
and other permutations.
,
, 
are perpendicular foot from 
to sides 
, prove that 
bisect segment
.
. Prajit then tries to place n points such that no four points are concyclic and no 
points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?
and 
Prove that
![$$ a+b+kc^2\geq\frac{3\sqrt[3]{k}}{4}$$](http://latex.artofproblemsolving.com/5/5/b/55be0378562556d1aa4400bef11d1510ff722854.png)
Where 
be an infinite sequence of positive integers. Prove that there exists a unique integer 
such that![\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]](http://latex.artofproblemsolving.com/9/4/f/94fc5a51b7e68588123c5b527fe75183bc4c4937.png)
such that
.
denote the number of positive integers less than that are relatively prime to . Prove that there exists a positive integer 
for which the equation 
has at least 
solutions in .
be the point such that 
is a parallelogram . Hence 
and 
Hence 
is a parallelogram.
. But 
. Hence 
is cyclic.
. 
lies inside triangle 
so that 
. The point is such that 
is a parallelogram. Prove that 
.
. Then points 
are concyclic. Using 
is a parallelogram we get
Similarly 
. Hence,
![[asy]
size(200);
pair B=dir(-160),C=dir(-20),E=dir(70),F=dir(135),A=extension(B,F,C,E),P=extension(B,E,C,F),Q=B+C-P;
draw(unitcircle,cyan);
fill(P--E--F--P--cycle,purple+grey+grey);
fill(B--Q--C--B--cycle,purple+grey+grey);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$E$",E,dir(E));
dot("$F$",F,dir(F));
dot("$P$",P,dir(-90));
dot("$Q$",Q,dir(Q));
draw(A--B--C--A,royalblue);
draw(B--E^^C--F,red);
draw(P--A--Q,green);
draw(B--Q--C^^E--F,brown);
[/asy]](http://latex.artofproblemsolving.com/c/3/a/c3a5e6bde8cb758fbb2214e5c8fa14940fbd8308.png)
denote homothety at 
with scale 
followed by reflection in internal angle bisector of 
. Note 
and 
. Thus 
is a point such that 
Hence 
. As also consists of reflection in internal angle bisector of , so 
follows. 
and 
. Since these are parallel pairs of lines, 
are the points at infinity for those pencils of parallel lines respectively. Note that 
by 
and 
. Thus, 
is isogonal to 
wrt. 
. Hence by DDIT on complete quadrilateral 
, there exists a projective involution swapping 
, 
, 
. This is taking the isogonal, and so 
are isogonal. 
Putting them together, we get![\[\frac{\sin(\angle QAC)}{\sin(\angle BAQ)} = \frac{CQ}{BQ}= \frac{BP}{PC} = \frac{\sin(\angle BAP)}{\sin(\angle PAC)}\]](http://latex.artofproblemsolving.com/2/1/9/219c9710c194c69c02102a61f129a2c27089f106.png)
and since 
, we have that 
and we're done. .
. Note that 
. Also note that 
. Both angle equalities are there since is a parallelogram. So, consider homothety 
under follow by reflection along the angle bisector of with scale 
. Note that sends 
to 
and 
to 
. Then sends to 
such that 
. So, it also implies that the line 
is a reflection of the line 
along the angle bisector of , which implies . So, we're done.
be the isogonal conjugate of 
Consider the lines through parallel to 
These lines intersect 
at points 
with 
between and 
and they intersect 
at points 
with between and 
Since 
we have that 
is cyclic. Also, since 
we have 
is cyclic, and similarly 
is cyclic, so 
is cyclic, and by Reim's theorem we have 
parallel. Now, take a homothety at sending 
to 
It is not hard to see that this takes to 
so 
are collinear, and we have 
as desired. 
be the isogonal conjugate of in , the angle condition implies that is on the perpendicular bisector of 
. Use barycentric coordinates with respect to and let 
. Then 
and 
. We have
where the last equality holds because is on the perpendicular bisector of . Hence are collinear, so 
. Done.
and 
. Then 
is cyclic and is a parallelogram, so![\[\angle PXY = \angle BCP = \angle CBQ, \quad \angle PXY = \angle CBP = \angle BCQ.\]](http://latex.artofproblemsolving.com/c/4/8/c48139ea59c1b9b90c2540bb70d5bc11e2b8d5eb.png)
as well as 
, so the quadrilaterals 
and 
are also similar with opposite orientation, which implies the desired. 
be the reflection of across the midpoint of 
. It is clear that 
is a parallelogram. Then, notice that 
, and 
. Hence, is a parallelogram, meaning that 
.![\[\angle ABQ = \angle PBQ+\angle ABP =\angle PCQ+\angle PCA = \angle ACQ.\]](http://latex.artofproblemsolving.com/3/1/f/31f97b897eac7a55612a5c680e47def8dfe46b2c.png)
, so is cyclic. Thus,![\[\angle BAP = \angle ABR = \angle AQR = \angle CAQ. \ \square\]](http://latex.artofproblemsolving.com/1/6/a/16a8865ddddc47799abcab7efec03a458234fb5f.png)
be the point such that 
is a parallelogram. First note that 
since is a parallelogram as well. Additionally, this gives 
so 
is also a parallelogram. Then 
so 
is cyclic.
and 
so 
Finally, 
as desired.
, 
, and .
is a parallelogram since 
and 
. Similarly, 
is a parallelogram. Then we have 
so 
is cyclic. Then 
so we are done.
, 
, 
, 
. The Sine Law in 
and 
gives 
, i.e. 
. Analogously 
. However, 
and 
from the parallelogram , thus 
. Hence 
, equivalent to 
, i.e. 
, as desired.
from point .
, and 
, so we get that , 
, and are involutions.
, so and are both involutions around the angle bisector of . So it follows that and are isogonal as desired.
over the midpoint of be 
We see that 
is a parralelogram. We show 
is cyclic because then 
Let 
Then 
Then note that 
Therefore 
as desired
until is the orthocenter of , then is the antipode so it's obvious.
to be parralelogram and trivial by similarity of 
and .
, 
being isogonal. From here Isogonal Line Lemma (or DDIT if you wish) gives the required.
, then 
is cyclic and 
.
and its reflection 
the 
angle bisector. is sent to and to and as this transformation preserves similarity, it sends to , hence and are isogonal.
.
and 
Prove that
,
,
,
and comparing them we get that
.
and 
then the equation reduces to 
be its reflection over 
and 
are obviously parallel. 
are antiparallel wrt 
since 
form a parallelogram, so done.![[asy]
unitsize(1cm);
pair A, B, C, P, Q, X;
A=(2,3sqrt(5));
B=(0,0);
C=(8,0);
P=(2,4sqrt(5)/5);
Q=B+C-P;
X=A+P-B;
draw(A--B--C--A--P--B--Q--C--P--A--X--P--A--Q--C--X);
draw(circumcircle(A,C,P));
label("$A$", A, NW);
label("$B$", B, W);
label("$C$", C, SE);
label("$P$", P, dir(-105));
label("$Q$", Q, S);
label("$X$", X, N);
[/asy]](http://latex.artofproblemsolving.com/7/0/4/704b5f7f0625f0924ab5d52d0476cee1554d07cc.png)
be the reflection of 
and 
,
is a paralellogram. Therefore, 
,
is cyclic. Therefore, 
.
and 
to meet 
;
and that 
.
