[/quote]
,
that satisfies![\[ f(x+yf(x))f(y)=f(x)(1+yf(y+1)) \]](http://latex.artofproblemsolving.com/f/7/a/f7a138fd2dc76f034525491a10f62f1e83de294d.png)
for all 
.
that satisfy the equation
This is really difficult.
,
satisfy that for any indices 
,
,
.![\[ \frac{a_1 a_2 \cdots a_n}{(a_1 + a_2 + \cdots + a_n)^n}. \]](http://latex.artofproblemsolving.com/8/8/1/881d5857b39895608db56e8f41264d79cff7850c.png)
Proposed by Wu Zhuo and Zhao Tingwei
such that![\[ 1^k + 2^k + 3^k + 4^k + 5^k + 6^k + 7^k + 8^k + 9^k = 3^m. \]](http://latex.artofproblemsolving.com/0/0/7/007105d7a32074b85e0305e91d9a97611f0ebd44.png)
Proposed by Wu Zhuo
Prove that
Where 
denote the set of positive integers. A function 
is said to be bonza if![\[
f(a)~~\text{divides}~~b^a-f(b)^{f(a)}
\]](http://latex.artofproblemsolving.com/7/b/b/7bb6b2b4fcde900ee6f7ef90b906525d61312796.png)
for all positive integers 
and 
.
such that 
for all bonza functions 
and all positive integers 
.
be an acute triangle with 
,
be the circumcircle of 
and 
lie on 
and 
meet on the external angle bisector of 
.
and 
,
and 
intersect 
and 
,
be the centre of the excircle of triangle 
opposite the vertex 
bisects 
,
be the circumcenter and orthocenter of 
respectively, and let 
be the reflection of 
.
lies on 
through 
at 
.
meets 
at 
.
,
Prove that:
When does the equality hold ?
be real numbers such that 
Prove that![\[2(ab+ bc + ca + da + db + dc) + 4\left(\frac{a^2 + b^2 + c^2 + d^2}{a + b + c + d}\right)^2 \geqslant 3(a^2 + b^2 + c^2 + d^2) .\]](http://latex.artofproblemsolving.com/3/4/c/34cbee97b986221240deaff2573572c9d5a570ab.png)
![\[2k^2(ab+ bc + ca + da + db + dc) + \left(\frac{a^2 + b^2 + c^2 + d^2}{a + b + c + d}\right)^2 \geqslant k(2-k)(a^2 + b^2 + c^2 + d^2) .\]](http://latex.artofproblemsolving.com/b/f/6/bf6e8960d20cfbb230ba8f783bb1300e293c17f8.png)
such that the following inequality holds:![\[
x^k y^k z^k (x^3 + y^3 + z^3) \leq 3
\]](http://latex.artofproblemsolving.com/2/3/0/230cc1b4378093eed4a28308584cc716337cd395.png)
for all positive real numbers 
satisfying the condition 
.
Y by amar_04, magicarrow, Purple_Planet, CaptainLevi16, Funcshun840
Let 
be a scalene triangle with circumcenter 
, incenter 
, and incircle 
. Let touch the sides 
, 
, and 
at points 
, 
, and 
respectively. Let 
be the projection of to 
. The line 
intersects the circumcircle of again at point 
. The circumcircles of 
and 
intersect again at points 
and 
respectively. Prove that the lines 
, 
, and 
are concurrent.
Proposed by MarkBcc168.
be a scalene triangle with circumcenter 
, incenter 
, and incircle 
. Let touch the sides 
, 
, and 
at points 
, 
, and 
respectively. Let 
be the projection of to 
. The line 
intersects the circumcircle of again at point 
. The circumcircles of 
and 
intersect again at points 
and 
respectively. Prove that the lines 
, 
, and 
are concurrent.Proposed by MarkBcc168.
This post has been edited 1 time. Last edited by MarkBcc168, Apr 28, 2020, 7:08 AM
and 
be the radical center of 
.
.
also lies on the polar of 
where 
so it is enough to prove 
.
is the radical axis 
of 
and 
so 
.
.
be the second intersection of 
.
are collinear whence ![\[-1 = (A, I; E, F) \stackrel{R}{=} (RA \cap EF, T; E, F) \]](http://latex.artofproblemsolving.com/d/b/e/dbec3949fc311f496a92113889659ff6f4b721a2.png)
so 
are collinear. Hence, ![\[\text{Pow}(M, \odot(ABC)) = MQ \cdot MA = ME \cdot MF = \text{Pow}(M, \omega) \]](http://latex.artofproblemsolving.com/b/d/0/bd00439b7b87a4c616a9457d7d7227d53da99172.png)
so 
also and we are done.
be the radical axis of 
,
and 
meet on 
,
concur on 
meet on 
.
be the point on 
such that 
.
,
lies on line 
,
.
,
.
.
is equivalent to proving that 
.
,
,
and the nine point circle of 
are coaxial. In other words, 
,
where we use 
and 
.
This completes the proof of the Lemma. 
be the foot of perpendicular from 
to 
respectively. Since 
passes through 
is cyclic, and hence, by Power of Point, we get 
.
,
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we get that 
concurs at a point 
.
.
WRT 
and as 
WRT 
WRT 
is the Polar of 
WRT 
are concurrent at 
.
is the Nine Point Circle of 
and 
Midpoint of 
around 
.
.
then 
and as 
is a cyclic quadrilateral and 
.
bisects 
.
.
WRT 
are concurrent at 
.
has equal Power WRT 
and also 
has equal Power WRT 
is the Radical Axis of 
.
.
are concurrent. 
be the midpoint of arc 
,
its antipode. Let 
be the antipode of 
with 
are collinear and that 
are collinear. Let 
be the 
satisfies 
by radical axis on 
.
is a well-known mixtilinear fact. Now define 
,
.
.
where 
,
is the pole of 
)
is tangent to the incircle and 
it follows that 
wrt incircle. Moreover, 
are collinear because 
and 
are isogonal where 
is the reflection of 
is cyclic with diameter 
.
is cyclic. Moreover
is tangent to 
,
cyclic 
,
are collinear.
and the center of the image of 
are collinear, hence 
be the image of 
of 
;
as well since 
are the intersections of 
.
concur where 
is the midpoint of 
is cyclic since 
where 
,
are collinear.
,
,
the second intersection of 
and similarly define 
,
lies on the Euler line of 
and 
(See 
lies on 
(
where 
is the Bevan Point of 
bisects minor arc 
,
bisects major arc 
at 
intersect 
intersect 
,
,
,
are cyclic.
,
.
,
bisects 
,
,
and 
are inversly similar,
are colinear.
and 
(Claim 1)
bisects 
bisects 
are colinear.
are colinear, we are done.
are colinear.
bisects 
also bisects 
are colinear, we are done.
are colinear.
,
,
,
and 
concurrent at 
,
,
,
concurrent at the orthocenter of 
,
and 
bisects 
,
,
intersect 
,
intersect 
.
are con-cyclic.
,
,
.
concurrent.
.
be the intersection between 
are cyclic, and 
tangent 
are cyclic,
,
are concurrent.
are concurrent.
are colinear.
.
,
is tangent to 
is tangent to 
colinear 
is the polar of 
=
is tangent to 
is tangent to 
is the polar of 
.
,
.
,
are concurrent.
.
are projective. Let 
and 
be the images of these points under inversion about 
,
is projective from 
to 
and 
are clearly projective, so 
be the second intersection point of 
with 
that maps 
.
is projective, it suffices to prove that two projective maps depending on 
.
and 
,
,
is tangent to 
,
concur. Let this intersection point be 
;
,
,
.
,
,
.
.
form an orthocentric system, so 
as well. This means that 
the points 
are collinear. Moreover, Pascal on 
yields that 
are collinear. Let 
,
,
,
is the radical axis of 
.
,
,
is collinear, as desired.
is well known.
, 
, 
are collinear.
and 
so 
radical center of 
are collinear.
are collinear.
and 
is cyclic.
are on radax of 
and 
are collinear.
, we get 
is the radax so 
.
is tangent to the incircle as that implies that 
inversion.
is then 
.
,
,
collinear.
is mapped to 
.
the 
is tangent to 
,
,
(mixtlinear excircle properties) and 
.
gives that the tangents at 
and 
,
is tangent, and so after undoing the inversion we are done.
be the excenters of 
are concurrent.
,
and 
which implies that 
passes through the intersection of 
,
,
which means that 
is tangent to 
.
, the point 
.
.
as well, so 
and ![[asy]
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are homothetic with center 
points 
and 
line on one line.
and 
is also on the polar of 
and 
,![[asy]
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[/asy]](http://latex.artofproblemsolving.com/1/9/f/19fcf5102ef781ac92c5750c3c9eec5d01c40d01.png)
and let 
s
are concurrent, we see that 
.
are collinear
and 
,
.
are collinear. Since 
and 
by Brocard, we see that 
is an orthogonal system which implies 
thus, 
are collinear. We have 
.
.
which yields the collinearity of 
and 
the 
e
e
and 
are cyclic.![\[\measuredangle TXP = \measuredangle AXP = \measuredangle CEP = \measuredangle EFP = \measuredangle TFP \]](http://latex.artofproblemsolving.com/4/1/4/414f9d6c5eaae6e45faeebcff351fbb33440efef.png)
which implies that 
and 
it follows that lines 
, 
and 
concur at a point. We wish to show that this point also lies on 
and 
are cyclic.
.
b
.
and 
it follows that 
lies on 
which implies that 
.
to 
.
(well known) and 
)
and 
intersect for the second time at the second intersection of 
where 
.![\[\measuredangle AI_cP = \measuredangle CEP = \measuredangle EFP\]](http://latex.artofproblemsolving.com/6/9/f/69f3cb7d73e59a13f02cbdc6d36e60efc5794d43.png)
which since 
implies that points 
which is well known to lie on 
which finishes the proof of the claim.
is tangent to 
is tangent to 
is tangent to 
and 
where 
and 
are the 
e
,
and 
,
and 
are coaxial.
to 
be the complete quadrilateral of 
are collinear.
implies 
are collinear. Radax on 
implies 
are collinear.
is radax of 
Clearly note that 
Thus, 
However, 
being inverses and 
Thus, 
are collinear. Hence, 
which implies 
finishing our result.
However, by our claim, we know 
which implies that 
are collinear. This implies the result.
Rename 
.
be 
,
,
.
is self polar so 
so 
is radical axis of 
and 
so we are done.
and so 
collinear so pole of 
,
collinear.
is cyclic.
and 
cyclic and 
tangent to incircle. 
which proves the claim.
are collinear
which implies the collinearity.
concur
.
where the last equality follows from 
.
intersect 
collinear because then Pascal on 
would finish. So it suffices to show 
.
intersect at 
is on 
,
collinear. Apply incircle inversion, then since 
collinear and 
we see 
are inverses. Also well-known 
passes through midpoint 
are collinear. Then by radaxes we have that 
.
so 
is cyclic. We know 
,
and finally 
which finished the problem.
be the excentral triangle of 
and 
onto 
,
collinear.
and 
.
the image of 
be the line through 
and 
.
and 
and 
,
lies on 
,
collinear. Similarly 
collinear. Also by homothety 
and 
concyclic, by Reim's theorem, 
is concyclic. Thus by pop, 
.
collinear, so 
.
and 
so we are done. 
