[/quote]
,
board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 
,
to be the largest value in row 
,
.
to be the largest value in column 
.
?
steps and has to reach 
is valid but 
is not valid.
,
,
,
and 
according to the diagram. The edges from points 
,![\[[ABCD]^2+[ABFE]^2+[ADHE]^2=[BCGF]^2+[CDHG]^2+[EFGH]^2,\]](http://latex.artofproblemsolving.com/c/f/b/cfb6dc6726af4711bcfe7827b6531471a75270fb.png)
where ![$[XYZW]$](http://latex.artofproblemsolving.com/9/c/2/9c25ef1b5cfb129136334ed811fb699a84f48a13.png)
denotes the area of quadrilateral 
.
of an acute triangle 
and 
.
and 
be the points on 
and 
,
and 
.
and 
is on the circumference of triangle 
be a function satisfying![\[
f(f(x)) = x^2 - 6x + 6
\quad \text{for all} \quad x \in \mathbb{Z}.
\]](http://latex.artofproblemsolving.com/8/a/e/8aebed1e260316d675ca7d104d83fabcd8b5717e.png)
Given that![\[
f(i) < f(i+1) \quad \text{for} \quad i = 0, 1, 2, 3, 4, 5,
\]](http://latex.artofproblemsolving.com/9/f/9/9f92574f9fe14e2e30c3e51e194c661fbd4ece07.png)
find the value of![\[
f(0) + f(1) + f(2) + \cdots + f(6).
\]](http://latex.artofproblemsolving.com/4/0/7/4075d7efa78dda4c46a5cf764fb8852e8c0fb09b.png)
such that for all 
the followings hold:
lie in the interiors of the sides 
respectively, such that 
is perpendicular to 
and 
is perpendicular to 
.
lies on the line 
and 
on that line.)
be the circumcentre of triangle 
.
and 
?
of positive integers, such that every sum
(with 
)
for all 
;
for all 
?
denotes the greatest common divisor of 
,
.
such that ![\[ f(x^2 + yf(z)) = xf(x) + zf(y) , \]](http://latex.artofproblemsolving.com/2/b/a/2baee1d8464992fc38a16561b48b18f4b35807bd.png)
for all 
.
,
,
be an infinite sequence of positive integers. Suppose that there is an integer 
such that, for each 
,
is an integer. Prove that there is a positive integer 
for all 
.
,
,
be positive real numbers such that 
.![$$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1.$$](http://latex.artofproblemsolving.com/2/0/7/207934bca53047858a3820f37705171ff0abfd45.png)
and 
and 
Prove that![\[\frac{a}{(b+c-a)^2} + \frac{b}{(c+a-b)^2} + \frac{c}{(a+b-c)^2} \geq \frac{3}{(abc)^2}.\]](http://latex.artofproblemsolving.com/0/a/b/0ab4b4e6c28a1ae67788525642661dbcd04d0bd8.png)
,
.
and 
,
is acute, and let 
be its orthocentre. Points 
and 
and 
and 
Y by Adventure10, ItsBesi
Let 
be an acute-angled triangle with circumcenter 
. Let 
be a circle with center on the altitude from 
in , passing through vertex and points 
and 
on sides 
and 
. Assume that
![\[BP\cdot CQ = AP\cdot AQ.\]](//latex.artofproblemsolving.com/1/c/2/1c22443c125d448bf3f48d2729ae4f2eb375a405.png)
Prove that is tangent to the circumcircle of triangle 
.
be an acute-angled triangle with circumcenter 
. Let 
be a circle with center on the altitude from 
in , passing through vertex and points 
and 
on sides 
and 
. Assume that![\[BP\cdot CQ = AP\cdot AQ.\]](http://latex.artofproblemsolving.com/1/c/2/1c22443c125d448bf3f48d2729ae4f2eb375a405.png)
Prove that is tangent to the circumcircle of triangle 
.This post has been edited 4 times. Last edited by djmathman, Apr 19, 2017, 3:55 AM
,
to 
,
is cyclic.
.
,
is the angle bisector of 
.
,
.
.
,
is tangent to 
and 
It's known that 
are isogonals WRT 
thus the inversion 
followed by symmetry WRT the angle bisector of 
swaps 
and 
Thus 
of 
on a same side 
So denoting 
and 
the points at infinity of 
by involution properties, we get then 
and 
But 
lies on the A-symmedian of 
and 
. From the condition (the center of 
lie on A-altitude) we know 
is anti-parallel to 
WRT 
. From 
we get there exist a point 
on 
and 
. Notice that 
pass through the midpoint of 
, hence we get the construction of 
such that 
.
and 
on A-symmedian of 
is an parallelogram 
and 
is cyclical, If 
is Miquel point of 
(Points 
)
and 
Then 
and 
and 
is a parallelogram. Suppose that 
.
and 
are tangents
are 
)
and how the last equality ( 
helps us to determinate that 
is the point of tangency?
is the circumcenter of 
i.e. 
and 
are tangents (If we drawing a line tangent to 
Note that 
Let 
be the 
be the centres of 
and 
are collinear and 
we have 
so 
are collinear, establishing the tangency.
for convenience
Since triangles 
and 
are similar, by the above ratio, we know that 
and 
and 
are similar
,
are cyclic
are cyclic as well (proof will be put later at the bottom)
be the line tangent to circle 
as 
which is equivalent to proving that 
or
Also, let 
lies on 
Also, note that
Thus,
Thus, the two circles are tangent at point 
as well
Hence, 
wrt to 
meet at 
.
meet 
again at 
It follows that![\[\angle APY = \angle YQC = 180 - \angle AQY\]](http://latex.artofproblemsolving.com/3/a/2/3a259cc941286257d1d5bcdf32be63cf0317e8e9.png)
thus 
are concyclic.
We have![\[\angle XYA = \angle XAY = \angle KTY = \angle KYT\]](http://latex.artofproblemsolving.com/8/0/e/80ee0cfa71735103e3f332b2a879fd7690033dd8.png)
which implies that 
are collinear, and the desired result follows.
be the midpoints of 
.
.
and thus 
.
is cyclic with diameter 
,
.![\[\frac{KM}{KN} = \frac{\delta(K, AB)}{\delta(K, AC)} = \frac{AB}{AC} = \frac{AB(\tfrac12 - t)}{AC(\tfrac12 - t)} = \frac{PM}{QN}\]](http://latex.artofproblemsolving.com/9/7/0/970be5b348c007575039e81e54798b185287b5e5.png)
where 
is 
by the length condition, as desired.
be the center of 
be the ![\[\angle O_1KA = \angle O_1AK = \angle BB'K = \angle B'KO_2\]](http://latex.artofproblemsolving.com/5/9/4/59482ee627807a669a394aa74254f5c4c4b2806f.png)
where 
are collinear so 
tangent at 
and 
due to ratios. This finishes.
to 
,
is a symmedian in 
,
tangent to 
(as well as the intersection of 
tangent to 
is cyclic. I now claim that 
(where 
in general, i.e. 
.
be (distinct) parallel lines. Let 
,
.
be tangent to 
at 
be tangent to 
at 
are tangent at 
be the centers of 
,
.
which implies the desired result. 
be the center of 
be the foot of perpendicular from 
simplifying which 
we find 
which is the sum of the 2 radii
are points on 
with 
and 
then 
is tangent to the reflection of 
the reflection of 
then we see 
so it suffices to show that 
collinear. To do this, consider the midpoints 
of 
It suffices to show that 
collinear.
so 
This rearranges to 
or 
or 
Thus from 
we finish by Menelaus.
the A-dumpty point, note that 
,
,
,
,
is symedian thus 
,
,
,
and 
share the same 90 degree angle we have that 
are colinear thus 
a parallelogram. (To see this, just note that for this point 
and 
are similar.)
,
,
.
and note that 
,
and 
are similar, so 
,
,
,
.
so the tangent to 
such that 
is a parallelogram:
gives us 
,![\[\angle QRC + \angle PRQ + \angle BRP = \angle B + \angle A + \angle C = 180.\]](http://latex.artofproblemsolving.com/7/d/2/7d25dc9fadb2cabd76d4221d32461c00ccd0f7d9.png)
.
,![\[\angle BMC = \angle BPR + \angle CQR = 2 \angle A = \angle BOC.\]](http://latex.artofproblemsolving.com/c/1/e/c1eaaa62a98b187930ff9754ddbda16218e24eaf.png)
,![\[\angle AMP = \angle AQP = \angle B = \angle RMP.\]](http://latex.artofproblemsolving.com/9/3/a/93a71a00c8515c3eb44f07fa80896eef30b28129.png)
be the circumcircle of 
,
be the circumcircle of 
.
.
,
,
,
.
.
bisects 
,
.
Similarly,
so that 
![\[\angle PXA = 180^{\circ} - \angle PAX - \angle XPA, \text{ and } \angle QXA = 180^{\circ} - \angle XQA - \angle QAX = \angle XQC - \angle QAX.\]](http://latex.artofproblemsolving.com/0/9/6/096eb85cf83e4612339a2e664fc346b3a4480fc2.png)
Hence,
as desired. 
,
are parallel, and further as 
are by definition collinear, then
so that 
are collinear as 
and 
which implies our claim.
is a parallelogram. Let 
.
be the point on 
.
,
which gives us that
Thus, we require 
and thus 
is indeed a parallelogram.
which means that 
and thus 
So clearly 
.
such that 
which in turn gives
Therefore, 
be a point on 
is the diameter of 
Points 
and 
and 
and 
are concyclic. Let 
Points 
be t the 
are similar 
and 
we get:
, combining this with 
we get that triangles 
are similar 
we get that:
Points 
and 
Points 
and 
is tangent to 
by miquel's theorem, 
![\[ \angle BXC = \angle BXR + \angle CXR = \angle BPR + \angle CQR = 2A \]](http://latex.artofproblemsolving.com/0/f/8/0f86dd398f74c6db1acc0f5b4d80528edc8be952.png)
Also, 
;
are concyclic.![\[ \angle AXP = \angle AXQ = B = 180 - \angle RXP \]](http://latex.artofproblemsolving.com/6/b/a/6bafc24e3dd87f72955709cb790ba8f6c56ab33e.png)
Hence 
bisects 
,
is 
,
.
,
WRT 
.
,
and 
.
inversion sends 
,
is a parallelogramm. Then, it's clear that 
is cyclic. Let 
and 
be the image of 
.
,
is tangent to 
is tangent to 
,
the centers of circles 
.
and 
.
,
are concyclic, so, by PoP we have 
.
,
.![$[AK]$](http://latex.artofproblemsolving.com/3/2/a/32a180e681de5ca6aeffe6bfa17cf23f043b9361.png)
and the 
are collinear.
:
.
.
.
.
.
.
.
.
.
,
,
,
.
,
and 
,
and the center of this circle is the midpoint of 
,
.
.
.
.
.
.
,
as 
bisector and we get that 
and so 
are collinear since 
