[/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 
.
?
and in 
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 
.
be on segment 
of an acute triangle 
such that 
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 
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
Let 
be a cyclic quadrilateral inscribed in a circle 
such that 
. Let 
be a point on the segment 
such that 
. The line 
intersect again at 
. The chords 
and 
meet at 
. Let 
be the symmetric point of 
about . Prove that 
and are parallel.
be a cyclic quadrilateral inscribed in a circle 
such that 
. Let 
be a point on the segment 
such that 
. The line 
intersect again at 
. The chords 
and 
meet at 
. Let 
be the symmetric point of 
about . Prove that 
and are parallel.This post has been edited 1 time. Last edited by Blastzit, Dec 5, 2020, 9:32 AM
are concyclic.
be the point on 
are concylic. Such point is unique.
,
.
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/6/2/0/6208efd20ab1d590b120d8c8b715bed13ab47834.png)
![\[\angle{ABD}=\angle{ACD}=\angle{MCE}=\angle{BFA}=\angle{MFE}\]](http://latex.artofproblemsolving.com/7/5/c/75c7abf7089926f59353eb3a6ee3fe0810221bbc.png)
![\[\frac{Sine(\angle{EFD})}{DE}=\frac{Sine(\angle{EDF})}{EF} \qquad (1)\]](http://latex.artofproblemsolving.com/0/2/8/028043b946c77173a7106aeb59353c739636cb94.png)
![\[\frac{Sine(\angle{BFC})}{BC}=\frac{Sine(\angle{CBF})}{CF} \qquad (2)\]](http://latex.artofproblemsolving.com/b/f/7/bf79a959d5763b58fe24c264607dfc16f2c79bc2.png)
dividing 
,
,
and using the "Ratio lemma" we get that 
bisects 
.
.
is the midpoint of 
lies on 
)
be a point on 
.
.
.
.
across the perpendicular bisector of 
are collinear.)
is cyclic.
Let 
.
.
and
Dividing last two, we have that
On the other hand,
hence 
and we are done.
be a point on 
.
.
and 
,
.
.
.
and 
.
aswell, so 
and the conclusion 
follows.
.
through point 
and 
.
is cyclic and so 
.
and 
and so 
.
and 
so 
.
so 
.
be the intersection of 
to be the point on 
.
is isosceles and an easy angle chase shows that![\[\angle EFB=\angle AFB=\angle ADB=90^\circ-\frac{\angle BAD}{2}=\frac{\angle DCB}{2}=\angle CD'B=\angle ED'B\]](http://latex.artofproblemsolving.com/e/a/f/eaf6b1d0c74a1651e09e2922ff0681281197f93c.png)
so that 
is cyclic (note that we did not use the definition of ![\[SC \cdot SD=SB \cdot SF=SE \cdot SD'\]](http://latex.artofproblemsolving.com/7/1/8/71875ee7ed0df553369644263a67e6c3eea7d2ae.png)
and using that 
,
as desired.
,
bisects 
,![\[\frac{NE}{DE}=\frac{NF}{DF}=\frac{NC}{CB}\]](http://latex.artofproblemsolving.com/c/f/6/cf626e253e8dc453a87a2c352017e41130bb81f4.png)
and as 
,
Law of sines on triangles 
and 
respetively gives
and
Dividing gives 
(so by arcs it holds that 
)
.
such that 
,
and 
in 
is equal to arc 
which gives 
cyclic and Reims finishes.
is tangent to 
which means that 
so our claim is proven.
are equal in 
is cyclic, also note that by Claim 2 we get 
which gives us that the arcs 
are equal in 
ISLsusceles trapezoid so arcs 
Isusceles so 
are tangent to 
and since arcs 
are equal we get 
isogonal w.r.t. 
so 
and by the parallels 
thus we are done 
