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.
be positive real numbers.![$$\sqrt[3]{\left(\frac{a^{2022}-a}{b}+\frac{2021}{a^{\frac{a}{b}}}+1\right)\left(\frac{b^{2022}-b}{c}+\frac{2021}{b^{\frac{b}{c}}}+1\right)\left(\frac{c^{2022}-c}{a}+\frac{2021}{c^{\frac{c}{a}}}+1\right)}$$](http://latex.artofproblemsolving.com/c/4/9/c4923b16be8a8faf86422477a6ed6b4b60eb2552.png)
satisfying ![$$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$$](http://latex.artofproblemsolving.com/f/6/e/f6ed10f7fff1cc94edd8f451e75718a0916a8bfa.png)
holds for all 
+1 w
black and 
red cards. You pull out 
cards, one after another, and check their colour. If both cards are the same colour, then a black card is added to the deck. However, if the cards are of different colours, then a red card is used to replace them. Once the cards are taken out of the deck, they are not returned to the deck, and thus the number of cards keeps reducing. What is the likelihood the last card left in the deck is black?
be positive real numbers such that 
Prove that we have![\[(a+b)(b+c)(c+a)\geq 8.\]](http://latex.artofproblemsolving.com/2/a/6/2a63f19de634fce70ff4ebc02ad13d442a20c378.png)
such that the equality 
holds for all pairs of real numbers 
.
such that 
and 
.
be a given positive real and 
be the set of all positive reals. Find all functions 
such that ![\[f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.\]](http://latex.artofproblemsolving.com/b/0/6/b069e7b7ec1e277f4a4ce85a99434fdb54eb02f3.png)
of the vertices of the regular polygon 
we write the number 
and in the remaining ones we write the number 
Let 
be the number written on the vertex 
A vertex is good if ![\[x_i+x_{i+1}+\cdots+x_j>0\quad\text{and}\quad x_i+x_{i-1}+\cdots+x_k>0,\]](http://latex.artofproblemsolving.com/b/f/4/bf4cf984af7f310c1eb90b6b37126bfca9181305.png)
for any integers 
and 
such that 
Note that the indices are taken modulo 
Determine the greatest possible value of such that, regardless of numbering, there always exists a good vertex.
be a scalene triangle, 
be the median through 
, and 
be the incircle. Let touch 
at point 
and segment 
meet for the second time at point 
. Let 
be the triangle formed by lines and and the tangent to at . Prove that the incircle of triangle is tangent to the circumcircle of triangle .
in which 
is parallel to 
is inscribed in a circle of radius 
and centre 
The non-parallel sides 
and 
are extended to meet at 
while diagonals 
and 
intersect at 
Prove that 
.
. Let 
, so we want to show that
This would follow immediately if 
.
, 
and are collinear.
is an isosceles trapezium, as it's cyclic (well known) then 
and lie on the same line by symmetry.
intersect the circle at 
. Then
As 
is a diameter (passes through ), we have 
by Thales theorem.
, let 
be the intersection of and 
and 
be the intersection of and 
. We have 
as shared 
and right angle, thus 
. due to 
and shared angle 
, which implies the result.
be the point of infinity along . It's well known that the inverse of with respect to 
is the foot of the perpendicular from to 
. By symmetry, 
, implying 
meaning must be the inverse of . This immediately gives the conclusion.
and 
Prove that
is an obvious solution. Now, let 
.
or 
.
.
,
.
we have 
,
.
and bisector of 
,
gives away that there's a clean inversion soln.... If only I knew how to do that
and 
are tangent at 
,
,
.