be a natural number. At a mathematical olympiad training camp the same 
courses are organised every day. Each student takes exactly one of the 
from the set of positive integers to the set of positive integers such that, for all positive integers 
and 
,![\[ a, f(b) \text{ and } f(b + f(a) - 1).\]](http://latex.artofproblemsolving.com/4/a/b/4ab97da27e05d0f9e7762f74d0d0bd2dc8b9842f.png)
.
and 
be the incircle and circumcircle of the acute triangle 
,
so that all of its sides are tangent to 
,
are both on 
.
and 
,
and 
,
and 
intersects on 
,![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
larger than 
,
,
)![\[ \begin{cases} \displaystyle x_1+ \phantom{2^2} x_2+ \phantom{3^2} x_3 + \cdots + \phantom{n^2} x_n &= n+2, \\ x_1 + 2\phantom{^2}x_2 + 3\phantom{^2}x_3 + \cdots + n\phantom{^2}x_n &= 2n+2, \\ x_1 + 2^2x_2 + 3^2 x_3 + \cdots + n^2x_n &= n^2 + n +4, \\ x_1+ 2^3x_2 + 3^3x_3+ \cdots + n^3x_n &= n^3 + n + 8. \end{cases} \]](http://latex.artofproblemsolving.com/0/8/f/08fc55e0cc886c4c44f678e75eb07e4d814c4a77.png)
,
of the triangle, construct outwardly three squares with centres 
,
,
respectively. Let 
.
lies on 
.
be an integer, and let 
and 
be sequences of non-negative integers. For each integer 
,
and 
,
is the integer such that 
is divisible by 
.
equals the following arithmetic mean: ![\[a_i = \frac{1}{2b_i+1} \sum_{s=-b_i}^{b_i} a_{i+s}.\]](http://latex.artofproblemsolving.com/1/1/7/117ca4963c8e443fd3cbfb55a907bee4be295151.png)
Suppose that neither 
nor 
of the numbers 
are zero.
is given on the plane. Let the points 
,
and 
be on the sides 
and 
)
are taken such that 
is an equilateral hexagon. Suppose that the circumcenters of 
are coincident. Then determine the least possible value of:
Note: 
area of 
Y by Mathematicalprodigy37, mithu542
If you have the quadratic 
, call the roots 
and 
with 
. Then by Vieta's, 
and 
. If we can find 
, we can find 
. Note that 
and 
. Therefore, 
, and 
. Now all we have to do is solve the following system:
and 
. Solving for by elimination, we get 
, so 
, which is the quadratic formula. Substituting this into our first equation, we get 
, and solving for yields 
,
, call the roots 
and 
with 
. Then by Vieta's, 
and 
. If we can find 
, we can find 
. Note that 
and 
. Therefore, 
, and 
. Now all we have to do is solve the following system:
and 
. Solving for by elimination, we get 
, so 
, which is the quadratic formula. Substituting this into our first equation, we get 
, and solving for yields 
,This post has been edited 1 time. Last edited by Soupboy0, Mar 28, 2025, 3:33 PM
