[/quote]
,
be the circumcircle of triangle 
.
intersects 
,
is the point on the segment 
such that 
.
meets 
.
intersects segments 
,
at 
,
.
and 
be two isogonal conjugate with respect to triangle 
.
and 
be two points lying on the circle 
such that 
and 
are perpendicular and parallel to bisector of 
and 
bisect two arcs 
containing 
and not containing 
.
and prove 
satisfying 
,
denotes the set of all real numbers.
with integer coefficients such that, for any positive integer 
has an integer root.
,
satisfying the equation 
.
Let 
and 
Show that![\[\sum_{k=1}^n\frac{1}{1+\sum_{j\neq k}x_j}\le\frac n{1+(n-1)\sqrt[n]{x_1x_2\cdots x_n}}.\]](http://latex.artofproblemsolving.com/e/4/c/e4ca07e5003fad8937af3c44dcbc9906fa48e5b4.png)
![\[ (abcde)^2 = a^2+b^2+c^2+d^2+e^2+f^2. \]](http://latex.artofproblemsolving.com/c/c/5/cc5979d6737e2009a4b61a5eb8061023e0ec59ea.png)
in integers.
,
,
.
.
: 
-is integer.
,![$P_{i}(x)\in \mathbb{R}[x], \forall i=\overline{1,n}.$](http://latex.artofproblemsolving.com/0/5/1/0514f85850830af9a9e1a4dd5e7a4e8e68b2dff9.png)
.
polynomial with only one root.
are real numbers such that 
prove inequality 
Y by
A sequence 
is called good, if 
are non-negative integers, and 
is either 
or 
for all 
.
Fix a positive integer
, and Ivan has a whiteboard with some ones written on it. In each step, he may erase any good sequence that appears on the whiteboard, and then he writes the number 
such that 
Suppose Ivan starts with the least possible number of ones to obtain 
after some steps, determine the minimum number of steps he will need in order to do so.
Proposed by Ivan Chan Kai Chin
is called good, if 
are non-negative integers, and 
is either 
or 
for all 
.Fix a positive integer
, and Ivan has a whiteboard with some ones written on it. In each step, he may erase any good sequence that appears on the whiteboard, and then he writes the number 
such that 
Suppose Ivan starts with the least possible number of ones to obtain 
after some steps, determine the minimum number of steps he will need in order to do so.Proposed by Ivan Chan Kai Chin
This post has been edited 2 times. Last edited by navi_09220114, Saturday at 4:14 PM
for 
and 
good sequence where 
.
denote the least number of ones obtainable from 
denote the least number of moves to obtain 
,
ones minimum. Immediately, we get the recursive formula![\[f(n)=\min(f(a_1)+f(a_2)+\cdots+f(a_m))\text{ over all good sequences $2^{n-1}<2^{a_1}+2^{a_2}+\cdots+2^{a_m}\le 2^n$}\]](http://latex.artofproblemsolving.com/2/3/4/234ea297eed1ed9f49351386fdb08a7986549069.png)
Henceforth, call a 
.
-
.
,
.
and 
so there exists 
such that 
by discrete IVT. Then![\[f(n)>f(a_1)+f(a_2)+\cdots+f(a_d)\ge f(n-1)\]](http://latex.artofproblemsolving.com/0/8/3/083552d962fa4ad32d1cfcc10dbe232583ddf5e4.png)
hence proven. 
for all 
![\[2^{n+1}\rightarrow 2^{n-1}, 2^n\]](http://latex.artofproblemsolving.com/c/2/d/c2d847341ecbf1f692b160629ce1a1becf268a5b.png)
implies 
.
is 
,
times if 
and at most 
times if 
.
,
exists, we take the last instance of it, so it is not followed by 
.![\[3f(k)>f(k)+f(k+1)\]](http://latex.artofproblemsolving.com/6/2/b/62b4fa7ea68b3e9bca6c0a1c177c246cf5ca219c.png)
we have reduced the value of 
exists, we can replace with 
.
.
.
for all 
.![\[\left(1+\frac12+\cdots+\frac1{2^{m-1}}\right) 2^{a_m}=2^{a_m+1}-2^{a_1}\]](http://latex.artofproblemsolving.com/d/f/6/df6d1ba63f1ec237653c849174448134f28271ad.png)
Since 
,
.
,
works so if any previous terms exist we can delete them, contradiction. Hence the 
.
.![\[2^{a_1}+\cdots+2^{a_{d-2}}+\left(1+\frac12+\cdots+\frac1{2^{m-d}}+\frac1{2^{m-d}}\right) 2^{a_m}=2^{a_1}+\cdots+2^{a_{d-2}}+2^{a_m+1}\]](http://latex.artofproblemsolving.com/c/a/4/ca414a4d16be69d8a18696f5700cd9a974a719c0.png)
So clearly 
.
,
,
.
,
works. Hence the sequence is 
.
,
into 
will reduce 
.
.
which are trivial. If 
,
.![\[f(k-1)+f(k)+f(k)+f(k+1)+\cdots+f(n-2)=f(k)+f(k+1)+f(k+1)+\cdots+f(n-2)=\cdots=f(n-3)+f(n-2)+f(n-2)=f(n-2)+f(n-1)\]](http://latex.artofproblemsolving.com/7/6/a/76aa3e7030dab3c958546781d1d412a828bb482c.png)
as well as 
.
.![\[g(n)=\min(g(a_1)+g(a_2)+\cdots+g(a_m))+1\text{ over all great sequences $2^{n-1}<2^{a_1}+2^{a_2}+\cdots+2^{a_m}\le 2^n$}\]](http://latex.artofproblemsolving.com/1/a/3/1a31e26b90c44615dbb8e7d8bc6653c6ee465875.png)
We will show that 
for all 
.
,
,
.![\[g(3)+g(4)+g(4)+g(5)+\cdots+g(n-2)+1=\cdots=g(n-3)+g(n-2)+g(n-2)+1=g(n-2)+g(n-1)+1\]](http://latex.artofproblemsolving.com/d/4/f/d4ff55ec9fda7f732e0dc02bac9b4bef08e93e71.png)
and we also have![\[g(3)+g(4)+g(4)+\cdots+g(n-2)+1=g(2)+g(3)+g(3)+\cdots+g(n-2)+\mathbf{2}\text{ (!!!)}=g(1)+g(2)+g(2)+\cdots+g(n-2)+1\]](http://latex.artofproblemsolving.com/0/0/6/006a64eca77a9001281d5f1ac9986174a2d95913.png)
![\[=g(0)+g(1)+g(1)+\cdots+g(n-2)+1=g(0)+g(0)+g(0)+g(1)+\cdots+g(n-2)+2\]](http://latex.artofproblemsolving.com/f/c/4/fc4b0ed88ac87eb263d99f69b9d18a22a4b28b0d.png)
hence taking either sequence 
or 
gives the result 
for 
and 
