Maximum number of nice subsets
by FireBreathers, Apr 23, 2025, 10:27 PM
Given a set 
of natural numbers with 
elements with odd number. A nonempty subset 
of is called 
if the product of the elements of divisible by the sum of the elements of , but not by its square. It is known that the set itself is good. Determine the maximum number of subsets (including itself).
of natural numbers with 
elements with odd number. A nonempty subset 
of is called 
if the product of the elements of divisible by the sum of the elements of , but not by its square. It is known that the set itself is good. Determine the maximum number of subsets (including itself).
be an integer greater than zero. Then, the value of the sum of the reciprocals of the cubes of the roots of the equation![\[
mx^4 + 8x^3 - 139x^2 - 18x + 9 = 0
\]](http://latex.artofproblemsolving.com/a/3/a/a3a9bca412e927632e4921c51f2a6593409031d9.png)
is equal to:
,![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
be triangle, inscribed in parabola. Tangents in points 
forms triangle 
.
.
)
for which there exists an even natural number 
such that the number![\[
(a - 1)(a^2 - 1)\cdots(a^n - 1)
\]](http://latex.artofproblemsolving.com/9/c/a/9caf4eeb82ff46b5ba55ab4b6bc28f0cace586ec.png)
is a perfect square.
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
is the square of an integer![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
.
be its cirumcircle, 
its orthocenter, and 
the foot of the altitude from 
.
.
be the point on 
such that 
and let 
be the point on 
.
,
,
and 
are tangent to each other.![[asy]
/* IMO 2015 Problem 3, free script by liberator, 14 July 2015 */
import olympiad;
import cse5;
unitsize(3.2cm);
pen ddd = rgb(0.4,0.6,0.8);
pointpen=black;
defaultpen(fontsize(10pt));
pair A = dir(75);
pair B = dir(205);
pair C = dir(-25);
pair O = origin;
pair H = orthocenter(A,B,C);
pair F = foot(A,B,C);
pair D = IP(L(H,F,0,2015),unitcircle);
pair M = midpoint(B--C);
pair Ap = B+C-H;
pair Q = IP(L(M,H,0,2015), unitcircle);
pair Qp = A+Ap-Q;
pair K = IP(L(Qp,H,-0.1,2015), unitcircle);
pair P = circumcenter(D,H,K);
D(unitcircle, heavygreen);
D(circumcircle(K,Q,H), red+linewidth(1));
D(circumcircle(F,K,M), heavymagenta);
D(circumcircle(D,H,K), grey+linetype("0 2")+linewidth(1));
D(A--B--C--cycle, ddd+linewidth(1));
D(A--D, ddd);
D(A--Qp, ddd);
D(Q--Ap, ddd+dashed);
D(K--Qp, ddd+dashed);
D(H--P--K, ddd);
D(rightanglemark(C,F,A,2), ddd);
D("A",A,dir(A)*2);
D("B",B,dir(B));
D("C",C,dir(C));
D("D",D,dir(-110));
D("F",F,dir(F));
D("H",H,dir(-65)*2);
D("K",K,dir(-15)*1.6);
D("M",M,dir(-70));
D("P",P,dir(P));
D("Q",Q,dir(Q));
D("A'",Ap,dir(Ap));
D("Q'",Qp,dir(Qp));
D("\Gamma",O,dir(160)*28);
[/asy]](http://latex.artofproblemsolving.com/a/a/6/aa6ae4ba390a833b56195faffc62aaf837297443.png)
be the antipodes of 
respectively w.r.t 
be the reflection of 
are collinear. Also observe that 
are collinear, since 
be the circumcenter of 
,
:
.
to 
at 
;
from rectangle 
,
,
,
is tangent to 
.
,
is tangent to 
:
,
at 
respectively. Denote 
and 
,
as the intersection of the lines 
and 
.
is the excenter of 
,
and 
are coaxal.
denote the second intersection of circles 
and 
.
.
is cyclic, from perpendicularity of tangents.
is tangent to 
.
and 
,
.
and 
,
.
at 
,
.
,
and 
.
be the points on 
and 
,
and 
.
and 
is on the circumference of triangle 
October 2018
April 2018