AoPS Academy
is given. Let 
be set of all 
segments of real line of type ![$[i, j]$](http://latex.artofproblemsolving.com/a/6/4/a644edd67caa59eaded765c417b6d9822f782ec2.png)
, where 
and 
are integers, 
. A subset 
is said to be valuable if the intersection of any two segments from 
is either empty, or is a segment of nonzero length belonging to . Find the number of valuable subsets of set .
rows of coins. There is 1 coin in the first row, 2 coins in the second row, 3 coins in the third row, and so on, up to coins in the th row. Initially, all of the coins show heads (H). Carley plays a game in which, on each turn, she chooses three mutually adjacent coins and flips these three coins over. To win the game, all of the coins must be showing tails (T) after a sequence of turns. An example game with 4 rows of coins after a sequence of two turns is shown. for which it is possible to win the game starting with rows of coins.
satisfying 
, prove that![\[ \frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 1- ab-bc-ca \]](http://latex.artofproblemsolving.com/f/e/8/fe85cf52b883c214806ab837a2eb25cebb511d7b.png)
the incircle is tangent to 
at 
. Let 
be a point on 
different from and 
, and let 
and 
be incenters of 
and 
respectively. Suppose that the circumcircle of 
cuts 
again at 
. Prove that 
.
the incircle is tangent to at . Let be a point on different from and , and let and be incenters of and respectively. Suppose that the circumcircle of cuts again at . Prove that .
![[asy]
size(8cm);
defaultpen(fontsize(10pt));
pair A = dir(120), B = dir(210), C = dir(330), I = incenter(A,B,C), D = foot(I,A,B), P = B+dir(B--C)*abs(B-C)*0.65 , K = incenter(A,B,P), L = incenter(A,C,P), Q = A+dir(A--P)*abs(A-D), E = foot(I,B,C), M = (K+L)/2, N = (E+P)/2, R = foot(K,B,C), S = foot(L,B,C);
dot("$A$", A, dir(120));
dot("$B$", B, dir(210));
dot("$C$", C, dir(330));
dot("$D$", D, dir(150));
dot("$P$", P, dir(285));
dot("$K$", K, dir(180));
dot("$L$", L, dir(345));
dot("$Q$", Q, dir(60));
dot("$E$", E, dir(255));
dot("$M$", M, dir(90));
dot("$N$", N, dir(270));
dot("$R$", R, dir(270));
dot("$S$", S, dir(270));
draw(A--B--C--A--P--K--L--P);
draw(circumcircle(K,P,L));
draw(Q--K--R^^M--N^^L--S^^D--K--E, dotted);
draw(A--K--B, dashed);
[/asy]](http://latex.artofproblemsolving.com/1/6/4/164bf5f9dd0390bd7d0331f776d1b00fbe4d6317.png)
be the point where the incircle touches 
;
(the other case is analogous). Define 
,
as midpoints of 
,
respectively; 
,
as projections of 
.
,
.
,
.
,
by SAS. Now 
so 
by AAS and the conclusion follows. 
![\[\angle KPL = \angle KPA+\angle LPA = \frac{1}{2}\angle APB + \frac{1}{2}\angle APC = \pi/2,\]](http://latex.artofproblemsolving.com/f/a/5/fa5e3c7c14c71e608b3a286107701b217a331f6d.png)
so 
.
be the projection map onto line ![\[Q=2\phi(M)-P.\]](http://latex.artofproblemsolving.com/a/3/a/a3ac969421ce243f4f6c235409817de54de17668.png)
Since 
,![\[Q=\phi(K)+\phi(L)-P.\]](http://latex.artofproblemsolving.com/1/6/d/16d886faad0b8b262e6067e548a34b8e040ee40f.png)
We see that![\[|P\phi(K)|=\frac{AP+BP-AB}{2}\]](http://latex.artofproblemsolving.com/f/9/9/f99db20149d010976df4c886ab215535c897a8c0.png)
and![\[|P\phi(L)|=\frac{AP+CP-AC}{2},\]](http://latex.artofproblemsolving.com/0/a/b/0abac91393f66856087d4ad780e0c6732988ea3a.png)
so![\[PQ = |P\phi(K)| + |P\phi(L)| = AP-\frac{AB+BC-AB}{2} = AP-AD,\]](http://latex.artofproblemsolving.com/4/a/7/4a77fbdb00c6328273f8f0275d8a34d88c65572e.png)
so 
,