[/quote]
,
be the set of all positive integers. Determine all subsets 
of 
such that
and 
intersect at 
and 
.
;
,
and 
intersects 
on 
.
and 
meet at 
.
.
?
be a positive integer. A Japanese triangle consists of 
circles arranged in an equilateral triangular shape such that for each 
,
,
,
row contains exactly 
circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of 
,
such that in each Japanese triangle there is a ninja path containing at least 
,
and 
.
is prime, 
with 
and 
be a function such that for every regular 
-
we have 
.
for all reals 
.
)
are the altitudes of the triangle. The bisector of 
intersects 
at 
.
and 
.
passes through the midpoint of 
.
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
Y by tiendung2006, Mogmog8, ImSh95, Rama12
Let 
be a scalene triangle with incenter 
and let 
meet the circumcircle of triangle again at 
. The incircle 
of triangle is tangent to sides 
at 
, respectively. Let 
be the circumcenter of triangle 
and let 
be the intersection of and the altitude from 
to 
so that and lie on the same side with respect to 
. Denote by 
a circle centered at and passing through , and let 
meet again at 
.
Prove that the lines
and 
meet on the circumcircle of triangle 
.
be a scalene triangle with incenter 
and let 
meet the circumcircle of triangle again at 
. The incircle 
of triangle is tangent to sides 
at 
, respectively. Let 
be the circumcenter of triangle 
and let 
be the intersection of and the altitude from 
to 
so that and lie on the same side with respect to 
. Denote by 
a circle centered at and passing through , and let 
meet again at 
.Prove that the lines
and 
meet on the circumcircle of triangle 
.This post has been edited 1 time. Last edited by Olympiadium, Apr 24, 2022, 12:55 AM
![[asy]
size(8cm);
import geometry;
import olympiad;
pair A,B,C,I,D,E,M,O,L,N,X,F;
A=(6.2,7.2);
B=(-4,-3);
C=(9,-3);
I=incenter(A,B,C);
D=foot(I,A,B);
E=foot(I,A,C);
M=intersectionpoints(line(A,I),circle(A,B,C))[0];
O=circumcenter(B,D,E);
L=intersectionpoints(line(A,foot(A,B,C)),incircle(A,B,C))[0];
N=intersectionpoints(line(A,foot(A,B,C)),circle(O,length(segment(O,L))))[1];
X=extension(B,M,D,L);
F=intersectionpoints(line(B,M),circle(B,D,E))[0];
draw(circle(A,B,C));
draw(incircle(A,B,C));
draw(circle(B,D,E),darkgreen);
draw(circle(L,N,E),orange+dotted);
draw(circle(O,length(segment(O,L))),red);
draw(A--B--C--cycle);
draw(A--N);
draw(E--F);
draw(L--X);
draw(F--X);
draw(A--M);
draw(L--E);
draw(F--N);
draw(D--E);
dot(A^^B^^C^^I^^D^^E^^M^^O^^L^^N^^X^^F);
label("$A$",A,NE);
label("$B$",B,dir(235));
label("$C$",C,dir(330));
label("$I$",I,NW);
label("$D$",D,dir(280));
label("$E$",E,dir(210));
label("$M$",M,dir(250));
label("$O$",O,SE);
label("$L$",L,dir(70));
label("$N$",N,dir(270));
label("$X$",X,dir(200));
label("$F$",F,dir(265));
[/asy]](http://latex.artofproblemsolving.com/3/7/b/37bf37b398e461bc968e329b00297c17f04c24a1.png)
lies on 
the line through 
Notice 
so 
is the reflection of 
Hence, 
is a cyclic isosceles trapezoid. 
,
,
)
,
,
,
.
is equivalent to 
,
)
.
are concyclic as desired.
and 
and 
such that 
.
be the incenter of the triangle 
and assume that the angle bisector of 
intersects 
at 
and the altitude from 
the second intersection of line 
lies on the same circle, we only have to prove 
denote the intersection of line 
and the line through 
.
.
and 
.![\[\measuredangle BL'E = 90 + \measuredangle L'BC = 90 + \measuredangle MBC = \measuredangle 90 + \measuredangle IAC = \measuredangle ADE\]](http://latex.artofproblemsolving.com/9/a/0/9a0d7a622e1bff6761eaa0bf8b0dc43aea2017e0.png)
which implies that 
which implies that 
.
,
is an isosceles trapezoid, which proves the claim.![\[\measuredangle LPL' = \measuredangle ABM + \measuredangle LDA = \measuredangle ABC + \measuredangle CAM + \measuredangle LDA \]](http://latex.artofproblemsolving.com/2/c/3/2c3447a46c3a9f8ac29df0ad390c3dddf1a1d1ad.png)
and![\[\measuredangle L'EL = \measuredangle AEL + \measuredangle L'EC = 90 + \measuredangle BCA + \measuredangle AEL\]](http://latex.artofproblemsolving.com/d/b/6/db61100b0c31100f9baec51b21b3c8d1c2c2ad29.png)
Thus,![\[\measuredangle LPL' + \measuredangle L'EL = (\measuredangle ABC + \measuredangle CAM + \measuredangle LDA) + (90 + \measuredangle BCA + \measuredangle AEL) = \measuredangle BAC + \measuredangle CAI + \measuredangle DLE + \measuredangle CAB + 90 = 0 \]](http://latex.artofproblemsolving.com/b/f/e/bfefd367b7d95b276845d8b53ac1ed4e247a005f.png)
which implies that points 
