[/quote]
,
.
?
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 
.
be a point such that 
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)
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
and 
.
Equality holds when 
or 
and 
,
cookies from one of the piles, of which he eats 
for which Marica has a winning strategy.
Y by Adventure10, Mango247
Let 
be a cyclic quadrangle, let the diagonals 
and 
cross at 
, and let 
and 
be the incentres of the triangles 
and 
, respectively. The line 
crosses the segments 
and 
at 
and 
, respectively. Prove that the triangle 
is isosceles.
be a cyclic quadrangle, let the diagonals 
and 
cross at 
, and let 
and 
be the incentres of the triangles 
and 
, respectively. The line 
crosses the segments 
and 
at 
and 
, respectively. Prove that the triangle 
is isosceles.This post has been edited 1 time. Last edited by parmenides51, Sep 8, 2018, 5:07 PM
Reason: name edit
Reason: name edit
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(23.cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.52, xmax = 18.48, ymin = -4.84, ymax = 6.3; /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0.);
draw((5.24,2.8)--(10.84,2.82)--(11.08,-0.78)--cycle, linewidth(2.) + zzttqq);
draw((5.24,2.8)--(5.111826910108407,-0.9522349243668591)--(11.08,-0.78)--(10.84,2.82)--cycle, linewidth(2.) + zzttqq);
/* draw figures */
draw((5.24,2.8)--(10.84,2.82), linewidth(2.) + zzttqq);
draw((10.84,2.82)--(11.08,-0.78), linewidth(2.) + zzttqq);
draw((11.08,-0.78)--(5.24,2.8), linewidth(2.) + zzttqq);
draw(circle((8.047086407998094,0.8258057605332061), 3.4317891830809644), linewidth(2.));
draw((5.24,2.8)--(5.111826910108407,-0.9522349243668591), linewidth(2.) + zzttqq);
draw((5.111826910108407,-0.9522349243668591)--(11.08,-0.78), linewidth(2.) + zzttqq);
draw((11.08,-0.78)--(10.84,2.82), linewidth(2.) + zzttqq);
draw((5.111826910108407,-0.9522349243668591)--(10.84,2.82), linewidth(2.));
draw((6.53,1.53)--(9.72,1.52), linewidth(2.));
draw((6.53,1.53)--(5.111826910108407,-0.9522349243668591), linewidth(2.));
draw((9.72,1.52)--(11.08,-0.78), linewidth(2.));
draw((9.72,1.52)--(8.159318418787835,4.2557592548047545), linewidth(2.));
draw((6.53,1.53)--(8.159318418787835,4.2557592548047545), linewidth(2.));
/* dots and labels */
dot((5.24,2.8),dotstyle);
label("$A$", (5.32,3.), NE * labelscalefactor);
dot((10.84,2.82),dotstyle);
label("$B$", (10.92,3.02), NE * labelscalefactor);
dot((11.08,-0.78),dotstyle);
label("$C$", (11.16,-0.58), NE * labelscalefactor);
dot((5.111826910108407,-0.9522349243668591),dotstyle);
label("$D$", (5.2,-0.76), NE * labelscalefactor);
dot((8.124522937395426,1.031747925363762),linewidth(4.pt) + dotstyle);
label("$O$", (8.2,1.2), NE * labelscalefactor);
dot((9.72,1.52),dotstyle);
label("$I$", (9.98,1.58), NE * labelscalefactor);
dot((6.53,1.53),dotstyle);
label("$J$", (6.28,1.66), NE * labelscalefactor);
dot((8.159318418787835,4.2557592548047545),dotstyle);
label("$E$", (8.24,4.46), NE * labelscalefactor);
dot((7.315749969195022,1.5275368339523665),linewidth(4.pt) + dotstyle);
label("$M$", (7.4,1.68), NE * labelscalefactor);
dot((8.86998441801185,1.5226646256488652),linewidth(4.pt) + dotstyle);
label("$N$", (8.94,1.68), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/1/c/1/1c1c8f158a8a619707189eea254d64ad650da0d4.png)
be the intersection of sides 
and 
.
is cyclic.
.
. Now from simple angle chasing we get 
,
be the incenters of triangles 
;
be the arc midpoints of 
and 
.
is a kite, hence 
.
.
is cyclic and some angle chasing, 
is parallel to 
.
