[/quote]
,
and 
intersect at 
and 
.
;
and 
,
and 
intersects 
.
.
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)
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)
Y by Mathuzb, tenplusten, anantmudgal09, R8450932, nguyendangkhoa17112003, GeoMetrix, amar_04, Adventure10, Mango247, Rounak_iitr
Circle 
is tangent to sides 
,
of triangle 
at 
,
respectively, such that 
, 
and 
. 
,
lies on 
such that 
, 
. Let 
and 
meet at 
. 
lies on minor arc 
of , such that the tangent of to is parallel to . Prove that the incenter of 
lies on 
.
is tangent to sides 
,
of triangle 
at 
,
respectively, such that 
, 
and 
. 
,
lies on 
such that 
, 
. Let 
and 
meet at 
. 
lies on minor arc 
of , such that the tangent of to is parallel to . Prove that the incenter of 
lies on 
.
be a cyclic quadrilateral with circumcenter 
Denote 
and 
Let 
be the antipode of 
Then 
are collinear.![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
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[/asy]](http://latex.artofproblemsolving.com/1/1/d/11d4150a3f4c7b5c18cdf7b025c701af2da10712.png)
and let 
It is 
are collinear, and 
On the other hand, since 
and 
is a diameter of this circle, we have 
Therefore, 
be the incenter of 
Note that 
is the perpendicular bisector of 
;
and 
It follows that 
Therefore, 
is cyclic with circumcenter 
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(15cm);
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 */
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Label laxis; laxis.p = fontsize(10);
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yaxis(ymin, ymax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */
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[/asy]](http://latex.artofproblemsolving.com/4/4/c/44c7c9aa0ca5fd1a5cb27d411622e978ca666b05.png)
and 
meet at 
By the above observations, the sidelines of 
are parallel to the sidelines of the intouch triangle of 
By homothety, we see that 
lies on major arc 
of 
Therefore, 
Applying the lemma to cyclic quadrilateral 
are collinear. 
from 
as the reflections of 
in lines 
respectively. Let 
and 
.
be the touch-points of the incircle and 
hence 
is cyclic. Also 
.
.
is cyclic and is the image of the incircle under suitable homothety at 
and 
move with constant velocity. Hence we can find a spiral similarity with pivot 
and 
.
be point at infinity with 
.
is projective.
.
(likewise let 
for the final case). Then 
and 
hence 
and we're done. 
lies on 
is diameter of 
and 
hence 
are colinear and we are done.
,
,
.
,
,
be the contact points of the incircle on 
.
where 
.
,
the standard determinants give ![\begin{align*} t(u+t)(v+t) &= (v-t)\left[ (v+t)y-(w-t)z \right] \\ t(u+t)(w+t) &= (w-t)\left[ -(v+t)y+(w+t)z \right]. \\ \implies t(u+t)\left[ (v+t)(w-t)+(w+t)(v+t) \right] &= \left[ -(w-t)^2(v-t)+(v+t)(w-t)(w+t) \right]z \\ z &= \frac{t(u+t)(v+t)(2w)} {(w-t)\left[ (v+t)(w+t)-(w-t)(v-t) \right]} \\ &= \frac{w(u+t)(v+t)}{(w-t)(v+w)}. \end{align*}](http://latex.artofproblemsolving.com/6/4/9/649bf3b9add66aab1c4cbf10c8a7b34788f85798.png)
Thus we recover ![\[ K = \left( \frac{t(v+t)}{u+t} : \frac{v(w+t)}{v-t} : \frac{w(v+t)}{w-t} \right). \]](http://latex.artofproblemsolving.com/0/3/0/03077957e9acf524747244aeedd230e1f18cb1de.png)
On the other hand, by homothety we have ![\begin{align*} \vec L &= \left( 1 + \frac tu \right) [2 \vec I - \vec X] - \frac tu \vec A \\ \left( 2u+2t \right) \vec I - u \vec L &= \left( u+t \right) \vec X + t \vec A \\ &= t \vec A + (u+t) \frac{w \vec B + v \vec C}{w+v} \\ \end{align*}](http://latex.artofproblemsolving.com/9/7/c/97c7f85ac495ff84bfe21dbba35477078b76e5f0.png)
Consequently the point 
lies on 
and is distinct from 
)
,
.
is inversion image of 
,
.
are collinear. Done
.
and the intouch triangle of 
.
implies 
is cyclic; its center is the intersection of the 
bisectors of 
,
be the second intersection of 
with 
respectively. Obviously, the incenter 
of 
is the circumcenter of 
so simple angle chasing yields 
On the other hand, by Reim's theorem we get 
so 
and 
are homothetic 
are collinear.
.
,
,
.
lie on a circle centered at 
lies on 
is tangent to 
as well as to the tangent to 
,
.
,
.
are collinear. 
,
are reflections of 
is cyclic and center is 
,
This completes the proof of claim. 
,
are collinear. 
![[asy]
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I=incenter(A,B,C);
D=(A+4B)/5;
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F=reflect(B,I)*D;
G=reflect(C,I)*EE;
K=extension(D,G,EE,F);
X=extension(D,F,EE,G);
L=extension(K,I,X,X+(0,1));
M=foot(X,I,K);
filldraw(circumcircle(D,F,G),tfil,tri);
filldraw(circumcircle(D,EE,L),sfil,sec);
filldraw(incircle(A,B,C),fil,pri);
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draw(L--M--X,tri);
draw(D--G,sec); draw(EE--F,sec);
filldraw(A--B--C--cycle,fil,pri);
dot("$A$",A,N);
dot("$B$",B,SW);
dot("$C$",C,SE);
dot("$D$",D,W);
dot("$I$",I,W);
dot("$E$",EE,dir(30));
dot("$F$",F,SW);
dot("$G$",G,SE);
dot("$K$",K,dir(110));
dot("$X$",X,S);
dot("$L$",L,N);
dot("$M$",M,dir(240));
[/asy]](http://latex.artofproblemsolving.com/9/a/6/9a686ae1b8b915662487760ca1c063108443ce30.png)
and 
.
and the contact triangle are homothetic at 
is a diameter of 
,
,
by construction, 
.
,
.
are 
and 
respectively, and the coordinates of 
are 
,
respectively. Computing a determinant, the equations of 
Solving, we get that point 
Also, if we consider point 
,
.
.
into a determinant, we get a cubic in 
,
.
points. We will choose 
.
At 
we have 
and we get a similar result for 
.
,
as desired.
![[asy]
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dot((-5.546903657206548,-6.2449680793904125),linewidth(4pt) + dotstyle);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/a/3/3/a33b17e24eee806bba80a64f07246c0535f629c9.png)
and 
.
be the touchpoints of the incircle with 
.
and 
are homothetic and so taking a homothety centred at 
.
is a collinear triplet 
is the 
-
is the centre of 
.
are inverses w.r.t 
is a collinear triplet. By brocard we have that 
.
is a collinear triplet as desired
,
is cyclic. Inversion around 
with radius 
gives 
.
such that 
are tangent to 
but so is 
where 
,
.
must be the tangents of 
to 
are collinear.
must lie on 
.
.
are collinear.
Since these involutions overlap in two pairs, these are in fact the same involutions, hence 
,
.
by parallel lines. To finish,
and 
is 
are pairs of an involution on 
.
is the midpoint of 
and 
,
,
.
WRT 
,
is the Miquel Point of 
of 
is the perpendicular bisector of 
.
.
.
are the 
angle bisectors of 
.
Let 
.
and 
.
.
and 
.
is cyclic 
implies 
.
,
,
.
and 
is diameter. Consider the miguel point 
.
,
intersect at 
intersect 
passes through 
also passes through 
,
for some real number 
.
,
.
.
linearly. In particular, taking 
,
collinear.
denote the linear map that sends 
,
denote the linear map that sends 
.
coincide for two distinct values of 
,
intersect on 
are concyclic.
,
are concyclic. 
.
so that
-
,
,
,
.
sending the 
to 
.
it remains to compute 
and solving we get 
Now to show the determinant ![$[IKL]=0$](http://latex.artofproblemsolving.com/6/d/a/6dad4b4a6bbdbddf1c78f88515849dd1eef4eaa8.png)
:
.
,
where everything cancels (yay!)
,
.
so 
is cyclic.
,
must be collinear, so it suffices to show that 
goes through 
,
is the antipode of 
,
lie on circle with center as 
hence 
cyclic.Note 
and 
we have 
also from fact that 
with 
.
is center of 
cyclic.
hence 
are collinear.
.
.
(well know property)
is diameter 
,
hence 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/f/d/d/fdd0afbfdd53127bd43f1cd713aa091264506b38.png)
and 
and 
,
with center 
and contact triangle are homothetic with center 
.
,
because 
,
and we are done!
and 
.
from the definition of ![[asy]
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dot("$L$", L, dir(90));
dot("$X$", X, dir(90));
dot("$Y$", Y, dir(60));
dot("$M_A$", MA, dir(270));
[/asy]](http://latex.artofproblemsolving.com/8/3/5/835c6ae89047b96724f5ab87613808e21663970a.png)
and 
with 
and 
.
and 
,
and 
.
and 
bisects 
,
.
.
and 
.
and 
.
is cyclic; then 
and 
.
,
,
and 
are isosceles trapezoids. Hence 
and 
.
,
. This implies 
.
and on 
.
.
.
,
which implies 
.
are collinear and we are done.
