[/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 
.
be a triangle 
)
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 Mathuzb, AopsUser101, Adventure10
Let 
be the intersection of 
and 
in isosceles trapezoid 
(
, 
) . The circumcircle of triangle 
inersects 
for the second time at 
. Point 
lies on 
such that 
. Prove that 
.
be the intersection of 
and 
in isosceles trapezoid 
(
, 
) . The circumcircle of triangle 
inersects 
for the second time at 
. Point 
lies on 
such that 
. Prove that 
.This post has been edited 1 time. Last edited by Etemadi, Apr 26, 2018, 10:21 AM
is cyclic. So from 
we conclude that 
.
is Cyclic. Let 
& 
are cyclic. Now, Note: 
and since 
is isosceles trapezoid, then 
.
.
.
is exterior bisector of 
.
is the ex-center of triangle 
with respect to 
.
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(9.080071469145949cm);
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draw(B--A, linewidth(1.2));
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draw(D--C, linewidth(1.2));
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draw(X--B, linewidth(1.2));
draw(A--X, linewidth(1.2));
draw(Y--A, linewidth(1.2));
draw(D--Y, linewidth(1.2));
draw(Y--C, linewidth(1.2));
draw(C--A, linewidth(1.2));
draw(D--B, linewidth(1.2));
draw(Y--P, linewidth(1.2));
draw(circle((-0.5487676153490452,0.23096704811575783), 0.5953920330094988), linewidth(1.2));
draw(P--X, linewidth(1.2));
/* dots and labels */
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label("$A$", (-0.641252415982058,0.8572858397873546), NW * labelscalefactor,qqqqcc);
dot(D,linewidth(4.pt) + qqqqcc);
label("$D$", (-1.0720851531544535,-0.23246755423693838), SW * labelscalefactor,qqqqcc);
dot(B,linewidth(4.pt) + qqqqcc);
label("$B$", (0.6222822501790012,0.8500449534483228), NE * labelscalefactor,qqqqcc);
dot(C,linewidth(4.pt) + qqqqcc);
label("$C$", (1.0205309988257534,-0.16005869084661992), NE * labelscalefactor,qqqqcc);
dot(P,linewidth(4.pt) + qqqqcc);
label("$P$", (-0.051120179350961574,0.35404423922464123), SW * labelscalefactor,qqqqcc);
dot(X,linewidth(4.pt) + qqqqcc);
label("$X$", (0.21317217202370123,1.689987768776017), NE ,qqqqcc);
dot(Y,linewidth(4.pt) + qqqqcc);
label("$Y$", (-1.2386255389521863,0.2418105009696476), NE * labelscalefactor,qqqqcc);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/7/e/2/7e24761a449ebc5236576edde22a24896d9915b5.png)
cyclic and 
it is trivial that 
is cyclic.
so it is enough to show 
and 
also 
!
.
and 
.
.
.
is external angle bisector of 
and 
.
of 
and the result follows immediately.
and 
and 
to be a cyclic quadrilateral so 
is cyclic.![\[\measuredangle AYD = \measuredangle XYD = \measuredangle YXC = \measuredangle AXB = \measuredangle APD\]](http://latex.artofproblemsolving.com/0/c/1/0c132c8526f8184003af27296d7bef9bc5ee21dc.png)
Thus, it is clear that 
![\[\measuredangle XYP = \measuredangle AYP = \measuredangle ADP = \measuredangle PCB = \measuredangle PCY\]](http://latex.artofproblemsolving.com/7/1/0/71000c95a6693c95ac5b939510208870f3ead62a.png)
and![\[\measuredangle PXY = \measuredangle PXA = \measuredangle PBA = \measuredangle BAP = \measuredangle BXP = \measuredangle CXP\]](http://latex.artofproblemsolving.com/a/c/2/ac2292a4e223f686f49443a672f87ad1481e658e.png)
Further, 
and 
share a side - 
which implies that they are indeed congruent as desired.![\[\measuredangle YDA = \measuredangle YPA = \measuredangle YCP + \measuredangle PCY = 2\measuredangle YCP = 2\measuredangle YCA\]](http://latex.artofproblemsolving.com/5/5/b/55b020ec072dc9502d113c2f0858b155612f7b5d.png)
as was needed.
