[/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 
.
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
Given a parallelogram 
. The line perpendicular to 
passing through 
and the line perpendicular to 
passing through 
intersect at point 
. The circle centered at point and radius 
intersects the line 
at point 
, (
) and the line 
at point 
, (
). Prove that the line 
passes through the point .
. The line perpendicular to 
passing through 
and the line perpendicular to 
passing through 
intersect at point 
. The circle centered at point and radius 
intersects the line 
at point 
, (
) and the line 
at point 
, (
). Prove that the line 
passes through the point .
,
.
page.
.
which gives us that AC and CP are perpendicular.
,
,
,
the circumcircle of our circle. First of all because 
we have that 
be the other tangent line to 
,
.
,
is the midpoint of 
,
are colliniar.![[asy]
import graph; size(8cm);
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pen dotstyle = black; /* point style */
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draw((-0.4471684078444506,2.5348300714317697)--(-0.7919984792762205,2.2876616635873193)--(-0.54483007143177,1.9428315921555495)--(-0.2,2.19)--cycle, linewidth(0.6));
draw((-1.8911245634895693,3.0233168610331784)--(-1.5625184116915543,2.7549551703981328)--(-1.2941567210565084,3.083561322196148)--(-1.6227628728545236,3.3519230128311936)--cycle, linewidth(0.6));
/* draw figures */
draw((-5.92,-1.91)--(-1.26,-1.33), linewidth(0.6) + wrwrwr);
draw((-5.92,-1.91)--(-0.2,2.19), linewidth(0.6) + wrwrwr);
draw((-4.86,1.61)--(-1.26,-1.33), linewidth(0.6) + wrwrwr);
draw(circle((-1.3085923842821308,3.7366215702667773), 1.9028965699960712), linewidth(0.6) + wrwrwr);
draw((-5.92,-1.91)--(0.46969661092200743,4.413898179665534), linewidth(1.6) + dotted + wrwrwr);
draw((-3.045525745709047,4.513846025662389)--(-0.2,2.19), linewidth(0.6) + wrwrwr);
draw((-3.045525745709047,4.513846025662389)--(-5.92,-1.91), linewidth(0.6) + wrwrwr);
draw((-4.86,1.61)--(-5.92,-1.91), linewidth(0.6) + wrwrwr);
draw((-5.92,-1.91)--(-1.3085923842821308,3.7366215702667773), linewidth(0.6) + wrwrwr);
draw((0.46969661092200743,4.413898179665534)--(-1.26,-1.33), linewidth(0.6) + wrwrwr);
draw((-4.86,1.61)--(-0.2,2.19), linewidth(0.6) + wrwrwr);
draw((-1.3085923842821308,3.7366215702667773)--(-0.2,2.19), linewidth(0.6) + wrwrwr);
draw((-1.3085923842821308,3.7366215702667773)--(-3.045525745709047,4.513846025662389), linewidth(0.6) + wrwrwr);
/* dots and labels */
dot((-5.92,-1.91),dotstyle);
label("$A$", (-6.32,-1.87), NE * labelscalefactor);
dot((-1.26,-1.33),dotstyle);
label("$B$", (-1.08,-1.63), NE * labelscalefactor);
dot((-0.2,2.19),dotstyle);
label("$C$", (0.06,1.93), NE * labelscalefactor);
dot((-4.86,1.61),linewidth(4pt) + dotstyle);
label("$D$", (-5.22,1.81), NE * labelscalefactor);
dot((-1.3085923842821308,3.7366215702667773),linewidth(4pt) + dotstyle);
label("$P$", (-1.34,4.11), NE * labelscalefactor);
dot((-2.004239071286051,1.9654380555051696),linewidth(4pt) + dotstyle);
label("$Y$", (-2.06,1.59), NE * labelscalefactor);
dot((0.46969661092200743,4.413898179665534),linewidth(4pt) + dotstyle);
label("$X$", (0.68,4.29), NE * labelscalefactor);
dot((-3.045525745709047,4.513846025662389),linewidth(4pt) + dotstyle);
label("$Z$", (-3.6,4.47), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/7/c/f/7cfd187361feacfacd11679044fdfabd156c5b45.png)
is equal to 
!
is equal to 
.
,
are collinear which is what you need to prove.
meet 
.
meet 
at 
and let 
be the circumcenter of 
.
and 
be the midpoints of 
respectively.. Finally let 
be the projection of 
onto 
and 
to calculate 
and 
.
to get 
.
.
.
and thus 
.
,
and 
.
.
with 
.
,
The coordinates now are 
.
. The equation of line 
Clearly 
.
,
.
Since 
,
intersect 
Because 
are collinear, furthermore 
are collinear.
