[/quote]
,
and 
are natural numbers such that
is a prime number. Find all possible values of 
.
be a positive integer. Aslı and Zehra are playing a game on an 
grid. Initially, 
stones are placed on some of the unit squares of this grid.
![\[\frac{x^3+1}{(y-1)(z+1)}+\frac{y^3+1}{(z-1)(x+1)}+\frac{z^3+1}{(x-1)(y+1)}\]](http://latex.artofproblemsolving.com/5/c/6/5c6894b617b990ec69c62ed1cf7d062d4660af53.png)
where 
are reals.
and let 
be positive integers such that 
.
and determine when equality holds.
equals to a positive integer power of 
.
![\[\frac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C} = 2\]](http://latex.artofproblemsolving.com/9/2/a/92a524bbdca92e9dd4091275714bf346bad96cd7.png)
is a pentagon whose vertices lie on circle 
where 
.
and 
intersect at 
,
meet 
at 
.
is the midpoint of arc 
on 
.
holds, then what is the value of 
?
.
be arbitrary points on sides 
and 
.
.
satisfy the equality, 
prove that the triangle must have a right angle.
candidates and 
is an odd integer. Each candidate is evaluated by each judge as either pass or fail. Suppose that each pair of judges agrees on at most 
candidates. Prove that ![\[{\frac{k}{m}} \geq {\frac{n-1}{2n}}. \]](http://latex.artofproblemsolving.com/3/8/7/387c8824b28f6422306e3c25b34fccb0a841f4cd.png)
be real numbers such that![\[ a^2+x^2=b^2+y^2=c^2+z^2=(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2 \]](http://latex.artofproblemsolving.com/7/e/b/7ebfb2f663d390714310d3c43b9c68e1b101d632.png)
.
,
and 
,
intersects 
is tangent to the circles with diameters 
and 
.
of non-negative real numbers such that the following three conditions are all satisfied:
;
;
of 
[/list]
is a permutation of 
,
.
for which there exist real numbers 
satisfying 
,
and
for 
.
Y by rkm0959, anantmudgal09, baopbc, ThE-dArK-lOrD, guangzhou-2015, buratinogigle, nguyendangkhoa17112003, Ya_pank, Adventure10, Rounak_iitr
Let 
be a non-cyclic convex quadrilateral. The feet of perpendiculars from 
to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. The feet of perpendiculars from 
to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. Let the orthocenter of 
be 
. Prove that the common chord of circumcircles of 
and 
bisects 
.
be a non-cyclic convex quadrilateral. The feet of perpendiculars from 
to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. The feet of perpendiculars from 
to 
are 
respectively, where 
lie on segments 
and 
lies on 
extended. Let the orthocenter of 
be 
. Prove that the common chord of circumcircles of 
and 
bisects 
.
be the isogonal conjugate of 
WRT 
respectively. Clearly, 
and 
so one of the intersection 
of 
and 
lies on 
Let 
be the second intersection of these two circles, then note that 
and 
are concyclic we get 
hence 
Let 
be the midpoint of 
Since 
are concyclic (the 9-point circle of 
)
be the orthocenter of 
.
concur.
.
is cyclic on circle 
.
so 
.
.
and 
,
,
,
are collinear. But since 
is the 
,
as desired, and the lemma is proved. 
be the midpoints of 
.
,
are collinear. It remains to prove that this line passes through the other intersection of 
.
be the unique right circumhyperbola of 
of 
as well. Thus, 
w.r.t 
,
so 
are collinear. Thus, 
bisects 
be a non-cyclic quadrilateral and 
be the orthocenter of triangle 
.
be the pedal circle of 
wrt 
and 
be the pedal circle of 
.
passes through the mid-point of 
.
be the projections of 
respectively. Let 
meet the sides 
respectively.
.
.
is the common chord of 
.
be the isogonal conjugate of 
;
be the isogonal conjugate of 
are symmetric in 
.
lies on the pedal circle 
and 
hence 
all lies on the circle with diameter 
.
be the midpoint of 
and 
meet at 
and 
meet at 
.
,
and 
hence 
.
and 
are both tangent to 
and 
meet at 
.
by Brokard's theorem.
,
.
,
,
.
is cyclic, and similarly, 
is cyclic. Let 
,
are tangent to circle 
w.r.t. 
,
and 
.
be the orthocenter of 
.
and 
are concurrent one of the intersections of 
.
,
.
,
.
and 
are perspective and the concurrency follows. Now it is left to show that this point lies on both circles. Since 
and 
are cyclic, by Miquel's theorem it follows that 
it follows by homothety that 
passes through 
.
be the center of this hyperbola. Then by the fundamental theorem, this must be the other intersection point of 
with radical axis of circles 
,
.
the circle passing through 
.
,
lies on line 
.
is one of the intersections of circles 
lies on the radical axis of 
.
,
with respect to 
,
