AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
such that a positive integer power of 
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 
. Let 
and 
intersect at 
, 
meet 
at 
. 
is the midpoint of arc 
on , not containing 
. If 
holds, then what is the value of 
?
be a cyclic quadrilateral and let the intersection of and 
be 
. Let the points 
be arbitrary points on sides and respectively which satisfy the conditions 
and 
. Prove that 
.
satisfy the equality, 
prove that the triangle must have a right angle.
candidates and judges, where 
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)
.
, inscribed in the circle 
, satisfies 
and 
, and is the intersection point of the diagonals and . The circle with center 
and radius 
intersects in two points and . Prove that the line 
is tangent to the circles with diameters 
and 
.
of non-negative real numbers such that the following three conditions are all satisfied:
;
;
of such that 
[/list]
is a permutation of 
, and they are both permutations of 
. Note that any list is a permutation of itself.
for which there exist real numbers 
satisfying 
, 
and
for 
.
for some real number 
and ![\[ a_n=na_{n-1}+(n-1)(n!(n-1)!-1) \]](http://latex.artofproblemsolving.com/4/1/f/41f0cc7b041a8bba056882b5c794503195c374fd.png)
for integers 
. Given that 
, and given that 
, where 
and 
are positive integers whose greatest common divisor is 
, compute 
and 
are in the coordinate plane such that 
. Let 
denote the locus of all points 
in the coordinate plane satisfying 
, and let be the midpoint of . Points 
and 
are on such that 
and 
. The value of 
can be expressed in the form 
, where and are relatively prime positive integers. Find 
.
moves.
. Then suppose 
and 
correspond to the same state (some distinct values of 
and 
must satisfy this by PHP). Therefore, applying the algorithm 
times to a state returns that state back to itself. This is true no matter what state we are currently at, since every piece is unique on the cube. This means that after algorithms at the beginning, we will return to the solved state.
moves.
and then 
then trivially the puzzle gets resolved after applying this finite sequence of legal moves.
Prove that
Let 
Prove that
