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.
with circumcenter 
and orthocenter 
, let 
be an arbitrary point on the circumcircle of triangle such that does not lie on line 
and that line 
is not parallel to line 
. Let 
be the point on the circumcircle of triangle such that 
is perpendicular to , and let 
be the point on line 
such that 
. Let 
and 
be the points on the circumcircle of triangle such that 
is a diameter, and 
and 
are parallel. Let 
be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
be a sequence defined by 
. Then the product of four consecutive terms in can be written as the product of two terms in . Find 
if 
.
with 
, a tangent to the circumcircle of at point 
intersects the extension of 
at point 
. 
is the midpoint of 
, and 
intersects the circumcircle of at 
. Prove that 
.
be positive real numbers such that 
. Prove that![\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]](http://latex.artofproblemsolving.com/6/5/5/65558feba82e32266d3d3cbdfea85e079483403f.png)
with integer coefficients such that for all positive number 
and prime 
satisfying 
, we have 
.
that satisfy 
, where 
, then the number of integers in the range of 
is __
, since scaling up area by some amount will scale perimeter by the square root of that amount, if the perimeters are unequal, they will still be unequal after a scale. Then the perimeter of a triangle that has side length 
is 
. If we can prove that the derivative is positive or negative for the entire interval 
, we will be done.
. We can now just prove 
, both of which are trivial given . Therefore, there exists no noncongruent right triangles with the same area and perimeter.
and 
with 
and 
, then 
and 
.
.
and one angle 
and simplified using those conditions and got this equation
.
, where 
is the hypotenuse.
and 
respectively where 
and 
are the hypotenuses.
----(i)
-----(ii)
----(iii)
, which implies 
and perimeters are same, we get 
. This implies 
. Assuming 
and 
and . Adding and subtracting these two we get 
and 
.
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
be reals such that 
P
satisfying 
,
,
and the other right triangle has legs 
,
.![\[a+b+\sqrt{a^2+b^2}=c+d+\sqrt{c^2+d^2}\]](http://latex.artofproblemsolving.com/8/a/2/8a23f74cf8f85dc0e4ccff1714feec3ef1121d21.png)
and ![\[ab=cd\]](http://latex.artofproblemsolving.com/7/2/6/7260dcc8367798a67fc5b8c4f786038a85a09738.png)
looking for a contradiction. WLOG, 
so we have must have 
because lengths are non-negative and the function 
is strictly increasing over non-negative numbers. This implies 
because 
.
because the square-root function is stricting increasing. Putting this together, 
,
.
.
Using the quadratic formula: 
Therefore, 
or 
