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.
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 __
be a triangle and 
denote its inradius and ex-radii opposite to the vertices 
respectively. If 
and 
, then which of the following is/are true?
is obtuse
is acute
, where 
is semi perimeter
, where is semi perimeter
satisfy that: 
and 
. Find the maximum: 
be a trapezium in which 
and 
. Let 
be then midpoint of the diagonal 
. If ![$[ABCD] = n \times [CDE]$](http://latex.artofproblemsolving.com/e/9/3/e9347ba2c8e9396adfe716159e711408a3b9dc69.png)
, what is the value of ?![$[t]$](http://latex.artofproblemsolving.com/8/a/8/8a86702dc60b9db916798b1115a6b6822bf828b9.png)
denotes the area of the geometrical figure
.)
be reals such that 
P
satisfying 
![$ a,b\in[\frac{1}{2},1] $](http://latex.artofproblemsolving.com/0/d/2/0d24f742e1f7db553b3a8795c061c796f07a897f.png)
.
Let ![$ a,b\in[\frac{1}{2},2] $](http://latex.artofproblemsolving.com/a/8/d/a8d4c8205bac0e34129f582a8f255355eda12e0b.png)
.
and 
Prove that
altitude of trapezium, and 
.![$[ABCD] = \frac{3x+x}{2} \cdot h = 2xh,$](http://latex.artofproblemsolving.com/a/e/5/ae5f28beb8fea580969324426d0aec130bb4644a.png)
![$[BCD] = \frac{1}{2} xh,$](http://latex.artofproblemsolving.com/6/b/b/6bbe5aeb136750d6fff6cded78dcc4d91c20879b.png)
![$[CDE] = \frac{1}{2} [BCD]$](http://latex.artofproblemsolving.com/1/9/7/197ca7a07e6cec75f53e5b5b87599caa0adf2706.png)
.![$\frac{[ABCD]}{[CDE]} = \frac{[ABCD]}{\frac{1}{2} [BCD]} = \frac{2[ABCD]}{[BCD]} = \frac{4xh}{\frac{1}{2} xh} = \boxed{8}$](http://latex.artofproblemsolving.com/b/6/0/b60180720412a5be300340066e407a08a5aad75e.png)
![$[ABCD]=2CD\cdot h$](http://latex.artofproblemsolving.com/f/9/b/f9be5365db06272049135b856680ceb3288fdee9.png)
and ![$[\triangle CDE]=\frac{1}{2}[\triangle BDC]=\frac{1}{2}CD\cdot h$](http://latex.artofproblemsolving.com/9/9/f/99fccb8b236e142ad94bbf814051925a34645e34.png)
so answer is 
.