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.
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 
.
, 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 square, and 
the circle whose diameter is 
Let 
be an arbitrary point on the segment 
We know that 
meets on and 
meets it on 
Also 
and intersect in 
show that belongs to 
)!
and 
. Also let Y be the center of the square. Since 
, and since AB = DA, right triangles QDA and SAB are congruent. This means that S is the image of Q when rotated 90 degrees around Y. By similar arguments, T is the image of Q when rotated around Y 90 degrees the other way. So TQS is an isosceles right triangle. Let 
, so that 
.
. (We have used the fact that 
which implies 
.) Likewise, 
. We then have 
, and after some calculation this comes out to 
.
and 
, so 
.
, EFBM is cyclic, so M is on the circle as desired.
. in the plane and points 
on it s.t. the line and 
meet on . let 
be an arbritary point of . let be the third point of intersection of 
with and let Q be the third points of intersection of 
with . then 
and 
concur on .
by the first condition. furthermore we know that 
where 
is the third point of intersection of with ; this is clear since both sides are the zeroes of some linear functions(the intersections with lines). Thus we have 
. defining 
analogously, we get 
. Thus we have, since 
that 
and this means that 
(this is the crucial step; this implication holds for all curves of genus at least 1). as the union of the and the line 
. it is clear that and meet on . therefore we immediately get the result.
, they are symmetric to each other with respect to the parallel to the side AB of the square through its center O. Of course, the diagoanls AC and BD are also symmetric to each other with respect to this parallel. Since the point O lies on this parallel, it thus follows from symmetry that < QOC = < ROB.
lies on the line CD.
is the point Q and also lies on the line CD., and 
are collinear. Since the points and both lie on the line CD, it follows that the third point must also lie on the line CD. In other words, the lines BO, ME and CD concur. But we know that the lines BO and CD intersect each other at the point D; thus, the line ME passes through the point D. In other words, the point M lies on the line DE. Similarly, the point M lies on the line CF. Thus, the problem is solved.
, where M is the point of intersection of the lines CF, DE and OR.
. But since the quadrilateral BRQC is a rectangle (in fact, it has three right angles: < RBC = < ABC = 90°, < BCQ = < BCD = 90° and < QRB = 90°), we have BR = CQ, and similarly AR = DQ, and thus this equation becomes 
. This completes our proof.
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 
![$ 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