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: 
is circumscribed around the circle with centre 
. Let points and 
be the midpoints of sides 
and 
respectively and let 
. Prove that is either a trapezoid or a parallelogram.
is a triangle. WLOG AC'D'>90 (why?). Now, translate the ratio 
into a fact about this new triangle, and use the fact that AC'D'>90 to show that I is the unique point on segment AD' so that BM/BN'=IM/IN' (the circle of Appolonius helps here). Therefore, you can show that I is the reflection of B about MN'. Thus, BIA is 90, and this implies the required parallel condition.
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
)
of 
.
,
,
,
at 
.
for which 
is perpendicular to 
is purely imaginary, or![\[\left(\frac{z-\alpha}{\alpha}\right) + \overline{\left(\frac{z-\alpha}{\alpha}\right)}=0\,.\]](http://latex.artofproblemsolving.com/5/b/4/5b45fa86e1066e586c21ea8cd1806c2045cbaab2.png)
,
.![\[\frac{z}{\alpha^2}+\bar{z}=\frac{2}{\alpha}\,.\]](http://latex.artofproblemsolving.com/6/8/5/685d6638ec3f5c331a3bbe36dc4c0e5d2ed0471c.png)
![\[\frac{z}{\beta^2}+\bar{z}=\frac{2}{\beta}\,.\]](http://latex.artofproblemsolving.com/c/6/8/c684216cd7f7c176412585430f91e5758d588fc8.png)
is the intersection of ![\[B = \frac{2\alpha\beta}{\alpha+\beta}\,.\]](http://latex.artofproblemsolving.com/4/0/b/40b7cf11f38f32d8b17956fefcf36bb254a69b8f.png)
![\[C = \frac{2\beta\gamma}{\beta+\gamma}\,,\]](http://latex.artofproblemsolving.com/c/0/d/c0dbd286ae9900e19666db22cd1a61c4a367c54e.png)
![\[D = \frac{2\gamma\delta}{\gamma+\delta}\,.\]](http://latex.artofproblemsolving.com/6/7/7/677eb71513c18d1b754ffbbaf35ac1a1d14a8a39.png)
![\[A = \frac{2\delta\alpha}{\delta+\alpha}\,.\]](http://latex.artofproblemsolving.com/c/7/6/c7626d72e7370022994a25f8353beec748725e6c.png)
![\[M=\frac{A+B}{2} = \alpha\left(\frac{\beta}{\alpha+\beta}+\frac{\delta}{\alpha+\delta}\right)\]](http://latex.artofproblemsolving.com/f/3/c/f3c430cd2c652f950f2a00614446cde447c033ce.png)
![\[N = \frac{C+D}{2} = \gamma\left(\frac{\beta}{\gamma+\beta}+\frac{\delta}{\gamma+\delta}\right)\,.\]](http://latex.artofproblemsolving.com/4/8/a/48a45aaceced9199dbcd8bc90378a3a77700d606.png)
![\[|MI| = \left|\frac{\beta}{\alpha+\beta}+\frac{\delta}{\alpha+\delta}\right| = \frac{\big|\alpha(\beta+\delta)+2\beta\delta\big|}{|\alpha+\beta||\alpha+\delta|}\]](http://latex.artofproblemsolving.com/1/3/8/138c66a5e03e01b94aad5e3b268155fa277b7bdd.png)
![\[|NI| = \frac{\big|\gamma(\beta+\delta)+2\beta\delta\big|}{|\gamma+\beta||\gamma+\delta|}\,.\]](http://latex.artofproblemsolving.com/1/7/9/1794d2f8437fb1e95405fe4101b7cc0c3c8499c5.png)
![\[|AB| = \left|\frac{2\alpha\beta}{\alpha+\beta}-\frac{2\delta\alpha}{\delta+\alpha}\right| = \frac{2|\beta-\delta|}{|\alpha+\beta||\alpha+\delta|}\]](http://latex.artofproblemsolving.com/f/5/8/f5849e15e927a373c9a22d3029339974f5a055ff.png)
![\[|CD| = \frac{2|\beta-\delta|}{|\gamma+\beta||\gamma+\delta|}\,.\]](http://latex.artofproblemsolving.com/a/1/2/a1240920c1ea33c2ff758ffe63c438e47bb66c45.png)
holds if and only if![\[{\big|\alpha(\beta+\delta)+2\beta\delta\big|}= \big|\gamma(\beta+\delta)+2\beta\delta\big|\;\;\;\;\;\text{...(*)}\]](http://latex.artofproblemsolving.com/2/6/d/26dd660fdd2c9ccbd86467ea0cdad797be305c10.png)
,
.
,
be the intersection of 
be the image of reflection of ![\[y=-x=-\frac{2\beta\delta}{\beta+\delta}\,.\]](http://latex.artofproblemsolving.com/3/e/7/3e73f32addba702ada7d54d58b94987df571b7ca.png)
![\[|\alpha-y| = |\gamma-y|\,,\]](http://latex.artofproblemsolving.com/1/2/5/125116af4988fb0b12093c57cbbdb93427b2e837.png)
of the segment 
.