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.
, where 
, then the number of integers in the range of 
is __
is inscribed in circle 
. Let 
denote its incenter and 
its 
-excenter. Let 
denote the midpoint of arc 
. Line 
meets a second time at 
. The perpendicular to 
at meets sides 
and 
at 
and 
respectively. The circumcircle of 
meets 
a second time at 
, and the circumcircle of 
meets 
a second time at 
. Prove that 
passes through the antipodal to on .
. Prove that if the leading coefficient of the polynomial equals then the greatest common divisor of the other coefficients also equals .
be a triangle with circumcenter 
. Point 
is the intersection of the parallel line from to with the perpendicular line to from 
. Let 
be the point where the external bisector of 
intersects with . Let 
be the projection of onto 
. Prove that the lines 
have a common point.
dwarves, whose weigths are 
, who want to cross a river. They have a small boat which can lift at most 
each time without sinking. For each journey of the boat a non-empty subset of the dwarves to be taken to the other side is chosen and one of these dwarves is chosen as the 
for that journey. Since return journeys are counter-current, no dwarf is able to do the rower for more than one return journey. Is it possible for all the dwarves to reach the other side of the river?
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 
.
and 
. We have 
and 
. Dividing the two expressions we get 
. But 
and thus 
. Because both angles are less than 
we get 
. and , which are equal. But then 
and because 
the result follows.
and . We have and . Dividing the two expressions we get . But and thus . Because both angles are less than we get . and , which are equal. But then and because the result follows.
be a parallelogram; since clearly 
, from 
we get 
, done.
. I have a decent diagram, and intuitively I agree with this statement, but could someone please provide a proof for this?
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)
satisfying 
and 
Show that 
be opposite of 
so 
but 
by symmetry so 
now by symmetry 
and 
.
and cevian 
,![\[AD=AE=\sqrt{a^2-2b^2}.\]](http://latex.artofproblemsolving.com/3/5/e/35e5faeb1107188881b853ac350dc18d347a8432.png)
By the Ratio Lemma on ![\[\frac{\sin\angle{BAD}}{\sin\angle{DAE}}=\sqrt{1-\frac{2b^2}{a^2}}<1,\]](http://latex.artofproblemsolving.com/5/c/7/5c73f2bd03afc02116abab050cf2ccf265813f3b.png)
so ![\[\sin\angle{BAD}<\sin\angle{DAE}.\]](http://latex.artofproblemsolving.com/8/7/7/877217fa7aa04b23d316aef78f5c2552396bf746.png)
Both of these angles are in ![$\left[0,\frac{\pi}{2}\right]$](http://latex.artofproblemsolving.com/8/5/3/8539f1d70f8e7db15780e94c1dad183ce02f2044.png)
and sine is increasing in this interval, so 
.
is a parallelogram. Since 
,
.
done. 
must be less than