AoPS Academy
, let 
be the biggest divisor of 
except itself. Does there exists a positive integer 
such that the equality 
has exactly 
solutions?
be an acute angled triangle with orthocenter 
and 
. Let 
be any other point on the arc 
of the circumcircle of 
and let line 
intersect line 
at 
and let line 
intersect line 
at 
. Let the circumcircles of 
and intersect again at 
(
). Let the lines 
intersect the circumcircle of 
again at 
(
). Prove that the lines 
concur.
be a degree 6 polynomial with integer coefficients. Prove that there exists a real number ![$a \in [0,3]$](http://latex.artofproblemsolving.com/c/e/0/ce048e201e2d842cb846be9a98a64da6bd47a364.png)
such that![\[
|P(a)| \geq \frac{45}{64}.
\]](http://latex.artofproblemsolving.com/1/7/5/175d3c5451574d61d0d4923d2b9bc240ec3935fd.png)
). And indeed, the 
should be 
; mind you, when it works for the latter, it will a fortiori work for the former!
of the quadrilateral 
is 
, by the known fact that the area of a convex quadrilateral is half the product of its diagonals with the sine of their angle, followed by the fact that the sine of an angle is at most 
, followed by Ptolemy's inequality.
, so if we prove that 
, i.e. 
, that will be enough. But this is a simple vectorial computation.
and 
, so 
.
, by the triangle inequality.
. Then 
, 
, 
, so the inequality is equivalent to 
, which, as far as I remember, is the triangle inequality, expressed in the Euclidean norm (in any dimension) by Minkovski's inequality. The fact it is not appearing in its usual form
,
(having as last recourse the triangle inequality - what else?); otherwise it is difficult to foresee a change in the chain of formulae.
Prove that:
Where does the equality holds ?
be a convex quadrilateral and points 
and 
on sides 
such that![\[\tfrac{AB}{AE}=\tfrac{CD}{DF}=n\]](http://latex.artofproblemsolving.com/b/4/2/b42cd6305fab848a60c79689f17bdb55e1ba0d31.png)
![\[\tfrac{AB}{AE}=\tfrac{CD}{CF}=n\]](http://latex.artofproblemsolving.com/f/8/2/f82128eb770639038420fce432811dd37da6243d.png)
and 
than we have 
hence a mistake in question?
instead of 
it seems.
such as 
are parallelograms.
collinear 
such as 
.
.
.
and 
be perpendicular lines on 
,
so 
.
inequality is equivalent with 
.
.
and 
.
.