be an acute triangle with 
,
its orthocenter and 
it's circumcenter. Let 
be a point on the parallel through 
to 
such that 
and 
are on different sides of 
.
the second intersection of the circumcircles of 
,
the midpoint of 
,
the intersection of 
and the parallel through 
,
the intersection of 
and the perpendicular through 
and 
a point on 
,
is the midpoint of 
.
lie on a line.
is inscribed in a circle 
.
— the circle inscribed in a segment 
circle of 
— the centers of circles 
and 
are parallel.
be cyclic quadrilateral. Let 
and 
intersect at 
intersect at 
.
.
with the diagonals of 
and 
are tangent at a fixed point.
,
.
and 
,
is acute, and let 
and 
,
and 
and 
and 
![$f \in C([0,1];[0,1])$](http://latex.artofproblemsolving.com/4/7/5/475518b7db0b787dd90b93f602c7b912c663fba3.png)
bijective, 
and ![$(a_k) \in [0,1]^\mathbb N$](http://latex.artofproblemsolving.com/a/d/8/ad8a80d847c60ee8e916a1f9b08b7b03ec17c814.png)
with 
converge.
converge?
,
.
satisfies the quadratic equation 
and deduce that the value of 
.
.
,
is 
.
![$f\in C^1[a,b]$](http://latex.artofproblemsolving.com/8/4/8/848da12534d564560fe1cb016142159234be98b4.png)
,
and suppose that 
,
,
for all ![$x\in [a,b]$](http://latex.artofproblemsolving.com/7/8/d/78d9dcd969a2c45f7cc9234aa1377b6ec1c2d5d3.png)
.
for all 
converge?
is called perfect if the sum of its positive divisors 
is twice 
.
is a perfect number since the sum of its positive divisors is 
,![\[
\sum_{p|n} \frac{1}{p + 1} < \ln 2 < \sum_{p|n} \frac{1}{p - 1}
\]](http://latex.artofproblemsolving.com/0/c/0/0c08f527e87f7aae4ca5226f19bd084249551fe3.png)
where the sums are taken over all prime divisors 
of 
![$f\in C[0,b]$](http://latex.artofproblemsolving.com/8/9/8/898aec0300888e562794b7a551be34f5c2676236.png)
,
and let 
be periodic with period 
.
has a limit as 
and
where 
,
,
,
and 
for 
.
and give an example where 
.
Y by
Let 
be polygons, no two of which are similar. Show that there are polygons 
where 
is similar to 
so that for no 
does contain a polygon that's congruent to 
.
Note. Here, the word "contain" means for the construction we have, we cannot select a size for so that is wholly contained in , and so it does not intersect the edges of at all.
be polygons, no two of which are similar. Show that there are polygons 
where 
is similar to 
so that for no 
does contain a polygon that's congruent to 
.Note. Here, the word "contain" means for the construction we have, we cannot select a size for
so that is wholly contained in , and so it does not intersect the edges of at all.This post has been edited 1 time. Last edited by somebodyyouusedtoknow, Apr 26, 2025, 11:41 PM
Reason: Added clarity for the notes
Reason: Added clarity for the notes
.
is obvious; for fixed 
consider the largest copy of 
inside it and slightly enlarge it to obtain 
.
.
with 
for all 
we seek for appropriate 
.![$i\in [n]:= \{1, \ldots, n\}$](http://latex.artofproblemsolving.com/5/5/7/5572574af6c66362b508aebea9989529caa5811f.png)
define 
as the largest copy of 
inside 
as the smallest copy of ![$i\in [n]$](http://latex.artofproblemsolving.com/e/e/5/ee51c5c5eb09111cc774a00e9970074ad9f9058c.png)
.![$\max_{i\in [n]} R_i^- \subsetneq \min_{i\in [n]}R_i^+$](http://latex.artofproblemsolving.com/5/0/1/501e5e0ead06031f5c49050fed96583c031780da.png)
.
.
