[/quote]
,
be positive reals. Prove that there is a cyclic permutation 
of 
such that the inequality:![\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]](http://latex.artofproblemsolving.com/5/5/3/553527fc40a1b7d29dbdfb4b5e63635308f46c85.png)
and 
.
and circumcenter 
.
,
the midpoints of 
,
.
with diameter 
,
at a point 
.
meets line 
at 
.
and 
intersect at a point 
on 
.
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of 
is the sum of 
and incenter 
and let 
,
,
are selected on sides 
,
such that 
,
,
.
intersects 
other than 
and 
meet on 
,
,
,
,
,
initially contains one coin. The following operations are allowed
,
,
;
,
,
and 
.
coins.
Prove that
its circumradius. Let 
,
across the line 
,
across the line 
.
be a real number and let 
be arbitrary real numbers in 
.
and 
are then defined recursively by
.
for all sufficiently large 
.
is a prime number
is inscribed in circle 
.
its 
.
meets 
at 
and 
meets 
a second time at 
meets 
a second time at 
.
passes through the antipodal to 
intersects the lines 
,
at points 
,
,
of the segments 
,
,
,
.
be a positive integer. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
,
.
Y by Mango247, Mango247, Mango247
Let 
be an acute triangle. Let 
and 
be the midpoints of sides 
, respectively 
. Let 
and 
be the the midpoints of the arcs 
and respectively of the circumscriberd circle of triangle . Let 
be the intersection of the straight line 
and the perpendicular to 
through 
. Define 
and 
similarly. Show that the straight line 
and 
intersect at one point.
be an acute triangle. Let 
and 
be the midpoints of sides 
, respectively 
. Let 
and 
be the the midpoints of the arcs 
and respectively of the circumscriberd circle of triangle . Let 
be the intersection of the straight line 
and the perpendicular to 
through 
. Define 
and 
similarly. Show that the straight line 
and 
intersect at one point.This post has been edited 1 time. Last edited by parmenides51, Dec 8, 2020, 1:10 PM
.
,
,
is the angle bisector of 
.
Therefore, by ceva's theorem on 
,
and 
is perpendicular bisector of 
is a parallelogram we can easly conclude the point of concurrency will in fact be the Isotomic Conjugate of the incenter of 
,
.
