[/quote]
,
specifies the greatest common divisor of 
and 
.
such that: 
and 
.
and 
be on sides 
and 
respectively in 
such that 
.
be the circumcenter of 
and let 
.
![\[ a_n=2^n+3^n+6^n -1,\ n\geq 1. \]](http://latex.artofproblemsolving.com/6/4/9/6499d8f992f3e72765cab1f3afe15df99d01b12a.png)
that satisfy the equation
be a triangle with its 9-point center 
and excentral triangle 
.
-
,
,
as 
,
and 
for the 
-
-
,
,
.
be the radical center of the circles 
,
,
.
,
,
are concurrent.
,
and incentral triangle 
.
intersect 
again at 
.
and 
.
and 
be the 9-point centers of 
,
,
are collinear.
be an isogonal conjugate pair. Suppose the line through 
and perpendicular to 
intersects 
again at 
.
.
and perpendicular to 
intersects 
again at 
.
.
and 
be the orthocenters of 
and 
,
and 
.
and 
be the isogonal conjugates of 
with respect to 
are concurrent.
with 
.
lies on Euler line of 
be the angle bisector of 
.
at 
.
at point 
.
be positive real numbers such that : 
. Prove that : 
such that 
.![\[
P = \frac{a}{a^2 + b^3} + \frac{b}{b^2 + c^3} + \frac{c}{c^2 + a^3}
\]](http://latex.artofproblemsolving.com/1/3/9/139529da9f360d82378a2fac5e756daba4183bbf.png)
Y by
The diagram below shows two circles, Circle 
and Circle 
. The centre 
of Circle lies on the circumference of Circle , and the centre 
of Circle 2 lies on the circumference of Circle . The two circles intersect at points 
and 
. The length of the line joining and is cm, rnd the length of the line joining and is 
cm. Find the area common to both circles (shaded).
and Circle 
. The centre 
of Circle lies on the circumference of Circle , and the centre 
of Circle 2 lies on the circumference of Circle . The two circles intersect at points 
and 
. The length of the line joining and is cm, rnd the length of the line joining and is 
cm. Find the area common to both circles (shaded).This post has been edited 2 times. Last edited by parmenides51, Feb 26, 2021, 11:37 PM
.
via special triangles and then 
.
Probably, since 30-60-90 is a special triangle 
is equilateral, and then 
from with you can find circular segment 
by taking Sector 
with respect to 
Then double this to obtain an answer of 
is equilateral gives us the fact that 
which gives us the area of the 
sector from which we can subtract 
and we get half the area of the desired region
condition is extraneous.
