be positive real numbers such that 
.![\[\frac{1}{a^2+155}+\frac{1}{b^2+155}+\frac{1}{c^2+155}\le \frac{11}{780}\]](http://latex.artofproblemsolving.com/4/2/7/427e12cbfdbbba6f26407dbb5c2430aee26c15eb.png)
When does the equality hold ?
and 
Prove that
where 
,
,
be non-negative real numbers, no two of which are zero. Prove that :
weights, each weighing 
and another 
weights with 
each. Write a algorithm that determines the ways in which a scale can be balanced with a weight 
.
and 
.
![$ a,b,c\in [0,1] $](http://latex.artofproblemsolving.com/2/1/b/21b329d0a92c18c6a2861742177969d5977d4558.png)
. Prove that
.
.
from the domain 
and put those numbers into the given function f(x)=1/x.It gives us 1=1/2.But it's not true.So ,it can't be one-one function.But in the answer,it is one-one function.Would anyone enlighten me where is my fault? Thank you!
Y by den_thewhitelion, aopser123, Adventure10, Mango247
At the 
triangle the midpoints of 
are respectively 
and the triangle tangent to the incircle at 
, 
and 
in the same order.The midpoint of 
is 
. 
and 
intersect at point 
. The 
centered circle passing through 
cuts the 
ray at point 
. The line passing through and parallel to the 
and 
meet at 
. 
and intersect at the 
point. There is 
point chosen at incircle. 
is tangent to incircle at point . Prove that 
are cyclic.
triangle the midpoints of 
are respectively 
and the triangle tangent to the incircle at 
, 
and 
in the same order.The midpoint of 
is 
. 
and 
intersect at point 
. The 
centered circle passing through 
cuts the 
ray at point 
. The line passing through and parallel to the 
and 
meet at 
. 
and intersect at the 
point. There is 
point chosen at incircle. 
is tangent to incircle at point . Prove that 
are cyclic.This post has been edited 3 times. Last edited by KereMath, Sep 15, 2017, 10:43 AM
be given and it's incircle 
.
,
at 
,
and 
,
midpoints of same sides, respectively. Let 
and 
.
with radius 
,
lies on tangent from Feuerbach point 
onto 
be foot of altitude from 
be intersection point of tangent from 
be harmonic conjugate of 
i.e. 
But from 
we have
And by Menelaus we are done! 
,
,
and 
so writing the equation of 
We deduce that 
and the equation of the line parallel to 
.
and the equation of line 
and we know that 
,
deducing 
,
.
,
and 
.
for any 
,
,
.
