Prove that 
be real numbers such that 
Prove that 
Where 
.
and 
respectively denote the midpoints of the arcs 
denote the midpoint of arc 
)
be the incenter of 
be the circle that is tangent to 
be the circle that is tangent to 
,
friends, 
people are seated around a circular table in a fixed order. Ali places 
Y by
In triangle 
, points 
, 
, 
are the midpoints of sides 
, 
, 
, respectively. Points 
, 
, 
are the midpoints of broken lines 
, , 
, respectively (point is called the midpoint of a broken line if it belongs to a broken line and divides it into two broken lines equal length). Prove that lines 
, 
, 
pass through one point.
, points 
, 
, 
are the midpoints of sides 
, 
, 
, respectively. Points 
, 
, 
are the midpoints of broken lines 
, , 
, respectively (point is called the midpoint of a broken line if it belongs to a broken line and divides it into two broken lines equal length). Prove that lines 
, 
, 
pass through one point.This post has been edited 1 time. Last edited by parmenides51, Apr 23, 2022, 8:26 PM
.
,
and 
.
and then deduce the midpoint of broken line 
be the midpoint of 
.
is the midpoint of 
and 
is the midpoint of ![\[
A_3 = 2M_A-A_4 \text{ hence } A_2 = 2A_3-A = 2(2M_A-A_4)-A = 4M_A-2A_4-A.
\]](http://latex.artofproblemsolving.com/4/4/d/44d1749a106c9bd504bcaeb2624c9789fd3b808e.png)
So we need to compute 
,
and 
.
.![\[
A_4 = \left(\frac{b+c}{2c},\frac{c-b}{2c},0\right) \implies A_2 = \left(-\frac{b}{c},\frac{b}{c},1\right)
\]](http://latex.artofproblemsolving.com/e/a/3/ea39f990c854a38c8e3e1949f9e0882888e3b7da.png)
Similarly, one has that (cyclically repeating notation):![\[
B_4 = \left(\frac{c-a}{2c},\frac{c+a}{2c},0\right)\implies B_2 = \left(\frac{a}{c},\frac{-a}{c},1\right)
\]](http://latex.artofproblemsolving.com/2/1/4/214edd5f86ab5fa8ecc43f271bd80ec4f2ff2c4f.png)
and![\[
C_4 = \left(0,\frac{a-b}{2a},\frac{a+b}{2a}\right) \implies C_4 = \left(1,\frac{b}{a},\frac{-b}{a}\right).
\]](http://latex.artofproblemsolving.com/9/a/1/9a1f61d1a46a31dd609766731f8d8c81f25e9f98.png)
Now Ceva with triangle 
,
and 
finishes the job 
.
