Hint

Posted by

Jorjia 6 Oct 2018 19:50

I can't solve it, yet.

But i think that, it solviable as fermat point geometry construction (see wiki), and using angles, as Shen Yang suggested

<AOB = acos((c1*c1-c2*c2-c3*c3)/(2*c2*c3)),

<AOC = acos((c2*c2-c1*c1-c3*c3)/(2*c1*c3) and

<BOC = acos((c3*c3-c1*c1-c2*c2)/(2*c1*c2)) .

(c1,c2,c3 - are prices).

In wiki fermat point: constructed triangles ABC' , BCA' and CAB'

where

<ABC' = 60, <BAC' = 60,

<CBA' = 60, <BCA' = 60, and

<ACB' = 60, <CAB' = 60.

Fermat point X - is intersection of AA' and BB' and CC' lines.

In there, we must construct triangles ABC' , BCA', and CAB' , with

<ABC' = <BAC' = <AOB / 2

<BCA' = <CBA' = <COB / 2

<ACB' = <CAB' = <AOC / 2

and intersection of AA' , BB' and CC' - will be ans, iff it's in ABC triangle. otherwice A, B, or C will be ans.