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.