Problem A. 386. (December 2005)
A. 386. Consider one of the two arcs of a rectangular hyperbola centred at O, and choose an arbitrary point P on it. Let Q and R denote the intersections of the hyperbola and the circle of radius 2OP drawn about P. Prove that QPR=120^{o}.
(5 pont)
Deadline expired on January 16, 2006.
Solution. Place the hyperbola into the Cartesian coordinate system such that its equation is xy=1. Draw the other branch of the hyperbola as well. The circle and the second branch have another 2 intersections: S and P' which is the reflection of P to O.
Let P=(a,b) and compute the intersections of the hyperbola and the circle. A point (x,y) lies on both curves if and only if
xy=1
and
(xa)^{2}+(yb)^{2}=4(a^{2}+b^{2}).
Then
x^{4}2ax^{3}3(a^{2}+b^{2})x^{2}2bx+1=0.  (1) 
Let P'=(a,b), Q=(q_{1},q_{2}), R=(r_{1},r_{2}) and S=(s_{1},s_{2}). The roots of the quartic equation (1) are exactly a, q_{1}, r_{1} and s_{1}.
From Viéte's formulas, the sum of roots is 2a. Therefore
Similarly,
Equations (2) and (3) together yield that P is the centroid of triangle QRS. But P is the circumcenter as well. This implies that triangle QRS is equilateral, point P is its center anf QPR=120^{o}.
Statistics:
14 students sent a solution.  
5 points:  Bogár 560 Péter, Erdélyi Márton, Estélyi István, Hujter Bálint, Jankó Zsuzsanna, KisfaludiBak Sándor, Kónya 495 Gábor, Nagy 224 Csaba, Paulin Roland, Szakács Nóra, Tomon István. 
4 points:  Dücső Márton. 
2 points:  1 student. 
0 point:  1 student. 
