# 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 2*OP* drawn about *P*. Prove that *QPR*=120^{o}.

(5 pont)

**Deadline expired on January 16, 2006.**

**Solution.** Place the hyperbola into the Cartesian co-ordinate 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

(*x*-*a*)^{2}+(*y*-*b*)^{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 2*a*. Therefore

q_{1}+r_{1}+s_{1}=3a. | (2) |

Similarly,

q_{2}+r_{2}+s_{2}=3b. | (3) |

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, Kisfaludi-Bak 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.

Problems in Mathematics of KöMaL, December 2005