Mathematical and Physical Journal
for High Schools
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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 \angleQPR=120o.

(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







Let P'=(-a,-b), Q=(q1,q2), R=(r1,r2) and S=(s1,s2). The roots of the quartic equation (1) are exactly -a, q1, r1 and s1.

From Viéte's formulas, the sum of roots is 2a. Therefore




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 \angleQPR=120o.


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