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=120o.
Deadline expired on 16 January 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 QPR=120o.
|Statistics on problem A. 386.|
|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