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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

xy=1

and

(x-a)2+(y-b)2=4(a2+b2).

Then

(x-a)^2+\left(\frac1x-b\right)^2=4(a^2+b^2)

x4-2ax3-3(a2+b2)x2-2bx+1=0.(1)

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

q1+r1+s1=3a.(2)

Similarly,

q2+r2+s2=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 \angleQPR=120o.


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.

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