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Irish Mathematical Olympiad

Geometry Difficulty 5.1 AIME, harder Prove it Ireland

A fixed point OO is joined to any point PP on a given line which does not contain OO. On OPOP a point QQ is taken such that ∣OP∣⋅∣OQ∣|OP| \cdot |OQ| is constant. Show QQ lies on a fixed circle if PP varies. Examine all possible cases.

Solution

Let RR be the foot of the perpendicular from OO on the given line and let SS be chosen on OROR such that ∣OR∣⋅∣OS∣=∣OP∣⋅∣OQ∣|OR| \cdot |OS| = |OP| \cdot |OQ|, i.e. SS is a point on the locus. The value of the constant determines the position of SS, either between OO and RR or not. Then P,Q,SP, Q, S and RR lie on a circle in all cases (see formulation of the hint). Hence ∠PQS=90∘\angle PQS = 90^\circ since ∠PRO=90∘\angle PRO = 90^\circ. This implies that QQ lies on the circumference of the circle on OSOS as diameter which is a fixed circle. In every case QQ can be taken on either side of OO on the line OPOP. The analysis in every case (see drawings below) is similar to above.
Figure 1
Figure 2

Figure 3

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