olymmath

NMC 1995 Problem 1

    Let ABA B be a diameter of a circle with centre OO. We choose a point CC on the circumference of the circle such that OCO C and ABA B are perpendicular to each other. Let PP be an arbitrary point on the (smaller) arc BCB C and let the lines CPC P and ABA B meet at QQ. We choose RR on APA P so that RQR Q and ABA B are perpendicular to each other. Show that ∣BQ∣=∣QR∣|B Q|=|Q R|.

    Figure 7 .

    Dataset text NuminaMath olympiads. Quoted with source.

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