Through a point $P$ exterior to a given circle pass a secant and a tangent to the circle. The secant intersects the circle at $A$ and $B$, and the tangent touches the circle at $C$ on the same side of the diameter through $P$ as the points $A$ and $B$. The projection of the point $C$ on the diameter is called $Q$. Prove that the line $QC$ bisects the angle $\angle AQB$.
Problem
Source: Baltic Way 2004 Problem 16
Tags: projective geometry, geometry proposed, geometry
sprmnt21
21.11.2004 10:41
O==center(ABC); AO = BO ==> <BAO = <ABO; PC^2 =PQ*PO = PA*PB ==> AQOB cyclic ==> <AQP = <ABO = <BAO = <BQO = a. <CQA = 90-a = <CQB.
Valentin Vornicu
25.10.2005 20:29
Let the lines CQ and AB meet at X. From the definition of the line CQ, it directly follows that the line CQ is the polar of the point P with respect to our circle. Hence, by the harmonic property of a polar, the point X, where this line CQ meets the line AB, is the harmonic conjugate of the point P with respect to the segment AB. In other words, the points P and X divide the segment AB harmonically. Hence, the rays QP and QX divide the angle formed by the rays QA and QB harmonically. But on the other hand, we know that the rays QP and QX are perpendicular (in fact, the ray QP is the diameter of the circle through the point P, and the ray QX is the ray CQ). It is well-known that if two perpendicular rays divide an angle harmonically, then these two rays are the two angle bisectors (internal and external) of the angle. Hence, the rays QP and QX are the two angle bisectors of the angle AQB. Since the ray QX is the ray QC, it follows that the ray QC bisects the angle AQB. Proof complete. This is in fact Darij's solution, re-posted.
Virgil Nicula
26.10.2005 14:09
This easy and nice problem is a immediate consequence of the properties of a harmonical division. See and other problems to http://www.mathlinks.ro/Forum/viewtopic.php?t=46146