L is a tangent to the circle C and M is a point on L. Find the locus of all points P such that there exist points Q and R on L equidistant from M with C the incircle of the triangle PQR.
Solution
Answer: Let X be the point where C meets L, let O be the center of C, let XO cut C gain at Z, and take Y on QR so that M be the midpoint of XY. Let L' be the line YZ. The locus is the open ray from Z along L' on the opposite side to Y.
mainly by Gerhard Wöginger, Technical University, Graz (I filled in a few details)
Let C' be the circle on the other side of QR to C which also touches the segment QR and the lines PQ and QR. Let C' touch QR at Y'. If we take an expansion (technically, homothecy) center P, factor PY'/PZ, then C goes to C', the tangent to C at Z goes to the line QR, and hence Z goes to Y'. But it is easy to show that QX = RY'.
We focus on the QORO'. Evidently X,Y' are the feet of the perpendiculars from O, O' respectively to QR. Also, OQO' = ORO' = 90. So QY'O' and OXQ are similar, and hence QY'/Y'O' = OX/XQ. Also RXO and O'Y'R are similar, so RX/XO = O'Y'/Y'R. Hence QY'·XQ = OX·O'Y' = RX·Y'R. Hence QX/RX = QX/(QR - QX) = RY'/(QR - RY') = RY'/QY'. Hence QX = RY'.
But QX = RY by construction (M is the midpoint of XY and QR), so Y = Y'. Hence P lies on the open ray as claimed. Conversely, if we take P on this ray, then by the same argument QX = RY. But M is the midpoint of XY, so M must also be the midpoint of QR, so the locus is the entire (open) ray.
Gerhard only found this after Theo Koupelis, University of Wisconsin, Marathon had already supplied the following analytic solution.
Take Cartesian coordinates with origin X, so that M is (a, 0) and O is (0, R). Let R be the point (b, 0) (we take a, b >= 0). Then Q is the point (2a - b, 0), and Y is (2a, 0). Let angle XRO be θ. Then tan θ = R/b and angle PRX = 2θ, so tan PRX = 2 tan θ/( 1 - tan2θ) = 2Rb/(b2 - R2). Similarly, tan PQX = 2R(b - 2a)/( (b - 2a)2 - R2).
If P has coordinates (A, B), then B/(b - A) = tan PRX, and B/(b - 2a + x) = tan PQX. So we have two simultaneous equations for A and B. Solving, and simplifying slightly, we find A = -2aR2/(b2 - 2ab - R2), B = 2b(b - 2a)R/(b2 - 2ab - R2). (*)
We may now check that B/(2a - A) = R/a, so P lies on YZ as claimed. So we have shown that the locus is a subset of the line YZ. But since b2 - 2ab - R2 maps the open interval (a + √(a2 + R2), ∞) onto the open interval (0, ∞), (*) shows that we can obtain any value A in the open interval (-∞,0) by a suitable choice of b, and hence any point P on the ray (except its endpoint Z) by a suitable choice of R.
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
8 Sep 1999