### 35th Putnam 1974

**Problem A2**

C is a vertical circle fixed to a horizontal line. P is a fixed point outside the circle and above the horizontal line. For a point Q on the circle, f(Q) ∈ (0, ∞] is the time taken for a particle to slide down the straight line from P to Q (under the influence of gravity). What point Q minimizes f(Q)?

**Solution**

Answer: Let X be the lowest point of C. Q is the (other) point at which PX intersects C.

Let k be the angle which PQ makes to the horizontal. Then the acceleration of the particle along PQ is g sin k. Hence the time squared is proportional to PQ/sin k. So we minimise the time by minimising PQ/sin k. Let PQ meet the circle again at R and let PT be tangent to the circle. Then PQ·PR = PT^{2}, a constant. So we minimise the time by maximising PR sin k. But PR sin k is just the vertical distance of R below P. That is clearly maximised by taking R = X.

35th Putnam 1974

© John Scholes

jscholes@kalva.demon.co.uk

18 Aug 2001