1st ASU 1961 problems Problem 6 (a) A and B move clockwise with equal angular speed along circles center P and Q respectively. C moves continuously so that AB=BC=CA. Establish C's locus and speed. *(b) ABC is an equilateral triangle and P satisfies AP=2, BP=3. Establish the maximum possible value of CP. Solution (a) Represent A, B as complex numbers z 1 + w 1 e it , z 2 + w 2 e it . Then C is (z 1 + w 1 e it ) + (z 2 + w 2 e it - z 1 - w 1 e it ) e i π/3 , which is also of the form z + w e it . However, there is one subtlety. There are actually two circles possible for C depending on which side of AB we place it. The continuity requirement means that C is normally confined to one of the circles. However, if A and B ever coincide then C may be able to switch to the other circle. If we regard "moves continuously" as allowing a discontinuous velocity, then a switch is always possible (provided A and B coincide). (b)
: 5. P must be the opposite side of AB to C (or we could increase CP, whilst keeping AP and BP the same, by reflecting in AB). Similarly it must be on the same side of AC as B, and on the same side of BC as A. For any P in this region the quadrilateral APBC is convex and hence satisfies Ptolemy's inequality CP·AB ≤ AP·BC + BP·AC, with equality iff APBC is cyclic. But AB = BC = CA, so we have CP ≤ AP + BP = 5 with equality iff P lies on the arc AB of the circle ABC. Note that there is just one such point, because the locus of P such that BP=1.5 AP is a circle which cuts the arc just once. Ptolemy's inequality for 4 points A, B, C, D: AB·CD + BC·AD ≥ AC·BD with equality iff ABCD is a cyclic quadrilateral (meaning A, B, C, D lie on a circle in that order). Proof Take E inside ABCD such that ∠DAE = ∠CAB and ∠ADE = ∠ACB. Then ADE and ACB are similar, so DE/CB = AD/AC and hence BC·AD = AC·DE. It also follows that AE/AB = AD/AC. But we also have ∠EAB = ∠DAC and hence AEB and ADC are also similar. So EB/AB = DC/AC, and hence AB·CD = AC·EB. Adding, we have: AB·CD + BC·AD = AC(BE + ED) ≥ AC·BD with equality iff E lies on BD, or equivalently ABCD is cyclic. This glosses over one point. It only follows that ∠EAB = ∠DAC if ABCD is convex. For the convex case, we have that ∠EAB = ∠CAB + ∠EAC and ∠DAC = ∠DAE + ∠EAC, or ∠EAB = ∠CAB - ∠EAC and ∠DAC = ∠DAE - ∠EAC. Either way ∠EAB = ∠DAC. But in the non-convex case, we can have ∠EAB = ∠CAB + ∠EAC and ∠DAC = ∠DAE - ∠EAC (or - ... +) and hence the angles ∠EAB and ∠DAC are not necessarily equal. 1st ASU 1961 © John Scholes jscholes@kalva.demon.co.uk