A1, A2, ... , An are points in the plane. A non-zero real number ki is assigned to each point, so that the square of the distance between Ai and Aj (for i ≠ j) is ki + kj. Show that n is at most 4 and that if n = 4, then 1/k1 + 1/k2 + 1/k3 + 1/k4 = 0.
Solution
Suppose we have four points A, B, C, D with associated numbers a, b, c, d. Then AB2 = a + b, AC2 = a + c, so AB2 - AC2 = b - c. Similarly, DB2 - DC2 = b - c, so AB2 - AC2 = DB2 - DC2. Let X be the foot of the perpendicular from A to BC, and Y the foot of the perpendicular from D to BC. Then AB2 - AC2 = (AX2 + XB2) - (AX2 + XC2) = XB2 - XC2. Similarly for D, so XB2 - XC2 = YB2 - YC2. Hence X = Y, so AD is perpendicular to BC. Similarly, BD is perpendicular to AC, and CD is perpendicular to AB. Hence D is the (unique) orthocenter of ABC. So n <= 4.
Suppose n = 4, so we have four points A, B, C, D with associated numbers a, b, c, d. We have AB2 + AC2 - BC2 = (a + b) + (a + c) - (b + c) = 2a. But by the cosine formula it is also 2 AB AC cos BAC. Hence a = AB AC cos BAC. Similarly for A, B, D etc. Hence ab/cd = (AB AC cos BAC)(BA BD cos ABD)/( (CA.CD cos ACD)(DB DC cos BDC) ) = (AB2/CD2) (cos BAC/cos BDC) (cos ABD/cos ACD).
Take ABC to be acute with D inside. Then angle ABD = angle ACD ( = 90o - angle BAC), and angle BDC = 90o + angle ACD = 180o - angle BAC. So cos BAC/cos BDC = -1. Thus ab/cd = - AB2/CD2 = - (a + b)/(c + d). Hence ab(c + d) + cd(a + b) = 0, so 1/a + 1/b + 1/c + 1/d = 0.
© John Scholes
jscholes@kalva.demon.co.uk
22 Oct 2000