From: Pierre Bornsztein Subject: Re: geometry question Date: Wed, 11 Aug 1999 23:10:45 +0200 Newsgroups: sci.math To: Jerome Leo Terry Keywords: Ptolémée's theorem on quadrilaterals Jerome Leo Terry wrote: > > The following is a question from the Euclid Contest, a contest open to > undergraduate math students. I am preparing to enter such a contest this > October and I am hoping that someone can give me a hint how to solve it. > > Consider a quadrilateral ABCD inscribed in a circle of diameter d with > center 0. Points A and D are the left and right end points of the diameter > and points B and C lie above the diameter. Let AB = a, BC = a, and CD = b. > If a,b are integers with a not equal to b, > > (a) prove that d cannot be a prime number. > (b) find the minimum value of d. > > From Ptolémée's theorem, the convex quadrilateral ABCD is inscriptible if and only if BD*AC = BC*AD + AB*CD (1) From Pythagore's theorem : AC^2 = d^2 - b^2 and BDç2 = d^2 - a^2 Then (1) can be written as : ( d^2 - a^2)(d^2 - b^2) = a^2(d+b)^2 <=> d(d-b)=2a^2 (2) a) Since d is a diameter we have d>b and d>a. Suppose, for a contradiction, that d is a prime. Then, from (2), d divides 2a^2 it follows that d divides 2 or d divides a. -if d divides a : then d =< a . Contradiction. - if d divides 2 : then d = 2. thus 1 =< a,b < 2 = d with a,b distinct integers. Contradiction. then, d cannot be a prime. b) From above (the if and only if part), the minimal value of d is given by (2), with a,b distinct positive integers. (2) can be written as : d^2 -bd -2a^2 = 0 It is a quadratic equation. Since d>0, we easily find that : d = (b+ sqrt(b^2+8a^2))/2 (3) If d is not supposed to be an integer, the minimum is achieved for a=1 and b=2 (they have to be distinct). and then d = 1+sqrt(3) = 2,73205... If d has to be an integer, then b^2+8a^2 has to be a square. And from a), d cannot be equal to 1,2,3,5,7. - if d = 4 : then, 1 =< a,b =< 3. It's easy to verify that there is no solution in this case. - if d = 6 : then, 1 =< a,b =< 5 No solution again -if d = 8 : we find a = 2 and b = 7 A direct computation gives a = 2 and b = 7 and then d = 8 is the minimum value of d. Pierre.