From: Pierre Bornsztein <pbornszt@club-internet.fr>
Subject: Re: geometry question
Date: Wed, 11 Aug 1999 23:10:45 +0200
Newsgroups: sci.math
To: Jerome Leo Terry <v57jlt@morgan.ucs.mun.ca>
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.