Given lengths R and r with R > 2r. Construct a triangle with side lengths in arithmetic progression and R and r its circumradius and inradius respectively.

let a,b,c be the sides of the triangle with a<b<c

now since a,b,c are in arithmetic progression we have that

(1) a=b-k c=b+k

now there is a little equation relating R,r,a,b,c and it is

(2) R=(abc)/(2*p*r)  where p=perimeter=(a+b+c)

substituting (1) into (2) we get

R=[(b-k)(b)(b+k)]/[2*(3*b)*r]

R=(b^2-k^2)/(6r)

now we also require that R>2r so

(b^2-k^2)/(6r)>2r

b^2-k^2>12r^2

now we have that

r=(1/2)Sqrt[(b^2-4k^2)/3]

r^2=(b^2-4k^2)/12

so we have

b^2-k^2>b^2-4k^2

thus

4k^2>k^2

4>1

Thus it would appear that any triangle with sides in arithmetic progression satisfies the requirement that R>2r

now all we need is to figure out restraints on b,k for a,b,c to form a valid triangle

a+b>c  b-k+b>b+k  b>2k

a+c>b  b+k+b-k>b  2b>b  2>1

b+c>a  b+b+k>b-k  2b+k>b-k  b+k>0

thus the only restraint is that b>2k  thus an example for such a triangle would be with b=3  k=1  and thus a=2, c=4

Edited on January 30, 2007, 3:58 pm

Posted by Daniel on 2007-01-30 14:06:08