Prove that if a right triangle has all sides of integral length, then it has at most one angle bisector of integral length.
Am I right in thinking the right angle bisector is the geometric mean of the legs?
I derived formulae for the lengths of the other two bisectors.
Call the triangle ABC with right triangle at C and legs a and b and hypotenuse c.
Construct the bisector of angle A. Call its length x. We have
cos(A) = b/c
and
cos(A/2)= b/x
Applying the double angle formula for the first:
b/c = 2[cos(a/2)]^2  1
substituting
b/c = 2[b/x]^2  1
solve for x and simplify
x=(b/(b+c))*sqrt(2bc+2c^2)
Similarly the other bisector
y=(a/(a+c))*sqrt(2ac+2c^2)
Unless my algebra is faulty (a real possiblity) these formulae should work.
The only thing is for a {5, 12, 13} triangle they give irrational results
2sqrt(15)
(5/3)sqrt(13)
(12/5)sqrt(26)

Posted by Jer
on 20050304 18:50:00 