 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Sastry's Evil Bisectors (Posted on 2005-03-04) Prove that if a right triangle has all sides of integral length, then it has at most one angle bisector of integral length.

 No Solution Yet Submitted by owl Rating: 4.2000 (5 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Doesn't seem to work. | Comment 1 of 10

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 2005-03-04 18:50:00 Please log in:

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