Home > Shapes > Geometry
| 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: 4.4000 (5 votes)
|
|
Jer's arithmetic needs work...
|
 | Comment 3 of 10 | 
|
Jer, simplifying
b/c = 2(b/x)^2 - 1
I get:
x = b[2c/(b+c)]^0.5
|
|
Posted by vertigo
on 2005-03-04 19:15:46 |
|
|
Please log in:
Forums (0)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
|
