Problem:
Let be a triangle with . Points and lie on sides and , respectively, such that and . Segments and meet at . Determine whether or not it is possible for segments to all have integer lengths.
Solution:
Solution from Zuming Feng: The answer is no, it is not possible for segments to all have integer lengths.
Assume on the contrary that these segments do have integer side lengths. We set and . Note that is the incenter of triangle , and so . Applying the Law of Sines to triangle yields
by the addition formula (for the sine function). In particular, we conclude that is rational. It is clear that . By the subtraction formulas, we have
from which it follows that is not rational. On the other hand, from right triangle , we have , which is rational by assumption. Because cannot not be both rational and irrational, our assumption was wrong and not all the segments , can have integer lengths.
Alternate Solution from Jacek Fabrykowski: Using notations as introduced in the problem, let , and . The angle bisector theorem implies
and the Pythagorean Theorem yields . Both equations imply that
and since is rational, is rational too (observe that to reach this conclusion, we only need to assume that , and are integers). Therefore, is also rational, and so is . Let now (similarly to the notations above from the solution by Zuming Feng) and where . It is obvious that and are both rational and the above shows that also is rational. On the other hand, , which is a contradiction. The solution shows that a stronger statement holds true: There is no right triangle with both legs and bisectors of acute angles all having integer lengths.
Alternate Solution from Zuming Feng: Prove an even stronger result: there is no such right triangle with having rational side lengths. Assume on the contrary, that have rational side lengths. Then is rational. On the other hand, in triangle . Applying the law of cosines to triangle yields
which is irrational. Because cannot be both rational and irrational, we conclude that our assumption was wrong and that not all of the segments can have rational lengths.
This problem was proposed by Zuming Feng.
The problems on this page are the property of the MAA's American Mathematics Competitions