USAMO, 2010

The answer is no, it is not possible for segments $AB$, $BC$, $BI$, $ID$, $CI$, $IE$ to all have integer lengths.

Suppose on the contrary that these segments do have integer side lengths. Set $\alpha = \angle ABD = \angle DBC$ and $\beta = \angle ACE = \angle ECB$. Note that $I$ is the incenter of triangle $ABC$, and so $\angle BAI = \angle CAI = 45^\circ$. Applying the Law of Sines to triangle $ABI$ yields
$$\frac{AB}{BI} = \frac{\sin(45^\circ + \alpha)}{\sin 45^\circ} = \sin \alpha + \cos \alpha,$$
by the sine addition formula. In particular, we conclude that $s = \sin \alpha + \cos \alpha$ is rational. It is clear that $\alpha + \beta = 45^\circ$. By the sine and cosine subtraction formulas, we have
$$s = \sin(45^\circ - \beta) + \cos(45^\circ - \beta) = \sqrt{2} \cos \beta,$$
from which it follows that $\cos \beta$ is not rational. On the other hand, from right triangle $ACE$, we have $\cos \beta = AC/EC$, which is rational by assumption. Therefore, $\cos \beta$ is both rational and irrational, a contradiction. Hence, our assumption was wrong, and not all the segments $AB$, $BC$, $BI$, $ID$, $CI$, $IE$ can have integer lengths.

We proceed again by contradiction; suppose it were possible for $AB$, $BC$, $BI$, $ID$, $CI$, $IE$ to all have integer lengths. Write $BD = m$, $AD = x$, $DC = y$, $AB = c$, $BC = a$ and $AC = b$. The angle bisector theorem implies
$$\frac{x}{b-x} = \frac{c}{a}$$
and the Pythagorean Theorem yields $m^2 = x^2 + c^2$. Both equations imply that
$$2ac = \frac{(bc)^2}{m^2 - c^2} - a^2 - c^2,$$
and since $a^2 = b^2 + c^2$ is rational, $a$ is rational too (observe that to reach this conclusion, we only need to assume that $b, c$, and $m$ are integers). Therefore, $x = \frac{bc}{a+c}$ is also rational, and so is $y$. As in the previous solution, now define $\angle ABD = \alpha$ and $\angle ACE = \beta$ where $\alpha + \beta = \pi/4$. It is obvious that $\cos \alpha$ and $\cos \beta$ are both rational, and the above shows also that $\sin \alpha = x/m$ is rational. On the other hand, $\cos \beta = \cos(\pi/4 - \alpha) = (\sqrt{2}/2)(\sin \alpha + \cos \beta)$, which is a contradiction.

We prove an even stronger result: There is no such right triangle with $AB$, $AC$, $IB$, and $IC$ having rational lengths. Suppose the contrary, that $AB$, $AC$, $IB$, and $IC$ have rational lengths. Then $BC^2 = AB^2 + AC^2$ is rational. On the other hand, in triangle $BIC$, $\angle BIC = 135^\circ$. Applying the law of cosines to triangle $BIC$ yields
$$BC^2 = BI^2 + CI^2 - \sqrt{2} \cdot BI \cdot CI,$$
which is irrational. Because $BC^2$ cannot be both rational and irrational, we conclude that our assumption was wrong and that not all of the segments $AB$, $AC$, $IB$, and $IC$ can have rational lengths.