Problems of Triangles

We first solve for the tangent segments for each of the three sides of the triangle, as shown below on the left.

We have $a + b = 2, a + c = 3, b + c = 4$ ; half the sum of all three gives $a + b + c = \frac{9}{2}$ , then subtracting each original equation yields $a = \frac{1}{2}, b = \frac{3}{2}$ and $c = \frac{5}{2}$ .

Now the added parallel tangent line creates a smaller triangle which is similar to the original triangle, as shown above on the right. Then

$\begin{array}{rlccrl} \dfrac{\frac{5}{2} - \alpha}{\alpha + \beta} &= \ \dfrac{3}{2} & & & \dfrac{\frac{5}{2} - \beta}{\alpha + \beta} &= \ \dfrac{4}{2} \\ \\ 5 - 2\alpha &= \ 3(\alpha + \beta) & & & 5 - 2\beta &= \ 4(\alpha + \beta) \\ \\ 5\alpha + 3\beta &= \ 5 & & & 4\alpha + 6\beta &= \ 5 \\ \\ \\ & & \alpha = \ \dfrac{5}{6}; & \ \beta = \ \dfrac{5}{18} \end{array}$

so that the length of XY is $\alpha + \beta = \boxed{\dfrac{10}{9}}$