IMO Shortlist, 2007

Let $N$ be the midpoint of segment $B T$ (see Figure 1).
Line $X N$ is the axis of symmetry in the isosceles triangle $B X T$, thus $\angle T N X = 90^{\circ}$ and $\angle B X N = \angle N X T$.
Moreover, in triangle $B C T$, line $M N$ is the midline parallel to $C T$; hence $\angle C T M = \angle N M T$.
Due to the right angles at points $M$ and $N$, these points lie on the circle with diameter $X T$. Therefore,
$$\angle M T B = \angle M T N = \angle M X N \quad \text{and} \quad \angle C T M = \angle N M T = \angle N X T = \angle B X N.$$
Hence
$$\angle M T B - \angle C T M = \angle M X N - \angle B X N = \angle M X B = \angle M A B$$
which does not depend on $X$.

Figure 1

Let $S$ be the reflection of point $T$ over $M$ (see Figure 2).
Then $X M$ is the perpendicular bisector of $T S$, hence $X B = X T = X S$, and $X$ is the circumcenter of triangle $B S T$.
Moreover, $\angle B S M = \angle C T M$ since they are symmetrical about $M$.
Then
$$\angle M T B - \angle C T M = \angle S T B - \angle B S T = \frac{\angle S X B - \angle B X T}{2}.$$
Observe that $\angle S X B = \angle S X T - \angle B X T = 2 \angle M X T - \angle B X T$, so
$$\angle M T B - \angle C T M = \frac{2 \angle M X T - 2 \angle B X T}{2} = \angle M X B = \angle M A B,$$
which is constant.

Figure 2