48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions

Let $N$ be the midpoint of segment $BT$ (see Figure 1). Line $XN$ is the axis of symmetry in the isosceles triangle $BXT$, thus $\angle TNX=90^{\circ }$ and $\angle BXN=\angle NXT$. Moreover, in triangle $BCT$, line $MN$ is the midline parallel to $CT$; hence $\angle CTM=\angle NMT$. Due to the right angles at points $M$ and $N$, these points lie on the circle with diameter $XT$. Therefore, $$\angle MTB=\angle MTN=\angle MXN\text{and}\angle CTM=\angle NMT=\angle NXT=\angle BXN.$$ Hence $$\angle MTB-\angle CTM=\angle MXN-\angle BXN=\angle MXB=\angle MAB$$ which does not depend on $X$.

Figure 1

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Alternative solution.

Let $S$ be the reflection of point $T$ over $M$ (see Figure 2). Then $XM$ is the perpendicular bisector of $TS$, hence $XB=XT=XS$, and $X$ is the circumcenter of triangle $BST$. Moreover, $\angle BSM=\angle CTM$ since they are symmetrical about $M$. Then $$\angle MTB-\angle CTM=\angle STB-\angle BST=\frac{\angle SXB-\angle BXT}{2}.$$ Observe that $\angle SXB=\angle SXT-\angle BXT=2\angle MXT-\angle BXT$, so $$\angle MTB-\angle CTM=\frac{2\angle MXT-2\angle BXT}{2}=\angle MXB=\angle MAB,$$ which is constant.

Figure 2