Let \(\triangle ABC\) be an acute angled triangle. Let \(A_1, B_1, C_1\) be the midpoints of \(BC,CA,AB\) respectively. The internal angle bisector of \(\angle C_1A_1B_1\) intersects lines \(AB\) and \(AC\) at points \(A_2, A_3\) respectively. Let \(M_A\) and \(N_A\) be the circumcenters of \(\triangle BCA_2 \) and \(\triangle BCA_3\) respectively. Points \(M_B, N_B, M_C, N_C\) are defined analogously. It turns out that lines \(M_AN_A, M_BN_B, M_CN_C\) are concurrent at a point \(X\) within \(\triangle ABC.\) Given that \(\angle ABC = 30^{\circ}, \angle BCA= 45^{\circ}\) and \(k=\dfrac{AB+BC+CA}{XA+XB+XC},\) find \(\left \lfloor 100 k \right \rfloor.\)

**Details and assumptions**

- The floor function \(\left \lfloor x \right \rfloor \) denotes the largest integer \(\leq x.\) For example, \(\left \lfloor 4.3 \right \rfloor = 4, \left \lfloor \pi \right \rfloor= 3.\)

- You might use a scientific calculator.

- A picture will accompany soon.

- GeoGebra users will be prosecuted.

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