# Find The Angle

Geometry Level 5

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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