【Question 19】
There are two isosceles triangles with a leg length of 1, one with a vertex angle of $30^{\circ}$ (an acute isosceles triangle), and the other with a vertex angle of $90^{\circ}$ (a right isosceles triangle). As shown in the figure below, these two types of triangles are placed around a point in a clockwise direction, following these rules: the 1st and 2nd are acute isosceles triangles, the 3rd is a right isosceles triangle; the 4th and 5th are acute isosceles triangles, the 6th is a right isosceles triangle, and so on. If the $n$-th triangle exactly overlaps with the 1st triangle, find the smallest value of $n$.
【Analysis and Solution】
(Method One)
By drawing, we can get that the minimum value of $n$ is 23.
(Method Two)
(1) If $n=3 k+1$ (where $k$ is a natural number), then $2 k \times 30^{\circ}+k \times 90^{\circ}=150 k^{\circ}$ is an integer multiple of $360^{\circ}$;
The minimum value of $k$ is 18;
At this point, the minimum value of $n$ is $3 \times 18+1=55$.
(2) If $n=3 k+2$ (where $k$ is a natural number),

then $(2 k+1) \times 30^{\circ}+k \times 90^{\circ}=30(5 k+1)^{\circ}$ is an integer multiple of $360^{\circ}$;
The minimum value of $k$ is 7;
At this point, the minimum value of $n$ is $3 \times 7+2=23$.
In conclusion, the minimum value of $n$ is 23.