The problem is translated from another language and is an 11th grade problem.
ABC is an equilateral triangle with side length $a$. Another equilateral triangle DEF fits inside ABC with its sides forming the angle $\alpha$ with the sides of ABC, as shown in the diagram.
I) Find the side length of the triangle DEF in terms of $a$ and $\alpha$.
II) If DE || BC, what is the side length in terms of $a$ and explain the geometric significance of the result.
Let $b$ be the side length of the small equilateral triangle. And let x and y be the side segment lengths of the large triangle, with $x+y=a$.
(I) Use the sine rule for one of the corner triangles to establish,
$$\frac xb = \frac{\sin\alpha}{\sin 60},\>\>\>\>\>\frac yb = \frac{\sin(120-\alpha)}{\sin 60}$$
Then,
$$x+y=a=b\>\frac{\sin\alpha+\sin(120-\alpha)}{\sin60}$$
Solve for the side length of the small triangle $b$,
$$b=\frac{a\sin60}{\sin\alpha+\sin(120-\alpha)}=\frac{a}{\sqrt3\sin \alpha+ \cos\alpha}$$
(II) In the case where DE || BC, $\alpha =60$ and $ b=\frac a2$. Geometrically, the vertexes of the small equilateral triangle bisect the sides of the large triangle. And the large triangle is portioned into 4 congruent equilateral triangles.