Problem 2. On each side of an equilateral triangle, a point is taken. Each side of the triangle with vertices at these points is perpendicular to one of the sides of the original triangle. In what ratio does each of the taken points divide the side of the original triangle? What is the ratio of the areas of the original and the formed triangles?
Solution. Let points $K, L, M$ lie on the sides $A B, B C$, and $A C$ of an equilateral triangle $A B C$, respectively, such that $K L \perp B C, L M \perp A C, M K \perp A B$.

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1) Then $\angle M K L=180^{\circ}-\angle A K M-\angle L K B=180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}$. Similarly, we can show that $\angle K M L=60^{\circ}$, so triangle $K L M$ is equilateral. Right triangles $A K M, B L K$, and $C M L$ are congruent by a leg and an acute angle. Since $C M=A K=1 / 2 A M$ (as the leg opposite the $30^{\circ}$ angle), then $C M: A M=1: 2$. Similarly, $A K: K B=1: 2, B L: L C=1: 2$.
2) Let $C M=A K=B L=a$. Then the side of the original triangle is $3 a$, and the side of the inscribed triangle is $a \sqrt{3}$, i.e., $\sqrt{3}$ times smaller. Since the areas of equilateral triangles are proportional to the squares of their sides, the area of the inscribed triangle will be 3 times smaller.

Answer: 1) $1: 2$. 2) $3: 1$.