3. On the surface of the cube $A B C D E F G H$, a broken line composed of four identical segments is constructed in the faces $A B F E, B C G F, C D H G$ and $G H E F$, which starts from vertex $A$ and ends at vertex $E$. Determine in what ratio this broken line divides the edge $C G$.
SOLUTION. Let $K, L, M$ be the points of the given broken line, which lie in sequence on the segments $B F, C G, G H$. Let the length of the edge of the cube be one and the foot of the perpendicular from point $K$ to the edge $C G$ be $P$ (Fig. 1). Right triangles $A K B, K L P$ and $E M H$ coincide in their hypotenuses $A K, K L$ and $E M$ and in the unit legs $A B, K P$ and $E H$. Therefore, they are congruent by the $S$ theorem and it holds that $|B K|=|L P|=|M H|=u$. From the right triangles

![](https://cdn.mathpix.com/cropped/2024_04_17_eeefd9d1448a1390df7fg-2.jpg?height=436&width=637&top_left_y=102&top_left_x=386)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_04_17_eeefd9d1448a1390df7fg-2.jpg?height=431&width=635&top_left_y=104&top_left_x=1081)

Fig. 2

$L M G$ and $A B K(|G L|=1-2 u,|G M|=1-u)$, we express the squares of the lengths of their hypotenuses using the Pythagorean theorem and compare them: $1+u^{2}=(1-u)^{2}+(1-2 u)^{2}$.

The equation has a single root less than $1: u=\frac{1}{4}(3-\sqrt{5})$. The ratio $|C L|:|L G|=2 u$ : $:(1-2 u)=\frac{1}{2}(\sqrt{5}-1)$ is equal to the golden ratio.

For the position of point $L$ on the edge $C G$, there is another possibility, illustrated on the same part of the cube grid in Fig. 2. Clearly, then points $C$ and $L$ are identical and $u=|B K|=|G M|=$ $=|M H|=\frac{1}{2}$, while the ratio $|C L|:|L G|$ turns out to be zero.

## ADDITIONAL PROBLEMS: