Given an equilateral triangle $ABC$. Let the reflection of $A$ over $C$ be $B_{1}$, the reflection of $B$ over $A$ be $C_{1}$, and the reflection of $C$ over $B$ be $A_{1}$. In what ratio do the sides of the triangle $A_{1} B_{1} C_{1}$ bisect the sides of the triangle $ABC$?
I. solution. Let the line $A_{1} B_{1}$ intersect $A B$ at point $K$, and let the reflection of $B$ over $C$ be $D$. Then $B_{1} D$ is parallel to $A B$ - since $B_{1} D$ is the reflection of $A B$ over $C$ - and $B$ divides the segment $A_{1} D$ in the ratio $1:2$, so $K$ also divides $A_{1} B_{1}$ in the ratio $A_{1} K: K B_{1}=1: 2$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_27eb7c5e85e8684cdebeg-1.jpg?height=461&width=720&top_left_y=293&top_left_x=713)

The side $B_{1} C_{1}$ is constructed in the same way from the triplet of points $A, C, B$ as $A_{1} B_{1}$ was from $C, B, A$, so $B C$ also divides $B_{1} C_{1}$ in the ratio $1:2$, and similarly, $C A$ divides $C_{1} A_{1}$ in the same ratio.

We did not use the fact that triangle $A B C$ is equilateral, so our conclusion is valid for any triangle.

Dettai István (Pannonhalma, Bencés g. I. o. t.)

Angster Judit (Pécs, Nagy Lajos g. I. o. t.)

II. solution. Let the line through $C$ parallel to $A B$ intersect $A_{1} B_{1}$ at $L$. Then $C L$ is the midline of triangle $A B_{1} K$, and $B K$ is the midline of triangle $A_{1} C L$; hence $L$ bisects $B_{1} K$, and $K$ bisects $L A_{1}$: $B_{1} L = L K = K A_{1}$. Thus, $A_{1} K: K B_{1}=1: 2$. - We can reason similarly for the other sides.

Terlaky Edit (Kaposvár, Táncsics M. g. I. o. t.) Györy Jenö (Budapest, Berzsenyi D. g. I. o. t.)

III. solution. Let the line through $A_{1}$ parallel to $A B_{1}$ intersect $A B$ at $C_{2}$. Then $A_{1} C_{2}$ is the reflection of $C A$ over $B$, so it is equal to it, and thus to $B_{1} C$. The diagonals of the parallelogram $A_{1} C_{2} B_{1} C$ bisect each other, which implies that $A_{1} B_{1}$ is the line of one of the medians of triangle $A_{1} C_{2} C$. $C_{2} B$ is another median, so $K$ is the centroid of the triangle, and thus $A_{1} K$ is two-thirds of the median, and one-third of the length of $A_{1} B_{1}$.

Soós Miklós (Budapest, Fazekas M. gyak. g. II. o. t.)

IV. solution. The sought ratio $A_{1} K / K B_{1}$ is equal to the ratio of the distances from points $A_{1}$ and $B_{1}$ to the line $A B$. These distances are the altitudes of triangles $A B A_{1}$ and $A B B_{1}$ to their common side; thus, the ratio of the areas of the triangles also gives the sought ratio.

The area of triangle $A B B_{1}$ is twice the area of triangle $A B C$ because the median $B C$ bisects the area of triangle $A B B_{1}$. The area of triangle $A B C$ is equal to the area of triangle $A B A_{1}$ because the median $A B$ bisects the area of triangle $A A_{1} C$. Thus, the area of triangle $A B B_{1}$ is twice the area of triangle $A B A_{1}$, so $B_{1} K$ is twice the length of segment $K A_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_27eb7c5e85e8684cdebeg-1.jpg?height=344&width=483&top_left_y=1872&top_left_x=756)

$(x=2 / 5, \quad(x+y)=2 / 7)$

Figure 2

Remark. If on the extensions of the sides of triangle $A B C$ we measure the corresponding sides scaled by $\lambda$, then - as can be easily shown similarly to the above - (Figure 2):

$$
\frac{A_{1} K}{K B_{1}}=\frac{t_{A B A_{1}}}{t_{A B B_{1}}}, \quad t_{A B B_{1}}=t_{A B C}+t_{B B_{1} C}, \quad t_{B B_{1} C}=\lambda t_{A B C}, \quad t_{A A_{1} B}=\lambda t_{A B C}
$$

Thus,

$$
\frac{A_{1} K}{K B_{1}}=\frac{\lambda \cdot t_{A B C}}{(1+\lambda) \cdot t_{A B C}}=\frac{\lambda}{1+\lambda}
$$

In our case, $\lambda=1$, so the value of the ratio is $1 / 2$.