We know the following about five points in the plane:

a) Three points form an equilateral triangle.

b) The other two points are symmetric with respect to one axis of the triangle, the distance between them is 1 unit, and the distance from the point on the axis is $\sqrt{7}$ units.

c) The distance between one element of each of the symmetric point pairs is 2 units.

What is the side length of the equilateral triangle?
The five points can be arranged according to Figures 1, 2, and 3. Based on the conditions, $A D = A E = \sqrt{7}$, and $D E = 1$, further $B D = C E = 2$ or $C D = B E = 2$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e7d1134074860b7783e6g-1.jpg?height=569&width=330&top_left_y=244&top_left_x=884)

1. Figure
![](https://cdn.mathpix.com/cropped/2024_05_02_e7d1134074860b7783e6g-1.jpg?height=1118&width=588&top_left_y=913&top_left_x=758)
2. Figure

![](https://cdn.mathpix.com/cropped/2024_05_02_e7d1134074860b7783e6g-1.jpg?height=515&width=290&top_left_y=2144&top_left_x=901)

4. Figure

The situation in Figure 4 is not possible because in that case $E C > 2$. Using the notations in the figures, let $E^{\prime} C = x$ and $F F^{\prime} = E E^{\prime} = y$.

First, we examine the cases according to Figures 1 and 2. Based on the figures, $A F = \frac{3 \sqrt{3}}{2}$ and $A F^{\prime} = \frac{\sqrt{3}}{2}(2 x + 1)$. To solve the problem, it is sufficient to determine $x$. This can be done using the right-angled triangles $E E^{\prime} C$ or $E E^{\prime} B$ for each figure. In the first case,

\[
\begin{aligned}
x^{2} + y^{2} & = 4, \quad \text{and in the other case} \\
(x + 1)^{2} + y^{2} & = 4
\end{aligned}
\]

In both cases, $E E^{\prime} = y = \left|A F^{\prime} - A F\right| = \left|\frac{\sqrt{3}}{2}(2 x + 1) - \frac{3 \sqrt{3}}{2}\right| = \sqrt{3}|x - 1|$. Accordingly, from (1): $x^{2} + 3(x - 1)^{2} = 4$, and from (2): $(x + 1)^{2} + 3(x - 1)^{2} = 4$. Solving these quadratic equations, we get $x = \frac{3 \pm \sqrt{13}}{4}$, and from (2): $x = 0, x = 1$. Since $x \geq 0$, the solutions are $x = \frac{3 + \sqrt{13}}{4}, x = 0$, and $x = 1$. Therefore, the side $B C = 2 x + 1$ of the equilateral triangle can be $1, 3, \frac{5 + \sqrt{13}}{2}$.

In the case of Figure 3, the only difference is that $A F^{\prime} = \frac{\sqrt{3}}{2}(1 - 2 x)$ and $E E^{\prime} = y = A F - A F^{\prime} = \sqrt{3}(x + 1)$. From the right-angled triangles $E E^{\prime} C$ or $E E^{\prime} B$,

\[
\begin{aligned}
x^{2} + y^{2} & = 4, \quad \text{or} \\
(1 - x)^{2} + y^{2} & = 4
\end{aligned}
\]

Then, from (3) and (4), we get the following quadratic equations:

\[
\begin{aligned}
x^{2} + 3(x + 1)^{2} & = 4, \quad \text{or} \\
(1 - x)^{2} + 3(x + 1)^{2} & = 4
\end{aligned}
\]

Solving these equations, we get $x = \frac{-3 \pm \sqrt{13}}{4}$, and from (4): $x = 0, x = -1$. Since $x$ is non-negative, $x = \frac{-3 + \sqrt{13}}{4}$, and $x = 0$ are possible. Therefore, the side $B C = 1 - 2 x$ of the equilateral triangle can be $1$ or $\frac{5 - \sqrt{13}}{2}$.

Comparing the results, we see that the problem has four solutions: $1, 3, \frac{5 + \sqrt{13}}{2}, \frac{5 - \sqrt{13}}{2}$.

Based on the conditions, the five points can be easily constructed.

Based on the work of Balázs Kotnyek (Bp., Jedlik Á. Gymn., IV. grade)

Note. In the calculations for Figures 1 and 2, we could have allowed $F^{\prime} C - F^{\prime} E^{\prime} = x$ to be negative. This would have saved us the calculations for Figure 3.