In the plane, there are four different points. Among the six distances between the four points, four distances are one unit, and one is 1.2. What could be the length of the unknown sixth distance?
Solution. First, consider the case when three segments of unit length form a regular triangle. Let the vertices of this triangle be denoted by $A, B, C$. The fourth point is 1 unit away from $A$ and 1.2 units away from $B$. Points that are 1 unit away from $A$ lie on a circle, and similarly, points that are 1.2 units away from $B$ lie on another circle. These two circles intersect at two points, which we will call $D$ and $D'$, as shown in the figure. We need to determine the missing sixth distance in both cases.

![](https://cdn.mathpix.com/cropped/2024_05_02_969a1e36e705d6557702g-1.jpg?height=594&width=374&top_left_y=389&top_left_x=867)

In the first case, let the angle $\angle ABD$ be $\beta$, and apply the cosine rule in triangle $ABD$ for $\beta$:

$$
1^{2}=1^{2}+1.2^{2}-2 \cdot 1 \cdot 1.2 \cdot \cos \beta
$$

From this, $\cos \beta=\frac{1.44}{2.4}=0.6$ and $\beta \approx 53.13^{\circ}$. Next, apply the cosine rule in triangle $DCB$. Here, $\angle CBD = \angle CBA + \angle ABD \approx 60^{\circ} + 53.13^{\circ} = 113.13^{\circ}$.

$$
DC^{2}=1^{2}+1.2^{2}-2 \cdot 1 \cdot 1.2 \cdot \cos 113.13^{\circ}
$$

From this, $DC^{2} \approx 2.44 + 2.4 \cdot 0.3928 = 3.3827$, and $DC \approx 1.84$.

In the second case, we need to determine the distance $D'C$. Let the angle $\angle D'AB$ be $\alpha$. Apply the cosine rule in triangle $ABD'$:

$$
1.2^{2}=1^{2}+1^{2}-2 \cdot \cos \alpha, \quad \cos \alpha=\frac{0.56}{2}=0.28
$$

From this, $\alpha \approx 73.74^{\circ}$ and $\angle D'AC = \alpha - 60^{\circ} \approx 13.74^{\circ}$.

In triangle $ACD'$:

$$
CD'^{2}=1^{2}+1^{2}-2 \cdot \cos 13.74^{\circ}
$$

Thus, $CD' \approx 0.24$.

Finally, if the unit segments do not form a triangle, we get a quadrilateral with one diagonal of 1.2 units and all sides of 1 unit. Since the quadrilateral is a rhombus, its diagonals bisect each other at right angles. We can apply the Pythagorean theorem in the right triangle:

$$
x=\sqrt{1^{2}-0.6^{2}}=0.8 \quad \text{and} \quad BD=1.6.
$$

The missing sixth segment length, depending on the arrangement of the points, can be 1.84, 0.24, and 1.6 units.

![](https://cdn.mathpix.com/cropped/2024_05_02_969a1e36e705d6557702g-1.jpg?height=331&width=393&top_left_y=2144&top_left_x=861)