[ Examples and counterexamples. Constructions ]

Six identical parallelograms of area 1 were used to cover a cube with an edge of 1. Can we assert that all parallelograms are squares? Can we assert that all of them are rectangles?
Examples of such cube nets are shown in Figure 1.
![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-40.jpg?height=522&width=918&top_left_y=2188&top_left_x=44)

Fig. 1

## [Elementary (basic) constructions with a compass and straightedge] Problem 107725 Topics: [ Construction of triangles by various elements $]$ Difficulty: $3+$ Grades: $7,8,9,10$

Given a line and a point outside it, how can you construct a line parallel to the given line and passing through the given point using a compass and straightedge, while drawing the fewest possible lines (circles and lines), so that the last line drawn is the desired line? What is the minimum number of lines you were able to achieve?

## Solution

Given a line $a$ and a point $O$ (notations). Mark two arbitrary points $A$ and $B$ on the line. Draw a circle with center at point $B$ and radius $AO$, and a circle with center at point $O$ and radius $AB$. They will intersect at point $X$. The quadrilateral $AOXB$ is a parallelogram because its opposite sides are equal. Now we can draw the desired line $OX$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-41.jpg?height=400&width=780&top_left_y=812&top_left_x=-1)

Fig. 1

Expressing this solution in different words, we can say that we constructed the triangle $BOX$ in the standard way by two vertices ($B$ and $O$) and the lengths of two sides, equal to the lengths of segments $AO$ and $AB$. Clearly, $\triangle ABO = \triangle XOB$ (by three sides). Therefore, $\angle ABO = \angle XOB$, which are alternate interior angles for lines $a$ and $OX$ and the transversal $BO$. From the equality of these angles, it follows that $a$ and $OX$ are parallel. Another solution. Mark an arbitrary point $A$ on the line and draw a circle with center at point $A$ and passing through point $O$. This circle intersects the line at two points; denote them as $M$ and $N$. Next, measure (see explanation at the end of the problem) the segment $MO$ with the compass and draw a circle with center at point $N$ and radius $MO$. The desired line passes through point $O$ and point $B$, the intersection of the two constructed circles.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-41.jpg?height=452&width=806&top_left_y=1649&top_left_x=0)

Fig. 2

$\triangle MAO = \triangle NAB$ by three sides, so the heights of these triangles, drawn from vertices $O$ and $B$, are equal. The bases of these triangles ($MA$ and $NA$) lie on line $a$, so points $O$ and $B$ are at the same distance from line $a$.

A disadvantage of this solution is that if point $A$ happens to be the foot of the perpendicular dropped from point $O$, then points $O$ and $B$ coincide and do not define the required line. This same issue is "masked" in the first solution in the statement "Mark two arbitrary points $A$ and $B$ on the line." If the points are arbitrary, they can accidentally coincide (and then the construction will not work), and to mark two non-coincident points on the line, additional lines will need to be drawn.

Now let's prove that it is impossible to achieve the construction with only two lines. The second line must be the desired line. To draw it, we need to obtain a second point that is at the same distance from line $a$ as point $O$. However, after drawing one line, all points on this line, except for the points of intersection with line $a$, will be indistinguishable, and it is impossible to find a second point at the required distance from line $a$ by drawing only one line.

Explanation. In the solution, we mentioned a parallelogram, triangles, the transversal $BO$, and angles. However, for the construction, we only needed the points (vertices of the parallelogram and triangles, endpoints of the transversal segment, and endpoints of the segments forming the angles). The segments themselves were not needed for the construction, so we did not draw them and, of course, did not count them when tallying the number of lines drawn.

Clarification for the problem. In classical works on geometry, the question of which constructions with a compass and straightedge are possible in principle is discussed, but not the number of operations required for a particular construction. However, differences of opinion can arise on this issue. For example, in Euclid's "Elements," it is considered impossible to measure a distance with a compass and transfer it to construct a circle with an arbitrary center. However, in Theorem 2 of this book, it is proven that transferring a measured distance is possible, but not in one step, but through a certain construction performed in several steps.

After this theorem, one can forget about how the distance is transferred—whether in one step or several—unless the question is about the principle of the construction's possibility, not the number of necessary constructions. In modern geometry textbooks, it is generally accepted that no special constructions are required to transfer a distance. For example, in the well-known textbook by Pogorelov, it is stated that if the center and radius are given, the circle is considered constructed. In this case, the authors of the problem proceed from this point of view.

## Answer

3 lines.

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