Develop spatial imagination

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DEVELOP
SPATIAL
IMAGINATION

N. G. Fedin

Exams often offer stereometric problems, for which the most important thing is to imagine how the figures are located in space. Foreign messages suggest ideas for solving planimetric problems. In this article we will talk about some questions and tasks that should, to some extent, interest high school students and help them expand the horizons of their spatial imagination.

Crossbreeding primo

1. Let’s draw two crossing primaries $a$ and $b$. On figure 1 their image is given.
   Are you surprised that the primes are shown parallel? In vain. This is a correct drawing, although not visual. It is known that crossing primes $a$ and $b$ can be placed in parallel planes $a$ and $b$ so that the prime $a$ lies in the plane $\alpha$, and the straight line $b$ lies in $\beta$, and then project them onto the plane $\gamma$, intersecting the planes $\alpha$ and $\beta$ (Fig. 2). In this plane we will obtain images $a_1$ and $b_1$ of straight lines $a$ and $b$.

2. Let the given crossing primes $a$ and $b$ and points $A_1, A_2$ on the line $a$ and $B_1, B_2$ on the line $b$ (Fig. 3). Let’s carry out receptions $A_1B_1$ and $A_2B_2$.

Fig. 1
Fig. 2

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Can these lines intersect?

Of course, the drawing can derail us. But if we assume that the lines $A_1B_1$ and $A_2B_2$ intersect, then it turns out that the four points $A_1$, $A_2$, $B_1$, $B_2$ lie in the same plane, and therefore the original lines will also lie in the same plane, which cannot be the case. We come to the conclusion that the lines $A_1B_1$ and $A_2B_2$ cannot intersect.

3. Does a figure consisting of two intersecting lines have a center of symmetry?

It may seem that the center of symmetry of two crossing lines exists and coincides with the midpoint $M$ of their common perpendicular (Fig. 4). However, this is not true. In fact, if the center of the symmetry of these two were straight, then the straight line $a$ with the central symmetry would turn into a parallel straight line $b$, but $b$ is not parallel to $a$. Consequently, intersecting lines cannot have a symmetrine center.

4. Do two intersecting straight lines have an axis of symmetry?

This question is much more difficult than the question about the center of symmetry of intersecting lines. It turns out that two crossing lines have three axes of symmetry: $l_1$, $l_2$ and $l_3$. However, one of them is a common perpendicular to the original lines $a$ and $b$. The other two are mutually perpendicular, intersect in the middle $M$ of the segment $KP$ and are bisectors of adjacent angles formed by straight lines $a_1$ and $b_1$, corresponding to parallels $a$ and $b$ (Fig. 4). Try to prove this fact yourself.

Solving planetary problems using a stereometric method

Obviously planimetric problems are solved on a plane, and stereometric problems are associated with the image of a spatial figure on a flat drawing. Such direct solutions to problems or proofs of the theorem are not surprising; they are generally acceptable. But solving planimetric problems in a stereometric, spatial, three-dimensional way is already unusual and interesting. Of course, not all problems can be solved by this method, but only problems of a certain type.

5. There are three rays on the plane, emanating from one point (Fig. 5, $a$) and dividing the plane into three angles, each of which is less than $180^\circ$. In each of the three angles, one point is given: $A_1$, $B_1$ and $C_1$ - so that these points do not lie on the same straight line. Required to construct a triangle

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Fig. 5, a.

ABC, the vertices of which lie on the given rays, and the sides pass through the given points.

It is planned to solve this problem in a difficult way, but stereometrically it can be solved quite easily. Let us imagine (Fig. 5, b) that the rays form in space the edges of a trihedral angle with the vertex at point O. Let us imagine the points A₁, B₁, C₁ located in the faces of the trihedral angle. These points define a certain plane. Let us denote this plane by β; it will intersect the faces of the trihedral angle along the required sides BC, CA and AB. It remains to construct the line of intersection of the planes of the faces of the trihedral angle with the plane β.

To do this, we project these points onto the xOy plane parallel to Oz. Points A₁, B₁, C₁ will move to points A₀, B₀, C₁ respectively (point C₁ will remain in place).

Thus, we have reduced the problem to a projection one, which in school is often called the problem of cutting a polyhedral angle with a plane. We find the point x₀ of intersection of the line A₁B₁ with the plane xOy, this will be the point of intersection of the lines A₁B₁ and A₀B₀. Points x₀ and C₁ lie in both the β plane and the xOy plane, therefore, the cutting plane β intersects the xOy plane along the straight line x₀C₁.

Then we find the intersection points

Fig. 5, b.

straight line x₀C₁ with rays (edges) Ox, Oy, we get points A and B. Wires BA₁ and AB₁, we get the third point at the intersection with the axis (ray Oz) - vertex C of the desired triangle ABC. The spatial representation helped us find an idea for solving the planimetric problem.

6. Two triangles ABC and A₁B₁C₁ are given on the plane; also, lines AA₁, BB₁ and CC₁ intersect at one point S, and lines AB and A₁B₁, BC and B₁C₁, AC and A₁C₁ intersect at points x, y and z. Prove that points x, y and z lie on the same line l.

This problem is known in geometry under the name Desargues’ theorem*).

Here is the idea of ​​its proof. Let’s imagine that point S is the vertex of a paramid with base ABC and cutting plane A₁B₁C₁ (Fig. 6). Then the lines AB, BC and AC lie in the plane ABC, and the lines A₁B₁, B₁C₁ and A₁C₁ lie in the plane A₁B₁C₁. Therefore, points x, y, z lie both in the plane of triangle ABC and in the plane of triangle A₁B₁C₁, that is, they are located on the line of intersection of these planes, on straight line l.

Conduct a rigorous proof of Desargues’ theorem on your own.

*) Gerard Desargues - French mathematician and engineer (1593-1662).

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Fig. 6.

The condition that lines $AA_1$, $BB_1$ and $CC_1$ intersect at one point $S$ is used in which place: lines $AB$ and $A_1B_1$ (and others) do not intersect in space, but intersect at point $x$. In further reasoning, only the planes of triangles are involved; you can forget about nature.

I wonder if this theorem is valid for quadrilaterals $ABCD$ and $A_1B_1C_1D_1$, in which the lines $AA_1$, $BB_1$, $CC_1$ and $DD_1$ intersect at one point $S$? Will the conditions for the quadrilateral intersection of the continuations sides $AB$ and $A_1B_1$, $BC$ and $B_1C_1$, $CD$ and $C_1D_1$, $DA$ and $D_1A_1$ lie on the same straight line? Generally speaking, this is not the case. For two quadrilaterals, a similar theorem will be true under some additional conditions.

Try to find these conditions yourself.

Indicated and e. Consider the straight line connecting the points of intersection of the diagonals of the quadrilaterals.

It is interesting that neither the length of the segments nor the magnitude of the angles appeared in the task conditions. In these problems, the solution was only about the points of intersection of lines and planes, about whether the points belonged to straight lines. We considered a planometric figure as a projection of some spatial figure onto a plane. This method of projection or projection

Fig. 7.

characteristic of a special branch of geometry called projective geometry.

Intersection and image of figures

Let’s consider several problems.

7. Given a triangular pairida $ABCD$ with vertex $D$. On the continuations side $CA$ and $CB$ the points $K$ and $P$ are taken as well as $AC = 2AK$ and $BC = 2BP$; $M$ is the middle of edge $CD$. A cutting plane is drawn through points $K$, $P$, $M$. In what ratio will this plane divide the volume of the paramid?

To solve the problem, let’s see what the plane $KPM$ represents for the paramida, that is, the plane defined by the three points $K$, $M$ and $P$ (Fig. 7).

Triangles $ABC$ and $KCP$ are similar, hence $KP \parallel AB$. Then $KP$ is parallel to the plane $ABD$, $KP = \frac{3}{2} AB$. The cutting plane will intersect the face $ABD$ along the segment $A_1B_1$, parallel to $KP$, and therefore parallel to $AB$. Let’s draw the middle line $ME$ of the triangle $BCD$, $CE = EB = BP$, $ME \parallel BD$. Note that $BB_1$ is the midline of triangle $MPE$, from which $BD = 2ME = 4BB_1$, therefore $A_1B_1 = \frac{3}{4} AB$ (triangles $ABD$ and $A_1B_1D$ are similar). Areas of similar

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Fig. 8.

figures are related as squares of similar sides, therefore

$$S_{\triangle A_1B_1D_1} = \frac{9}{16} S_{\triangle ABD}.$$

Let $V$ be the volume of pyramid $ABCD$. When calculating the volume $V_1$ of the pyramid $A_1B_1MD$, it is advisable to take face $A_1B_1D$ as its base. then the height $H_M$ will be half the height $H_C$, since $M$ is the midpoint of the segment $CD$. We have:

$$V_1 = \frac{1}{3} S_{\triangle A_1B_1D} H_M = \frac{1}{3} \cdot \frac{9}{16} S_{\triangle ABD} \times \frac{H_C}{2} = \frac{9}{32} \left( \frac{1}{3} S_{\triangle ABD} H_C \right) = \frac{9}{32} V.$$

The volume of the bottom of the pyramid $V''$ will be $\frac{23}{32}$ of the volume of the pyramid $ABCD$. Now there are objects at the top and bottom of the pyramid: $V' : V'' = 9:23$.

8. On the plane of the table there are four balls of radius $R$, touching each other (the centers of the balls form a square). Then, into the “hole” formed by these balls,

Exercises

1. On the plane of the table there are three balls touching each other and having a radius $R$. Then four balls of the same radius are placed in the “hole” formed by these balls. Naty is the distance from the top point of the fourth ball to the plane of the table **).

2. Given a truncated cone with axis $OO_1 = h$ and base radii $r$ and $R$. On the circle of bases we take all possible points $A$ and $B$ such that $A$ and $B$ lie on different circles and $OO_1$ and $AB$ are intersecting segments. Naty boundaries change the distance between $OO_1$ and $AB$.

3. The vertex $S$ of a regular triangular pyramid $ABCS$ moves along the height $OS$ to infinity, and then approaches the plane of the base in height. Within what limits will the linear angle of the dihedral angle at the edge $AS$ change?

*) An interesting question in physics: what should be the friction force between the balls and between the balls and the table so that this entire structure does not fall apart?
**) The same question about physics.