Problem 1093

122. a) Centrally project the plane $\pi$ from Fig. 28, a onto another plane $\pi^{\prime}$ so that the line $p$ becomes the distinguished line of the plane $\pi$. In this case, the lines $AM$ and $BC$ will transform into parallel lines $A^{\prime} M^{\prime}$ and $B^{\prime} C^{\prime}$; similarly, we have $B^{\prime} N^{\prime} \| A^{\prime} C^{\prime}, C^{\prime} L^{\prime} \| A^{\prime} B^{\prime}$. The lines $AS, BT$, and $CR$ will transform into the medians $A^{\prime} S^{\prime}, B^{\prime} T^{\prime}$, and $C^{\prime} R^{\prime}$ of the triangle $A^{\prime} B^{\prime} C^{\prime}$ (Fig. 285), which intersect at a single point $P^{\prime}$; therefore, the lines $AS, BT$, and $CR$ also intersect at a single point $P$.

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Fig. 285.

b) Centrally project the plane $\pi$ from Fig. 28, b onto another plane $\pi^{\prime}$ so that the line $RS$ becomes the distinguished line of the plane $\pi$. Then, on the new drawing, we will have $K^{\prime} L^{\prime} \| A^{\prime} B^{\prime}, K^{\prime} M^{\prime} \| A^{\prime} C^{\prime}$ (Fig. 286, a). Project the parallelogram $A^{\prime} B^{\prime} C^{\prime}$ into an equilateral triangle $A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$ (Fig. 286, b). In this case, $A^{\prime \prime} K^{\prime \prime}$ and $B^{\prime \prime} L^{\prime \prime}$ intersect on the axis of symmetry $C^{\prime \prime} D$ of the triangle; $A^{\prime \prime} K^{\prime \prime}$ and $C^{\prime \prime} M^{\prime \prime}$ intersect on the axis of symmetry $B^{\prime \prime} E$.

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c)

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g)

Fig. 286.

In this case, $A^{\prime \prime} K^{\prime \prime}$ and $B^{\prime \prime} L^{\prime \prime}$ intersect on the axis of symmetry $C^{\prime \prime} D$ of the triangle; $A^{\prime \prime} K^{\prime \prime}$ and $C^{\prime \prime} M^{\prime \prime}$ intersect on the axis of symmetry $B^{\prime \prime} E$ of the triangle. Therefore, the point $Q^{\prime \prime}$ must coincide with the point of intersection of both axes of symmetry, i.e., with the center of the triangle, and the lines $L^{\prime \prime} K^{\prime \prime}, K^{\prime \prime} M^{\prime \prime}$, and $M^{\prime \prime} L^{\prime \prime}$ are the midlines of the triangle.

From the fact that $L^{\prime \prime} M^{\prime \prime} \| B^{\prime \prime} C^{\prime \prime}$ and the property of parallel projection (see p. 18), it follows that $L^{\prime} M^{\prime} \| B^{\prime} C^{\prime}$. In the context of the property of central projection, this means that the point $T$ of intersection of $LM$ and $BC$ also lies on the distinguished line of the plane $\pi$, i.e., that the points $R, S$, and $T$ lie on the same line.

1) Note that in the general projection of the plane $\pi$ onto the plane $\pi^{\prime}$, the triangle $ABC$ does not necessarily transform into the triangle $A^{\prime} B^{\prime} C^{\prime}$; for example, this is clearly not the case if the line $p$ intersects the sides of the triangle $ABC$ (in central projection, a triangle can transform into a rather complex figure; see Fig. 20 on pp. 34-35). However, the three points $A, B$, and $C$ will transform into three new points $A^{\prime}, B^{\prime}$, and $C^{\prime}$; the lines connecting the points $A, B$, and $C$ pairwise will transform into lines connecting the points $A^{\prime}, B^{\prime}$, and $C^{\prime}$ pairwise. Only in this sense should the statement that Fig. 28, a transforms into Fig. 285 under projection be understood. For simplicity, we do not fully draw the lines $AB, AC$, etc., and $A^{\prime} B^{\prime}, A^{\prime} C^{\prime}$, etc., but only some segments of these lines. However, in solving the problem, we rely on the fact that, for example, the line $AB$ transforms into the line $A^{\prime} B^{\prime}$; the statement that the segment $AB$ transforms into the segment $A^{\prime} B^{\prime}$ could be simply false. This remark applies to the solutions of several other subsequent problems as well.

In addition, the reasoning in this problem contains another, more significant, clarification. The point $P^{\prime}$ in Fig. 285 may lie on the distinguished line of the plane $\pi^{\prime}$; in this case, the lines $AS, BT$, and $CR$ will not intersect at a single point but will be parallel. Such inaccuracies are present in the solutions of most subsequent problems; this is specifically mentioned on p. 51 and following.