By Pogorelov, A.V.
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Problem 10(b) is, clearly, a generalization of Problem 7(a). t Here we have in mind either one of the points of intersection of the line 1 with the circle S. DISPLACEMENTS s 23 o Figure 14 11. Find a point X on the circumference, such that the chords A X and BX cut off on the chord CD a segment EF whose midpoint is J (Figure 14). 12. The strip formed by two parallel lines clearly has infinitely many centers of symmetry (Figure 15). Can a figure have more than one, but only a finite number of centers of symmetry (for example, can it have two and only two centers of symmetry)?
T Thus we see that ij thejigures F and F’are relalcd by a rotalion through an angle a, then Corresponding segments of these jigwes are equal and make an angle a with each other. Let us show that, conversely, if to each point of the jigwe F there corresponds a point ojanothjigure F‘, and thcscjFgurcs arc such tkat corresponding segments are equal and mzke a% angle a with each other (so that the segments of the figure F become parallel to the corresponding segments of the figure F’ when they are turned through an angle u in the chosen direction),then F and F’ are related by a rotation through an angle a about some center.
We shall soon use this fact. For the details of this construction, see, for example, Bwgwim Ploblcn, Book I in this series, Problem 1895/2, Note. i Further, let A and A' be any other corresponding points of the figures F and F'. Consider the triangles OMA and OM'A'. One has OM = OM' (by construction of the point 0), MA = M'A' (this was given); in addition, QOMA = +OM'A', because the angle between OM and OM' is equal to the angle between MA and M'A', that is, the points M,M', 0, and K(K is the point of intersection of AM and A'M') lie on a circle and the inscribed angles OMA and OM'A' cut off the same arc.
Analytical geometry by Pogorelov, A.V.
by Anthony
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