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51. A Parabola as an Envelope On one side of an arbitrary angle, mark off n equally spaced points 1, 2, . . . , n from the vertex (0), and on the other side n equally spaced points (not necessarily with the same spacing on the first side) n " 1, n " 2, . . . , 2, 1, 0 from the vertex (n). Prove that the lines joining points with the same number on both sides envelope (i.e. are tangent to) a parabola. A=12 1 2 3 4 7 6 5 9 8 10 11 S 12 11 4 3 2 B=1 Figure 1 The proof uses the Theorem of Apollonius. If two parabola tangents SA and SB, with the points of tangency A and B intersect a third parabola tangent at P and Q with O the point of tangency on SP QO QS . the parabola, then PA OP SQ A P S O Q B Figure 2 The proof of this theorem follows from the known property of parabolas that the point of intersection of two parabola tangents lies on a line parallel to the parabola’s axis, and passing through the midpoint of the chord joining the points of tangency. (This is easily established from the equation y 4p1 x 2 of the parabola with focus F0, p , directrix y "p and axis x 0. ) It follows from this theorem that if A and B are two points on a parabola, A and B their U U 1 projections on the directrix, M the midpoint of segment AB, and M the projection of M on the directrix, then M is the midpoint of A B , i.e., AM M B. U U U U U U A M F A' B M' d ire c trix B' Figure 3 Proof of the Theorem of Apollonius Let AP a, PS ), BQ b, QS *, OP p and OQ q. Let a , ) , b , . . . be the projections of AP, PS, BQ, . . . on the directrix. Then U U U (1) p U (2) q U (3) b * U (4) ) * U U U a U b U a ) U p q U U U From (1), (2) and (4), we get ) * a b , and then from (3), ) b and * a . Then U U U U U SP PA ) a ) aU bU aU , QO OP q p qU pU bU aU , BQ QS * b bU bU aU , which proves the Theorem of Apollonius. U * U U U U R Proof of the envelope. Let S be the vertex of the given angle, e and f the length of the segments marked off n times on each side of the angle, and A and B those points of the sides so that SA ne and SB nf, with A numbered n and B numbered 0. (Figure 1 shows the case n 12. ) Consider the parabola that is tangent to SA and SB at A and B. (The reader can use analytic geometry, for instance, to show that a unique such parabola exists.) The tangent line joining the point P numbered 7 on SA SP BQ meets the other side in a point Q so that PA by Apollonius’ Theorem. But QS SP v n"v , and it follows that Q is the point onf SB numbered 7. Thus all the lines PA R drawn connecting like numbered points are tangent to a parabola. Note 1. This is the idea behind a lot of "string art". Note 2. It’s a good exercise to use some sort of geometry software program to 2 implement this "construction". Note 3. One can use Apollonius’ Theorem to construct the point of tangency on each QO SP line. (From Figure 2, OP PA n"v v . ) 3