Newsgroups: rec.puzzles Date: 29 Jul 1998 12:13:20 +1000 > Three straight lines are parallel to each other, and the distance > between any two lines is arbitrary. > _________________________________ > > > _________________________________ > > _________________________________ > > Your task is to draw an equilateral triangle such that on each of > the above three lines, one, and only one, apex falls. > > The tools you have are a pencil, a pair of compasses, and a ruler > without scales. [ If you enjoy this sort of problem, I recommend reading Geometric Transformations I (and II and III, for that matter) by I. Yaglom. ] SPOILERS AHEAD Choose a point X on the middle line. Rotate the bottom line 60 degrees about point X. Let Y be the point of intersection of this rotated line with the top line. Then X, Y and the preimage of Y are the vertices of an equilateral triangle. In more detail: 1) Construct the circle with centre X and tangent to the bottom line. Let T be the point of tangency. 2) Construct the equilateral triangle XTV. V will lie on the circle. 3) Construct the tangent to the circle at V. This is the rotated line. Y is the appropriate point of intersection. 4) Construct the perpendicular bisector of XY. This intersects the bottom line at Z, the desired preimage of Y. XYZ is equilateral. In extreme detail: 1a) With the compasses set to a larger distance than that of X from the bottom line, draw a circular arc centred on X intersecting the bottom line at two points. 1b) Construct the perpendicular bisector of the segment formed. It passes though T and X. T is the point of intersection with the bottom line. 1c) Draw the circle with centre X and radius XT. 2a) Draw the circle with centre T and radius TX. This intersects the other circle at V (either choice of V will do). 3a) Construct the perpendicular bisector of TV. This goes through X and intersects the bottom line at a point W. The line WV is the desired tangent. (Or you could just construct the perpendicular to XV at V.) The last required step (constructing perpendicular bisectors) is left as an exercise for the reader. :) (It's a pretty fundamental operation, so anyone attempting this problem should know it anyway.) Cheers, Geoff. ------------------------------------------------------------------------------- Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@cs.usyd.edu.au | Gameplayer by vocation. ------------------------------------------------------------------------------- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Problem: Express the area of an equilateral triangle having one vertex on each of the three parallel lines in terms of the distances a and b. _ ______________C__________________ | b _|_____A___________________________ a _|________________B________________ Trigonometric Solution: [1] Let s = AB = BC = AC, and phi the angle between the horizontal line through A and the side AB. a = s * sin(phi) b = s * sin(pi/3 - phi) = s * (sin(pi/3) * cos(phi) - cos(pi/3) * sin(phi)) = s * (sqrt(3)/2 * cos(phi) - 1/2 * sin(phi)) = sqrt(3)/2 * s * cos(phi) - a/2 s * cos(phi) = ( a + 2b ) / sqrt(3) s*s = (s*sin(phi))^2 + (s*cos(phi))^2 = a^2 + ( a + 2b )^2 / 3 = 4/3 (a^2 + a*b + b^2) area = (a^2 + a*b + b^2)/sqrt(3) Problem: Given a refence triangle and three parallel lines. Contruct a triangle ABC which is similar to the refence triangle where the points A, B, and C are on the three lines. _ ____________C____D_______________ _ _____A___________________________ _ _____________________B___________ Solution: [3] Analysis: Let ABC the triangle with the requiered properties. Let D be the point where the circumcircle of triangle ABC cuts the C line a second time. Then by the inscribed angle theorem we have beta = angle(ABC) = angle(ADC) and gamma = angle(ACB) = angle(ADB). The supplement angle of triangle BDC at D is of course alpha = angle(BAC). Konstruction: Select an arbitrary point D on the top line. Shift the angles alpha, beta, and gamma of the reference triangle to point D. _____________D_____________ beta / \ alpha /gamma\ The intersection of the two rays with the lower parallel lines determine the points A and B. Then construct a circle through the three points A, B, and D. The other intersection gives C. References: [1] Liang-shin Hahn; Complex Numbers and Geometry, MAA 1994 (Spectrum Series), ISBN 0-88385-510-0 - p62-63 the area of the equilateral triangle is (a^2 + ab + b^2)/sqrt(3) where a and b are the distances of the middle line to their neighbor lines. - p114-115 excercise 19 Gives a trigonometric method for computing the side length of the triangle. Then a "slick proof" is described. This slick on is invalid, which you have to show. [2] Eckard Specht; geometria - scientiae atlantis. 300+ Aufgaben zur Geometrie und zu Ungleichungen insbesondere zur Vorbereitung auf Mathematik-Olympiaden, Otto-von-Guericke-Universität Magdeburg, 2001, ISBN 3-929757-39-7 Problem A.87 http://www.math4u.de/ (online version) [3] R\"udiger Thiele; Mathematische Beweise, Deutsch Taschenb\"ucher Band 25, Harri Deutsch Verlag, 1979 - Section 3.2, p65-68 using the inscribed angle theorem (Umfangswinkelsatz) [4] Brian Bolt; Even more mathematical activities, Cambridge Univ. Press, 1987 (german: Die dritte mathematische Fundgrube, Klett Verlag, 1993, ISBN 3-12-722730-2) Sect 71: Parallele Grenzen, p76, 177-178 - equilateral triangle with vertices on three lines. - square with vertices on a parallelogram. -- http://www.mathematik.uni-bielefeld.de/~sillke/ mailto:Torsten.Sillke@uni-bielefeld.de