van Aubel's Theorem: -------------------- Given a quadrilateral. Erect squares on its sides. The segments between opposite centers of opposite squares have equal length and are perpendicular. Proof: like [Engel, 12.1.3 E11] Let the corners of the quadrilateral be represented by the complex numbers w, x, y, z in anti clock wise order. The centers of the squares: (w+x)/2 + i*(w-x)/2 center of the square on side wx. (x+y)/2 + i*(x-y)/2 center of the square on side xy. (y+z)/2 + i*(y-z)/2 center of the square on side yz. (z+w)/2 + i*(z-w)/2 center of the square on side zw. The two vectors between the opposite centers: p = (-w-x+y+z)/2 + i*(-w+x+y-z)/2 vector center_square(wx) to center_square(yz) q = (+w-x-y+z)/2 + i*(-w-x+y+z)/2 vector center_square(wx) to center_square(yz) The theorem follows from i*p = q. - - - - - - - - - - - - - - - - - - - - - - - - - - - - Special case: quadrilateral = parallelogram Thébault's Theorem: The centres of squares placed on any parallelogram form a square. +---------------+ | | | | | | + | P | / \ | | / \ | | / \ | | / Z---------------Y / Q / / \ + / / \ \ / / \ \ / / + \ / / Q' / W---------------X / | | \ / | | \ / | | \ / | P' | + | | | | | | +---------------+ Given a parallelogram WXYZ. Erect squares on its sides. Show: The centers of the squares form a square. The area of this square is area(PQP'Q') = area(WXYZ) + (area(square_P) + area(square_Q))/2. area(WXYZ) = sqrt(area(square_P)*area(square_Q)) * sin(angle(XYZ) Further the parallelogram and the sqares of type P and Q tile the plane such that the center of the squares form a square lattice. Gutierrez showed how van Aubel's theorem follows from Thébault's theorem. Proof: Let WXYZ by a quadrilateral. Let Q, P, Q', P' be the midpoints as shown in the figure. Let V be the midpoint of WY. Then VQ, VP are of equal length and perpendicular by Thébault's theorem. Similarly, VQ', VP' are of equal length and perpendicular by Thébault's theorem. Now triangles PVP' and QVQ' are congruent as a rotation at V of a right angle transforms PVP' into QVQ'. Therefore PP' = QQ' and they are perpendicular. - - - - - - - - - - - - - - - - - - - - - - - - - - - - The Tiling: (special case: quadrilateral = rhombic) +-------+_ / \ / '-,_ / \ / +-------+_ / \ / x / \ / '-,_ +-------+_ / \ / +-------+_ | | '-,_ / \ / x / \ / '-,_ | o | _+-------+_ / \ / +-------+_ | | ,-' | | '-,_ / \ / x / \ / '-,_ +-------+_ | o | _+-------+_ / \ / +-------+ / \ / '-,_| | ,-' | | '-,_ / \ / x / \ / / \ / +-------+_ | o | _+-------+_ / \ / / \ / x / \ / '-,_| | ,-' | | '-,_ / \ / +-------+_ / \ / +-------+_ | o | _+-------+ | | '-,_ / \ / x / \ / '-,_| | ,-' | | | o | _+-------+_ / \ / +-------+_ | o | | | ,-' | | '-,_ / \ / x / \ / '-,_| | +-------+ | o | +-------+_ / \ / +-------+ / \ / '-,_| | ,-' | | '-,_ / \ / x / \ / / \ / +-------+_ | o | _+-------+_ / \ / / \ / x / \ / '-,_| | ,-' | | '-,_ / \ / +-------+_ / \ / +-------+_ | o | _+-------+ | | '-,_ / \ / x / \ / '-,_| | ,-' | | | o | _+-------+_ / \ / +-------+_ | o | | | ,-' | | '-,_ / \ / x / \ / '-,_| | +-------+ | o | +-------+_ / \ / +-------+ '-,_| | ,-' | | '-,_ / \ / x / \ / +-------+_ | o | _+-------+_ / \ / '-,_| | ,-' | | '-,_ / \ / +-------+_ | o | _+-------+ '-,_| | ,-' | | +-------+_ | o | '-,_| | +-------+ The center (marked o and x) of the squares form a square lattice. The area of the small ox-squares = 1 square + 2 triangles. The area of the small o-square = 2 squares + 4 traingles. - - - - - - - - - - - - - - - - - - - - - - - - - - - - Application: The regular 12-gon can be tiled with 12 regular triangles and 6 squares. The o-square above have the same side length as the circumference of the 12-gon. So area of the regular 12-gon is 3*r*r with r the distance between A and B. _+-------+_ ,-' | | '-,_ +_ | | _+ / '-,_| | ,-' \ / +-------+ \ / / \ / \ \ +_ / \ / \ _ B | '-,_ / \ / \ ,-' | | _+-------A-------+_ | | ,-' \ / \ / '-,_| +- \ / \ / + \ \ / \ / / \ _+-------+_ / \ ,-' | | '-,_ / +_ | | _+ '-,_| |_,-' +-------+ Puzzle: Trigg Q268 Draw three lines AA, BB, CC. Show they intersect in O. _B-------C_ ,-' | | '-,_ A_ | O | _A / '-,_| | ,-' \ / +-------+ \ / / \ / \ \ +_ / \ / \ _ + | '-,_ / \ / \ ,-' | | _+-------+-------+_ | | ,-' \ / \ / '-,_| C- \ / \ / B \ \ / \ / / \ _+-------+_ / \ ,-' | | '-,_ / +_ | | _+ '-,_| |_,-' +-------+ References: Alexanderson, G.L.; Seydel, Kenneth K\"urschak's tile. Math. Gaz. 62, 192-196 (1978). Zbl 0397.51012 - DODECAGON INSCRIBED IN UNIT CIRCLE D. DeTemple, S. Harold; A round-up of square problems, Mathematics Magazine (1996) 15-27 - proof of Van Aubel by the Finsler-Hadwinger theorem Arthur Engel; Problem-solving strategies, Problem Books in Mathematics. New York, NY: Springer. x, 403 p. (1998) ISBN 0-387-98219-1/hbk Zbl 887.00002 - Chap. 12.1.3 complex numbers, E11 van Aubel Alfinio Flores; Tiling with Squares and Parallelograms (proof by picture), College Mathematics Journal 28:3 (1997) 171 Martin Gardner; Mathematical Circus, Alfred A. Knopf (1979) New York (german: Mathematischer Zirkus, Ullstein, 1984) chapter 14: Simplicity - van Aubel's Thm (problem without proof) Antonio Gutierrez; Van Aubel's Theorem: Quadrilateral with Squares http://agutie.homestead.com/files/vanaubel.html (Geometric proof of van Aubel using Thébault.) Geometry: Triangle and Squares Problem 1 http://agutie.homestead.com/files/triangle_squares1.html (Geometric proof og Thébault.) Richard I. Hess; PME 558 The Pi Mu Epsilon Journal 7 (1983) 615 (problem by R. I. Hess) The Pi Mu Epsilon Journal 7 (1984) 669 and 8 (1984) 41 (corrections) The Pi Mu Epsilon Journal 8 (1985) 183 (solution by Leon Bankoff) - van Aubel configuration puzzles I. M. Jaglom (Yaglom); Geometric Transformations, Random House (1962) Problem 24b, p95-96 - proof of van Aubel's Theorem by symmetry operations Paul J. Kelly; Von Aubel's Quadrilateral Theorem, Mathematics Magazine (Jan. 1966) 35-37 - vector proof and generalizations to nonconvex figures E. Kitchen; Dörrie Tiles and Related Miniatures. Mathematics Magazine 67, 128-130, 1994. Michael de Villiers; Dual generalisations of Van Aubel's theorem, The Mathematical Gazette 82 No 495 (Nov. 1998) 405-412 Michael de Villiers; More on dual Van Aubel generalisations, The Mathematical Gazette 84 No 499 (Mar. 2000) 121-122 Michael de Villiers; Generalizing Van Aubel Using Duality, Mathematics Magazine 73:4 (Oct. 2000) 303-307 http://mzone.mweb.co.za/residents/profmd/aubel.pdf KRS Sastry; Problem 2133, Crux Mathematicorm 23:4 (May 1997) 249-250 (solution of Bradley) - similar rectangles are placed on the sides of a parallelogram. Dick Tahta; Edmund Van Aubel (1900-?): discovered various triangle theorems, as well as one about the squares drawn on the sides of a quadrilateral. http://s13a.math.aca.mmu.ac.uk/Geometry/TriangleGeometry/HistoricalNotes.html http://s13a.math.aca.mmu.ac.uk/Geometry/TriangleGeometry/LemoinePoint/LemoinePtOrthLpedal1.html Charles W. Trigg; Mathematical Quickies, New York 1967, (reprint: Dover Publ., 1985) Q 268. Dissected Dodecagon a nontrivial tripple-crossing of diagonals Alex Bogomolny; - Thébault's Theorem: The centres of squares placed on any parallelogram form a square. http://www.cut-the-knot.org/Curriculum/Geometry/Thebault1.shtml - Thébault's Theorem: equilateral triangles on a square http://www.cut-the-knot.org/Curriculum/Geometry/Thebault2.shtml - A similar name - a different theorem; http://www.cut-the-knot.com/triangle/VanObel.html Nick Hobson; Four Squares on a Quadrilateral, Nick's Mathematical Puzzles 62, www.qbyte.org/puzzles/p062s.html Solution (Engel's solution with complex numbers.) Mathworld; www.mathworld.wolfram.com/vonAubelsTheorem.html Hontuka; http://www.auemath.aichi-edu.ac.jp/graduate/1998/ogura/hontaku.htm -- http://www.mathematik.uni-bielefeld.de/~sillke/ mailto:Torsten.Sillke@uni-bielefeld.de