Torsten Sillke, 1995-12-31 Trapezoid Question: last update, 2002-01-16 given AD, BC, CA, and BD. These are the lateral sides and the diagonals. (The diagonals have different lengths.) Construct the trapezoid. c D-------C / \ d/ \ b / \ A-------------------B a Problem: Find a geometric construction ^^^^^^^^^ Annotation: The related problem (given AB, CD, CA, and BD) is simple. ---------------------------------------------- An algebraic solution ---------------------------------------------- Let O be the crossing point of the diagonals AC and BD. Solution I (with vectors): Let the center of the coordinate system be O. The triangles AOB and COD are similar. Let A, B, C, and D be vectors in this system, then C = -z * A D = -z * B with z = c/a. The distances: a = |AB|, b = |BC|, c = |CD|, and d = |DA| and e = |AC| and f = |BD|. Then the squares of the distances are scalar products: b^2 = < B - C, B - C > = < B + z*A, B + z*A > d^2 = < A - D, A - D > = < A + z*B, A + z*B > e^2 = < A - C, A - C > = < A + z*A, A + z*A > f^2 = < B - D, B - D > = < B + z*B, B + z*B > b^2 - d^2 = ( - ) * (1 + z)(1 - z) f^2 - e^2 = ( - ) * (1 + z)^2 If != that means e != f we have b^2 - d^2 1 - z --------- = ----- = y f^2 - e^2 1 + z Resolving the equation gives 1 - y f^2 - e^2 - b^2 + d^2 z = ----- = --------------------- = c/a (*) 1 + y f^2 - e^2 + b^2 - d^2 On the other hand we have the scalar product identity < A - C, A - C > + < B - D, B - D > - < B - C, B - C > - < A - D, A - D > = 2 < B - A, C - D >. Application of the equation to the trapeziod yields f^2 + e^2 - b^2 - d^2 = 2ac (**) Combining (*) and (**) yields 2a^2 = (f^2 + e^2 - b^2 - d^2)*(f^2 - e^2 + b^2 - d^2)/(f^2 - e^2 - b^2 + d^2) 2c^2 = (f^2 + e^2 - b^2 - d^2)*(f^2 - e^2 - b^2 + d^2)/(f^2 - e^2 + b^2 - d^2) Solution II: (calculate a,c by equal areas) The Triangles ABD and ABC have equal area. Using the formula of Heron we get (a + d + f)(a - d - f)(a + d - f)(a - d + f) = (a + b + e)(a - b - e)(a + b - e)(a - b + e), that is (a^2 - (d+f)^2)(a^2 - (d-f)^2) = (a^2 - (b+e)^2)(a^2 - (b-e)^2) which is a linear equation in a^2. 2a^2 = ((d^2 - f^2)^2 * (b^2 - e^2)^2)/(f^2 - e^2 - b^2 + d^2) = (f^2 + e^2 - b^2 - d^2)*(f^2 - e^2 + b^2 - d^2)/(f^2 - e^2 - b^2 + d^2) Solution III: (calculate c/a or (a-c)/(a+c) by equal angles) Using the cosine law we get b^2 = a^2 + e^2 - 2a*e*cos(angle(BAC)) for triangle BAC, d^2 = c^2 + e^2 - 2c*e*cos(angle(ACD)) for triangle ACD. As angle(BAC) = angle(ACD) we get b^2 c - d^2 a = ac(a-c) - e^2(a-c). The same calculations for the triangles BDC and ABD yields b^2 a - d^2 c = ac(c-a) - f^2(c-a). Adding these two equations results in (b^2 - d^2)(a + c) = (f^2 - e^2)(a - c), and therefore (b^2 - d^2)/(f^2 - e^2) = (a-c)/(a+c) = (1-z)/(1+z) with z = c/a. References: - C. Krediet; Een trapezium te construeeren als gegeven zijn de opstaande zijden en de diagonalen (Ein Trapez zu konstruieren, wenn die nichtparallelen Seiten und die Diagonalen gegeben sind). Wisk. Tijdschr. 4, 241-244. Published: (1908) JFM 39.0563.01 Verf. gibt hierfür drei Konstruktionen. - W. Kroll; P 1042: Trapezkonstruktion PM 42:6 (2000) proposed by W. Kroll PM 43:4 (2001) 199-201 solutions Man finde eine m\"oglichst elegante Konstruktion (mit Zirkel und Lineal) eines Trapezes aus seinen beiden Schenkeln b, d und den Diagonalen e, f. - Theo K\"uhlein; Ebene Trigonometrie II, Mentor-Repetitionen, Mentor Verlag, 1962, Berlin (7th ed. 1974) ISBN 3-580-63140-3 - Sect 50: Das Trapez Problem 103 is the trapezoid question: given b, d, e, f. The translation trick is used to derive the relation (b^2 - d^2)/(f^2 - e^2) = (a-c)/(a+c) with the Pythagorean Thm. c D-------C e = AC / /| \ f = CG d/ d/ |h \ b a+c = AG / / | \ a-c = BF A-------F---E-------B-------G c q p c First look at the right triangles GEC and AEC. h^2 = f^2 - (p+c)^2 = e^2 - (q+c)^2 f^2 - e^2 = (p+c)^2 - (q+c)^2 = (p+c+q+c)*(p-q) = (a+c)*(p-q) Now look at the right triangles BEC and FEC. h^2 = b^2 - p^2 = d^2 - q^2 b^2 - d^2 = p^2 - q^2 = (p+q)*(p-q) = (a-c)*(p-q) Division gives (b^2 - d^2)/(f^2 - e^2) = (a-c)/(a+c) = (1-z)/(1+z) with z = c/a. - Eckart Schmidt; Berechnungsgrundlagen zu P 1042 Trapezkonstruktion PM 43:6 (2001) 293-294 - Hermann Schmidt; Die Inversion und ihre Anwendungen, Oldenbourg Verlag (1950) Muenchen Chapter VI. Inversoren, Section 2. Der Inversor von Hart The Hart's Inversor consists of a the special trapezoid c=d and e=f. The trapezoid is flexible iff e=f iff the trapezoid is cyclic. - Kurt Sieber; amg-info 1/74 (AMG = Arbeitskreis fuer Mathematik und Grenzgebiete) (Lsg einer Trapezaufgabe von Rung) - L. Vautré; Sur le trapèze. J. de Math. élém. (4) III. 99-107. Published: 1894 JFM 25.0904.05 Sätze und Aufgaben über das Paralleltrapez. - H. Chr. Espensen; Geometriske Konstruktionsopgaver (trekant og trapez) med løsninger. (Danish) København. 48 S. (1927). JFM 54.0663.10 references: lines parallel to a and c - American Mathematical Monthly 54 (1947) 537-538 m = 2 a c / (a+c) (length of a parallel section through O.) m = harmonious_mean(a, c) - Archimedes 15:8 (1963) 126 m = geometric_mean(a, c) (parallel section m gives similar trapezoids) m = harmonious_mean(a, c) (parallel section m goes through O.) m = square_mean(a, c) (parallel section m halves the area) -- mailto:Torsten.Sillke@uni-bielefeld.de http://www.mathematik.uni-bielefeld.de/~sillke/