Halving the angles in the right 3-4-5 triangle: Torsten Sillke, 1998-10 - L O O K ! - . . . . . . . . . . . . . . . . . . . . B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . o . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . o . . . . . . . . . . . . Z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . o . . . . . . . . . . . . . . . . . . . . . . . . . . . I . . . . X . . . . . . . . . . . . . . . . . . . . . . . . . o . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A . . . . o . . . . o . . . . Y . . . . C Halving the angle: angle(CAB) = 2*angle(YAI) => arctan(3/4) = 2*arctan(1/3) angle(ABC) = 2*angle(ZBI) => arctan(4/3) = 2*arctan(1/2) Halving the area: area(ABC) = 2*area(AYIZ) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Halving the angles in the triangle 3-4-5 with complex numbers or there geometric interpretation: (2+i)^2 = 3+4i . . . E . . . . . . . . . . . . . . . . . . . . D . . . . . . . C . . . . . . . A . B F . . . . . . AB/BC = 2, AD/DE = 2, AF/FE = 3/4 (3+i)^2 = 8+6i . . . . . . . . E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D . . . . . . + . . . . . . C . . . . . . A . . B . . . . F . AB/BC = 3, AD/DE = 3, AF/FE = 4/3 (2+i)(3+i) = 5+5i . . . . . E . . . . . . . . . . . . . . . . . . . . D . . . . . . . + . . . . . . . C . . . . . . . A . B . . F . . . . AB/BC = 2, AD/DE = 3, AF/FE = 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - The halve angles of the 3-4-5 triangle give a proof of Charles Trigg's three square problem [Gardner] +---+---+---O | | | | A---B---C---D angle(OAD) + angle(OBD) = angle(OCD) arctan(1/3) + arctan(1/2) = arctan(1) Another Proof: [Conway, Guy] E---+ | | +---+---+---O | | | | A---+---C---D | | +---B angle(OAD) + angle(CAB) = angle(OAB) As ABOE is a square. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - A relation of Lewis Carroll: ---------------------------- arctan(1/(n+c))+arctan(1/(n+d))=arctan(1/n). This relation is valid iff cd = n^2 + 1. [Conway, Guy] Proof: ((n+c) + i)((n+d) + i) = (n^2 + n(c+d) + cd - 1) + (2n + c + d)i Now: n^2 + n(c+d) + cd - 1 = n (2n + c + d) <=> cd = n^2 + 1. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Halving the angles and trisecting the right triangle via the incenter: Let A,B,C be the vertices, I the incenter and X,Y,Z the points where the incircle touches the sides. Let s_a = AY = AZ (congruent triangles AIY, AIZ), s_b = BX = BZ (congruent triangles BIX, BIZ), s_c = CY = CZ (congruent triangles CIY, CIZ). We have the relations a = s_b + s_c b = s_a + s_c c = s_a + a_b Adding this three equations gives the semiperimeter s. s = (a+b+c)/2 = s_a + s_b + s_c s_a = (-a+b+c)/2 = s - a = AY = AZ s_b = ( a-b+c)/2 = s - b = BX = BZ s_c = ( a+b-c)/2 = s - c = CY = CZ Now let the right angle be at point C. As the primitive Pythagorean triangle is of the form (a, b, c) = (odd, even, odd) the values s, s_a, s_b, s_c are all integral for each Pythagorean triangle (not only the primitive ones). The areas of the three components are related as Area : Area_A : Area_B : Area_C = s : s_a : s_b : s_c The radii of the inscribed and escribed circles of the right triangle are (rho, rho_a, rho_b, rho_c) = (s_c, s_b, s_a, s) This relation was shown by [Nagel 1836, Appendix] cited after [Baptist 1992, p257]. Therefore we can find such a trisection for every right triangles with matches. For every triangle we find dissection into right triangles Area = s * rho = s_a * rho_a = s_b * rho_b = s_c * rho_c This gives for the right triangle the additional relation Area = s * s_c = s_a * s_b. Determine the right triangle given two parameters: Given (2 of): a, s_b, s_c Determine: s_a From Area = (s_a + s_b + s_c)*s_c = s_a*s_b it follows s_a = s_c*a/(s_b - s_c) Given (2 of): c, s_a, s_b Determine: s_c From Area = (s_a + s_b + s_c)*s_c = s_a*s_b it follows s_c = (sqrt(c^2 + 4*s_a*s_b) - c)/2 Given (2 of): c, s_c, s [Nagel, 1827, problem 1: given a+b, c] Determine: Area, s_a, s_b Area = s * s_c (x - s_a)(x - s_b) = x^2 - c*x + Area = 0 Geometric contruction of a and b: Draw the incircle and the outer circle of C. Construct the common tangents of the circle. Given (2 of): a, s_a, s [problem: given a, b+c] Determine: b By the Pythagorean thm a^2 + b^2 = c^2 = (2s - a - b)^2. Simplifying gives ab = 2s(a + b - s) which is two times the area. b = 2s (s-a)/(2s-a). - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Halving the angle A of a right triangle ABC. Let a=BC, b=AC, and c=AB. D alpha := angle(BAC) = angle(DBA) + angle(BDA) |\ as angle(BAC) is an exterior angle of the triangle ABD | \ Let AB = AD then alpha/2 = angle(BDA). Then | \ | \ tan(alpha/2) = a / (b+c) A \ | \ \ | \ \ | \ \ | \\ C---------B Using tan(alpha/2) = (s - c) / (s - a), the incirle relation, we again conclude: s*(s-c) = (s-a)*(s-b) = a*b/2. Determine the right triangle given two parameters: Given (2 of): a, s_a, s Geometric contruction of the triangle: Draw the right triangle BCD with BC = a and CD = b+c = 2s - a. The mid perpendicular of BD intersects CD in point A. The triangle ABC has the sides a, b, and c. Appendix: A problem from [BWM 1986/II Problem 2] see [Baptist 1992, p257]. A triangle is right if and only if s + s_a + s_b + s_c = rho + rho_a + rho_b + rho_c Solution: Show one of the following equations: (rho_a - s)(rho_b - s)(rho_c - s) = s^2 (rho + rho_a + rho_b + rho_c - 2s) (-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2) = 8 Area^2 (2s - rho - rho_a - rho_b - rho_c)(2s + rho + rho_a + rho_b + rho_c) Thm: right triangle The following sentences are equivalent: (0) the triangle is right at C (1) a^2 + b^2 = c^2 (2) (a+b+c)*(a+b-c) = (-a+b+c)*(a-b+c) (3) s*(s-c) = (s-a)*(s-b) (4) Area = s*s_c (5) rho_c = s (6) cos(gamma) = 0 Proof: (0) <=> (1) by the Pythagorean thm (1) <=> (2) by simplification (2) <=> (3) by substitution (3) <=> (4) by Heron's formular Area^2 = s*s_a*s_b*s_c. (4) <=> (5) by Area = rho_c * s_c (0) <=> (6) by definition of cos at the unit circle. References: - Peter Baptist; Die Entwicklung der Neueren Dreiecksgeometrie, BI, Mannheim 1992, ISBN 3-411-15661-9 - Heinrich Hemme; Das Problem des Zw\"olf-Elfs, Vandenhoeck & Ruprecht, G\"ottingen, 1998, ISBN 3-525-40736-X 6. Die Halbierung des Dreiecks (3-4-5) - Kobon Fujimura; The Tokyo Puzzles, New York, 1978, p20, Problem 119 (japan. original: Tokio, 1969) - halving the right triangle (3-4-5) with two length 1 segments - John H. Conway, Richard K. Guy; The Book on Numbers, Copernicus an imprint of Springer Verlag, 1996 (german: Zahlenzauber, Birkh\"auser, 1997) - Chapter 9: some transcendent numbers Section: The number Pi. - Carroll's rule: arctan(1/(n+c))+arctan(1/(n+d))=arctan(1/n) <=> cd=n^2+1 - Martin Gardner; Mathematical Circus Alfred A. Knopf (1979) New York Chapter 11.3: Three Squares - an angle addition problem, elementary solutions with no trigs arctan(1/3)+arctan(1/2)=arctan(1), (3+i)(2+i)=5+5i [triangle (a,b,c)=(3,4,5) has (s,s_a,s_b,s_c)=(6,3,2,1)] - Heinrich Hemme; Die magischen Vierecke des Abul Wafa, Rätsel und Knobeleien wie aus 1001 Nacht, rororo 61969, 2004, ISBN 3-499-61969-5, - problem 39: Der geknickte Stab - Calvin T. Long; On the Radii of the Inscribed and Escribed Circles of Right Triangles - A Second Look, College Mathematics Journal 14:5, 1983, 382-389 - D MacHale; The 3-4-5 triangle again. Math. Gaz. (1989) v. 73(463) p. 14-16. - It is shown that the 3, 4, 5 triangle is the only one with integer-sides such that the semi-perimeter equals the area. - Stephen B. Maurer; MSJ 521, The Mathematics Student Journal 27:4 (1980) 3 (problem Stephen B. Maurer) The Mathematics Student Journal 28:2 (1981) 4 (solution Greg Slaughter, Chun-Nip Lee) In the right triangle XYZ, W ist the midpoint of hypotenuse XY, and the circle with diameter ZW intersects WX at V. Calculate the length of XZ if XY=50 and WV=7. Solution: angle ZVW is right (Thales circle). So XZ^2 = XV * XY = (XW-WV)*XY = (XY/2-WV)*XY = 30^2. So YZ^2 = YV * XY = (YW+WV)*XY = (XY/2+WV)*XY = 40^2. - Christian Heinrich von Nagel; De triangulis rectangulis ex algebraica aequatione construendis, Dissertation, T\"ubingen, 1827 - Christian Heinrich von Nagel; Untersuchungen \"uber die wichtigsten zum Dreieck geh\"origen Kreise, [inverstigations of the most importend circles of the triangle] Leipzig 1836 - Appendix concerning the right triangle - Ch Trigg; 54 different proofs, JoRM 4 (Apr 1971) 90-99 - proofs of arctan(1/3)+arctan(1/2)=arctan(1) - Ch Trigg; geometrical proof of a result of Lehmer's, Fib. Quar. 11 (1973) 539 - a generalization for n squares in a row. - R. Narth; Math. Gaz. (Dec 1973) 334-336 + (Oct 1974) 212-215 - proofs of arctan(1/3)+arctan(1/2)=arctan(1) - Ross Honsberger; Mathematical Morsels, The Dolciani Mathematical Expositions No. 3 The Mathematical Association of America, 1978 (german: Gitter-Reste-Wuerfel, Vieweg, 1984) problem 14: a surprising property ... - The incircle radius = (a+b-c)/2 for Pyth. triangles. AMM 1956 Problem E 1197, p493 (solver Leon Bankoff) - Baltic Way - 92; Mathematical Team Contest, Vilnius, 1992-11-5 to 1992-11-8. reprint: Crux Mathematicorum 23:8 (Dec 1997) 462-463 Problem 20. Let a<=b<=c be the sides of a right triangle, and let 2p be its perimeter. Show that p*(p-c) = (p-a)*(p-b) = S where S is the area of the triangle. Solution: p*(p-c) = ((a+b+c)/2)((a+b+c)/2-c) = ((a+b)^2-c^2)/4 = (a^2+b^2+2ab-c^2)/4 = ab/2 = S (p-a)(p-b) = ((a+b+c)/2-a)((a+b+c)/2-b) = (c+(b-a))(c-(b-a))/4 = (c^2 - (b-a)^2)/4 = (c^2 - (b^2+a^2) + 2ab)/4 = ab/2 = S - BWM = Bundeswettbewerb Mathematik Jahr/Runde [German Mathematical Competition year/round] -- mailto:Torsten.Sillke@uni-bielefeld.de http://www.mathematik.uni-bielefeld.de/~sillke/