Ladder-Box Problem (Leiter und Kiste Problem) Torsten Sillke Problem: A Ladder Problem, [Angela Dunn] A cubic box with 1-foot edges is placed flat against a wall. A ladder sqrt(15) feet long is placed in such a way that it touches the wall as well as the free horizontal edge of the box. Find at what height the ladder touches the wall. : |\ : 1/a | \ c given b and c, calculate a : |__\ : 1/b | |\ x = a + b : |__|_\ y = 1/a + 1/b : b a c_min = (b^(2/3) + b^(-2/3))^(3/2) : Solution-type for (b=1): T(rigonometric), C(onic section), : L(adder-symmetry), S(ymmetry a + 1/a), P(olynomSym in a), : N(umeric), D(iophantic Solutions) : rel: similar triangles: x/y = a/(1/b) = b/(1/a) : rel: 2 * triangle area: x*y = (1/b)*x + b*y : rel: Pythagorean thm: (a+b)^2 + (1/a+1/b)^2 = c^2 Ladder square box problem: We show a solution adopted to [Zerger 2nd]. |\ | \ | \ c/2 + x | \ | \ |____t\ | 1 |\ | 1| \ c/2 - x |_____|_t\ Use the Pythagorean theorem in its trigonometric form sin^2 t + cos^2 t = 1 Plug in the ratios for sin and cos to get (1/(c/2 - x))^2 + (1/(c/2 + x))^2 = 1 Simplifing gives (c/2 + x)^2 + (c/2 - x)^2 = (c^2/4 - x^2)^2 and further c^2/2 + 2 x^2 = c^4/16 - c^2 x^2/2 + x^4 This is a quadratic equation in x^2. x^2 = c^2/4 + 1 - sqrt(c^2 + 1). There are other interesting relations in this figure. |\ | \ | \ c1 a1| \ | \ are similar to |_____\ | 1 |\ |\ 1| 1| \ c2 1/c2 | \ 1 |_____|__\ |__\ 1 a2 1/c1 First we get the area relation from Euclid's supplementing rectangles. Second proof by similar triangles. (1) a1 a2 = 1 By Pythagorean thmeorem (2) c1^2 = a1^2 + 1 and c2^2 = a2^2 + 1. (2') c1^2 = a1 (a1 + a2) and c2^2 = a2 (a1 + a2). Then we have the curious relation (3) c1 c2 = a1 + a2 Proof (c1 c2)^2 = ( a1^2 + 1 )( a2^2 + 1 ) = ( a1^2 + a1 a2 )( a2^2 + a1 a2 ) = a1 a2 (a1 + a2)^2 = (a1 + a2)^2 By the trigonometric Pythagorean theorem we get (4) 1/c1^2 + 1/c2^2 = 1 By similar triangles we get (5) c1 = a1 c2 and c2 = a2 c1. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Ladder square box problem made easy: |\ | \ | \ a p| \ | \ |_____\ | h |\ a , '|\ h| h| \ b , ' h| \ b |_____|__\ '________|__\ h q p q You see that the ladder box problem is equivalent to another problem for right triangles. Given a+b and h determine the other parts. Determine c (that is p+q) in the right triangle. Start with the binomial (a+b)^2 = a^2 + b^2 + 2ab. Now a^2 + b^2 = c^2 by Pythagorean theorem. The area of the triangle is area = ab/2 = hc/2. Thus (a+b)^2 = c^2 + 2hc. Solving the quadratic equation we found one positive root c = -h + sqrt(h^2 + (a+b)^2). Determine p and q. (x - p)(x - q) = x^2 - c x + h^2 = 0. The roots p and q are p = c/2 + sqrt(c^2/4 - h^2), q = c/2 - sqrt(c^2/4 - h^2). Determine a and b. (x - a)(x - b) = x^2 - (a+b) x + hc = 0. The roots a and b are a = (a+b)/2 + sqrt((a+b)^2/4 - hc), b = (a+b)/2 - sqrt((a+b)^2/4 - hc). Let h = 1 and R some positive rational number. (a+b)^2 = 2(sqrt(R) + 1/sqrt(R))^2 = 2(R + 1/R) + 4 >= 8 c = sqrt((a+b)^2 + 1) - 1 p = (sqrt(2(R + 1/R) + 5) + sqrt(2R + 1) - sqrt(2/R + 1) - 1)/2 q = (sqrt(2(R + 1/R) + 5) - sqrt(2R + 1) + sqrt(2/R + 1) - 1)/2 p+h= (sqrt(2(R + 1/R) + 5) + sqrt(2R + 1) - sqrt(2/R + 1) + 1)/2 q+h= (sqrt(2(R + 1/R) + 5) - sqrt(2R + 1) + sqrt(2/R + 1) + 1)/2 a = (sqrt(2(R + 1/R) + 4) + sqrt(2R + 4) - sqrt(2/R + 4))/2 b = (sqrt(2(R + 1/R) + 4) - sqrt(2R + 4) + sqrt(2/R + 4))/2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - References: mainly ordered by time. - Thomas Simpson; A Treatise of Algebra, London 1745, p250 - Isaac Newton; Arithmetica universalis, 1707 edited by W. Whiston, 1722 edited by Newton, 1761 commented by Johannis Castillionei, Arithmetica universalis, sive de compositione et resolutione arithmetica. Cum commentario Johannis Castillionei. contains Newton's lecture notes of the years 1673-1683. problem XXIV: segment in a corner. - A. Cyril Pearson; The twentieth century standard puzzle book, George Routledge & Sons, London 1907, Part II, no. 102, p. 103 box = (15, 12), ladder = 52, -> (x, y) = (20, 48) (<- Euclides Ex 623) - Geoffrey Mott-Smith; Mathematical Puzzles, Dover Publ. (1954, 1946 repr.) Problem 102: The Bay Window box = (3,9), ladder = 20, -> (x, y) = (12, 16) [D] - A. Sutcliffe; A ladder and wall problem, Problem E1832, AMM 72 (1965) 1021 problem AMM 74 (1967) 325-326 solution [D] <- Schaaf, Bibliography of Rec. Math. II.3.12 - Angela Dunn; Mathematical Bafflers, Dover Publ. (1980, 1964 repr.), Problem: A Ladder Problem, p14, p41, [S] box = 1, ladder = sqrt(15). - Bild der Wissenschaft (Math. Kabinett) 10:1970 p1050 [T] - Bild der Wissenschaft (Math. Kabinett) 1:1972 p89 [S] - Bild der Wissenschaft (Math. Kabinett) 9:1978 p147-147 [C] - Bild der Wissenschaft (Math. Kabinett) 1:1979 p112 [S] - D. John Baylis; Mathematics Teaching, No 54 (1971) p24 problem Mathematics Teaching, No 57 (1971) p13 solution [S] - KOMAL F2432, Komal 67 (1983) 30 - Hugh Apsimon; Math. Byways in Ayling, Beeling, and Ceiling, Oxford Univ. Press 1984 (Recreations in Mathematics) (Chap. 1 [S?] & Chap. 2 [D]) - Monte Zerger; The ladder problem, Math. Magazine 60:4 (1987) 239-242 [T,L,(C)] - letter to the editor, Math. Magazine 61:1 (1988) 63 (->Zerger) [L] - section of circle and hyperbola, mathematiklehren 1 (1983) 50-54 [C?] - Euclides 66 (1990/91) 222 Ex 623 (2 refs) [???] - Alfons Grassl; Diskussion und Kritik. Zu: Neue Standards f\"ur das L\"osen geometrischer Berechnungsaufgaben ... MNU 51:1 (Jan. 1998) 51 [u = x*y] - Heinrich Hemme; Der Wettlauf mit der Schildkr\"ote, G\"ottingen, 2002, Vandenhoeck & Ruprecht, ISBN 3-525-40740-8 problem 26: Leiter und Kiste [u = x+y] - Herbert Pieper; Heureka, Ich hab's gefunden, Deutscher Verlag der Wissenschaften, Berlin, 1991, ISBN 3-326-00693-4 problem 40: Eine Gerade gegebener L\"ange in einem rechten Winkel - shows Newton's solution for the negative solution of the ladder and box problem. - Anna Maria Fraedrich; Die Satzgruppe des Pythagoras, Lehrb\"ucher und Monographien zur Didaktik der Mathematik Band 29, BI-Wiss.-Verl., 1995, ISBN 3-411-17321-1 - III.A.3.b Beispiel 3: p321-322 Konstruiere das rechtwinkliege Dreieck ABC aus der Hypotenuse c und der Winkelhalbierenden w_c. - K. Herterich; Die Konstruction von Dreiecken, Stuttgart 1986 p255 - Martin Mettler; Vom Charme der 'verblassten' Geometrie, Eurobit-Verlag Timisoara, (2000) Rumänien (ISBN 973-9441-97-1). http://www.mathematik.uni-mainz.de/monoid/VomCharmeDerVerblasstenGeometrie.html Section 9.4: Zum Leiterproblem, p144-145, Monoid Aufgabe 366 - rec.puzzles FAQ ==> geometry/ladder.and.box <== [???] - David Singmaster: 6.L.1. LADDER OVER BOX - sio: general solution for b=1 and special problem parameter otherwise. b=2, c=10/3: (x,y) = (8/3, 2) or (2.949163, 1.553560) cubic root b=2, c=7/2: (x,y) = (3.249, 1.300) or (2.5178, 2.4312) with a1 = (-2 - Sqrt[5] + Sqrt[23 + 10 Sqrt[5]])/2 a2 = (-2 + Sqrt[5] + Sqrt[23 - 10 Sqrt[5]])/2 b=1: standard case a + 1/a = z = - 1 + sqrt( 1 + c^2 ) a1 = (z + sqrt(z^2 - 4))/2, a2 = (z - sqrt(z^2 - 4))/2 x1 = (z + 2 + sqrt(z^2 - 4))/2, x2 = (z + 2 - sqrt(z^2 - 4))/2 b=1: special case c = 2*r + 1/r for some rational r. You get the same solution if you transform r -> 1/(2r). a1 = (Sqrt[c^2 + 1] + Sqrt[4*r^2 + 1] - Sqrt[r^-2 + 1] - 1)/2 a2 = (Sqrt[c^2 + 1] - Sqrt[4*r^2 + 1] + Sqrt[r^-2 + 1] - 1)/2 x1 = (Sqrt[c^2 + 1] + Sqrt[4*r^2 + 1] - Sqrt[r^-2 + 1] + 1)/2 x2 = (Sqrt[c^2 + 1] - Sqrt[4*r^2 + 1] + Sqrt[r^-2 + 1] + 1)/2 b=1, c=3 (r = 1/2 or 1): (x,y) = (1.670, 2.492) or (2.492, 1.670) a1 = (Sqrt[10] + Sqrt[5] - Sqrt[2] - 1)/2 a2 = (Sqrt[10] - Sqrt[5] + Sqrt[2] - 1)/2 x1 = (Sqrt[10] + Sqrt[5] - Sqrt[2] + 1)/2 x2 = (Sqrt[10] - Sqrt[5] + Sqrt[2] + 1)/2 b=1, c=9/2 (r = 1/4 or 2): (x,y) = (1.302, 4.307) or (4.307, 1.302) a1 = (Sqrt[85] + 2*Sqrt[17] - Sqrt[5] - 2)/4 a2 = (Sqrt[85] - 2*Sqrt[17] + Sqrt[5] - 2)/4 x1 = (Sqrt[85] + 2*Sqrt[17] - Sqrt[5] + 2)/4 x2 = (Sqrt[85] - 2*Sqrt[17] + Sqrt[5] + 2)/4 b=1, c=11/3 (r = 1/3 or 3/2): (x,y) = (1.420, 3.381) or (3.381, 1.420) a1 = (Sqrt[130] + 3*Sqrt[10] - Sqrt[13] - 3)/6 a2 = (Sqrt[130] - 3*Sqrt[10] + Sqrt[13] - 3)/6 x1 = (Sqrt[130] + 3*Sqrt[10] - Sqrt[13] + 3)/6 x2 = (Sqrt[130] - 3*Sqrt[10] + Sqrt[13] + 3)/6 b=1, c=2Sqrt[2] (r = 1/Sqrt[2]): (x,y) = (2, 2) a = 1 x = 2 b=1, c=5/Sqrt[2] (r = Sqrt[2] or 1/Sqrt[8]): (x,y) = (1.449, 3.225) or (3.225, 1.449) a1 = (Sqrt[6] + 2)/2 a2 = (Sqrt[6] - 2) x1 = (Sqrt[6] + 4)/2 x2 = (Sqrt[6] - 1) We can write this in a more symmetric way. c^2 = 2(Sqrt[R] + 1/Sqrt[R])^2 = 2(R + 1/R) + 4 >= 8 a1 = (Sqrt[c^2 + 1] + Sqrt[2R + 1] - Sqrt[2/R + 1] - 1)/2 a2 = (Sqrt[c^2 + 1] - Sqrt[2R + 1] + Sqrt[2/R + 1] - 1)/2 x1 = (Sqrt[c^2 + 1] + Sqrt[2R + 1] - Sqrt[2/R + 1] + 1)/2 x2 = (Sqrt[c^2 + 1] - Sqrt[2R + 1] + Sqrt[2/R + 1] + 1)/2 c1 = (c + Sqrt[2R + 4] - Sqrt[2/R + 4])/2 c2 = (c - Sqrt[2R + 4] + Sqrt[2/R + 4])/2 Examples: R c^2 + 1 2R + 1 2/R + 1 1 9 3 3 1/2 10 2 5 1/3 35/3 5/3 7 2/3 28/3 7/3 4 - sio[S]: you can shear the box (rombic box) without disturbing symmetry (b=1) a + 1/a = z = cos(phi) - 1 + sqrt( (cos(phi)+1)^2 + c^2 ) a1 = (z + sqrt(z^2 - 4))/2 a2 = (z - sqrt(z^2 - 4))/2 KOMAL F2432: There is a vertical fence of height m in distance f from a vertical wall. A ladder of length h should be placed in such a way that it leans on the wall , the ground, and the top of the fence. If all of the data are in meters, calculate with cm accurency the distance of the leaning points of the ladder from the bottom line of the wall. (Numerical example: m=f=5, h=15.) (Note you can solve it with quadratics.) - Carnificina Calculatoris; Das Leiterproblem: Aufgabe für die Monate Dezember 2002/Januar 2003 http://schumann-marl.bei.t-online.de/carnificina/statistik.htm problem index http://schumann-marl.bei.t-online.de/carnificina/leiter/leiter.htm problem http://schumann-marl.bei.t-online.de/carnificina/leiter/loesung.htm solution - Mathe Zirkel; http://www.math.uni-goettingen.de/zirkel/loesungen/blatt08/ - Jim Wilson; EMAT 4600/6600 Problem Solving in Mathematics, The University of Georgia, Problem: Ladder and Box http://jwilson.coe.uga.edu/emt725/EquiPer/Ladder.Box.html - Ladder and Box http://mathcentral.uregina.ca/QQ/database/QQ.09.98/blade1.html - Das Leiter-Rätsel http://mitglied.lycos.de/sirlizium/start.php?page=22 - Zahlenjagd; http://www.zahlenjagd.at/loes0802.html - Jürgen Köller; Leiter Aufgaben, http://www.mathematische-basteleien.de/ladder.htm (english) http://www.mathematische-basteleien.de/Leiter-Aufgaben.htm (german) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - /*******************************************************************/ /* |\ */ /* | \ The Ladder and Box Problem */ /* | \ -------------------------- */ /* x' = 1/x | \ */ /* | A \ c given: y, c */ /* |_____\ what is: x, x' ? */ /* | |\ */ /* | | \ As the triangles A and B are */ /* 1/y | | \ similar, it is x' = 1/x. */ /* | | B \ 2 3____ 3____ 3 */ /* |_____|____\ c = ( V y*y + 1 / V y*y ) */ /* y x min */ /*-----------------------------------------------------------------*/ /* 2 2 2 */ /* ( x + y ) + ( 1/x + 1/y ) = c */ /*******************************************************************/ u = ( c*c - y*y - 1.0/(y*y) ) / 3.0; /*******************************************************************/ /* This gives the cubic equation: z^3 + 3*u*z^2 = 4*c^2 */ /* z = u*(-1 + 2*cos( arccos( 2*c*c/(u*u*u) - 1 ) / 3 ) ) */ /* = c / ( sqrt(u) * cos( arccos( c/sqrt(u*u*u) ) / 3 ) ) */ /* */ /* z >= 2 z is the only positive root if C >= C.min */ /* z = 2 iff y = 1 */ /*******************************************************************/ z = c / ( sqrt(u) * cos( acos( c/sqrt(u*u*u) ) / 3.0 ) ); s = sqrt( c*c + z - 1.0/(y*y) ); /*******************************************************************/ /* y*z - 2/y */ /* x*x + ( y - s ) * x + 1/2 ( z - --------- ) = 0 */ /* s */ /*******************************************************************/ x1= 0.5*( s - y + sqrt( (s - y)*(s - y) - 2*(z - (z*y-2/y)/s) ) ); x2= 0.5*( s - y - sqrt( (s - y)*(s - y) - 2*(z - (z*y-2/y)/s) ) ); -- http://www.mathematik.uni-bielefeld.de/~sillke/ mailto:Torsten.Sillke@uni-bielefeld.de