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Crossed ladders problem

From Wikipedia, the free encyclopedia
Mathematical puzzle

Thecrossed ladders problem is apuzzle of unknown origin that has appeared in various publications and regularly reappears in Web pages andUsenet discussions.

The problem

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Crossed ladders of lengthsa andb.h is half theharmonic mean ofA andB; equivalently, thereciprocals ofA andB sum to the reciprocal ofh (theoptic equation). Givena, b, andh, findw.

Two ladders of lengthsa andb lie oppositely across an alley, as shown in the figure. The ladders cross at a height ofh above the alley floor. What is the width of the alley?

Martin Gardner presents and discusses the problem[1] in his book of mathematical puzzles published in 1979 and cites references to it as early as 1895. The crossed ladders problem may appear in various forms, with variations in name, using various lengths and heights, or requesting unusual solutions such as cases where all values are integers. Its charm has been attributed to a seeming simplicity which can quickly devolve into an "algebraic mess" (characterization attributed by Gardner toD. F. Church).

Solution

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The problem description implies thatw > 0, thata >w, andb >w, thath > 0, and thatA >h,B >h, whereA andB are the heights of the walls where sides of lengthsb anda respectively lean (as in the above graph).

Both solution methods below rely on the property thatA,B, andh satisfy theoptic equation, i.e.1A+1B=1h,{\displaystyle {\tfrac {1}{A}}+{\tfrac {1}{B}}={\tfrac {1}{h}},}, which can be seen as follows:

Divide the baseline into two parts at the point where it meetsh{\displaystyle h}, and call the left and right partsw1{\displaystyle w_{1}} andw2{\displaystyle w_{2}} respectively. The angle wherea{\displaystyle a} meetsw{\displaystyle w} is common to two similar triangles with basesw{\displaystyle w} andw1{\displaystyle w_{1}} respectively. The angle whereb{\displaystyle b} meetsw{\displaystyle w} is common to two similar triangles with basesw{\displaystyle w} andw2{\displaystyle w_{2}} respectively. This tells us that
Bw=hw1andAw=hw2,wherew1>0,w2>0,{\displaystyle {\frac {B}{w}}={\frac {h}{w_{1}}}\quad {\text{and}}\quad {\frac {A}{w}}={\frac {h}{w_{2}}},\quad {\text{where}}\quad w_{1}>0,w_{2}>0,}
which we can then re-arrange (usingw1+w2=w{\displaystyle w_{1}+w_{2}=w}) to get
1A+1B=1h.{\displaystyle {\frac {1}{A}}+{\frac {1}{B}}={\frac {1}{h}}.}

First method

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Two statements of thePythagorean theorem (see figure above)

A2+w2=b2{\displaystyle A^{2}+w^{2}=b^{2}}

and

B2+w2=a2{\displaystyle B^{2}+w^{2}=a^{2}}

can be subtracted one from the other to eliminatew, and the result can be combined with1A+1B=1h{\displaystyle {\tfrac {1}{A}}+{\tfrac {1}{B}}={\tfrac {1}{h}}} with alternatelyA orB solved out to yield thequartic equations[2]

A42hA3+(hA)2(a2b2)=0,{\displaystyle A^{4}-2hA^{3}+(h-A)^{2}(a^{2}-b^{2})=0,}
B42hB3+(hB)2(b2a2)=0.{\displaystyle B^{4}-2hB^{3}+(h-B)^{2}(b^{2}-a^{2})=0.}

These can be solved algebraically or numerically for the wall heightsA andB, and the Pythagorean theorem on one of the triangles can be used to solve for the widthw.

Second method

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The problem may be reduced to thequartic equationx3(xc) − 1 = 0, which can be solved by approximation methods, as suggested by Gardner, or the quartic may be solved inclosed form byFerrari's method. Oncex is obtained, the width of the alley is readily calculated. A derivation of the quartic is given below, along with the desired width in terms of the quartic solution. Note that the requested unknownw does not appear directly in most of the derivation.

From1A+1B=1h,{\displaystyle {\tfrac {1}{A}}+{\tfrac {1}{B}}={\tfrac {1}{h}},} we get

(Eq. 1)AB=h(A+B).{\displaystyle {\text{(Eq. 1)}}\quad AB=h(A+B).}

Using thePythagorean theorem, we can see that

w2+B2=a2{\displaystyle w^{2}+B^{2}=a^{2}} andw2+A2=b2.{\displaystyle w^{2}+A^{2}=b^{2}.}

By isolatingw2{\displaystyle w^{2}} on both equations, we see that

a2B2=b2A2,{\displaystyle a^{2}-B^{2}=b^{2}-A^{2},}

which can be rearranged and factored into

(Eq. 2)a2b2=(B+A)(BA).{\displaystyle {\text{(Eq. 2)}}\quad a^{2}-b^{2}=(B+A)(B-A).}

Square (Eq. 2) and combine with (Eq. 1):

(a2b2)2=(B+A)2(BA)2,{\displaystyle (a^{2}-b^{2})^{2}=(B+A)^{2}(B-A)^{2},}
(a2b2)2=(B+A)2(B22AB+A2).{\displaystyle (a^{2}-b^{2})^{2}=(B+A)^{2}(B^{2}-2AB+A^{2}).}

Rearrange to get

(a2b2)2=(A+B)2(A2+B22AB).{\displaystyle (a^{2}-b^{2})^{2}=(A+B)^{2}(A^{2}+B^{2}-2AB).}

Then

(a2b2)2=(A+B)2(A2+B2+2AB4AB),{\displaystyle (a^{2}-b^{2})^{2}=(A+B)^{2}(A^{2}+B^{2}+2AB-4AB),}
(a2b2)2=(A+B)2((A2+2AB+B2)4AB),{\displaystyle (a^{2}-b^{2})^{2}=(A+B)^{2}{\big (}(A^{2}+2AB+B^{2})-4AB{\big )},}
(a2b2)2=(A+B)2((A+B)24AB).{\displaystyle (a^{2}-b^{2})^{2}=(A+B)^{2}{\big (}(A+B)^{2}-4AB{\big )}.}

Now, combine with (Eq. 1):

(a2b2)2=(A+B)2((A+B)24h(A+B)),{\displaystyle (a^{2}-b^{2})^{2}=(A+B)^{2}{\big (}(A+B)^{2}-4h(A+B){\big )},}
(a2b2)2=(A+B)2(A+B)((A+B)4h).{\displaystyle (a^{2}-b^{2})^{2}=(A+B)^{2}(A+B){\big (}(A+B)-4h{\big )}.}

Finally,

(Eq. 3)(a2b2)2=(A+B)3(A+B4h).{\displaystyle {\text{(Eq. 3)}}\quad (a^{2}-b^{2})^{2}=(A+B)^{3}(A+B-4h).}

Let

x=A+Ba2b2,{\displaystyle x={\frac {A+B}{\sqrt {a^{2}-b^{2}}}},}
c=4ha2b2.{\displaystyle c={\frac {4h}{\sqrt {a^{2}-b^{2}}}}.}

Then

x3(xc)1=0.{\displaystyle x^{3}(x-c)-1=0.} (same as Eq. 3 with the sides reversed)

The above fourth-power equation can be solved forx using any available method. The width of the alley is then found by using the value found forx: The identity

A+B=A+A2+(a2b2){\displaystyle A+B=A+{\sqrt {A^{2}+(a^{2}-b^{2})}}}

can be used to findA, andw can finally be found by

w=b2A2.{\displaystyle w={\sqrt {b^{2}-A^{2}}}.}

A quartic equation has four solutions, and only one solution for this equation matches the problem as presented. Another solution is for a case where one ladder (and wall) is below ground level and the other above ground level. In this case the ladders do not actually cross, but their extensions do so at the specified height. The other two solutions are a pair of conjugate complex numbers. The equation does not have the ladder lengths explicitly defined, only the difference of their squares, so one could take the length as any value that makes them cross, and the wall spacing would be defined as between where the ladders intersect the walls.

As the wall separation approaches zero, the height of the crossing approachesh=aba+b.{\displaystyle h={\frac {ab}{a+b}}.} This is because1A+1B=1h{\displaystyle {\tfrac {1}{A}}+{\tfrac {1}{B}}={\tfrac {1}{h}}} (proven at the start) impliesh=ABA+B,{\displaystyle h={\tfrac {AB}{A+B}},} and asw goes to zero,b goes toA anda goes toB according to the top diagram.

As the solutions to the equation involve square roots, negative roots are equally valid. They can be interpreted as both ladders and walls being below ground level and with them in opposing sense, they can be interchanged.

The complex solutions can be interpreted as wallA leaning to the left or right and wallB below ground, so the intersection is between extensions to the ladders as shown for the casea,b,h = 3, 2, 1. The laddersa andb anda2b2{\displaystyle a^{2}-b^{2}} are not as specified. The basew is a function ofA,B, andh, and the complex values ofA andB can be found from the alternative quartic

x42hx3+Dx22hDx+h2D=0{\displaystyle x^{4}-2hx^{3}+Dx^{2}-2hDx+h^{2}D=0}

withD beinga2b2{\displaystyle a^{2}-b^{2}} for one wall andb2a2{\displaystyle b^{2}-a^{2}} for the other (±5 in the example). Note that the imaginary solutions are horizontal and the real ones are vertical. The valueD is found in the solution as the real part of the difference in the squares of the complex coordinates of the two walls. The imaginary part = 2XaYa = 2XbYb (wallsa andb). The short ladder in the complex solution in the 3, 2, 1 case appears to be tilted at 45 degrees, but actually slightly less with a tangent of 0.993. Other combinations of ladder lengths and crossover height have comparable complex solutions. With combination 105, 87, 35 the short ladder tangent is approximately 0.75.

Integer solutions

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There are solutions in which all parameters are integers.[3] For example,[2] (a, b, A, B, w1,w2,w,h) = (119, 70, 42, 105, 16, 40, 56, 30). Such solutions involvePythagorean triples for the two right triangles with sides (A,w,b) and (B,w,a) and integer solutions of theoptic equation1A+1B=1h.{\displaystyle {\tfrac {1}{A}}+{\tfrac {1}{B}}={\tfrac {1}{h}}.}

Application to paper folding

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Folding a rectangular sheet of paper into thirds using the crossed ladders problem

The optic equation of the crossed ladders problem can be applied to folding rectangular paper into three equal parts:

1/1/2 +1/1 =1/h   ∴   2 + 1 =1/h   ∴  h =1/2 + 1 =1/3

One side (left in the illustration) is partially folded in half and pinched to leave a mark. The intersection of a line from this mark to an opposite corner (red) with a diagonal (blue) is exactly one third from the bottom edge. The top edge can then be folded down to meet the intersection.[4]

It is also exactly one third horizontally from the left edge; folding the right edge to meet the intersection lets the paper be folded into thirds lengthwise.

Similarly, folding the left side twice to get quarters lets one fold the sheet into five equal parts:

1/1/4 +1/1 =1/h′   ∴   4 + 1 =1/h′   ∴  h′ =1/4 + 1 =1/5

and folding it thrice to get eights lets one fold the sheet into nine equal parts, etc.:

1/1/8 +1/1 =1/h″   ∴   8 + 1 =1/h″   ∴  h″ =1/8 + 1 =1/9

Extended crossed ladders theorem

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The crossed ladders theorem was extended to crossed ladders within a triangle. In 2002, Harold Joseph Stengel (1947–2007), an American secondary school teacher of mathematics, proved the extended theorem.[5]

Let AC be the base of a triangle ABC. Let ladder (line) AD have its foot at A and intersect BC at D; likewise, let ladder CE have its foot at C and intersect AB at E. Let AD intersect CE at F. Extend parallel lines from the points E, B, F, and D, intersecting AC at the points I, G, J, and H, respectively. Then

1/EI +1/DH =1/FJ +1/BG

whence it follows that

1/area (△ AEC) +1/area (△ ADC) =1/area (△ AFC) +1/area (△ ABC).

See also

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  • Right trapezoid, the quadrilateral with vertices at the tops and bottoms of the two ladders

References

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  1. ^Gardner, Martin (1979).Mathematical Circus: More Puzzles, Games, Paradoxes and Other Mathematical Entertainments from Scientific American. New York:Knopf. pp. 62–64.ISBN 9780394502076.
  2. ^abWeisstein, Eric W."Crossed Ladders Problem".MathWorld – A Wolfram Web Resource.
  3. ^Bremner, A.; Høibakk, R.; Lukkassen, D. (2009)."Crossed ladders and Euler's quartic"(PDF).Annales Mathematicae et Informaticae.36:29–41.MR 2580898.
  4. ^Meyer, Daniel; Meyer, Jeanine; Meyer, Aviva (March 2000). "Teaching mathematical thinking through origami".Academic.Writing: Interdisciplinary Perspectives on Communication Across the Curriculum.1 (9): 1.doi:10.37514/awr-j.2000.1.9.41.; see in particular section "Dividing into thirds"
  5. ^Stengel, H. (2002–2003)."Letter to the Editor: The extended crossed ladders theorem"(PDF).Mathematical Spectrum.35 (1):18–20.

External links

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