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The Crossed Ladder Puzzle

The traditional "crossed ladder" puzzle begins by telling you there are two buildings, A and B, separated by a city street of width x.  There are two ladders, AD of length a and BE of length b, are positioned as shown.  The puzzle gives you f, a, and b and asks you to find x, the width of the street.

The solution depends on the "Crossed Ladder Theorem", which says 1/e + 1/d = 1/f.  This, together with the Pythagorean Theorem, gives you the following equations:

a2 = d2 + x2,
b2 = e2 + x2

Eliminating x gives you the following quartic equations:

e4 - 2fe3 + (f-e)2(a2-b2) = 0
d4 - 2fd3 + (f-d)2(b2-a2) = 0

The solutions to quartics are not easy, but numeric solutions aren't hard.  Also, in typical contrived examples, the quartics can be easily factored.'s example is a case in point: a=105, b=87, f=35.  These values give you the following quartics:

e4 - 70e3 + 3456e2 - 241920e + 4233600
d4 - 70d3 - 3456d2 + 241920d - 4233600

Synthetic division quickly gives e=60, d=84.  Then the Pythagorean Theorem gives you x=63.

The Crossed Ladder Theorem

The heart of the Crossed Ladder Puzzle is the fact that 1/e + 1/d = 1/f.  This is called the Crossed Ladder Theorem.


By similar triangles,

AF'/AB = f/d, and
BF'/AB = f/e

The sum of AF'/AB and BF'/AB is 1, so

1 = f/d + f/e

Dividing by f, the result follows:

1/f = 1/d + 1/e

A Triangle Area Puzzle

A triangle is divided into four regions by two straight lines.  The areas of three of the regions are given.  What is the area of the fourth region.  (Note: diagram is not drawn to scale)

The solution to the puzzle involves a variant of the Crossed Ladder Theorem, which I will call the "Extended Crossed Ladder Theorem", which states,

1/c + 1/f = 1/d + 1/e

For our puzzle, the values of e, f, and d in the diagram to the left are clear from the areas in the diagram above:

e = (5+10)/w
f = 10/w
d = (8+10)/w,

where w is (1/2)(AB).

From the Extended Crossed Ladder Theorem, w/c + w/10 = w/18 + w/15, so c=45/w, which means the area of the entire triangle is 45.  To find the indicated area, we subtract the other areas from 45, so the answer to the puzzle is 45 - 10 - 5 - 8 = 22.

The Extended Crossed Ladder Theorem

A.k.a the "Crossed Ladders in a Triangle" theorem

1/c + 1/f = 1/d + 1/e


A simple proof of the Extended Crossed Ladder Theorem works by drawing two additional lines, representing buildings, which are parallel to the four lines labeled c, d, e, and f.  We extend the two ladders and two sides of the triangle to meet these two new buildings.

We label the heights of the points where the extended ladders meet the buildings as a and b.  We label the heights of the points where the extended triangle sides meet the buildings as x and y.  Note x and y encompass the entire heights of the buildings.  In other words, part of line x overlaps a, and part of line y overlaps b.

Now there are four crossed ladders, crossing at points C, D, E, and F.  From the standard Crossed Ladder Theorem, we get the following identities:

1/c = 1/x + 1/y
1/f = 1/b + 1/a
1/d = 1/x + 1/b
1/e = 1/y + 1/a

By adding the first two and adding the second two, we get

1/c + 1/f = 1/x + 1/y + 1/a + 1/b
1/d + 1/e = 1/x + 1/y + 1/a + 1/b

So 1/c + 1/f = 1/d + 1/e, and the theorem is proved.

Final Remark

In the foregoing diagrams, the lines labeled a, x, b, y, c, d, e, and f all appear to be perpendicular to the base, AB.  This isn't actually necessary to the validity of either of the Crossed Ladder Theorems, and in fact, I've taken care to avoid saying that.  The only condition that is necessary for these theorems to be true is that the lines all be parallel to one another.  The reason for this is due to the similar triangles that result from dropping perpendiculars from each of these same points -- the lengths of the perpendiculars are all in proportion to the corresponding lines.

Internet references

Mathworld: Crossed Ladders Problem, Crossed Ladders Theorem The Crossed Ladder Problem 

Related pages in this website:

Summary of geometrical theorems 


The webmaster and author of this Math Help site is Graeme McRae.