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Ladder Puzzles

Date: 03/16/2001 at 14:38:56
From: Craig M Chamberlain
Subject: Triangle ladder puzzle

I found a very interesting math puzzle back when I was in high school, 
and every so often I return to it to see if I can solve it myself - 
but no such luck.

It was in Popular Science magazine circa 1980-1985, and I am not sure 
if they ever gave the answer so I am going to type the exact 
question and see what help you can provide. There was also a diagram 
which I have duplicated here very simply.

"Cussed Ladders"

"The January Calcu-letter introduced an ancient and tough ladder 
puzzle with a comparatively mild warm-up. You were told of two ladders 
leaning opposite ways between two buildings, one touching eight feet 
above the ground, the other twelve. At what height above the ground 
would they cross?

  |\         /|
  | \       / |
  |  \     /  |
  |   \   /   |
  |    \ /    |
8'|    /|\    | 12'
  |   / | \   |
  |  /  |x \  |
  | /   |   \ |
  |/    |    \|
     a     b

"The problem could have been solved using geometry, algebra, 
trigonometry, analytic geometry - or by selecting an arbitrary value 
for the distance between the buildings, since the height must be the 
same no matter what distance is used.

"Maybe the simplest way is to label the distance on the ground from 
each building to line x as a and b, which allows you to set up two 
simple ratios based on similar triangles:

      12     x       8     x
     ---- = --- and --- = ---
     a+b     a      a+b    b

"Multiplying both equations through and substituting for the common 
terms yeilds what may have been intuitively obvious: b = 3/2a. Now you 
can rewrite the above equations using 3/2a in place of b and 5/2a 
in place of a+b. Solving both equations for x (the a's cancel) gives 
the height above the ground at which the ladders intersect: 4.8 feet.

"Now you are ready to try the second variation of the cussed ladders. 
This time, ladder lengths are given as 40 and 30 feet, and they cross 
at a point 10 feet from the ground. How far apart are the buildings 
against which the ladders are leaning?

"Because in this puzzle you have three dimensions given instead of 
two, it may at first glance look to be the easier problem. It is not. 
This one will require that you use your calculator - as well as your 

So that is the problem as it was presented with no answer. My thought 
was again to use similar triangles, this time using 10' instead of x 
and again using a and b for the base. Now we have two similar 

     / | \             10     b        10     30
    /  |  \           ---- = ---- and ---- = ----
   /   |   \           a      10       b      a+b
  /    |10' \
 /     |     \
   a      b

Once I follow this through (I hope you can see my reasoning here), 
as in the first problem I substitute in b for the other ratio and 
proceed to solve for the other part of the base, yet when I use these 
to check my answer by using Pythagoris to get back 30 and 40 for the 
hypotenuse, it doesn't work. 

I am stuck!


Date: 03/17/2001 at 19:00:13
From: Doctor Floor
Subject: Re: Triangle ladder puzzle

Hi, Craig,

Thanks for writing.

Let's say that the walls are the lines x = 0 and x = D in a 
rectangular grid, and that the x-axis is the floor.

Let's say that the shorter, 30-foot-long ladder is placed at O(0,0). 
This ladder reaches x = D at point A(D, sqrt(900-D^2)) (where we used 
the Pythagorean theorem to calculate the y-coordinate).

[Note: of course 900-D^2 must be greater than 10, which shows that D 
must be smaller than sqrt(800).]

The point where this ladder reaches the 10-foot-high mark is

 P( 10D/sqrt(900-D^2) , 10 ).

The line through P and B(D,0) has gradient

            10 - 0            10sqrt(900-D^2) 
   ----------------------  = -------------------
   10D/sqrt(900 - D^2)-D     D(10-sqrt(900-D^2))

so the equation is

         10sqrt(900-D^2)       10sqrt(900-D^2)
   y = ------------------- x - ---------------
       D(10-sqrt(900-D^2))     10-sqrt(900-D^2)

(Point B now fits fine).

This means that the y-intercept is:


Since the second ladder is 40 feet long, this must be equal to


So we can solve D from the equation:

 ---------------- = sqrt(1600-D^2) 


 10sqrt(900-D^2) = sqrt[(900-D^2)(1600-D^2)] - 10 sqrt(1600-D^2)

and after regrouping

 10[sqrt(900-D^2) + sqrt(1600-D^2)] = sqrt[(900-D^2)(1600-D^2)]

After that, squaring gives

 100[2500-2D^2 + 2sqrt[(900-D^2)(1600-D^2)]] = (900-D^2)(1600-D^2)

and simplifying yields

 200sqrt[1,440,000 - 2500D^2 + D^4] = 1,190,000 - 2300D^2 + D^4

Squaring this again, we find an equation that is quartic in D^2. From 
there it is theoretically possible to solve it, but the computations 
are tedious.

For some information on how to solve quartic equations, see the Dr. 
Math FAQ:

  Cubic and quartic equations   

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum   

Date: 03/14/2001 at 09:55:59
From: David Sabath
Subject: Geometry

This is a problem for which I have never seen a satisfactory answer.  
I have worked on it on and off for quite a while, and have presented 
it to several people who should know how to do it. I searched your 
archives and found a similar problem.  

The similar problem was the problem of the week for June 6-10, 1994.  
Your problem of the week was as follows: Two ladders, one 20 feet 
and one 30 feet, are leaning on opposite walls of an alley 10 feet 
wide. A basketball hoop hangs at the intersection of the ladders and 
the question is how high the hoop is. This is a fairly straightforward 
problem and the answer is rather easily computed.

My problem is this: Still two ladders, one 20 and one 30 feet long, 
leaning on opposite walls of an alley. However, I want to know how 
wide the alley has to be in order for the intersection point of 
the ladders to be exactly 10 feet high.

This problem appears simple at first glance, but I can assure it is 
not.  I would appreciate any help you can give me.

David Sabath
Math Enthusiast

Date: 03/22/2001 at 15:21:11
From: Doctor Jackpo
Subject: Re: Geometry
Hi David,

Let's look at a diagram so we can talk more easily about the geometry 


I'll prove in a moment that 

  1/AC + 1/DB = 1/FG.  

If this is true, then it must be the case that

  1/AC + 1/DB = 1/10 

and also that 

  AC^2 + AB^2 = 900


  BD^2 + AB^2 = 400.

Solving these 3 simultaneous equations,

  you would get AC = 27.3572325237..
                BD = 15/7612871098...
    and finally AB = 12.3118572378...

So, why is it true that 1/AC + 1/DB = 1/FG ?

Begin by looking at triangles ABC and ABD.  We know that triangle FGB
is similar to CAB and triangle DBA is similar to FGA. Therefore, 

  GB / (GB + GA) = FG / AC


  GA / (GB + GA) = FG / DB

Adding these two equations gives us

  FG ( 1/DB + 1/AC ) = GA + GB / (GB + GA)

which simplifies to 

    FG (1/DB + 1/AC) = 1 

       (1/DB + 1/AC) = 1/FG

I hope this helps. If you have more questions, write back.

- Doctor Jackpo, The Math Forum   

Date: 03/23/2001 at 03:02:38
From: Doctor Floor
Subject: Re: Geometry

Hi, David,

In addition to the letter Dr. Jackpo wrote as an answer to your 
question, let me write you a way to (theoretically) find the exact 

I quote from Dr. Jackpo:

   so we know that 1/AC + 1/DB = 1/10       [1]
   and we also know that AC^2 + AB^2 = 900  [2]
   and BD^2 + AB^2 = 400.                   [3]

Now we can go on:

[2]-[3] gives

  AC^2 - DB^2 = 500                        [4]

[1] rewrites into 

  1/AC + 1/DB = 1/10
  1/AC = 1/10 - 1/DB
  1/AC = (DB - 10)/(10DB)
  AC = 10DB / (10-DB)

And when we substitute this into [4] we find:

  100DB^2/(10-DB)^2 -DB^2 = 500
  100DB^2 - DB^2(10-DB)^2 = 500(10-DB)^2

This reduces to a quartic equation.

A quartic equation can be solved in exact roots. It would be too 
tedious to do this now here. For a description of the method, see from 
the Dr. Math FAQ:   

I hope this addition is helpful to you.

Best regards,
- Doctor Floor, The Math Forum   
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry

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