A 3' * 3' packing case is placed against a wall.

geometry/ladder.and.box Added material

Problem

A 3' * 3' packing case is placed against a wall.  A 15' ladder is
placed such that it touches the floor, the edge of the packing case and
the wall.

The question is:

How high up the wall is the ladder?


    |
  ^ |\
    | \
    |  \
    |   \ 15
  ? |    \
    |     \
    |      \
    |_______\ 
    |   3   |\
    |      3| \ 
  v |_______|__\___________________

Show solution

Let me redraw your diagram:

     M
     |\
     | \
     |  \
     |   \
     |    \
     |     \
     |      \
   Q |_______\ N
     |   3   |\
     |      3| \ 
     |_______|__\___________________
             Q'  P

I take your diagram to mean that MP = 15, and NQ = NQ' = 3.  And we
want to find MQ+3.  Now fold rotate triangle MQN around N, so that we
get a new right triangle:

                   N
                   _
                __/|\
             __/   | \
          __/      |  \
       __/         |   \
   M  /____________|____\  P
                   Q

Define a = NP, b = MQ, c = QP, d = MP.  Then MN = 15-a, and we want to
find b+3.

Now, b is to 3 as 3 is to c, so d = b+c = b+(9/b).  Triangle MNP is a
right triangle, so

    a^2 + (15-a)^2 = d^2

Now, consider the area S of triangle MNP.  We can consider segment MP
to be the base, in which case we get

    S = 3d/2

Or we can consider segment NP to be the base, in which case we get

    S = a(15-a)/2

So we have

    a(15-a) = 3d

From this we can write

    15^2 = [a + (15-a)]^2
         = a^2 + 2a(15-a) + (15-a)^2
         = a^2 + (15-a)^2 + 2a(15-a)
         = d^2 + 6d

or

    d^2 + 6d - 225 = 0

which we can solve easily to get d = sqrt(234)-3.  (The other root
is negative and hence cannot be a length.)  Recall that

    d = b+(9/b)

We can rewrite this as

    b^2 - db + 9 = 0

which we solve to get b = (d + sqrt(d^2-36))/2.  The desired length
is then just b+3.

byron elbows
brian@isi.edu

Additional material

[ Editor: numerically, the solution above gives a height of about
14.52 feet.  The other root of b^2 - db + 9 = 0, b = (d - sqrt(d^2-36))/2,
gives the second (flat) position of the ladder that the questioner
expected: b+3 is about 3.78 feet.  The two positions are mirror images,
with the height and the distance from the wall interchanged. ]

[ Editor: an exact analytic solution was also posted for a version with
a 1 x 1 box and a ladder of length 6, inferred from its equations; the
original statement of that version is not in the collected postings; in the equations x is the
height of the top of the ladder, x-1 = y is its height above the box,
and by similar triangles the foot is 1/y + 1 from the wall. ]

Let x=y+1, now clearly

  (y+1)^2 + (1/y + 1)^2 = 36

  y^4 + 2y^3 + 2y^2 + 2y + 1 [Editor: the original has 2 here] = 36y^2

  y^4 + 2y^3 + 3y^2 + 2y + 1 [Editor: the original has 2 here] = (y^2 + y + 1)^2 = 37y^2

  y^2 + y + 1 = +- sqrt(37) y

  y^2 + ay + 1 = 0, where a=(1 +- sqrt(37))

  y = (-a +- sqrt(a^2 - 4))/2

4 roots, out of which the two with a = 1 - sqrt(37) are appropriate for
the problem [Editor: the original has "problems"] (these two yield positive y), so

  x = 1 - ((1 - sqrt(37)) +- sqrt((1 - sqrt(37))^2 - 4))/2
    = (1+sqrt(37) +- sqrt(34 - 2sqrt(37)))/2

-- Stein Kulseth (stein.kulseth@nta.no)

[ Editor: the posting had a constant term of 2 instead of 1 in the
second and third equations; corrected here.  The two answers are
x = 5.878... and x = 1.205.... ]

[ Editor: a version with a 1 x 1 x 1 metre box and a ladder of 4 metres
was posted to geometry.puzzles in August 1993: "A ladder of 4 Meters is
standing against a wall, also against the wall is a cubic box of 1 x 1
x 1 meters.  The ladder is touching the box at the side.  At what
height is the ladder touching the wall?" ]

       |
       |\
       | \
       |  \ Ladder
    x  |   \
       |    \ 4
       |_____\
       |     |\
    1  | Box | \
       |_____|__\________
           1    y

By the Pythagorean Theorem, (x+1)^2 + (y+1)^2 = 16.
So, x^2 + y^2 + 2x + 2y + 2 = 16.
By similar triangles xy = 1.  Replace 2 with 2xy, and factor to get
(x+y)^2 + 2(x+y) - 16 = 0.  From the quadratic formula,
x+y = sqrt(17) - 1.

Consider the equation u^2 - (sqrt(17) - 1) u + 1 = 0.  The sum of the
roots is sqrt(17) - 1, and their product is 1, so the roots are x and
y.  Consequently,
x = [(sqrt(17) - 1) +- sqrt(14 - 2 sqrt(17))] / 2.
The height at which the ladder touches the wall is
x + 1 = [(sqrt(17) + 1) +- sqrt(14 - 2 sqrt(17))] / 2.

-- David Radcliffe <radcliff@csd4.csd.uwm.edu>