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Buildings and Ladders
Submitted by Christopher Schoettle, 10 February 1997.
Original article by Valerio De Angelis, this article by Allen Stenger.
We sent our Field Team out to place two ladders as shown in the figure, and to
measure how high they reached on each building
(y and z in the diagram).
The ladders are 15 meters and 20 meters long
(r = 20 and s=15).
The Field Team came back and reported that they had placed
the ladders with no trouble, but they couldn't measure the heights. The longest
tape measure they have is 6 meters, and they reported that the ladders definitely reached
more than 6 meters, but they couldn't measure how high. They did measure very carefully the
height where the ladders crossed
(h in the diagram),
to a fraction of a centimeter,
and they say that h
is exactly 576 cm.
They explained, "There are lots of
triangles in this problem, we're sure there must be some way to figure
out y and z
from h....
No, we don't know how to do it, we forgot everything
right after the exam, but there must be some clever person here who can figure it."
Are you the clever person they are thinking of? Can you figure
y and z
from the information given? If not,
we can send the Field Team back out to measure some more stuff, as long as it
can be done with a 6 meter tape; is there any additional measurement that would be
helpful?
Hint 1
Yes, there really is enough information, but since you are reading this hint we'll
assume you don't know how to do it yet. Let's imagine we send the Field Team out
to make more measurements. The horizontal distance between buildings is definitely
more than 6 m too, but they can measure that because it's on the ground (they can
measure 6 m, mark the point, then move the tape and and continue measuring).
If we know the distance x,
how could we find y and z?
Hint 2
With the value of x
it is easy, because x
is the base of two right triangles in the figure,
and we know their hypotenuses.
Therefore the Pythagorean theorem would give us the missing
sides y and z.
In fact if we knew x we could even figure h.
Do you see how?
Hint 3
Consider the two triangles ABC and EFC.
Clearly these triangles are similar (they have the same angles). So we have
\frac{x}{y} = \frac{FC}{h}
Using the same argument for the similar triangles
BCD and BFE, we get
\frac{x}{z} = \frac{x - FC}{h}
Adding these equations cancels the FC,
and there is a common factor x
that we can cancel, leaving us with
\frac{1}{z} + \frac{1}{y} = \frac{1}{h}
which we can also divide by 2
to write as
\frac{1}{2} \left( \frac{1}{z} + \frac{1}{y} \right) = \frac{1}{2h}
Why divide? Because we recognized the harmonic mean in this expression!
No matter what the lengths or angles of the ladders,
the crossing height h
is half the harmonic mean of the
heights at the walls.
So if we know x,
we can figure y and z
from the Pythagorean theorem and then figure
h from the harmonic mean.
Now, back to the original problem! Suppose we don't know
x,
but we do know
h.
How can we figure
y
and
z?
Click here for rest of the solution.
The Rest of the Solution
We already have most of the ingredients we need. The unknowns are
y
and
z.
The harmonic mean gives us one equation involving these,
and the Pythagorean theorem will give us another:
\begin{eqnarray}
x^2 + y^2 &=& r^2 \\
x^2 + z^2 &=& s^2.
\end{eqnarray}
Subtracting these gives
y^2 - z^2 = r^2 - s^2.
Solving the harmonic mean for
y
gives
y = hz/(z - h)
and substituting in the previous equation gives
\left( \frac{hz}{z-h} \right)^2 - z^2 = r^2 - s^2,
which we can multiply out and simplify to get a quartic equation,
z^4 - 2hz^3 + (r^2 - s^2)z^2 -2h(r^2 - s^2)z +h^2 (r^2 - s^2) = 0.
Substituting in the known values for
r,s,h
gives the quartic equation
z^4 - \frac{288}{25} z^3 + 175z^2 -2016z + \frac{145152}{25} = 0.
Here's the graph of this equation:
It appears to have a root that is approximately 4,
and a root that is approximately 9.
The Field Team reported that the height was definitely
greater than 6 m, and in any case it must be higher than the crossing height of
5.76 m, so we can reject the root near 4.
Substituting 9 into this ugly equation
shows that 9 really is an exact root! So we know
z = 9
m,
and from the harmonic mean we figure
y = 16
m.
Harmonic Mean and the Trapezoid
You probably already know the theorem that the median of a trapezoid has length
the
arithmetic mean
of the bases:
(The median is defined as the line bisecting the two sides.)
A less well-known fact is that the line through the intersection of the
diagonals and parallel to the bases has length the
harmonic mean
of the bases!
See if you can prove this; we already have all the
ingredients in Hint 3.
(The ladders are the diagonals, and the trapezoid is lying on its side.)
References
The ancient Greeks methodically defined ten different kinds of means,
but seven of these sank without a trace,
and of the remaining three only the arithmetic mean gets much use today;
the other two survivors (geometric mean and harmonic mean) are occasionally useful.
-
This
Math Pages page
explains the ten Greek means.
-
This
Math Forum page
lists some applications of the harmonic mean.
-
The problem is discussed in: Martin Gardner,
Mathematical Circus (Knopf, 1979), "Elegant Triangles", pp. 62-64.
-
The problem is discussed in this Wikipedia article.
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