My Favourite Problem No.4 Solution
A
D
E
x
F
B
C
y
z
Labelling the points A, B, C, D, E and F and distances x, y and z as in the diagram it can be seen
that triangles EBF and DBC are similar and that triangles ECF and ACB are similar.
Therefore
x
400
y

and
y z
x
600

z
y z
.
Adding these two equations together (the left hand sides and the right hand sides) gives
x
400
Therefore
Or
So
x
400

x
600

x
600

y
y z

z
y z

y z
 1.
y z
 1.
3x
2x

 1.
1200 1200
5x
1
1200
and
x 
1200
 240 .
5
The most surprising thing about this problem is that it is possible to calculate the height at which
the cable cars cross at all. The power of the concept of similar triangles is illustrated very well.
It’s worth noting that the value of x is independent of how far apart the starting points of the cable
cars are on the ground.
This means that in these pictures the marked distance (the height of the crossing point) is always
the same.
600m
400m
600m
600m
400m
Further Investigation
Similar triangles have many important applications. They are the basis for trigonometry which is
used extensively in all mathematics.
One particular application of similar triangles is to measure distances indirectly. This means that
it’s possible to estimate the distance to the moon without having to go anywhere near the moon!
You can read more about this here:
http://www.astro.washington.edu/courses/astro211/CoursePack/cp03a_distance_sun.pdf
400m