ALL SWIMMED OUT
Suppose Sammy the
swimmer at the tip of pier H wants to swim to the tip of pier I. Pier
H is 2 km long and pier I is 1 km long. Since the swim is very long
from H to I, Sammy thinks he will need to stop off at the beach to
take a break (at point J). Sammy can stop at any point on the beach
between the two piers (drag point J around once you construct this in
a geometry software or view the java applet below). If Sammy takes the
break, where should he stop on the beach if he wants to swim the least
distance (the blue path) for the entire trip? What is the shortest
distance Sammy can swim for the entire trip?
This is the typical
shortest distance using two triangles problem. We are trying to
evaluate at which point on the common leg (beach) of the triangles
which the swimmer must touch in order to swim the shortest distance.
Solve/Investigate the Problem
I hypothesized that
the shortest distance for Sammy to swim would be the distance created
by the base legs of two similar right triangles. In this case, the
heights of the triangles have a ratio of 2 to 1. Therefore, if my
hypothesis is correct, the legs would also have a ratio of 2 to one,
and the angles would be congruent at the point of least distance.
Investigation/Exploration of the Problem
I constructed a
diagram similar to the one given in the Investigation. Once placed the
common point at the vertices of the two right angles, I began moving
the point and investigating both the measurement changes and the angle
variances from point to point. Although it is not exact due to the
measurement limitations of GeoSketchPad, it does appear that the least
distance from H to I running through J is indeed at the point that the
triangles have proportional relationships.
Extensions of the
To continue on with
a proof, you could construct transverse triangles and, using the
Pythagorean Theorem (as well as typical minimum distance theory) and
prove that indeed the minimum distance is at the point of similarity.
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