# A two mile pier and a one mile pier extend perpendicularly into the ocean with four miles of shore between the two piers. A swimmer wishes to swim from the end of the longer pier to the end of the...

A two mile pier and a one mile pier extend perpendicularly into the ocean with four miles of shore between the two piers. A swimmer wishes to swim from the end of the longer pier to the end of the shorter pier with one rest stop on the beach. Assuming that the swimmer recalls that, in any right triangle, the hypotenuse2 = leg2 + leg2, find the shortest possible swim.

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### 1 Answer

The swimmer begins on the end of the 2 mile pier, swims to the beach and then out to the end of the 1 mile pier, and we are asked to find the shortest possible swim.

The easiest solution is to reflect the 1 mile pier across the line formed by the beach. (See attachment.) Now draw a line from the end of the 2 mile pier to the reflected end of the 1 mile pier. Extending the 2 mile pier 1 mile "below" the beach and connecting this point to the reflection of the 1 mile pier creates a right triangle whose sides are 3 miles and 4 miles long. The hypotenuse of this triangle is 5 miles long.

If d1 is the distance from the 2 mile pier to the shore, and d2 is the distance from the shore to the 1 mile pier, and if we assume that the point on the beach is where the line from the 2 mile pier to the reflected 1 mile pier intersects the beach, then d1+d2=5 miles.

Claim: 5 miles is the shortest distance. Let A be the end of the 2 mile pier, B the end of the 1 mile pier, B' the reflected end, and X the intersection of AB' and the shore. Suppose Y is a point on the shore between the piers and Y is not X. Then the total path is AY+YB. But by construction the total path using X is AX+XB=AX+XB'. Now consider triangle AYB'. By the triangle inequality theorem, AY+YB'>AB'=AX+XB'. Thus AY+YB>AX+XB for any choice of Y not equal to X.

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The shortest swimming distance is 5 miles.

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You can use calculus to minimize the sum of the distance functions, or algebra to find the minimum, but the geometric argument is easier.