A problem from the qualification round of the 2004/2005 Swedish Mathematical Contest, via the April 2009 issue of Crux Mathematicorum:
The cities A, B, C, D, and E are connected by straight roads (more than two cities may lie on the same road). The distance from A to B, and from C to D, is 3 km. The distance from B to D is 1 km, from A to C it is 5 km, from D to E it is 4 km, and finally, from A to E it is 8 km. Determine the distance from C to E.
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This solution is by Titu Zvonaru of Comănești, Romania. The sum of the distances AB, BD, and DE (3 + 1 + 4) equals the distance AE (8), so cities A, B, D, and E lie on the same straight road. The square of distance AC (25) equals the sum of the squares of CD and AD (9 + 16), so by the converse of the Pythagorean theorem CD is perpendicular to AD. This means that CD and DE are the legs of a right triangle, and Pythagoras shows that hypotenuse CE is 5 km.
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