By H. Fine, H. Thompson
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2). Was the length π of that kind? This was certainly a good question for mathematicians . . and not only for crazy ones! The proof of the transcendence of π by Lindemann in 1882 provided a final negative answer to the Greek circle squaring problem (See Sect. 3). Let us stress the fact that Greek geometers had some good reasons to believe that squaring a circle with ruler and compass could be possible, after all they were able to square more complicated figures delimited by arcs of circles. For example, 20 2 Some Pioneers of Greek Geometry Fig.
2 r R Writing π for this last ratio, which is thus independent of the size of the circle, we obtain the famous formula A = πR 2 . However, let us stress once more that for Greek geometers, such a ratio π is not a number and such a formula πR 2 for the area would not have been considered. As observed above, the area of a regular polygon is equal to its perimeter multiplied by half the apothem. Repeatedly doubling the number of sides, one thus expects to recapture the result, already “known” to the Egyptians (see Sect.
5, find the equation of this hyperbola. 3 Determine the locus of those points P such that the distances from P to two fixed points A and B is in a constant given ratio. 1 Write down a proof of Pythagoras’ theorem based on the consideration of areas in Fig. 26. 2 Construct with ruler and compass a regular pentagon with prescribed side. 3 Construct with ruler and compass a regular star pentagon with prescribed side. 4 Construct with ruler and compass a regular pentagon inscribed in a prescribed circle.