By Heinrich Dorrie

ISBN-10: 0486318478

ISBN-13: 9780486318479

ISBN-10: 0486613488

ISBN-13: 9780486613482

"The assortment, drawn from mathematics, algebra, natural and algebraic geometry and astronomy, is awfully attention-grabbing and attractive." — *Mathematical Gazette*

This uncommonly attention-grabbing quantity covers a hundred of the main well-known ancient difficulties of trouble-free arithmetic. not just does the publication endure witness to the intense ingenuity of a few of the best mathematical minds of historical past — Archimedes, Isaac Newton, Leonhard Euler, Augustin Cauchy, Pierre Fermat, Carl Friedrich Gauss, Gaspard Monge, Jakob Steiner, and so forth — however it presents infrequent perception and concept to any reader, from highschool math scholar to specialist mathematician. this is often certainly an strange and uniquely priceless book.

The 100 difficulties are provided in six different types: 26 arithmetical difficulties, 15 planimetric difficulties, 25 vintage difficulties bearing on conic sections and cycloids, 10 stereometric difficulties, 12 nautical and astronomical difficulties, and 12 maxima and minima difficulties. as well as defining the issues and giving complete suggestions and proofs, the writer recounts their origins and historical past and discusses personalities linked to them. frequently he supplies now not the unique answer, yet one or less complicated or extra attention-grabbing demonstrations. in just or 3 situations does the answer imagine whatever greater than a data of theorems of effortless arithmetic; for this reason, it is a ebook with a really large appeal.

Some of the main celebrated and exciting goods are: Archimedes' "Problema Bovinum," Euler's challenge of polygon department, Omar Khayyam's binomial enlargement, the Euler quantity, Newton's exponential sequence, the sine and cosine sequence, Mercator's logarithmic sequence, the Fermat-Euler best quantity theorem, the Feuerbach circle, the tangency challenge of Apollonius, Archimedes' selection of pi, Pascal's hexagon theorem, Desargues' involution theorem, the 5 normal solids, the Mercator projection, the Kepler equation, selection of the location of a boat at sea, Lambert's comet challenge, and Steiner's ellipse, circle, and sphere problems.

This translation, ready particularly for Dover by means of David Antin, brings Dörrie's "Triumph der Mathematik" to the English-language viewers for the 1st time.

Reprint of *Triumph der Mathematik*, 5th version.

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**Extra info for 100 great problems of elementary mathematics: their history and solution**

**Sample text**

Determination of the binomial coefficient C gives us immediately the sought-for binomial expansion: Here α and β pass through all the possible integral non-negative values that satisfy the condition α + β = n. β! one usually writes and also abbreviates this coefficient nα (read as n sub α). The expansion then takes on a somewhat simpler appearance: The coefficient nv is known as the binomial coefficient to the base n with index v. The binomial theorem was probably discovered by the Persian astronomer Omar Khayyam, who lived during the eleventh century.

This gives the problem the following appearance; VII. Finally, since of all the multiples of only 5 = 627365 added to the division remainder 110177 of the third line gives a number containing a 7 in the third place, we get K = 5 and at the same time abΔcde = 627365 and AB7CDE = 737542, which gives us all of the figures missing from the problem. 5 Kirkman’s Schoolgirl Problem In a boarding school there are fifteen schoolgirls who always take their daily walks in rows of threes. How can it be arranged so that each schoolgirl walks in the same row with every other schoolgirl exactly once a week?

Similarly, for the eighth line we obtain 8 · 12547ε = 10037**, and consequently t = 0 and u = 3. Since λ = λ · 12547ε results in a seven-place fourth line and only 8 and 9 have seven places, λ is either 8 or 9. V. From t = 0 and X 1 (together with R = r = 1, S = s = 0) it follows that T 1, and from n = 8, N 9, it follows that T 1, so that T = 1. N is therefore equal to 9 and X = 1. Since X = 1 and 2 · > 200000 (line 9), it follows that v = 1 and also that Y = 2, Z = 5, x = 4, y = 7, and z = ε. With the results obtained at this point the problem has the following appearance: VI.

### 100 great problems of elementary mathematics: their history and solution by Heinrich Dorrie

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