By Martin Erickson

ISBN-10: 0883858290

ISBN-13: 9780883858295

Every mathematician (beginner, beginner, alike) thrills to discover uncomplicated, stylish options to possible tough difficulties. Such chuffed resolutions are known as ``aha! solutions,'' a word popularized via arithmetic and technology author Martin Gardner. Aha! suggestions are amazing, wonderful, and scintillating: they demonstrate the wonderful thing about mathematics.

This booklet is a set of issues of aha! strategies. the issues are on the point of the varsity arithmetic scholar, yet there may be whatever of curiosity for the highschool scholar, the instructor of arithmetic, the ``math fan,'' and an individual else who loves mathematical challenges.

This assortment comprises 100 difficulties within the parts of mathematics, geometry, algebra, calculus, chance, quantity thought, and combinatorics. the issues start off effortless and usually get tougher as you move in the course of the ebook. a couple of strategies require using a working laptop or computer. an immense function of the booklet is the bonus dialogue of similar arithmetic that follows the answer of every challenge. This fabric is there to entertain and let you know or aspect you to new questions. in the event you do not take note a mathematical definition or thought, there's a Toolkit at the back of the booklet that might help.

**Read Online or Download Aha! Solutions PDF**

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**Additional info for Aha! Solutions**

**Sample text**

Every symmetry of the cube permutes (interchanges) the diagonals, and conversely every possible permutation of the diagonals comes from a symmetry of the cube. Similarly, we can show that the symmetry group of a regular octahedron is S4. Now that we know the symmetry group of the cube (S4), we find that the symmetry group of the regular tetrahedron goes along for the ride. Every symmetry of the cube automatically gives a symmetry of the regular tetrahedron (with the tetrahedron inscribed in the cube), or it moves the vertices of the tetrahedron to the other four vertices of the cube.

The centers of the equilateral triangles, A", 8", C" , are given by a", b", e", respectively, where I a" = 3(a' +b +e), I b"=3(a+b'+e), I e" = 3(a + b + e'). 2 Geometry 51 To show that A", B", e" are the vertices of an equilateral triangle, we will demonstrate that M(b" - a") = e" - a". Here is the algebra: M(b" - a") = ~M(a + b' + e - a' - b - e) 3 I = -M(a - b 3 I I = -(e - b 3 I = - (a 3 +b I - e +e - +a - e +b - +b +e I I a) I a) I - a - b - e) = e" -a. " By the way, the result in Napoleon's theorem also holds if the equilateral triangles are constructed inwardly rather than outwardly.

On) = + S5 S2 + S4 - SI - S3 1- ••• ••• , 31 I. 3 Geometry where we set Sk = 0 if k > n, so that the series in the numerator and denominator are finite. Let's give an aha! proof of this formula. The proof uses complex numbers and Euler's formula cos 0 + i sin 0 = e jo . Setting 0 equal to the sum of our n angles, we obtain COS(OI + ... + On) + i sin(OI + ... •• e jOn , and hence the identity COS(OI + ... + On) + i sin(OI + ... • . (cos On + i sin On). The real part of the left side of this identity is COS(OI + ...