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Download e-book for kindle: A Mathematical Jamboree by Brian Bolt

By Brian Bolt

ISBN-10: 051141496X

ISBN-13: 9780511414961

ISBN-10: 0521485894

ISBN-13: 9780521485890

A Mathematical Jamboree follows the very winning puzzle books by way of a similar writer. It includes 114 puzzles and actions to problem humans of every age. there's a exact remark on the finish of the booklet, giving ideas and motives, including the occasional follow-up challenge.

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Sample text

However, the substandard balls were each 1 gram too light. The retailers were informed of this discrepancy. The boxes all arrived in packs of ten, each with one substandard box - but which one? Sidney Sixiron, the professional at a prestigious golf course, had just taken delivery of a large order so needed to sort them out quickly. He soon found a way to do this which required only one weighing for each batch of ten boxes, using a pair of scales and a set of weights. How did he do it? Note that he did not need to know what a golf ball should weigh.

Now this table was compiled without taking any account of the costs. However, the second table gives the cost, in tens of pounds, of moving individual cars between the centres, so by using it we can see the cost of the above redistribution: 3x£60 + 2x£50 + 4x£40 + £40 + 4x£50 + £80 + £30 + 3 x£40 + £70 + 3 x£50 = £1130 Now the viability of the firm depends on it always making these redistributions as cheaply as possible. How should the cars be moved to minimise the cost? 48 car imovements to P Q R S A 3 0 2 4 9 from B 1 4 0 1 6 C 1 3 1 3 8 5 7 3 8 distribution costs to P Q R S A 6 2 5 4 from B 4 5 3 8 C 3 4 7 5 90 Pentagonal magic!

AX is common, to triangles AXR and AXC angle RAX = angle QAX from the construction angle ARX = angle AQX = 90Q Hence AR=AQ (ii) So triangles BXR and CXQ are congruent. BX = CX, as PX is the perpendicular bisector RX = QX, from the congruence of the previous triangles angle BRX = angle CQX = 90Q Hence RB = QC So AB = AR + RB = AQ + QC = AC, thus proving that triangle ABC is isosceles! Continuing with the same approach, we can prove CA = CB, and hence the triangle is equilateral! Where is the flaw in the argument?

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A Mathematical Jamboree by Brian Bolt

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