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Curiouser and Curiouser Author(s): Helen Morris Source: Mathematics in School, Vol. 26, No. 2 (Mar., 1997), pp. 34-35 Published by: The Mathematical Association Stable URL: http://www.jstor.org/stable/30211826 . Accessed: 07/04/2014 14:20 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp . JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. . The Mathematical Association is collaborating with JSTOR to digitize, preserve and extend access to Mathematics in School. http://www.jstor.org This content downloaded from 184.145.136.227 on Mon, 7 Apr 2014 14:20:57 PM All use subject to JSTOR Terms and Conditions

Curiouser and Curiouser

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Curiouser and CuriouserAuthor(s): Helen MorrisSource: Mathematics in School, Vol. 26, No. 2 (Mar., 1997), pp. 34-35Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30211826 .

Accessed: 07/04/2014 14:20

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

The Mathematical Association is collaborating with JSTOR to digitize, preserve and extend access toMathematics in School.

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Curiouser and

Curiouser

Compiled by

Helen Morris

Captive Queen

A captive Queen and her son and daughter were shut up in the top room of a very high tower. Outside their window was a pulley with a rope round it, and a basketfastened at each end of the rope of equal weight. They managed to escape with the help of this and a weight they found in the room quite safely. It would have been dangerous for any of them to come down if they weighed more than 15 lb more than the contents of the lower basket, for they would do so too quickly, and they also managed not to weigh less either. The one basket coming down would naturally of course draw the other up. How did they do it? The Queen weighed 195 lb, the daughter 165 lb, and the son 90 lb. The weight was 75 lb.

Extension

The Queen also had in her room a pig weighing 60 lb, a dog weighing 45 lb, and a cat of 30 lb. These have to be brought down safely too, with the same restriction. The weight can come down any way, of course. There must be someone at each end to put the animals into and out of the baskets.

From: The Complete Stories of Lewis Carroll 1993, Magpie Books, London

Captive Queen Solution

The solution is given in the form of who is still in the tower. The following symbols have been

used: Q - queen, D - daughter, S - son, W- weight, P - pig, d - dog, C - cat. There are times when someone or something has to be returned to the tower.

QDSW QDS QDW QD QSW QS QW DS DW D SW S W

at

Extension

QDSWPCd QDSPCd QDWPCd QDPCd QDWd QDWC QDC QSWC QSC QWC DSC Qs Qw DS DW D SW S -

34 Mathematics in School, March 1997

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Maths and Poetry

The M0bius Band

A mathematician confided That a Mobius band is one-sided But you'll get quite a laugh

When you cut one in half For it stays in one piece when divided.

Twisting the bandjust once You'll get a hoop that is one But twist the band twice Then it will be twice as nice For two pieces are better than one

Mathematical Graph-iti

Addddition Subraction Rat:o Division

Recurrrring Percen%age Equ=tion ld.p 2.dp

TriAngle Index AXes Factor!al

(S, E, T) Dec.mal Milll,000,000n

The Cubic Controversy

This is a story of two men, both from Italy - Girolamo Cardano (1501-1576) and Niccolo Fontana (1499-1557). Other key players in this story - Luigi Ferrari, Antonio Fior, Scipione del Ferro.

The scene - 16th century Italy.

The event - the discovery of an algebraic solution of cubic equations.

One day, in 1515, a mathematics professor at Bologna University, called Scipione del Ferro was working on his mathematics when he managed to solve algebraically the cubic equation x3 + mx = n. He was very excited as this had never been able to be solved

algebraically before. He did not publish his solution but revealed it to one of his students, Antonio Maria Fior. Now, in 1535, Niccolo Fontana (known as Tartaglia, the stammerer, because of an affliction to his speech), claimed to have found an algebraic solution to the cubic equation x3 + px2 = n. Tartaglia had a reputation of not always being truthful - he had claimed pieces of work as his own even when they had already been published. So, Fior challenged Tartaglia to a public contest on solving cubic equations, because he did not believe him. Tartaglia won the contest easily, solving all of the equations set by Fior.

News of Tartaglia's triumph reached Girolamo Cardano, who then invited Tartaglia to tea in order to try to encourage him to publish his results. Tartaglia did not want to publish, as he was saving his findings to be his 'piece de resistance' in a book that he was writing on algebra. Unscrupulous Cardano managed to persuade Tartaglia to show him his solution. Cardano was sworn to

secrecy. In 1545, Cardano published his own algebraic book, Ars Magna. In it appeared Tartaglia's solution of the cubic. Tartaglia was not pleased! One of Cardano's pupils, Ludovico Ferrari (later poisoned by his own sister), argued that Cardano had received his information second hand from del Ferro himself, and accused Tartaglia of plagiarism. Much angry dispute followed. The real hero of the story is probably del Ferro, who had the method of solving cubic equations in the first place.

The solution of the cubic equation, x3 + mx = n, as given by Cardano.

Choose values of a and b such that 3ab = m and a3 - b3 = n, and let x be given by a - b.

Then (a - b)3 + 3ab(a - b) = a3 - b3

Solving the equations for a and b we find

a = 3 q [(n/2) + 4 {(n/2)2 + (m/3)3}] and b = 34 [- (n/2) + 4 {(n/2)2 + (m/3)3}] an so x is determined. -

Contributions to this column are most welcome and should be sent to: Helen Morris, Balshaw's High School, Church Road, Leyland, PR5 2AH

Mathematics in School, March 1997 35

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