1985 mathematics olympiadAuthor(s): Tom GriffithsSource: The Mathematics Teacher, Vol. 79, No. 2 (FEBRUARY 1986), pp. 82-83Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27964792 .

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i !

j I

3 ? PCI|^?|iOHS

Reactions to articles and points of view on teaching mathematics ;

Multiplying negatives?II After seeing "On Multiplying Negative Numbers" (Mary L.

Crowley and Kenneth A. Dunn, April 1985, 252-56), we felt that the readers of the Mathematics Teacher would appreciate a model that has been univer

sally effective with our stu dents for a number of years. It deals theoretically with taking movies (videos?) of a car in motion and then showing the film on a projector (VCR?). If the car was traveling forward

(positive) and the film is shown forward (also positive), the result is a picture of the car

going forward (positive). The readers can fill in the

other three combinations for themselves with as little diffi

culty as our students. Many of our students even remember this example years later?and

frequently use it themselves in later classes to explain a "sign error" to a classmate.

Lawrence M. Gilbert Jack Bigelow The Park School Old Court Road

Brooklandville, MD 21022

Right-triangle relationship revisited I noticed in "Reader Reflec tions" (January 1985) that

Morris Wilkenfeld stated a the orem about a square inscribed in a right triangle : The length of the side of the square is equal to the product of the legs over the sum of the legs. This theo rem is actually more general.

Because of space limitations, letters may be subject to abridgment. Al though we are unable to acknowledge those letters that cannot be published, we appreciate the interest and value the views of those who take the time to send us their comments. Readers who are commenting on articles are encouraged to send copies of their cor respondence to the authors. Please double-space all letters that are to be considered for publication.

Notice that the length of the side of the inscribed rhombus is

independent of the length of the third side AB of the triangle. The length of the side of the rhombus is also independent of m lC.

Theorem : The length of the sides of a rhombus inscribed in a triangle, sharing a vertex

with the triangle, is

ab

a + ?'

where a and b are the sides

meeting at the shared vertex.

Given : CXYZ is a rhombus

Prove :

ab r =

a +

where a = BC and b = AC.

Proof. We observe that ZY ! BC and XY \ AC, since

they are pairs of opposite sides in a rhombus. By marking cor

responding angles along trans

versals, we see that AAZY - AYXB. Therefore,

b ? r r

ab ? ar ? br + r2 = r2, ab ? ar ? br = 0,

ab = r(a + b), and

ab _ a + b~

David Mulkey-Vermehren Frankfurt International

School Oberursel/Ts. West Germany

1985 mathematics olympiad This year the twenty-sixth In ternational Mathematics Olym piad was held at a resort com

plex in Joutsa, Finland. Joutsa is a small town two hundred ki lometers north of Helsinki. The

jury met at Heinola, one hundred kilometers north of Helsinki, and the awards cere

mony took place at the Univer

sity of Helsinki.

Thirty-eight countries were

represented this year by 209

competitors, the largest number ever in both cases. A full team consisted of six com

petitors. This year for the first time

the Canadian team trained at the University of Waterloo for two weeks before departing for Finland. Also involved in the

training session were prospec tive candidates for the 1986 I.M.O. to be held in Poland.

Many thanks are due to the

many members of the Univer

sity of Waterloo mathematics

department who assisted in the

training of the team. Special thanks are due to professor R. Dunkley who directed the

training session. Much appreci ated also was the team uniform and other financial assistance

provided by the Canadian Mathematics Competition Foundation. All travel ex

penses for the team were paid by the Canadian Mathematics

Society, for which they express their gratitude.

The team members were chosen by a consideration of the results of various contests, including the Euclid, the Des cartes, the Canadian Olympiad, and the U.S. Olympiad, which several Canadian team mem bers were invited to write. The Finland competition was a second I.M.O. for two of the Canadian team, Minh Tue Vo and Frank D'Ippolito, and a third for Martin Piotte.

The Canadian team finished in twelfth position in 1985, their best showing since their

82- -Mathematics Teacher

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first entry of a team at the

Washington Olympiad in 1981, in which they placed seventh. The team also won one silver and four bronze medals, which was a most commendable show

ing. Medals are awarded as fol lows : the top twelfth of the contestants receive gold medals; the next sixth, silver; and the next quarter, bronze.

The team members were Frank D'Ippolito (Sudbury, Ont., bronze), Moses Klein (To ronto, Ont., bronze), Martin Piotte (Montreal, Que., bronze), Guisseppe Russo (Sud bury, Ont.), Eric Veach

(Sarnia, Ont., bronze), and Minh Tue Vo (Montreal, Que., silver).

The leaders were R. Scoins, University of Waterloo, and Tom Griffiths, University of Western Ontario.

The order of placement of the first fifteen teams was as follows: Romania, U.S.A., Hungary, Bulgaria, Vietnam, Soviet Union, East Germany,

West Germany, France, Great

Britain, Australia, Canada and Czechoslovakia tied, Poland, Brazil, and Israel.

Students from Romania and

Hungary achieved perfect scores on the extremely diffi cult examination. Readers in terested in the I.M.O. problems are referred to the November issue of the Mathematics Maga zine (MAA 1985, 308).

Tom Griffiths

University of Western Ontario

London, ON N6A 2X4

Note on parts I write in reference to the article " Generalized Integration by

Parts" (Budmon Davis, March

1985, 217-18). I was a graduate student at

Wayne State University in the

early 1960s and had the good for tune to meet Karl W. Folley, who was chairman of the mathematics

department during my first year of studies. Folley had published a beautiful expository note, "Inte

gration by Parts," in the American Mathematical Monthly (vol. 54

[1947] : 542-43). Most graduate stu dents at Wayne became acquainted with that note because it helped teaching integration by parts and because credits edified their de

partment's leader. Students and faculty alike used

this method in their textbooks in calculus. See, for example, section 4.2 in Calculus by Munem and Foulis (Worth 1978) and exercise

2C, page 297 in Calculus with Ana

lytic Geometry by Niles and Habo rak (Prentice-Hall 1971) or exercise

2C, page 339 (2d ed. 1982). Folley's original paper is reprinted in Select ed Papers on Calculus (MAA 1968, 283-85).

As a teacher of calculus and friend of Folley, I thank Davis for

bringing this method to the atten tion of the teaching community once again. Good ideas last!

George E. Haborak

College of Charleston

Charleston, SC 29424

Budmon Davis responds : Regard ing Haborak's letter, I was un aware of the text by Munem and Foulis. I did look through about ten calculus books before writing the article. I did not know Folley nor about the article in the MAA's Selected Papers on Calculus. Upon looking it up, I see the great simi

larity. However, I did not claim that this technique was original but rather that it should be more well known.

Referring to Budmon Davis's in

teresting article, I would like to add this observation.

In mathematics, if one suspects that a given expression can be stated in a certain manner, one can

always write the expected form. If one can then compute all undeter mined parts via permissible math ematical operations, this not only establishes the correctness of one's

original expectation but, fur thermore, attains the sought result.

This principle is, for instance, used when splitting up a fraction into partial fractions for integra tion. It can also be used for some of the integrals that are usually done

by repeated integration by parts. We stipulate

\f{x)eKx dx = g(x)eKx + C,

J f(x) sinKxdx =

gi(x) sinKx + g2(x) cosKx + C,

?f(x)cosKxdx = g^x) sinKx + g2(x) cosKx + C,

where the polynomials to be ob tained are of the same degree as the corresponding polynomials in the integrand. We then proceed to determine the coefficients of the

polynomials in question by first dif

ferentiating both sides and then

comparing coefficients of like terms.

Examples (taken from the arti

cle):

(1) ?x3e~2xdx = (ax3 + bx2 + cx + d)e~2x + C

By differentiation of both sides, 3 = -2ax3 - 2bx2 - 2cx

-2d + 3ax2 + 2bx + c.

By comparing coefficients of like

terms,

-2a

3a-26

26-2c

c-2d

= 1

= 0

= 0

= 0,

6 =

> or

d= - -

Thus : dx

V 2 4 4 8/ (2) $(x

2 ? 2x) cos3x dx

= (ax2 + bx + c) sin3x + (dx2 + ex + f) cos3x + C.

Hence,

(x2 -

2x) cos3jc = [(3ojc2 + 3?JC + 3c) + (2dx + e)] cos3x + [-(3dx2 + 3ex + 3/) + (2ax + 6)] sin3x.

Thus,

and

3dx2 + Sex + Sf = 2ax + 6 |

3ajc2 + (36 + 2d)x + (3c + e)

= jc2 - 2jc.

From this, 3d = 0, 2a = 3e, 6 = 3/, 3a = 1, 36 + 2d = -2, and 3c 4- e = 0. We immediately find a = 1/3, 6

= - 2/3, c

= - 2/27,

d = 0, e = 2/9, and /" = -2/9; and the integrated result (3) can be written as follows :

(3) sinrajccosnxdjc

This, too, can be done by the

"conjecture" method, albeit by no

means as elegantly as by Davis's theorem. Incidentally, I do get a

slightly different result (the dis

crepancy lies in the sign of the de

nominator). Considering the indefinite inte

gral, I would conjecture the follow

ing:

j sinmjccos/ix = asinmjcsinnx + bcosmxco&nx + C

Then,

February 1986 -?-83

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