Problem 21, December 1995Author(s): Antonio QuinonesSource: The Mathematics Teacher, Vol. 89, No. 6 (SEPTEMBER 1996), pp. 500-501Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27969855 .

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"Consistent rules") suggests that the two rules be taught more con

sistently as

and

&) =fg+fg'

j'g-fg' g2

I suggest a new form for these two rules, as follows:

and

7Yr

I think the latter perhaps is bet ter than the former.

Zhang Zaiming Yuxi Teachers' College Hongtashan Yuxi Yunnan

653100 China

Wang responds: Zhang made an interesting suggestion. Accord ing to his suggestion, the deriva tive of the product of three func tions can also be written easily as

(fghj =(fgh? f g h

After we find the derivative, we

usually need to simplify it, which points out a possible inconve nience of this form. Simplifying this form will return to

&) =fg+fg'

for the product rule and

j'g-fg' g2

for the quotient rule. This form could be assigned for homework. For example, have students prove

and generalize it for the product of three and four functions.

?I =[Tg

Another proof of the converse A proof of the converse of the Pythagorean theorem was given by Richard Hinthorn, Roberto Gonzales, and Guillermo Mar tinez in the January 1996 Mathe matics Teacher (p. 27). I too have worked on finding a proof of this theorem that is not typically given in geometry textbooks. The following proof combines theorems from geometry and trigonometry. Given: AABC, with sides a, 6,

and c, such that a2 + b2 = c2.

Prove: AABC is a right triangle.

According to the law of sines,

sin A sin ? sinC

is true for any triangle. Therefore,

csinA

sinC '

csinB

sinC '

?sinC

sin 2 ? 2 a 2 * 2 c sin A

[ c sin

sin2C sin2C

and

a = -

b =

c =

_bhm2C

sin2

sin2C

Since

(sin2 A + sin2

j

sin2 sin2C1

sin2C sin2B'

sin2 A + sin2B = sin2 C. Since A, B, and C are the angles of a trian gle, C = 180 - (A + B). Therefore,

sin2 A+sin2 = sin2(l80-(A

+ B))

fsinl80cos(A + B)

-cosl80sin(A + B)^

= sin2(A + ?)

= (sin

A cos + cos A sin j2

= sin2 A cos2 5

+2 sin A cos ? cos A sin

+ COS2 Asin2 ?;

sin2 A-sin2 A cos2 ?

+ sin2 - cos2 Asin2 ? = 2 sin A cos cos A sin ?,

sin2A|l-cos2?J +

sin2?(l-cos2Aj = 2 sin A cos ? cos A sin ?,

sin2A|sin2?J

+sin2?(sin2Aj = 2 sin A cos ? cos A sin ?,

2sin2?sin2A = 2 sin A cos ? cos A sin ?,

sin ? sin A = cos ? cos A, tanA = cot?.

Therefore, angles A and ? are

complementary and angle C is a

right angle. Robin S. Kalder Bedford Central Schools Bedford, NY 10506

Problem 22, November 1995 Here is another solution of the Calendar problem from Novem ber 1995.

Find k if

Then

fai4*"1

f x8Jfc-2 a a

6

and

8*-2 = 1,

kJ-. 8

Zhang Zaiming Yuxi Teachers* College Hongtashan Yuxi Yunnan

653100 China

Problem 25, November 1995 This problem was presented as follows:

A figure is created by begin ning with a 2 2 square and continually adding to the fig ure as drawn. What is the area of the figure?

The solution given, 16/3, was ob tained by using the standard for mula for a geometric series. See figure 1, which presents an alter native solution, without words.

2

3(22 + l2 + (1/2)2 + ?.?) = 16

Charlie Marion charliemath@delphi.com Lakeland High School Shrub Oak, NY 10588

Problem 21, December 1995 Another method follows for solv ing the problem.

Find the digit represented by each different letter.

ABC ABC

+ ABC BBB

It uses simple algebra and criti cal thinking.

Since ABC ABC

+ ABC BBB,

500 THE MATHEMATICS TEACHER

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then S(ABC) = BBB = 111(B).

SinceBe JandABC>100 (it is a three-digit number), then 3 >

> 9 and the possible values of ABC are as follows:

37(3) = 111 37(4) = 148 37(5) = 185 37(6) = 222 37(7) = 259 37(8) = 296 37(9) = 333

Adding the unit, or C, column yields that three times C must equal plus some multiple of 10, or 3C = ;(10) + B, where is any possible integer.

The only possible value of ABC that complies is 148, since 3C =

x(10) + B, or 3(8) = 2(10) + 4.

3(ABC) = BBB = 111(B); 3(148) = 444 = 111(4).

Antonio Qui?ones Academia Disc?pulos de

Cristo Box 1947 Bayamon, PR 00960-1947

Problem 7,

January 1996 While working on the following Calendar problem, a student in my class noticed a variation that we found exciting to pursue.

Given: 69x64 = 4416 96 46 = 4416

Why? Can you find other two digit numbers that share this phenomenon?

The equations 12 31 = 372 and 21 13 = 273 give interesting results. The first and last digits of the product of two two-digit numbers are reversed when the digits of the factors are reversed. We called the factors "Buffy num bers" for three-digit products, after their finder, Buffy Last. The pair of products we called "Buffy pairs."

Definition 1. For all a, b, c, d, , y, e {1, 2,3,4}, the two

digit numbers (10a + b) and (10c + d) are Buffy numbers iff for the three-digit product (10a + 6)(10c + d) = lOOx +

IO3/ + z,ac = x and bd = z.

Theorem 1. If (10a + ) and (10c + d) are Buffy numbers, then (106 + a) arca7 (l?d + c) are

?i//5fy numbers.

proof. (10a + ?)(10c + d) = lOOx +

lOy + and ac-x and bd = z.

100ac + 10?c+10ad+?d = 100x+10;y + 2:

lOOx + lOfcc+lOad + 2 = l00x + l0y + z

106c+10ad = 10;y lOOz + IO?c+lOad + x

= 100z + 103/+x 100?d + 10?c + 10ad + ac

= 1002 + 103/+ ;

(lO? + a)(lOd

+ c) = 100z + lOy +

definition 2. TAe product of Buffy numbers and the corre

sponding product of Buffy numbers by theorem 1 are called Buffy pairs.

Some other Buffy numbers and pairs are the following: 12 23 = 276 and 21 32 = 672 21 23 = 483 and 12 32 = 384 13 23 = 299 and 31 32 = 992 21 31 = 651 and 12 13 = 156 14 21 = 294 and 41 12 = 492

An example such as 17 10 = 170 and 71 01 = 071 is not allowed because all digits are

positive integers by definition 1.

Royce E. Huffman Milwaukee Lutheran

High School Milwaukee, WI53225

Problem 8, January 1996 In the January 1996 issue, the Calendar problem for 8 January was as follows:

Solve for x:

, xx2-9x+20 lac -5*+

5] =1

This solution was given: Either of the following restrictions must be present. The exponent must be zero.

2-9*+ 20 = 0

(x-5X*-4) = 0

Thus - 4 or = 5. The base must be lor-1.

2 - 5x + 4 = 0 or 2 - 5x + 6 = 0 (x-lXx-4) = 0 or (*-2)(jc-3) = 0

= 1 or = 4 or = 2 or = 3

The solution set is {1, 2, 3,4, 5}.

This answer to the problem is correct. However, some state

ments made in the solution deserve a closer look. My first objection is to the implication that when the base is -1, the solution is 1 regardless of the value of the exponent. This situa tion is true only if the exponent is even. If the exponent is odd, the solution is -1; and if the expo nent is fractional, the solution does not exist. A clear counter example is found in the solution to this problem:

Solve for x:

|x2-5jt + 5j*

+6*+8 = 1

If I used the foregoing method to solve this problem, I would be led to believe that = 3 is a solu tion because it makes the base -1. Plugging in = 3 gives -1 because the exponent is 35, an odd number. It is interesting that = 2 is still a solution because

the exponent is 24, an even num ber. In the original problem, when the base is -1, the exponent is even; therefore, the solution was not wrong, just incomplete. Since it probably took some effort to come up with this problem, it is a safe assumption that the author was aware of all the details and expected the reader to under stand that this case was special. I point out this incompleteness in the solution mainly as a contrast to my second objection. My second objection to the

solution was with the implication that when the exponent is zero, the solution is 1 regardless of the base. According to my high school algebra teacher, who also hap pens to be my father, zero raised to the power of zero is undefined. It seemed that the solution was

again incomplete. As before, this special case did not arise in the original problem, but it is a subtle point worth exploring.

Suppose, for example, that the problem had been as follows:

Solve for x:

[x2-9j =1

We would be interested in = 3 as a possible solution to the prob lem because it makes the expo nent zero. It also makes the base zero and, under the confines of algebra, we would be stuck. With some calculus we can test this solution.

Starting with the limit as goes to 3 gives

lim z->3| M"3

:1.

Taking the natural log of both sides of the equation yields

In lim [ 2-

* 3 =ln(l)

= 0.

Asserting that the limit can be taken out of the natural log oper ator gives

lim x->3|

In

M"3 = 0.

Performing algebra leads to

lim (x-3)ln(jc2-9j

ln(*2-9) lim *->3|

:0,

= 0.

L x-3

Using FHopital's rule gives

dx v ;

where f[a) and g(a) are both infi nite or both zero.

lim *-?3|

ln(x2-9) 1

x-3 J

= lim x-?3|

2x

x2-9

.(->)' Performing more algebra and plugging in for the limit result in

lim x->3|

2x ( - 2

x2-9

lim x->3

lim z->3

-2x(x-3) x + 3

-2x^+6x

x + 3

= 0,

= 0,

Vol. 89, No. 6 ? September 1996 501

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