Correction October 1991Author(s): Marshall GordonSource: The Mathematics Teacher, Vol. 86, No. 2 (FEBRUARY 1993), p. 184Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27968217 .

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1. AABD s AGFE by SAS, and so/3 = m LEEG.

2. In GAE, AG = V? = GE and AE = V??.

3. Draw GH LM, and so

m ?GE# = 45?. Finally, we also obtain 0 + = 90? - 45? = 45? = a.

Zhang Zaiming Yuxi Teachers* College

Yunnan

653100, China

GH = y?GE2-EH2

Correction October 1991 In my October 1991 article "Counterintuitive Instances Encourage Mathematical Think

ing," I incorrectly stated that "when two original lines have

slopes with different signs, the summed line does not fall between them." I apologize to the readers; it's not clear what I had in mind?evidently not much.

Marshall Gordon The Park School Brooklandville, MD 21022

Correction Feb. 1992 I graduated with a doctorate in education from Peabody of Van derbilt University last May. Now I am a professor of mathematics and mathematics education at Naresaun University in Thai land. In my Theory and Practical

Mathematics course, I assigned my students an article from your journal. One of my students, Som Boonda, read "A Circle Is a ... Rose," (February 1992, 114-15).

In his report, he noted a mis take in this article on page 115. On this page,

a2 V3 24 8

should change to

24 + 16

His calculations are as follows:

A=i a2cos20d0 2 Jo

=>? [6cos20d0 1 Jo

_a_2 p(l + cos2fl)rifl

"2J0 2

=^-2 P(l + cos20)d0 4 Jo

_a2lQ , sin2?\?

_a2/ , sin(??/3)\ " 4 l6 2 /

_ a2 a2V3 24 16

Let us know if he is correct. I look forward to hearing from you.

Bunpot Suwannaprasert Nareasuan University Phitsanuloke 65000, Thailand

Urban responds: First, thank you very much for using my article in your class. I appreciate the keen interest of your student, Som Boonda. I am glad he verified the

integration for a particular case, which I derived in general oh page 114. No matter how careful one is when preparing an article, there always looms the possibili ty of a printing error, which I made in transferring my rough draft to word processing. Som Boonda was able to catch one of these. I hope the other printing errors, which were not mine, where replaces the correct and where the is missing on

page 114, did not confuse him. Just as the meaning of a sen tence can be understood when the grammar and spelling is not

perfect, the relationships of this article still hold, despite the printing errors.

I applaud your dedication to mathematics in your deter mination to make it letter-and number perfect.

I also invite you to write me about your mathematics curricu lum in Thailand. I am very inter ested in education around the world.

I wish you good luck in your future endeavors.

A corrected portion of page 114 follows:

Also, when , is odd, the total area enclosed by the curve is found as 0 varies in the interval [0, ], since the same curve is again outlined from ( , 2 ). Using equation (1), I found the area from 0 to 2 . Using the for mula for area yields

-if "2 ja

-if."

r2de,

cos ( ))2a

cos2 ( 0) d0

[^ + ?sin(2^))

-{?a+TnS?R{2n4 Using a = 0 and ? = 2 , we obtain the following:

A=^i?^ + ̂sin(27i27t)) 2 L\2 4 /

- (1.0

+ ?sin(27i.O))

which simplifies to

A - a

A corrected sentence for page 115 follows:

For example, the area 0in [0, /6] is

a2 , q2V3 14+ 6~'

30 April 1992 I would like to suggest the follow

ing alternative method for solv

ing the "April Calendar" (1992) problem for 30 April. The prob lem reads as follows:

Find the number of digits in 412 ? 520 when written in usual base-ten form.

Solution

4i2.52O _

^Qlogy2.QQlog5)20 = ^Ql2-log4.^Q2(Mog5 _

2Ql2-log4 + 20-log5

^q21.2052

.*. 1021< 412.520 <1022

So there are twenty-two digits. *: By the way, since 412? 520 can be

found on any scientific calendar in scientific notation, the answer can also be found with just a few keystrokes.

Paul R. Huff West Bloomfield High

School West Bloomfield, MI 48033

Alternate solution 5 October 1992 An alternate solution employing common logarithms is more expe dient than the exponential method used by the contributor.

The problem: List from least to greatest:

2I2I g55 88

Using the law, log nm = m log n, we set up the quantities in loga rithmic form as follows:

2121 = 121 log 2 = 121 (0.30103) = 36.424 629

955 = 55 log 9= 55(0.954 242 5) = 52.483 338

788 = 88 log 7 = 88(0.845 098) = 74.368 628

We immediately note that the logarithm of 788 has the greatest characteristic of the three loga rithms, representing a number with seventy-three digits to the left of the decimal place. Thus, the quantities are arranged prop erly in the problem. We are not asked to determine these quanti ties but merely to rank them, and this is well because a school cal culator is inadequate owing to

rounding. A computer is required for an exact computation.

Emory S. Commander P.O. Box 153 Hartford, AL 36344

In MCTM journals Readers of the Mathematics Teacher might enjoy the follow ing articles in this month's Arith

metic Teacher, which focuses on connections:

"Making Connections: A Case for Proportionality," Kathleen Cramer and Thomas Post

"Angle Sense: A Valuable Con nector," Rheta N. Rubenstein, Glenda Lappan, Elizabeth Phillips, and William Fitzger ald

"Making Connections with Estimation," Joanne E. Lobato

184 THE MATHEMATICS TEACHER

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