4 October 1990Author(s): Albert WhiteSource: The Mathematics Teacher, Vol. 84, No. 4 (APRIL 1991), p. 336Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27967170 .

Accessed: 09/06/2014 13:51

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 support@jstor.org.

.

National Council of Teachers of Mathematics is collaborating with JSTOR to digitize, preserve and extendaccess to The Mathematics Teacher.

http://www.jstor.org

This content downloaded from 81.83.40.92 on Mon, 9 Jun 2014 13:51:11 PMAll use subject to JSTOR Terms and Conditions

READER REFLECTIONS Reactions to articles and points ot view on

teaching mathematics

(Continued from page 317)

many times larger?" also not

asked, should be "one time larger." The expressions times as large and times larger are not equivalent.

Carl E. Heilman 592 West Main Street Mount Joy, PA 17552

4 October 1990 Regarding the calendar problem for 4 October 1990,

An equilateral triangle 1 cm on a side is inscribed in a cir cle. What is the diameter of the circle?

here is a slightly different solu tion: In equilateral triangle ABC

(see fig. 1) draw altitude AD. Tri

angle A DB is a 30?-60?-90? trian

gle. Therefore BD = 0.5. Let de note the center of the circle. Since

points A, By and C divide the circle into three equal arcs, m ?. BTC =

120. Thus triangle BDT is a 30? 60?-90? triangle. Using the fact that triangle ADB and triangle

BDT are similar, we have that the radius TB equals V3/3, that is, the diameter is 2V3/3.

A

Fig. 1

Albert White St. Bonaventure University St. Bonaventure, NY 14778

15 December 1990 Problem 15 on the December 1990

calendar,

Find the ratio of the area of

the smaller square to the area of the larger square,

asks for "the ratio of the area of the smaller square to the area of the larger square" (my emphasis).

The given solution comes from the correct approach, but it doesn't

go far enough. The stated answer of 1:V2 is the ratio of the two sides of the squares, but not of the areas. The ratio of the two areas is 1:2.

This point is small, and I don't wish to quibble, but it makes a difference if one is trying visually to estimate the relative areas. The error caught my eye, since I am

working on some other interesting (to me) ratios of areas of squares in circles.

James E. Kessler Vermont Technical College Randolph Center, VT 05061

The Mastery of Mathematics...

Requires a Mastery of the Basic Tools...

Give Students The Tool a Promotes Math Proficiency

336

Give Students The Tool That Promotes Math Proficiency

MacNumerics...A dynamic, interactive math * ematics program created to promote visual and systematic understanding of the basic concepts re uired for college or university studies. tarting with basic Cartesian coordinates and points,

advancing through differential and integral calculus MacNumerics is unequaled as a teaching/learning tool.

FEATURES: complex numbers conic sections functions graph paper help and show steps linear systems lines

matrices parametric equations

points polar equations polynomials graphic printing scientific calculator unit circle x-y regression comprehensive manual standard user interface easy to use

-SITE AND NETWORKING LICENSES AVAILABLE

Requires: Macintosh Plus, Macintosh SE, Macintosh Classic, All Macintosh II Computers, One meg of Ram memory.

Expanding Educational Horizons

SPRING BRANCH SOFTWARE, INC. P.O. BOX 342 MANCHESTER, IOWA 52057 PHONE: 319-927-6537

Mathematics Teacher

This content downloaded from 81.83.40.92 on Mon, 9 Jun 2014 13:51:11 PMAll use subject to JSTOR Terms and Conditions