High School Mathematics-Physics SMILE Meeting
Mathematics: Geometry

13 October 1998: Porter Johnson [IIT]
He talked about a Geometry problem where twins drove 3 nails at random into a table and formed a triangle. If the nails happened to lie along a straight line, there would be no triangle [or else a very skinny triangle with no interior!]. The probabilities of a new nail lying on the right or left of the line are each 50%, and for a triangle there are three lines, and you have to lie "left" "left" and "left" as you circulate around the boundary [counterclockwise]. By this simple argument, you might expect the probability of being inside as being 1/8 = 0.125. However, in the historic words of the 20th Century Physicist Wolfgang Pauli, "nicht Einfach; aber Falsch"---it isn't simple but it is wrong!

It is convenient to use vectors to decide whether a point is inside a triangle. Let us choose one vertex as special, call the vectors from it to the other vertices r1 and r2, and call the vector from it to the point in question r. Then the criterion for being inside the triangle is

r = a r1 + b r2 .

where a and b are both positive, with a + b < 1.

To gain insight, he employed a pseudo-random number generator and applied the ***Monte-Carlo technique. In a run of 10 million shots there was a computer crash because one of the sets of points was accidentally linear. He modified the program and then safely ran to 100,000,000 shots, without incident [it took several hours on the 80 MHz 486PC], obtaining 7,637,924 hits. The corresponding hit probability is thus .076379 ± .000100, which is consistent with the number 11/144 = .076388888888... , and lots of other more complicated numbers, as well. However, no clever soul has yet appeared with the solution. A copy of the FORTRAN program was passed out, along with a pseudo-random number generator touted by Liam Coffey [Physics Faculty Member and Computational Physics guru], just in case you also want to waste vast amounts of time on this problem or put your lazy computers to work  For details see the website http://mypages.iit.edu/~johnsonpo/aapt0611.html.

Roy Coleman commented that some pseudo-random number generators are not very successful in producing "random" numbers, as you can see from plotting them in pairs. The "safest" random number sequences involve tabulations of radioactive decay times, but they are difficult to use. The two pseudo-random number generators used here gave similar results.

***Much of twentieth century science involves Monte-Carlo simulations of actual experiments, theoretical models, hypothetical problems, and even "useless and insignificant puzzles". The inventor of the idea was the mathematician Stanislaw Ulam, who was involved in the development of the atomic and hydrogen bombs. Ulam became ill after the war, and spent a lot of time in hospital playing the card game Solitaire. He realized that it is more straightforward to play a few games to "estimate" the probability of winning the game, rather than to try to calculate the odds of winning directly. Ulam realized that many problems too difficult to be solved analytically could be resolved by this technique using fast computers. This story, along with many others, appears in his autobiography, Adventures of a Mathematician [ISBN 0-520-07154-9].

10 February 1998 James Chichester [Lincoln Way HS]
He brought in a product called Odd Balls, with ordering information

```             Orbix Corporation
6329 Mori Street
McLean, VA  22101
(703) 356-0695
```
Here is a picture of them:  Source: Shop; National Gallery of Art:  [http://shop.nga.gov/]

The product is shipped as 3 balls of three different sizes, which become parts of spheres snapped together in the center by what looks like a doughnut. Each section is a different size. Unlike an orange, each slice is different.
• -used to construct figures to show different colors
• -spun to show additive colors
• -used to show center of gravity of different figures
• -used as an inverted pendulum (like a rocking chair) period varies with center of mass
• -parts of whole -- fractions

Designer: Dr Ben F Sherman [a retired Nuclear Engineer]

Porter's comments: What happens when 2 cylinders intersect? They produce an intersection region that is rounded, but essentially becomes a square when looked at in a certain plane. The volume of that region may be computed as a single integral, and p does not appear in it.

V = ò -a a dx (a2-x2) = 4 a3 /3

10 October 2000 Fred Schaal (Lane Tech HS)
showed us "The Occurrence of Concurrence".  He explained that if 3 straight lines in a plane intersect in a single, common point, it is called "concurrence." With the aid of a meter stick, he constructed a large, nice looking triangle on the white board. With a large compass having a marker pen attached at its "chalk" end, Fred used the compass to construct the line which bisected one of the angles of the triangle. He did this in a contrasting color. Then he constructed the bisectors of the other two angles the triangle. If the board had not been so slippery and the compass marker had made legible marks, the bisectors of the three angles would have intersected at a single point within the triangle: concurrence! Unfortunately, the construction was not precise, and it didn't work out. But you made your mark, Fred! Thanks for an interesting lesson.

27 February 2001 Fred Schaal (Lane Tech Park HS, Math)
made a presentation on finding the lateral surface area of pyramids with a regular polygonal base. He illustrated the point with a pyramid with a square base. The lateral surface consists of four equivalent triangles, of base b and height s.  The area of each triangle is ½ b s, where the height s is the slant height of the pyramid, and the lateral surface area of the pyramid is

Lateral Area = 4 ( ½ b s) =  ½ (4 b s).
This can be expressed in terms of the base perimeter p = 4 b and the slant height s as
Lateral Area = ½ p s.
For an n sided regular polygonal base of base length b and slant height s, the perimeter is p = n b and the lateral surface area is
Lateral Area =   ½ n b s =  ½ p s.
By taking the limit as the number of sides becomes infinite, one obtains the result
Lateral Area =   ½ n b s =  ½ p s.
for a (right circular) cone, where the base perimeter p  is the circumference of a circle of radius r, or p = 2 p r. Thus we obtain the standard expression for the lateral surface area of a cone,
Lateral Area = ½ p s = ½ 2 p r s = p r s .
You can also calculate the lateral surface area by rolling the cone about its apex, and computing the area of the sector so obtained, corresponding to opening angle q = 2 p r / s and radius s. The result is
Lateral Area = ½ s2q = ½ s2 (2 p r / s) = p r s
Very good, Fred. [Fred also reported a gigantic crane being installed on a construction site at the corner of Western Avenue and Lake Street.]

06 November 2001: Fred Schaal (Lane Tech HS, Mathematics)  Congruences of Plane Figures
Fred
passed out a statement of the following principles on Euclidean Isometries:

1. A composition of reflections in two parallel lines is a translation.
2. A composition of reflections in two intersecting lines is a rotation.
3. In a plane, two congruent figures can be mapped onto one another by a composition of at most three reflections.
4. There are only four isometries.  They are the following:
reflection ... translation ... rotation ... glide reflection

Fred sketched figures on the board, which provoked much discussion as to how to establish the third result.  Porter Johnson suggested reflecting first about the perpendicular bisector to two identical vertices of congruence, so that the "old figure" and "new figure" will have that vertex (X=X') in common. Then, reflect about the the bisector to the angle YXY', where vertices Y and Y' are identical. The result will either be a congruence if the figures were not initially space-reflected, or else can easily be brought into congruence if the figures were initially space reflected.

Relevant References:

Fred also called our attention to the fact that the planets Mercury and Venus are visible in the sky just before dawn.

23 April 2002: Walter McDonald (Bowen HS and Chicago Veterans Administration) -- Higher Dimensional Geometry
Walter
described his efforts at tutoring students on visualizations of spaces of various dimensions.  He presented the following table of characteristic figures [called simplexes by mathematicians] in various spatial dimensions:

 Number of Dimensions Characteristic Figure 0 Point 1 Line 2 Triangle 3 Tetrahedron 4 Figure with Tetrahedron Faces

Walter asked what is the difference in 4 dimensional (Euclidean) space and what physicists call "space-time"?  Porter Johnson commented that the time interval Dt between two events and the spatial interval DL between the same two events can be regarded in terms of a unified "space-time", and that because of the constancy of the speed of light, v, it is required to define the invariant-interval-squared  between two events as [DL]2 -[v Dt]2 . This space is called Minkowski space in Special Relativity, and which is a four dimensional (Euclidean) space, when expressed in terms of real space variables, with the time variable multiplied by i = Ö(-1). Incidentally, the development of non-Euclidean geometry in mathematics and its applications in physics are an outgrowth of the examination of whether Euclid's fifth postulate [parallel lines never meet] is a consequence of the other four.  Non-Euclidean geometry is the mathematical framework for General Relativity, our (classical) theory of the Gravitational Field.  For an interesting discussion of non-Euclidean geometry, see the St Andrews University web page: http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Non-Euclidean_geometry.html.:

Very good, Walter!

23 April 2002: Leticia Rodriguez (Ruben Salazar Bilingual Center) -- Tesselations; Mathematical Applications; Scientific Method
Leticia
passed around the following book, which contains various tesselations [which are regular periodic patterns, or periodic and quasi-periodic "tilings" of space]:  Tesselations Teaching Masters; Dale Seymour Publications, 1989; ISBN 0-88661-462-7.  Leticia's primary students color these tesselations to make elaborate designs, and use them as a means to learn elementary ideas in mathematics [shapes, patterns, graphs, counting, geometry, etc] and elements of the scientific method [observing, estimating, collecting data, predicting, classifying, investigating, comparing, contrasting, problem solving, inferring, and drawing conclusions].  For further details see her website on the SMART home page: http://mypages.iit.edu/~smart/Porter Johnson mentioned that intricate, symmetric patterns are employed in many religions to convey a sense of spirituality in their cathedrals, chapels, churches, mosques, pagodas, shrines, and temples.  One beautiful example of these patterns is the Baha'i Temple in Wilmette Illinois; see the websites:   http://members.core.com/~fphayes/bahai.htm.

Leticia also pointed out that teachers are entitled to a 15% discount on educational and school supplies for classroom use (with proper identification) from April 15 to May 31, 2002 at Amazing Savings stores, located in Morton Grove (Harlem & Dempster) , Wheeling (Elmhurst & Dundee), Chicago (McCormick & Lincoln), Broadview (17th and Cermak), and Bloomingdale (Springbrook Shopping Center).  Thanks, Leticia!

24 September 2002: Hoi Hyunh ( HS, Physics) Math Notation and Visual Geometry
Hoi
began by suggesting the mnemonic that D is really a D, which stands for Difference.  [Do you remember "trouble that begins with T, which rhymes with P, and stands for Pool".] Hoi then showed us an ordinary-looking, box-like configuration that she had constructed as follows:

• She first made 4 thin square sheets out of cardboard [20 cm by 20 cm], which had different colors [red and white] on the two sides of each sheet.
• Then she cut each of the square sheets along the diagonals, obtaining 4 right triangles from each original square. She then attached these pieces together with wide flexible tape.
• Finally, she attached four squares together to form the sides of a box [red on the outside and white on the inside], with the top and bottom open.
Amazingly enough, the device was quite flexible, and it could be twisted around to make a wide variety of shapes and figures.

Very nice work, Hoi!

08 October 2002: Maria Vinci [Evergreen Park HS, Mathematics]     Tiling and Tessellation
Maria
passed around the book The Graphical Work by the Dutch graphical artist M C Escher (1898-1972) [Taschen GmBH 1989; ISBN 3-8288-5864-1], which contained various patterns, tilings, and tessellations. [For more details on the life of Maurits Cornelis Escher and his works see the website M C Escher by Cordon Art BV [http://www.mcescher.com/].  Maria showed various tessellated figures that students made in her classes, using images of an elephant or a human face in making periodic tilings.  Although Escher was primarily a graphical artist, he understood mathematics rather well, and his work has had a profound influence on mathematicians; for details see the website Mathematical Art of M C Escherhttp://www.mathacademy.com/pr/minitext/escher/index.asp PJ comment: The preparation of periodic micro-crystalline samples of protein structures, such as DNA, is a crucial component in X-ray scattering to determine the atomic structure of these materials.  For example, the double helical structure was deduced by Watson and Crick upon the basis of analysis of X-ray scattering of micro-crystals of DNA.  Thus, tessellations are also important throughout modern science.  We get the picture, Maria!

08 October 2002: Walter McDonald [VA Hospital; Bowen HS]    Fractals:  How Long is the Coastline of Florida?
Walter
explained that the length of certain intricate curves is indeterminate, because the lengths depend upon the scale of resolution.  For example, a tourist brochure may advertise that the coast of the State of Florida is 6000 km [4000 miles] in length, but even this estimate is imprecise, since it would be impossible to follow all the nooks and crannies that separate the land from the sea.  As the scale of resolution of the measurement decreases, the length increases.  He showed some "self similar curves", for which the structure has the same form when viewed at various scales --- including one on which we measured the following lengths with various resolutions:

 L: Length R:    Resolution #1 3 2 #2 7 1 #3 20 0.5
He calculated the fractal dimension D from the conditions (R2/R1)D = (L1/L2and (R3/R2)D = (L2/L3).  For this case, D is around 1.3 - 1.4. Walter gave us the following web-based references on fractals:
You helped us to see fractals almost everywhere we look, Walter! Good job!

23 November 2004: Porter Johnson called attention to information on Fractals on the Yale University website [http://classes.yale.edu/fractals/].  Fractals occur frequently in everyday situations, such as Crumpled Paper [http://classes.yale.edu/fractals/FracAndDim/BoxDim/PowerLaw/CrumpledPaper.html] and Bean Bags [ http://classes.yale.edu/fractals/Labs/CrumpledPaperLab/BBProcedure.html].   Does this actually work??

05 November 2002: Fred Schaal [Lane Tech High School, Mathematics]      TI-92 graphics calculator; Railroads
Fred
described an experiment that he had done, in which he could lay out a triangle on his calculator, and then have it determine all interior and exterior angles, as shown in the diagram below:

```
```
Note that, from the diagram, a + A = 180°; b + B = 180°; c + C = 180°.  In addition, it is true that A + B + C  = 360°, as can be seen by shrinking the triangle to a point, as shown.  By adding the first three relations, and then subtracting the fourth, we obtain a + b + c = 180°, indicating that the sum of the angles in a triangle is equal to 180°.  In addition, by drawing a line parallel to the base line at the top vertex, we  obtain the figure:
```
```
Evidently, a + b = C, indicating that the exterior angle C is the sum of the two interior angles a and b. Similarly, we can show that  b +c = A and c + a = B.

Fred noticed in his extensive travels by rail this summer that the tracks were noticeably bumpy --- presumably because freight trains also travel on the same tracks.  In addition, he observed a lateral [transverse] vibration of the train cars, and wondered why.  The consensus of the group was that the slight "play" in the  "knuckle couplers" that attach one car to another permits such transverse vibration, which can be amplified as the train travels along the track. Very interesting, Fred!

09 September 2003: Bill Colson [Morgan Park HS, mathematics]        Flashy Doo-Dah's
Bill
showed us Tetra-tops, a set of flashing and colorful polyhedron tops, with trading cards. Bill bought them at Walgreens for about \$5.00.Tetra-tops http://www.tetratops.com/ are manufactured by Duncan Toys Company http://www.theyoyostore.com/duntettop.html, the famous YO-YO manufacturer.  This excerpt is taken from the website above:

"Tops have fascinated humans since they were first discovered. No one knows when or where they were first spun, but they have been found in nearly every culture on Earth. The variety of designs is endless, however there is one remarkable similarity to all of them. Traditional tops all have only ONE axis of spin! Unlike traditional tops, Duncan's new TETRA-TOPS™ all have multiple axes of spin!"
These spinning tops include the five Platonic solids [ http://www.math.utah.edu/~alfeld/math/polyhedra/polyhedra.html], the tetrahedron, cube, octahedron, dodecahedron, and icosahedron.  Nifty, Bill!

27 January 2004: Fred Schaal [Lane Tech HS, mathematics]        Innies and Outies
Fred freehandedly drew a circle on the board, marked five points roughly equidistant on its circumference, and connected them to form a five-sided polygon, a pentagon or 5-gon. He then measured the interior and exterior vertex angles with a large wooden protractor, obtaining the following results:

 Vertex Interior angle Exterior Angle A 104° 76° B 98° 82° C 116° 64° D 115° 65° E 115° 75° TOTAL 548° 352°
These totals are close to the expected results, 540° and 360°, respectively. Fred then relocated the vertex point C to a point close to the center of the circle, C', drew the new 5-gon, and obtained the following results:
 Vertex Interior angle Exterior Angle A 104° 76° B 55° 135° C' 210° 150° D 70° 110° E 115° 75° TOTAL 554° 546°

The expected relation no longer works, because the sum of the angles for vertex C' is now close to  540°, rather than 360°.  If we had taken the exterior angle for C' to be -30°, we would have obtained 346°, which is more consistent with expectations.

Finally, Fred asked whether the rubber wheels on the Montreal and Paris METRO rail systems represent a practical means of noise reduction, or in fact are they too inefficient because of increased friction. Does anybody know the answer?  [It was mentioned that rubber wheels were once used in Chicago years ago.]  For additional information see the website Rubber-tired Metrohttp://en.wikipedia.org/wiki/Rubber-tired_metroInteresting, Fred!

09 March 2004: Fred Schaal [Lane Tech, HS Mathematics]           When is there more crust than "pie"?
Fred carefully drew a full circle on the blackboard, and marked the center of the circle, as well as two points on its circumference. He drew radial lines (r) from the center of the circle to these two points A, B --- enclosing a sector, or a pie-shaped slice of circle. Then he drew a straight line (chord) connecting A and B.  The area between the chord and the arc of the circle represents the crust, or outer region of the slice  Here is a rough diagram.

```

```
The slice angle between the two radii is q.  In terms of area, how much crust is there, and how much pie, for a given radius r and slice angle q?  Fred explained that the slice area  is Aslice =½ r2 q, with the slice angle q measured in radians, since the slice area is proportional to q, and for q = 2p (full circle) we get an area of pr2Fred then measured the radius r of his circle and the slice angle q, and calculated the slice area Aslice.

How do you determine the area Atri of the central  "edible" portion of the pie -- the slice minus the crust? It is the triangular region of base b and altitude h; so that  Atri = ½ b h.  Fred measured the chord length b and triangle altitude h directly on the diagram, and calculated Atri.  Alternatively, we can compute them using h = r sin q/2 and b = 2 r cos q/2, so that Atri =  r2 sin q/2  cos q/2 = ½ r2 sin q. Therefore, the fraction of the slice area that is crust area is given by

Acrust / Aslice = 1 - Atri / Aslice = 1 - sin q / q

This result, depends only on the slice angle q, and is independent of the pie radius.  Why? This table gives its value for various ways of slicing the pie into equal pieces:.

 Number of Slices q: degrees q: radians sin q / q Acrust / Aslice 12 30° 0.523 0.956 0.044 8 45° 0.785 0.901 0.099 6 60° 1.047 0.827 0.173 4 90° 1.571 0.637 0.373 3 120° 2.094 0.414 0.586 2 180° 3.142 0.000 1.000

Thanks for sharing the pie for us --- figuratively speaking!  Good, Fred!

23 March 2004: Fred Schaal [Lane Tech, HS Mathematics]           Of All the Crust!
Fred continued his discussion of slicing pie, which was begun at the last SMILE meeting. He had shown that the ratio, R, of crust area to total area is given by R = 1 - (sin q) / q. (q in radians)

He programmed this formula on his TI-83 graphing calculator, and projected on the screen at the front of the room the graph of R versus q for various ranges.  In particular, he noted that for q greater than 180° or p radians R became greater than 1.  Here is a summary of results:

 q: degrees q: radians R = 1 - (sin q) / q 60° 1.047 0.173 120° 2.094 0.586 180° 3.142 1.000 270° 4.712 1.212 360° 6.284 1.000 450° 7.854 0.873 630° 10.995 1.090 810° 14.137 0.927

It seems that the ratio R is approaching 1 at large slice angle q. Do we get more pie by circling many times?!

You can have your  p-pie and eat it, too!  Good, Fred!

28 October 2004: Benson Uwumarogie [Dunbar HS, mathematics]           Special Parts of a Triangle
Benson used patty papers to show us how to construct the altitudes, median lines, angle bisectors, and perpendicular bisectors of  a triangle by drawing the triangle on the paper, and then folding the paper appropriately.  With these patty papers (presumably named after the paper sheets placed between hamburger patties by butchers), his students constructed these features of triangles for themselves.  Unfortunately, the constructions did not show up well on the overhead projector, because the paper was not transparent.  By showing his work at various states, Benson illustrated these geometrical concepts.   Bill Colson commented that the book Patty Paper Geometry by Michael Serra has been published by Key Curriculum Press.  For details see their website http://www.keypress.com/.

A useful approach, Benson!

07 December 2004: F J Schaal [Lane Tech, mathematics]           Spheres to Cubes
Fred reminded us that a cube of side b0 has 6 square faces, a total surface area S0 = 6 b02, and total volume V0 = b03.  Let us enlarge the cube uniformly until its volume is doubled: V1 = 2V0 = b13. The length of a side is therefore equal to b1=  21/3  b0 ~ 1.27  b0.  Correspondingly, let us double the surface area S2 = 2 S0 = 6 b22 .  We obtain b2=  21/2  b0  ~ 1.41  b0 and V2 = b23 = 2.82 V0.  These results are the same as those obtained by Fred at the last MP SMILE meeting for a solid sphere. Bill Shanks pointed out that, if  an inverted hollow cone filled halfway to the maximum height (with snow or ice cream -- pick your favorite), the volume of edible material is only 1/8 of that when it is filled to the top.  What a rip-off! Thanks for the ideas, Fred and Bill!

07 December 2004: Bill Shanks [Joliet Central, happily retired]          Various Topics
Bill first held in his hand a hexagonal socket used with a 3/8" (8 mm) square drive to fit a 14 mm spark plug. He struck the hexagonal end smartly against the palm of his hand several times.   Each time we heard a short "pop" sound with a certain pitch. Bill then asked us what pitch of sound would occur when  he hit his palm with the other end.  Our survey consisted of votes in all three categories --- lower pitch, same pitch, higher pitch.  Then he did it ---and we heard a "pop" sound of obviously higher pitch. Bill then referred to the wine jug instrument (Helmholtz Resonator) presentation made at the 25 February 2003 MP SMILE meeting by Don Kanner.

At the last MP SMILE meeting Bill measured the length of a little wooden cube (a give-away), and obtained 1.27 cm (corresponding to a half-inch).  He calculated the volume of the cube, obtaining (1.27 cm)3, or a little more than 2 cm3.  We earlier had guessed that the cube was 1 cm on a side, with a volume of 1 cm3How can the volume of the cube more than double when its sides change only by a "small amount"?  To explain this, Bill put x = 1.00  and Dx = 0.27 into the expansion formula for (x + Dx)3:

(x + Dx)3 = x3 + 3 x2 D x + 3 x (D)2 + (D)3
... or ...
(1.00 +0.27)3 = 1.00 + 3(0.27) + 3(0.27)2 + (0.27)3
... or ...
(1.27)3 = 1.00 + 0.81 + 0.2187 + 0.0020 = 2.0484

Neat! Thanks, Bill!

22 February 2005: Fred Schaal [Lane Tech HS, mathematics]              Pump Up the Volume
Fred
brought out a metallic box (appropriately decorated for holiday gifts of candy or sweets), along with a ruler.  Fred recruited an assistant, Charlotte Wood-Harrington, to make measurements of  its outside length L, outside width W, and height H. Charlotte obtained the following values:  L = 14.25 cm, W = 14.15 cm, and H = 4.50 cm.  We then treated the box as a Rectangular Parallelopiped, and computed its volume:

V = L ´ W ´ H = (14.25 cm) ´ (14.15 cm) ´ (4.50 cm) = 907.36875 cm3 » 910 cm3
Of course, a candy Aficionado or Gourmand would be more concerned about the interior volume of the box, which would be significantly less.  We estimated the difference to be 10% to 20%.  One could measure the interior volume of the box by seeing how much water it holds, using a graduated cylinder that holds up to 1000 cm3 (one liter) of water.

Porter Johnson mentioned that glasses in restaurants, cafes, and bars in Europe are commonly marked to indicate the volume of a given level of fluid.  These markings, which were a legal requirement in Germany, might be one of the following:

 0,3 L = 3 dL 3 deciLiters,  or 300 cm3 0,5 L = 5 dL 5 deciLiters,  or 500 cm3 1,0 L 1 Liter,  or 1000 cm3

Watch out for those one liter beer steins!  Thanks, Fred!

12 April 2005: Benson Uwumarogi [Dunbar HS, mathematics]              Circle, Radius p
Benson
used a cloth measuring tape to determine the circumference and diameter of various round objects, and then calculated the ratio C / D, with these results:

 Object C: circumference D: Diameter Ratio metallic cookie tin 79 cm 25 cm 3.16 plastic lid 32 cm 10 cm 3.20
The ratio is seen to be roughly independent of the size of the circle, and in fact, when properly determined it should be the same for all circles; namely p = 3.1415926535 ...

01 November 2005: Benson Uwumarogie (Dunbar HS, mathematics)       Patterns
Benson
presented an interesting geometric puzzle. He drew a series of five circles marking either 2, 3, 4, 5, or 6 points on their circumferences. He then connected the points in all possible ways by straight lines. For example, for the circle with two points, only a single straight line across the interior of the circle can be drawn; for the circle with three points, a triangle can be drawn in the interior of the circle, etc. Each circle was thus divided into a number of non-overlapping regions within its interior. For example, for the "two-point" circle there are two regions; for the three point circle, the are four regions; for the four point circle, there are eight regions; for the five point circle there are 16 regions; for the six point circle there are 32 regions. This pattern suggests the general formula for n points, corresponding to 2n-1 regions.  The general formula works! How come? What a neat way to discover the relationship between physical patterns and mathematical formulas that describe the patterns.  Good work.  Thanks, Benson.

29 November 2005: Nneka Anigbogu  (Jones College Prep)                  Random rectangles
Nneka
handed out a sheet with 100 rectangles of various dimensions displayed.  Each rectangle was composed one or more identical squares.  Our activity was to obtain an estimate of the average number of squares.  This is an activity to illustrate use of statistics. Nneka asked us to briefly look at the 100 rectangles and to shout out estimates of the average area of all 100 in "standard units" (easy because each had an area that was an integral number of the standard squares mentioned above. Estimated averages were 8, 9, 11, 17, 50. Then Nneka had us choose any 5 rectangles in a "continuous run" (e.g., rectangles 5-9) and estimate the average of those 5. We got averages of 1-12.8, for 10 samples of 5 rectangles, with an overall average of 8.82. This completed the "subjective estimate" of the mean.

Nneke also handed out a table of 450 five digit random numbers. We used the table to pick 5 numbers and use any consecutive two digits from that five digit number, match these with the numbers of the corresponding rectangles, and get another group of 5-rectangle averages. Ten averages ranged from 4.0- 8.6, with an overall average of 5.81, the "random estimate" of the mean.

It turns out that the actual average of the 100 rectangles is 7.42. Surprisingly, the subjective estimate was a bit closer to the actual than the random estimate. Nneka noted that in her class the random method gave a number almost identical to the actual average. How come ours was so far off?  Great stuff! Thanks, Nneka.

07 February 2006: Fred Schaal (Lane Tech HS)       Constructing Points on an Ellipse
Fred
showed us how to make points on an ellipse using the blackboard and only a (chalk) compass and a straight edge. An ellipse is defined as a geometric shape with two focal points (foci), so that the sum of the distances from any point on the ellipse to each  is always the same. Fred drew the two foci (F) on the board and then a third point (P), as shown.

```                          P
.
/ \
x  /   \  y
/     \
*       *
F       F
```
Distance between foci (*) =  2 e a
x + y = 2a
Eccentricity = e
Semi-major axis = a
This third point along with the two foci will allow us to unambiguously draw the ellipse which goes through this third point.  Fred used the compass to measure and lay out -- on a straight line -- the total distance (x + y) from the two foci to the point P. Then he used the compass to measure a new-x and a new-y which summed to the same, original distance 2a.  With the compass set at the new-x, Fred drew an arc at distance new-x from F.  Then he set the the compass at the new-y and drew an arc at distance new-y  from the other F. The point at the intersection of the arcs was thus located at distance x from F and at the same time at distance y from the other F -- so it was another point P on the same ellipse!  Repeating this process for different pairs of x + y = 2a located more points, all on the ellipse. And so the entire ellipse could be defined, point-by-point!  For details see the website  http://www.du.edu/~jcalvert/math/ellipse.htm. Neat, Fred!

07 March 2006: Fred Schaal (Lane Tech HS, math)               Parabolic Points
In an extension of his presentation at the last meeting, Fred used a similar procedure to trace out the points on a parabola using only his (chalk) compass, a meter stick and the blackboard. He chose a focal point  (focus) at random above a horizontal line (directrix).  He used the compass to draw a portion of a circular arc with an arbitrary radius, with the center at the focus . Two arcs are then made with the compass held at the same radius, with their centers on the line. A tangent to these two arcs intersects the first arc at two points, which lie on the parabola. The process is repeated using the same focal point but different radii, generating points to trace out a parabola.  For additional information see the interactive webpage The Parabola by Alex Bogomolnyhttp://www.cut-the-knot.org/ctk/Parabola.shtml,

Neat, Fred!

07 March 2006: Porter Johnson (IIT, Physics)                  Sangaku-Followup
Porter continued the discussion of the “Circle Inscribing Sangaku”, which was introduced at the last class by Walter McDonald. This problem is discussed on the Mathworld Website on the web page http://mathworld.wolfram.com/CircleInscribing.html. However, that discussion is incomplete, in that it does not prove that the inscribed circle centered at O3 is tangent to the isosceles triangle ACB.

According to the statement of the problem, the  large circle of diameter 1 (unity)  is centered at point O, and a smaller circle of diameter r is centered at O2.  The smallest circle, which is of radius a and centered at O3, is tangent to the other two circles, and its center lies on the line O3A that is perpendicular to the major diameter XB.  The Mathworld website uses the fact that the right triangles OO3A and O2O3A have a common side, O3A, to determine the  length y of  that side O2A, as well as the radius a of the inscribing circle.  Their results are
(1 + r) a = r (1 - r)   ;

(1 + r)  y = r Ö[2 (1 - r)]    .

Let the symbol j represent the angle ACD.  Because the point C lies on the largest circle, its distance to the center O is 1/2. Furthermore, the right triangle ADC, has these side lengths:

[AD, DC, CA] = Ö(1 - r) / 2 ´ [ Ö(1 - r) , Ö(1 + r) , Ö 2 ]  .
Thus, we can show that
sin j = Ö[(1 - r) / 2]  .
Because the alternate interior angles O3AC and ACD are equal, we can compute the distance from the center O3 of the inscribed circle to the straight line AC:
y sin j = a
Consequently, the inscribed circle, with radius a and centered at O3, is tangent to the straight line AC. The result is thus established.

Porter then told us about Morley’s Theorem. Start with any triangle and trisect all three angles. Pairs of the trisecting lines from adjacent angles will intersect to make three points inside the original triangle. Connection of these three points will always produce an equilateral triangle!! Fred then illustrated this by laying out a carefully drawn figure on the board. For more details see the website  http://www.cut-the-knot.org/Curriculum/Geometry/Morley.shtml, which contains an adjustable triangle showing the result. See also http://www.jimloy.com/geometry/morley.htm, which contains the following comment:

"One of the interesting side results of some of the proofs is that the side of the equilateral triangle is equal to 8R sin(A/3) sin(B/3) sin(C/3), where A, B, and C are the angles of the larger triangle, and R is the radius of the circumcircle."
Fascinating, Porter.

04 April 2006: Karlene Joseph (Lane Tech)              Go Figure
Karlene brought a children’s book called Go Figure by Johnny Ball: .  It is a book about numbers. One section asks the question, “What if we had no numbers?” We couldn’t report sports scores, TV listings, and a million other things! There is also a section on the pyramids (Karlene loves ancient cultures!). The dimensions of the pyramids are arranged so that several interesting relationships occur. Another interesting item is that about 100 years ago the State of Indiana tried to pass a law decreeing that p would be exactly 3.2!  For details see The Indiana Pi Billhttp://www.agecon.purdue.edu/crd/Localgov/Second Level pages/indiana_pi_bill.htm.

Here’s another one about p (we thank Archimedes for this one). A circle inscribed inside a square of side 1 has a diameter of 1 and a circumference of p. The perimeter of the square is 4, which means p must be less than 4. Repeat with a hexagon inscribed inside the circle (hexagon perimeter 3.0); now p must be greater than 3. Archimedes iterated this up to a 96-sided polygon and found that 223/71 < p < 220/70 (or 3.14084507 < p < 3.142857143).  For details see Archimedes Traps Pihttp://physics.weber.edu/carroll/archimedes/pi.htm.

Then Karlene gave us each a triangle cut from a piece of paper (random shapes and sizes). We then talked about the things we know about triangles. One is that the total of the three angles is 180 degrees (the same as a straight line). Karlene then had us tear off the three corners and use the little angles to tuck into one another and see if we got a straight line and we surely all did! One can do the same with a planar quadrilateral to obtain 360 degrees. Neat stuff!  Thanks, Karlene.