Sources mentioned in my presentation at Lexington ISBN Toastmasters on February 23:
The Online Encyclopedia of Integer Sequences which allows you to search for previously studied sequences by simply entering the first several numbers of the sequence: http://oeis.org. This is an indispensable resource for mathematicians and math students. You can find out a lot about Alcuin's sequence here.
The paper I used to prepare my presentation was by Donald J Bindner and Martin Erickson. It appeared in the February
2012 American Mathematical Monthly, which can be obtained at JSTOR:
http://www.jstor.org/pss/10.4169/amer.math.monthly.119.02.115
Cost is $12 unless you have a JSTOR subscription for this journal, and you can read the abstract and first page for free. It is very much worth reading for those wanting an in-depth treatment of Alcuin's Sequence.
The Toastmaster Project
I am using a puzzle problem to illustrate the way in which I teach math to adult students who may not have much experience in the subject, which is the purpose of my company, Math for the Rest of Us.
I will introduce the puzzle and some problems related to it in this blog, and then will talk about its solution during the meeting of Lexington (MA) Toastmasters at 12 noon on February 23. (The source of my presentation and information relating to the solution will be listed on the blog after February 23.)
A rich merchant died and left an inheritance to his three sons. The inheritance consisted of 30 identical silver boxes, of which 10 were full of gold, 10 were exactly half-full of gold, and 10 were empty. He instructed that the inheritance be divided equally, so that each of the three sons received the same amount of silver and the same amount of gold. For the purpose of the division, it is impossible to remove any gold from any box. How can the treasure be divided equally?
This puzzle goes back to the early middle ages. Before following the link, try your hand at solving the puzzle. It can be solved by a little trial and error, and some elementary algebra may be helpful.
E19. Cutting a square cake
I found the following puzzle in Jacob's Geometry: Seeing, Doing, Understanding (3rd ed.). Given a square sheet cake, 9" on a side, divide it into 5 pieces that have the same amount of cake and the same amount of icing.
To be more mathematically precise, the problem is: Given a square, a units on a side. find a dissection of the square into 5 polygonal pieces, each with area a2/5 and each containing the same length of boundary of the original square, namely 4a/5. I think this is not too hard, but I will post my answer if anyone asks.
Extra credit: To generalize and sharpen this result, show that we can replace 5 by n, where n is any integer greater than or equal to 3. Also show that all polygons can be triangles or convex quadrilaterals, and that even so for any fixed n there are an infinite number of essentially different such dissections. (Two dissections are essentially different if one contains a polygon that is not congruent to any polygon in the other dissection.)
To be more mathematically precise, the problem is: Given a square, a units on a side. find a dissection of the square into 5 polygonal pieces, each with area a2/5 and each containing the same length of boundary of the original square, namely 4a/5. I think this is not too hard, but I will post my answer if anyone asks.
Extra credit: To generalize and sharpen this result, show that we can replace 5 by n, where n is any integer greater than or equal to 3. Also show that all polygons can be triangles or convex quadrilaterals, and that even so for any fixed n there are an infinite number of essentially different such dissections. (Two dissections are essentially different if one contains a polygon that is not congruent to any polygon in the other dissection.)
The Birthday Problem
A well-known problem asks for the smallest number of people (N) who must be in a room before it is more likely than not that two share the same birthday. The answer, surprising to most people who have not heard the problem before, is N = 23.
I thought it would be interesting to modify the problem where we ask for people who share that same day of the month for their birthday. While the answer is not as surprising as the original problem, the computation is much easier. Direct computation for the first problem using factorials will result in overflow on scientific calculators such as the TI-83. Also, the answer to the day-of-month problem (N = 7) is more suitable for empirical testing in small classes. Simply ask each student for their birth day (1 - 31) and record on a large month calendar. For N = 11 the probability of a match increases to almost 88%.
The formula for the probability of one or more matches amongst a group of N people is
Prob = 1 - (31)(30)...(32 - N)/31N
= 1 - 31! /[(31 - N)! * 31N]
I thought it would be interesting to modify the problem where we ask for people who share that same day of the month for their birthday. While the answer is not as surprising as the original problem, the computation is much easier. Direct computation for the first problem using factorials will result in overflow on scientific calculators such as the TI-83. Also, the answer to the day-of-month problem (N = 7) is more suitable for empirical testing in small classes. Simply ask each student for their birth day (1 - 31) and record on a large month calendar. For N = 11 the probability of a match increases to almost 88%.
The formula for the probability of one or more matches amongst a group of N people is
Prob = 1 - (31)(30)...(32 - N)/31N
= 1 - 31! /[(31 - N)! * 31N]
Mathematics and Humor
"Time flies like an arrow. Fruit flies like a banana." -- The Flying Karamazov Brothers.
Have you ever told a joke to someone who "doesn't get it"? If you patiently explain the referents you may get them to "understand" the joke, but they will probably respond something like "So, why is that funny?"
In the simple example above, you probably found this funny if (1) you are familiar with the maxim "Time flies like an arrow", (2) your knowledge of the English language allows you to understand that "flies" can be a verb meaning "passes swiftly" or a plural noun referring to a type of insect and that "like" can mean both "as" and "enjoy" (3) your knowledge of writing style leads you to expect that when the same word appears in two successive short sentences, it will usually have the same meaning in both sentences.
I think we face the same problem when we try to teach mathematical understanding. A proof is most memorable to us when, like in getting a pun, we make a connection between two or more apparently unconnected thoughts, what is often called an "Aha!" moment. Without previous deep knowledge of the constituent thoughts, the student may be able to follow the step-by-step logic, and may be able to remember the proof for tomorrow's test, but the proof will not be memorable, and both the theorem and the proof will soon be forgotten. One implication for pedagogy is that the curriculum must be carefully planned so that, when a mathematical topic is introduced, the students will understand the constituent parts and be able to appreciate their connection. Otherwise, we are mostly wasting our time.
I recently came across a proof of the Pythagorean Theorem that was new to me that gave me an aha! moment. This was given in Sanjay Gulati's excellent "Mathematics Academy" blog as a Geogebra demonstration. He does not indicate the original source of the proof. The aha! moment comes for the connection between the Pythagorean Theorem and an apparently unrelated theorem that I always teach in my elementary geometry class, the "crossed chords" theorem. The aha! moment occurs from looking at the following picture.
Then the crossed-chords theorem tells us that (c + a)(c - a) = b2, or c2 - a2 = b2.
Have you ever told a joke to someone who "doesn't get it"? If you patiently explain the referents you may get them to "understand" the joke, but they will probably respond something like "So, why is that funny?"
In the simple example above, you probably found this funny if (1) you are familiar with the maxim "Time flies like an arrow", (2) your knowledge of the English language allows you to understand that "flies" can be a verb meaning "passes swiftly" or a plural noun referring to a type of insect and that "like" can mean both "as" and "enjoy" (3) your knowledge of writing style leads you to expect that when the same word appears in two successive short sentences, it will usually have the same meaning in both sentences.
I think we face the same problem when we try to teach mathematical understanding. A proof is most memorable to us when, like in getting a pun, we make a connection between two or more apparently unconnected thoughts, what is often called an "Aha!" moment. Without previous deep knowledge of the constituent thoughts, the student may be able to follow the step-by-step logic, and may be able to remember the proof for tomorrow's test, but the proof will not be memorable, and both the theorem and the proof will soon be forgotten. One implication for pedagogy is that the curriculum must be carefully planned so that, when a mathematical topic is introduced, the students will understand the constituent parts and be able to appreciate their connection. Otherwise, we are mostly wasting our time.
I recently came across a proof of the Pythagorean Theorem that was new to me that gave me an aha! moment. This was given in Sanjay Gulati's excellent "Mathematics Academy" blog as a Geogebra demonstration. He does not indicate the original source of the proof. The aha! moment comes for the connection between the Pythagorean Theorem and an apparently unrelated theorem that I always teach in my elementary geometry class, the "crossed chords" theorem. The aha! moment occurs from looking at the following picture.
Harold Jacobs' Geometry
I've been considering a new text for a course in Euclidean Geometry that I teach for middle school teachers. I've been using Essentials of Geometry for College Students by Lial et al. The students seem OK with it, but I find it very boring. I supplement it with lots of my own exercises using Geometer's Sketchpad, paper folding, MIRA(tm), etc. to keep things interesting.
In looking for a replacement, the best book I have found so far is Geometry: Seeing, Doing, Understanding by Harold R. Jacobs. The latest (3rd) edition was published in 2003. Although I will probably use this book, I will transform many of the problems I assign from pencil, paper, ruler, and protractor to Geometer's Sketchpad. I would love it if the publisher W. H. Freeman would commission an update.
This is a high school text, but it is more challenging than Lial. The applications to "real life" are the most realistic and compelling that I have seen anywhere. I keep finding things that I didn't know, and ways of looking at geometry problems that I hadn't considered.
In one example on page 503 Jacobs shows a closed smooth curve bounding a convex region and consisting of circular arcs. One student said that the sum of the arc measurements must be 360 degrees, and the other doubts it because the curve is not a circle. From the nature of Jacobs' construction, it is easy to show that the sum of the arc measures is indeed 360 degrees. A good teacher could connect this with the fact that the sum of the exterior angles of a convex polygon is 360 degrees.
In another example, Jacobs gives an "Area Puzzle" where he guides students to prove a curious fact about triangle areas. If each vertex of a triangle (ABC in the figure below) is connected to a point 1/3 of the way from the next vertex (in CCW order, say) to the following vertex, and the intersections of these 3 segments (Cevians) are connected, an inner triangle (DEF) is formed. The area of DEF turns out to be 1/7 of the area of ABC. I have known this for some years, and even published a paper (with my brother Marshall and my nephew Michael) generalizing it to quadrilaterals and to ratios other than 1/3. The proof I used involved using analytic geometry to establish the result for a right triangle with vertices (0, 0) (1, 0), and (0, 1) and then arguing that the area ratio is preserved by affine transformations, so the result holds for all triangles.
Jacobs presents a neat synthetic proof that clearly shows where the strange ratio 1:7 comes from. He constructs 6 more triangles, each a translate of the central triangle, and then guides the student to show that the triangles can be dissected and reassembled to fill the original triangle. See the diagram below.
In looking for a replacement, the best book I have found so far is Geometry: Seeing, Doing, Understanding by Harold R. Jacobs. The latest (3rd) edition was published in 2003. Although I will probably use this book, I will transform many of the problems I assign from pencil, paper, ruler, and protractor to Geometer's Sketchpad. I would love it if the publisher W. H. Freeman would commission an update.
This is a high school text, but it is more challenging than Lial. The applications to "real life" are the most realistic and compelling that I have seen anywhere. I keep finding things that I didn't know, and ways of looking at geometry problems that I hadn't considered.
In one example on page 503 Jacobs shows a closed smooth curve bounding a convex region and consisting of circular arcs. One student said that the sum of the arc measurements must be 360 degrees, and the other doubts it because the curve is not a circle. From the nature of Jacobs' construction, it is easy to show that the sum of the arc measures is indeed 360 degrees. A good teacher could connect this with the fact that the sum of the exterior angles of a convex polygon is 360 degrees.
In another example, Jacobs gives an "Area Puzzle" where he guides students to prove a curious fact about triangle areas. If each vertex of a triangle (ABC in the figure below) is connected to a point 1/3 of the way from the next vertex (in CCW order, say) to the following vertex, and the intersections of these 3 segments (Cevians) are connected, an inner triangle (DEF) is formed. The area of DEF turns out to be 1/7 of the area of ABC. I have known this for some years, and even published a paper (with my brother Marshall and my nephew Michael) generalizing it to quadrilaterals and to ratios other than 1/3. The proof I used involved using analytic geometry to establish the result for a right triangle with vertices (0, 0) (1, 0), and (0, 1) and then arguing that the area ratio is preserved by affine transformations, so the result holds for all triangles.
Jacobs presents a neat synthetic proof that clearly shows where the strange ratio 1:7 comes from. He constructs 6 more triangles, each a translate of the central triangle, and then guides the student to show that the triangles can be dissected and reassembled to fill the original triangle. See the diagram below.
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