Friday 30 January 2009

Cat's Eyes Removed Ahead

I have written earlier about the strange fascination of British Signs and public postings... This one is like one I saw and mentioned in an earlier post.... Cats eyes are the British term for the little reflective things along the center of the highway.

Patterns in Iso-perimetric Problems - 3

In my last post on iso-perimeteric problems I finished with the question "If you have 360 feet of fencing and want to build a pen along a river or barn, the rectangular pen above is NOT the most efficient way to use three straight sections... assume you are limited to four fence posts, so you put two somewhere on the river, and can put two more along the fencing to hold straight sections of fence; how do you place them?

Now extend that problem to four sections of fence (and five fence posts)... and then generalize the results... "

The sweet mathematical idea to make all these types of problems easy is symmetry . Imagine that you had found the perfect solution to the fence problem with three sections and used it on your side of the river. Then the guy on the other side of the river decided to copy your master plan on his side...
NOW make the river really narrow..... no..more narrow... like a line.. Now erase the river completely...
What do you see when you look at your neighbors fence and your fence together? It must be a six sided figure.. and it must contain the most area it could contain for any six sided figure, since both sides hold the maximum possible... you couldn't make the total bigger without making one side or the other you must have the most efficient six sided figure possible, a regular hexagon.
So what does your half look like? It must be half of a regular hexagon divided across two opposite vertices.
The solution would be to divide the 360 feet of fencing into three sections, and make the angles between the fence sections 120o and slide the two ends up against the river.
To solve the three sided non-symmetric figure, we used 1/2 of a six sided symmetric figure. And the general solution from there, I hope, is trivial.

Actually with a little handwaving, this blog could have been much shorter..but I remember Pascal once wrote, "I have made this longer than usual, only because I have not had the time to make it shorter." [Lettres Provinciales, No. 16 (1657)]

Tuesday 27 January 2009

A Variation on a Harmonic Theme

One of the Blog sites I drop in on regularly is Sol Lederman's Wild About Math. He has a contest there about every two weeks, alternating with another site so that they produce a problem each week. The latest one is about the probability of hearing all five songs in a set if they are played randomly and you listen for eight times.. I know a way to do it..but the contest is not quite over, so I won't give that away..

But it did remind me of a related problem... Suppose you decided to set it out and wait for the band to play all five tunes. I just heard a nice talk at Gresham College by the current Gresham Lecturer in Geometry, and head of the Cambridge Dept of Math and Theoretical Physics, John Barrow about just this type of problem. The answer, somewhat nicely, is related to the harmonic sequence I wrote about before , 1/1 + 1/2 + 1/3 + ... etc...

Here is how it works... The average number of times it takes to succeed at something that has a probability of P, is 1/P... for example, the probability of rolling a one on a regular die is 1/6... if you set down repeatedly and counted how many times it would take until you rolled a one, on average it would take six trys... sometimes you might get it on the first roll, and sometimes you might have to roll 10 or 20 times...but on average, it would take six rolls...
Now lets look at the music problem... with five songs, you are going to hear one on the first that is pretty easy... but what is the probablity that the next song is new?
Well, if you have only heard one song, then 4/5 of the songs will be new to you, so it will take 5/4 or an average of 1.25 songs until you hear a second song... so for two songs, on average it will take 1 + 1.25 = 2.25 songs to hear both of them...
The probability of a third new song is 3/5, so it should take another 5/3 songs to hear a new one.. so for three songs we would have to listen to 1 + 1.25 + 1.667 = 3.917 songs on average...
If we write these out another way, we notice a pattern... the first took 5/5, the second took 5/4, the third took 5/3.. so for all five songs it would take 5/5 + 5/4 + 5/3 + 5/2 + 5/1 which is 137 /12 or about 11.4 songs to hear them all... and factoring out a five that is 5 (1/1 + 1/2 + 1/3 + 1/4 + 1/5)... the harmonic sequence..

Ok, how does that help.. well for five songs, not much... but what if there are 100 songs in your MP3 player... or 1000 song on your computer ... and you wonder... Hmmm how long would it take on a random shuffle to hear ALL the songs on my MP3 Player... well, 100 (1/1 + 1/2+ ... + 1/100) and that might take a little time to add up ... except for the incredible Euler... What Euler did was come up with a really good approximation for 1/1 + 1/2 + 1/3 + ..... + 1/n for any n... It turns out that as n gets bigger and bigger, the sum gets closer and closer to ln(n).. and Euler came up with a really good estimate of how wrong it would be. Today we often call the number gamma, or Euler's constant, but it is about .577... so if you want to know how long it will take to hear all 100 songs, just multiply 100 times (ln(100) + .577) .... I got about 518...
Ok, quick, let's check that with the answer we got for five songs.. 5 (ln (5) + .577) = 5 (1.609+.577) = 5(2.186) which is about 10.932....... compared to the actual 11.416.... NOT BAD for such a small number..

so for a thousand songs???? Well we leave that as a problem for the reader... good luck

Sunday 25 January 2009

Rheticus, and The Names of Trigonometric Ratios

Spending lots of time lately reading old English journal articles (1825-45) sent me by Dave Renfro who trys to help me stay up on the history of math. It is kind of great reading and watching the actual history of ideas unfold as they did in the old journals.... I came across an interesting letter from Agustus De Morgan about the protege of Copernicus, George Joachim of Rhaetia, also called Rheticus. It was Rheticus who managed to convince Copernicus to publish his long withheld manuscript.
I didn't realize for some years of teaching that in the early days the trigonometric functions were conceived to be lengths of segments in a circle of a given diameter (or radius) rather than the more modern view of ratios. I did not know until I read this article, that apparently it was Rheticus who first developed this approach. In fact, the tables he created to include in his publication of the trigonometric sections of De Revolutionibus were the first tables to include cosines (although he did not use these names). Here is the way De Morgan wrote it:
" Modern teachers (he writes in 1845) of trigonometry have pretty generally abandoned the system of independent lines, which used to be called sines, tangents, &c.; and have substituted, for the meaning of these words, the ratio of the sides of right-angled triangles. It appears that they have antiquity in their favor; indeed so completely has the idea of representing the ratios of the sides of triangles taken possession of the mind of Rheticus, that he abandons the use of the word sine. He dwells on the importance of the right-angled triangles, without any reference to the circle: his maxim expressed in the dialogue, is Triquetrum in planicie cum angulo recto, est magister Mathesos . It would also seem as if his choice of the semi-quadrantal arrangement with double descriptions was dictated merely by the convenience of heading one division with majus latus, and the other with minus latus. [Rheticus had labeled the top of his table with perpendiculum and basis, then the bottoms of these columns were reversed, much as Sine and Cosine were reversed at the top and bottom of tables used in my youth before calculators]........ The names cosine, cotangent, and cosecant are the consequence, not the cause, of this duplicate system of arrangements.......The introduction of the terms sine of the complement, complemental sine, and cosine, &c., followed after an interval of more than half a century."
De Morgan points out that one of the reasons it is so hard to find copies of much of Rheticus' work is that ", In the Index Expurgatorius, it is not Copernicus who is forbidden to be read generally; the prohibition only extends to the work De Revolutionibus, and is accompanied with a nisi corrigatur. But Rheticus is wholly forbidden to be read in any of his works. "
I think the difference in the two mens treatment in the Index may be because of the fact that Rheticus was Protestant, and in fact, was at Wittenburg, the very University where Luther had taught, and burned the Papal Bull.

An interesting anecdote told about Rheticus while he was, "puzzling himself about the motion of Mars, he invoked his genius or guardian angel to help him out of the difficulty: the angel accordingly lifted him up by the hair of his head to the roof and threw him down upon the pavement saying with a bitter laugh, 'That's the way Mars moves.' "

addendum James asked about the phrase "semi-quadrantal"... this just means he only went from 0 degrees to 45 degrees (1/2 of a quadrant) and then put Sin-Cos (he didn't use these words) at the top of the columns and Cos-Sin in the reverse order at the bottom... so that from 45 to 90 degrees was simply read up from the bottom....My old CRC tables were arranged the same way, and many textbooks did as well in the Fifties-sixties
 Giving Sin and Cosine as ratios was still pretty new when this was written by De Morgan. It appears that Peacock had initiated the practice in his lectures at Cambridge around 1830, and by 1837, according to De Morgan, it had become the accepted way to define the terms.

Saturday 24 January 2009

Patterns in Iso-perimetric Problems-2

In my first post on iso-perimetric problems, I ended with some questions, all related to, " A farmer has 360 feet of fencing, and wants to build " where in each case the farmer wanted to build a number of rectangular pens, and in one case no fence was needed along one edge as that was contained by a "straight river" (this is often replaced in textbook problems by making the pens against a barn or house).

In the first case, for two congruent pens, the maximal area I found was when each pen was 45 ft. along the sides that are not shared, and 60 ft in the direction dividing the two pens. for a total area of the two pens of 5400 sq. ft.

In the second case, if the river or barn replaces the fence on one side of two of the adjacent pens, the maximal area occurs when the sides parallel to the river are each 90 feet (twice as big as the last problem) and the remaining three sections perpendicular to this are each still 60 feet. There is a clue to a generalization in the fact that the new solution moves all the fence that would have been used on the river side of the pens to the opposite side and none to the ends.

In the case where there are three adjacent pens in a row, the solution is for each of the pens to be 30 feet along the sides that runs along all three pens, and 45 feet in the direction that divdes the pens.

Is there anything we could observe about these solutions that might answer the two remaining cases with four pens? "Yes!" he answers emphatically. Imagine each of the problems above with fences running north-south and east-west. In each solution, the amount of fencing used in the north-south direction is exactly equal to the amount used in the east-west direction, 180 feet, or one-half of the 360 feet available.

If we assume that will work in the two cases with four pens, then the one with four adjacent pens will have five fence lengths in the direction dividing the line of pens, and two long sections across both ends of the four pens... 180/8 gives 22.5 feet, and 180/5 = 36 feet, so each of the four pens will be 22.5 by 36 feet, and the total area of all the pens is 36 x 90 = 3240 square feet.
In the case of four pens arrainged in a square (don't you just know the four smaller pens will all be squares?), there are six fence sections along the east-west sides, and six along the north-south sides, so each fence was 30x30= 900 square feet, for a total of 360 square feet. A slightly better use of fencing than the four in a row...

In general, and that is where we are going... to maximize rectangular pens with M rows of N pens, and a total perimeter of P then the area of each pen will be P/2 divided by M(N+1) and in the other direction P/2 divided by N(M+1) so that half the fencing is used in each direction.

And as a foot note, as we noted in the case of four pens, 2pens x 2 pens is better than 4x1, and if we have R pens, we want to divide R into two factors of R that are closest together.

OK, so Onward to another relationship... If you have 360 feet of fencing and want to build a pen along a river or barn, the rectangular pen above is NOT the most efficient way to use three straight sections... assume you are limited to four fence posts, so you put two somewhere on the river, and can put two more along the fencing to hold straight sections of fence; how do you place them?

Now extend that problem to four sections of fence (and five fence posts)... and then generalize the results... Any takers?

More later... but I think to give you time in between, my next note will be about history, and in particular the reason that you don't know about George Joachim of Rhaetia.... admit it, you are excited already..

Thursday 22 January 2009

Patterns in Iso-perimetric Problems-1

Way back in my younger days I was a regional director for the Michigan Council of Math Teachers, and gave a lot of talks ... somewhere in that time I wrote some stuff about the type of optimization problems that mathematicians call iso-perimetric problems.. Often they involved a given perimeter (a farmer has 360 feet of fence).. and the object is to find the maximum area under some given conditinos (the area is a rectangle, etc..). Over the years I noticed some patterns to the types of solutions that often made the solution transparent with very little actual calculus or advanced math required. I wanted to write a couple of blogs to explain and, hopefully, organize my thoughts on them... but first, some introductory notes I wrote a long time ago about the problems for folks who are not familiar with the term isoperimetric, and to give a little history.

The word isoperimetric literally means "same perimeter". It is usually used in mathematics to refer to to figures having the same perimeter but different shapes. Mathematicians use the term Isoperimetric Problem to describe problems relating to finding which of two figures with equal perimeter have the greatest area. The problems date back at least to Zenodorus , a 2nd century BC Greek mathematician who wrote On Isometric (same measure) Figures. In it he showed that of all isoperimetric polygons having the same number of sides, the regular polygon had the greatest area. [For example of all quadrilaterals having a perimeter of 16, the 4x4 square has the greatest area.] He also showed that the area of a circle was greater than any regular polygon with the same perimeter as the circumference as the circle.

A related isoperimetric problem is called Dido's problem. Dido's problem is to find the maximum area for a figure with a given perimeter and bounded against a strait line. For example, if you had 100 feet of fence and wanted to enclose the maximum area with one side of the property along a straight river, or the side of a barn; what shape would enclose the maximum area.

Dido was a Phoenician princess in Virgil's Epic tale, THE AENEID. The story tells of her founding of Carthage. When she fled from her brother, Pygmalion, who was trying to kill her (sibling problems seemed pretty dramatic in the Greek classics) she landed along the coast of Northern Africa and tried to buy land from the local king, King Jarbas. She was told she could have as much land as could be enclosed by a bull's hide. Dido had her followers cut the hide into tiny strips which were then strung together to make a great length, and enclosed a great area against the sea. That's where the strait line on one side comes from in the problem. Dido's solution was a semicircle. (pay attention here, this is one clue about generalizing the solution of isoperimetric problems when one side is against a natural straight boundary)

Dido's problem is sometime called the Problem of Hengist and Horsa for a similar conclusion to an English story about two wizards/lords who come from Germany to defend against a Saxon invasion in the fifth century. The two turn out to be the antagonists of the young King Arthur and Merlin. Even Dido's story is related to England in a round-about way. Her brother Pygmalion would create a beautiful statue and then fall in love with it. Aphrodite brought it to life for him. This transformation of a stone statue found its way through a George B. Shaw play into the famous English musical, My Fair Lady

Perhaps one more small side-note to bring all this full circle to my interest in language, the model for Professor Henry Higgins in the play was a real live linguist named Henry Sweet who was instrumental in the development of hte International Phonetic Alphabet in 1897.

Ok, enough for now, but hopefully soon I can exaimine one of the patterns I have noticed. Maybe you can find it too... here is a guided discovery tour...

1) A farmer has 360 feet of fencing, and wants to build two congruent rectangular pens that share a common fence. Maximize the total area (this is a common textbook problem... ) and find the dimensions

2) step two is to do the same thing except one side of the area is bounded by a straight river.. so that side does not need to be fenced...

3) step three is to extend the original problem to three congruenct rectangular pens

4) Now there are two ways to arrange four pens, in a row or in a 2x2 square... try both ways...

Now look at all the answers and pay close attention to the dimensions... I will return to this soon.. so work fast...

Wednesday 21 January 2009

Mental Binary to Octal and Another Clock Problem Solution

Two brief items of interest, both very different. In my last blog I wrote about the problem sent by my good friend from Japan, Al Harmon.

How long after 7:00 will the second hand bisect the angle formed by the other two hands?
I gave the first answer, time wise, bisecting the obtuse angle between the hands. Al reminded me that there was, of course, another bisecting time after the second hand passed the hour hand... and supplied his answer, which is below if you want to work the problem out yourself.

A second problem came up from a question asked at the Math Forum. I occasionally answer questions on the T2T (Teacher to Teacher) web site at the Math Forum and came across an interesting question the other day which asked me how to convert a binary number to base eight. As a brief example, convert the binary number 1010010111011 to base eight... wait... mentally, no calculator, no pencil and paper... start now!

The Clock Problem
Al sent me his answer of 47.9327 seconds after the hour. I worked it out this way.... (and fortunately got the same answer.)

With the hour hand to the second hand as one of the two congruent angles and the angle from the minute hand passed the 12 to the minute hand... the first angle would be simply the angle of the second hand minus the hour of the minute hand, or in terms of T seconds, 6T- (210+T/120). The angle from the second hand to the 12 position on the clock would be 360-6T degrees, and the minute hand would be another T/10 degrees beyond the 12 position where it had started. Adding these two gives the angle between minute and second to be 360 - 59T/10
When we set these equal to each other and simplify we have 360 - 59T/10 = 719T/120 - 210. Transposing the like terms we get 570 = 1427T/120 and multiplying by the reciprocal of the fractional coefficient I got T=47+ 1331/1427
seconds ... checking the angles... at 7:00,47.932726 the hour hand is at 210 +47.932726 /120 = 210.3994394 degrees. The second hand is at 6(47.932726 ) = 287.596356 degrees so the angle between them is 77.1969166 degrees. The minute hand has moved T/10 degrees so it is at 4.7932726 degrees positive from the 12 position, and the second hand is 360- 287.596356=72.403... adding the minute hand gives the same angle as the first to six digits, so I'm calling that a solution.

Convert a binary number to base eight, mentally.
I realized that this conversion was really easy and the same idea could be used for any conversion from one base to some base that was an integer power of it... I don't remember the actual probelm, but suppose it was 1010010111011 (typing semi randomly)... The trick is to simply divide it into periods of three (starting from the right) 1 010 010 111 011 . Now for each period or part of a period (like the first one) simply convert that three digits to base ten (which for numbers less than eight, and all three digit binary numbers must be, is also the value in base eight)... and replace directly... so
1 010 010 111 011 is replaced by
1 2 2 7 3 which is the base eight value for the same number. and of course you could go back from base eight to base two by just reversing the directions, replace each digit of the base eight number with a binary equal.

You could do hexadecimal by marking off periods of four digits in the same way...

If the task was to convert base 3 to base 9 you would just take groups of two digits
2012 in base three is (20 12 ) 20 in base three just means two times three, so we replace 20 with a six.. and 12 is a three with two ones, so we replace it with a five to get 65 (base nine) = 2012 base three...
If you are a little rusty in converting bases, here is the conversion of any base to base ten ...
the place values in base ten are 1, 10, 100, 1000, etc working from right to left... you should recognize these as powers of 10... ie 10^0, 10^1, 10^2, etc... for any other base it works the same way... for base two the place values are 1, 2, 4, 8, 16... and for base three it would be 1, 3, 9, 27, 81, .... Base eight used above would be 1, 8, 64, 512...

To find the decimal expression for a number in another base, just multiply each digit times its place value and add the products together. 2012 base three is 2(27)+0(9)+1(3)+2 = 59 ; so I should get the same thing if I check 65 in base nine... 6(9) + 5 = 59.... (one right in a row!!!)

To express 12273 base eight in base ten we do 1(8^4) + 2(8^3)+ 2(8^2) + 7(8)+ 3... a big number, and I leave it to you to check that that is equal to what you get if you convert the binary 1010010111011 to decimal.

Thursday 15 January 2009

Another Clock Problem, and a Joke

Al sent me another clock problem, so I will put it here, and then to keep a little space between the problem and the answer, (for those who try to solve it themselves) I will post a joke that Jeannie sent me today at work...

First, the problem...How long after 7:00 will the second hand bisect the angle formed by the other two hands?

and now to distract you, a joke:

"While walking along the sidewalk in front of his church, our minister heard the intoning of a prayer that nearly made his collar wilt. Apparently, his 5-year-old son and his playmates had found a dead robin. Feeling that proper burial should be performed, they had secured a small box and cotton batting, then dug a hole and made ready for the disposal of the deceased.
The minister's son was chosen to say the appropriate prayers and with sonorous dignity intoned his version of what he thought his father always said: 'Glory be unto the Faaather, and unto the Sonnn , and into the hole he goooes.' "

It reminds me of when my grandaughter was young and would sob "It's not there!" when she didn't get her way... meaning, of course, "It's not FAIR."...

OK, here is my attempt at the solution: The hour hand starts at 210 degrees and moves 1/120 degrees per second... The minute hand starts at zero degrees and moves at 1/10 degree per second; and the second hand moves at 6 degrees per second from a zero degree start.

When the second hand bisects the other two, the angle between the hour and second hand will equal the angle between the second and minute hand (with the second hand in the middle)...
so I set hour-second = second - minute or...
210 + 1/120 T (in seconds) - 6T = 6T- 1/10 T and simplify to get

210 - 719/120 T = 59/10 T and collect T on right side to get
210 = 1427/120 T multiply by the reciprocal of the coefficient of T and

T= 17.6594 seconds afer 7 o'clock...
Checking the angles,, the hour hand is at 210.1471619 degrees amd the minute hand is at 1.76594 degrees, with the second hand at 105.9565522 degrees...The angle between hour and second is 210.147-105.957 = 104.19.......the hour between second and minute is 105.957-1.766 = 104.191... I'm calling that a winner...

Ok, for a second try... do the same thing on a 24 hour clock....

Tuesday 13 January 2009

A Fistful of Dollars

A Fistful Of Dollars: The Story of a Loan from Kieran Ball on Vimeo.

I have talked before about my enthusiasm for micro-loans to impact on world poverty, and particularly about Kiva, my personal method of making micro-loans. To those I have never explained this clearly to, try this video by Kieran Ball.

This video follows the path of a $25 loan from London, England to Preak Tamao village, Cambodia. is a website that allows internet users like you or I to lend money to people that need it in developing countries, with the aim of empowering them to lift themselves out of poverty. I encourage you to see if it might be a way for you to "lend a hand".

Click here to read Kieran's blog about the video. And if you want to get on board, contact Kiva today.

Monday 12 January 2009

What's Wrong with This Clock?

I used the picture above in a blog and mentioned the 24 hour clock in the Duomo (cathedral) in Florence. To me, what is wrong with it is that it goes the wrong moves mathematically clockwise.
The earliest timekeepping devices were probably shadow clocks. A vertical gnomen erected to cast a shadow as the sun moves. The problem with ancient shadow clocks were that they did not move through the same angle per hour each day. In the summer the angle distance between noon and 3 pm was different than in the winter. To resolve this difference the ancients resorted to cutting hemi-spherical depressions in the face of the sun-dial and making markings for time in several different seasons. It seems that Ptolemy may have known that this could be corrected by pointing the gnomen in the correct angle (the angle of latitude for the dial) but this idea was not put into common practice until the Islamic mathematicians of the Middle Ages made some advances in spherical geometry. Sundials can also be mounted vertically as seen in this 1749 dial I photographed in the Cathedral Close at Salisbury Cathedral.

The dial was actually erected three years before the creation of the Gregorian calendar. The lines drawn across the face mark the sun's path at different seasons of the year. One of the characteristics of vertically mounted sundials (in the northern hemishphere) is that the shadow moves counter-clockwise or in a mathematically positive rotation. In the Basilica di Santa Maria del Fiore (this is the famous one that has the dome by Brunelleschi) in Florence there is an unusual clock dating to the 1400's by Paulo Uccello. The clock not only rotates in a positive (anti-clockwise) direction, it is a 24 hour clock. This may be the oldest mechanical clock in the world to meet both those conditions. I have not seen another major clockwork with both conditions, and would love to hear from anyone who knows of others.

The clock is one-handed, showing only the hour. Originally it used the 24 hours of the hora italica (Italian time), a period of time ending with sunset at 24 hours, which was used until about the 18th century. The clock now shows the standard time of the region.

There is another much more modern twenty-four hour clock at the Shephard Gate of the Royal Observatory at Greenwich. The Wikipedia article states, "The clock, an early example of an electric clock, was a slave mechanism controlled by electric pulses transmitted by a master clock inside the main building. The 'network' of master and slave clocks was constructed and installed by Charles Shepherd in 1852. The clock by the gate was probably the first to display Greenwich Mean Time to the public". Note that it moves in the standard clockwise or mathematically negative rotation.

I have come across several other 24 hour clocks, but none that go in a mathematically positive rotation, and no others as old as the one in Florence. There is one in Piazza San Marco in Venice that has midnight at the "three O'clock" position; probably also for liturgical times using the old Italian method of beginning and ending the day at sunset. The Wells Cathedral mechanical clock (now in the Museum of Science in London) is a 24-hour clock that really counts to twelve twice around the clock face. It also travels in a standard clockwise rotation.

If you are willing to overlook the flaw of negative rotation, you can download a 24 hour clock widgit from Yahoo Widgets.

Saturday 10 January 2009

SOLUTION to Al's Clock Problem

Here's my solution to Al's Problem about the Concert Clock. My strategy was to find the angle of the hour hand between six and seven, and set it equal to the angle of the minute hand between nine and ten. I found that the concert starting at 6:49 1/11 and ending at 9:32 8/11.

Here is how I came up with that... The number of degrees rotation from straight up (12) on the clock for the minute hand is 6T where T is in minutes past the hour, for any hour. The hour hands move T/2 degrees per minute so after nine oclock the angle of the hour hand is at 270 + T/2 degrees for any time T, and between six and seven it is at 180 + T/2. My next step was to express these hour angles in terms of the minute angles...
Using the equation 180+T/2 =H (angle measure of hour hand) and 6T=M (angle measure of the Minute hand) we can solve together to eliminate T and get H=180 + M/12... This is the hour hand angle between six and seven... If we set this equal to the minute hand angle between nine and ten, we will find the angle when they are the same.
Setting 180 + M/12 = 6M and solving I get 196 4/11 degrees.

Now we can use the M = 6T and set it equal to the 196 4/11 and solve for T, The time turns out to be 32 8/11 minutes after the hour, so the ending time is 9:32 8/11.

To find the beginning time, we must set the hour hand angle for nine oclock agains the minute hand for six oclock, to get 270 + M/12 = M. This gives an angle of 294 6/11 degrees. Using the minute hand formula, M = 6T we see that the number of minutes after six oclock is 49 1/11, giving us 6:49 1/11 for the beginning time.

Hope I made that clear

I'm having fun with this, and hope you all are, so maybe I'll post another in a few days... Thanks again to Al for the nice problem.

Thursday 8 January 2009

Al's Revenge

My good friend, Al Harmon, figured out the puzzle I posted, and decided to reply in kind with a problem of his own. I post it here for a challenge to all my readers...

" Robin went to a concert that began between 6 PM and 7 PM and ended between 9 PM and 10 PM. The positions of the clock hands at the end of the concert were switched from their positions at the beginning (hour hand where the minute hand was, and vice-versa). What were the exact starting and ending times of the concert?" I assume he means a regular 12 hour clock, not the kind I have on the image at top... which looks like a clock I saw in Florence, and will mention later.

While your working on that, you may need a good chuckle, so here is a little document I pulled out of the file cabinet of unknown origin. I think it reflects the frustration all teachers feel when kids seem determined not to use their brain. Every teacher has one that drives them especially crazy... I remember a long argument among some mathteachers over how much should be counted off if a kid, given the problem "simplify the fraction 16/64" responds with

Two days ago I asked a student to find the square root of two... she pressed buttons on her calculator and said "One point four one four."... Close enough, so I ask for the square root of three... again the calculator and an answer... finally I ask what is the square root of four... she picks up the calculator, pushes the buttons, and then starts to turn red.... Ok... I set her up for that one... but if you want to send your own "pet peave student answer" I will collect and display after awhile... in the meantime, here is the page I found in the file cabinet today:

DAY 1: Teach them that (a+b)/c is (a/c) + (b/c)

DAY 2: Teach them that a/(b+c) is NOT (a/b) + (a/c)

DAY 3: Teach them that x / ln(x) is NOT "1 / ln"

DAY 4: Teach them that you can't solve (sin(kx)) = 1 by saying "x = 1/sin(k)"

DAY 5: Remind them that a/(b+c) is NOT (a/b) + (a/c)

DAY 6: Show them a movie of a student sitting in a field, writing "(a+b)^2 = a^2 + b ^2" and then getting HIT BY A TRAIN!

DAY 7: Remind them that a/(b+c) is NOT (a/b) + (a/c)

DAY 8: Teach them that if the domain of the a function f is the reals, the graph of y = f(x) is NOT a blank pair of axes, that perhaps they should adjust the "window"

DAY 9: Teach them that x/(y+z) is NOT (x/y) + (x/z)

DAY 10: Group work: Bring a trout to class. Have them solve sin(kx) = 1. If they get x = 1/sin(k), hit them with the trout. Make it a big trout.

Wednesday 7 January 2009

Duck, Here is the Cannonballs Solution

My recent blog asked to solve the problem below... if you haven't tried it yet and don't want to spoil the fun, then go to the original post now.... (the problem)
Several detachments of artillery divided a certain number of cannon balls. The first company took 72 and 1/9 of the remainder; The second detachment 144 and 1/9 of the remainder. The third company took 216 and 1/9 of the remainder; the fourth company took 288 + 1/9 of the remainder; and so on; ---- finally it was discovered by the commanding officer commanding the brigade of guns, the the shot had been equally divided. Determine the number of detachments and the number of balls in the pile.

My Solution approach The first person, as I explained will take 72+ (T-72)/9 or 64+T/9 cannonballs. The second will take 144 + [T-144 - (64-T/9)]/9 . This second expression simplifies somewhat to 1088/9 +8T/81. If the shares are to be equal, these must be the same quantity. The solution is pretty direct as shown here.

Once we know the total number of cannonballs, we just calculate the number that the first division takes in its turn, and then we can determine how many such shares there will be to distribute. The Total will expire after eight batteries have removed their share, each totaling 576 cannonballs, if the 72, 144, 216, pattern continues..Here is the way it looked on a spread sheet

Tuesday 6 January 2009

More on Al's Problem

There were a few questions about the problem in my last blog... As I stated I read it in a very old math prize competition, so I am as subject to misinterpret them as anyone else...but since it is MY blog... here is how I interpret it...
The first group walks up to a pile with T cannonballs, removes 72, then takes one-ninth of the (T-72) remaining cannonballs. Subsequent groups do the same with the remaining piles and the numbers assigned.

I will leave the question hanging another day ..but for those who need more... here is a second question if you have solved the first, and await furthur challenges... I think it should be possible to change the pattern of values given to the first few people and change the number of divisions that select cannon balls to any value less than the denominator of the fractional part (9 in this case)..


I did the simplest case, two divisions, and found it can be done if the first division takes 63 cannonballs plus 1/9 of the remainder. Then the second division will take what is left (which for the number I used for the total works out to two even shares)... This is NOT the smallest possible number of cannonballs, but that gives away too much.

How about three? Well working backwards from the solution for two, I realized that three would work if the first took 54 + 1/9 of the remainder, and the second took 63 + 1/9 of the remainder... the same pattern for larger numbers of detachments could start easily at 27, 36, 45, etc... and it seems as if it would be possible (and actually somewhat simple if I really understand) to generate a pattern for any number of divisions and any fraction we care to use in place of 1/9. Maybe if there is interest, I will address it later

Sunday 4 January 2009

A Problem for Al

I went into math education later than many, and was disappointed to find that math teachers do not, in general, love to do math problems. So I was fortunate that when I landed in Misawa, Japan to teach, they popped me right next door to an old veteran who loved problems as much as I did, and was very demanding about a clear proof and explanation. Today, after twenty years and thousands of miles between us, Al Harmon is still one of my math correspondents and an educational inspiration. He frequently sends me problems he is working on that he doesn't see a solution, or sometimes just doesn't think he has the most elegant solution. I am always flattered that he would think I could solve a problem that he couldn't,

I have another great internet advisor, Dave Renfro, who is no longer employed as a professor, but who still teaches many of us with his posts and references to old journals. Along the way he took to sharing with me the journals he thought might help me learn a little more about math, education, and their history. A few weeks ago he sent me a stack of reprints from the Philosophical Magazine from around 1825-1830. one insert included the "Examination of the scholars at Wyke-House" , a school in Middlesex, England. The scholars were tested on questions ranging from arithmetic to geometry and trigonometry, and took six days to answer the seventy-four questions. The image at top shows the prize medal presented to the winner. The first question this year asks for the student to "Numerate and "point off in periods, half-periods, etc... the numeral 123456789012345678901234567890 (raise your hand if you know what a half-period is.... and if not, look here).... and the last question asks for a proof that x+1/x = 2 cos(theta) could only have solutions when theta was a multple of 180o....several asked for proofs of geometric or trigonometric identities.

One caught my eye as the type of problem Al might love, so I wanted to share it here, with a thanks to both Al and Dave for their continuing contribution to my education.

Problem 51, from the 1827 examination: Several detachments of artillery divided a certain number of cannon balls. The first company took 72 and 1/9 of the remainder; The second detachment 144 and 1/9 of the remainder. The third company took 216 and 1/9 of the remainder; the fourth company took 288 + 1/9 of the remainder; and so on; ---- finally it was discovered by the commanding officer commanding the brigade of guns, the the shot had been equally divided. Determine the number of detachments and the number of balls in the pile.

Ok, So the first answer (with explanation) wins a pat on the back and high praise....... I noticed an interesting pattern when solving this problem, so there is a nice generalization to any number of regiments... For instance I could write a similar problem for any given number of regiments.... . Will provide that later, if needed.

Friday 2 January 2009

The Earth and the Turtle

A recent computer science blog reminded me of the wonderful story about what holds the earth in place. There are many versions, but the one I like best, as told by the folks at Wikipedia, comes from Stephen Hawking:

The most widely known version appears in Stephen Hawking's 1988 book A Brief History of Time, which starts:

“ A well-known scientist (some say it was Bertrand Russell) once gave a public lecture on astronomy. He described how the earth orbits around the sun and how the sun, in turn, orbits around the center of a vast collection of stars called our galaxy. At the end of the lecture, a little old lady at the back of the room got up and said: "What you have told us is rubbish. The world is really a flat plate supported on the back of a giant tortoise." The scientist gave a superior smile before replying, "What is the tortoise standing on?" "You're very clever, young man, very clever," said the old lady. "But it's turtles all the way down!"

Whether you see it as deeply metaphysical, or just poking fun at physics, you can still read it and smile.

Thursday 1 January 2009

WHY, we Flip and Multiply

A recent anonymous comment to my blog on "Division of Fractions by the Alien Method" wrote:

"This is really cool! And you are correct...I (like most other 5th graders a long time ago) memorized a method to divide fractions...I believe the mantra was "Yours is not to reason why, just invert and multiply"....or another one that students seem to use is "Keep, change, re-arrange". Although I can certainly perform the operation, and can even ask the question posed by 2/3 divided by 5/7 (If we think of it as a piece of wood with length 2/3, then I believe the question is how many 5/7's are there in the piece of wood). Unfortunately, that doesn't tell me "why" inverting the second fraction and then multiplying works....I've asked several very bright people and have never gotten an answer that sticks...can you enlighten????"

I want to make one comment about division of fractions that seems harder to visulaize than for general division, and then I hope to explain in simple terms just why "invert and multiply" works.

For every multiplication problem, there are two associated division problems; A x B = C begets C/A=B and C/B=A. Elementary teachers call these a "family of facts for C" (or did in the recent past.. educational language changes too fast for firm statments by a non-elementary teacher). So if we add units to one or both factors, appropriate units must be appended to the product. So how does this effect operations with fractions? Well if we have length, as in ANON's comment, then the division problem he states, "If we think of it as a piece of wood with length 2/3, then I believe the question is how many 5/7's are there in the piece of wood" he is dividing length by length to get a pure scaler counting how many pieces (or fractions of a piece) will fit into another. In the case he gives, the answer would be only 14/15 of a piece... becuase the 2/3 unit length is not quite enough to provide a 5/7 unit length piece...

The multiplication associated with this operation is then 14/15 of 5/7 units = 2/3 units... What about the other division in this family of facts... 2/3 units divided by 14/15 (a scaler here, not a length)will give 5/7 units length. What is this sitution describing? This seems the one most difficult for teachers and students alike. We all know what it means to divide a length into (by?) two pieces, but what sense does it make to divide it into 1/2 a piece.

We might try to make this clear to students by taking some common length (12 inches?) and see what happens if we divide it into (by) 8 pieces, then four, then two, then one, (each division is by half the previousl number)and look at the pattern of lengths. 12/8=3/2; 12/4 = 3; 12/2 = 6; 12/1= 12... I am confident most students could identify the next numbers in the sequence, 12/ (1/2) = 24, and 12/(1/4) = 48.

At this point, using whole numbers as divisors, the pattern for "invert and multiply" seems obvious, but this is far from a why for all fraction problems.

Let's look at one more case where we sneak in a related idea at the elementary level. Given a problem like 3.5 divided by .04, the student is taught to "move the decimal places enough to make the divisor (.04) a whole number. What we do is another problem (350 divided by 4) that has the same answer (87.5)as the original. Another why does that work that is not often explained.

What do the two operations have in common.... multiplication by one. In each case we have a division (fraction) operation and we simply mulitiply the fraction by a carefully chosen version of one that will make it easier to do. If we view 3.5/.04 as a fraction, then every fifth grader knows that multipliying it by one will not change its value. This is the core of what we do to find equivalent fractions... to get 3/5 = 6/10 we multiply by one, but expressed as 2/2... The decimal division problem uses the same approach... we multiply 3.5/.04 by 100/100 to get another name for the same fraction, 350/4.

Now to explain "invert and multiply" we just use the same idea... dividing fractions is simply fractions which have fractions instead of integers in the numerator and denominator. We want to multiply by one in a way that the division problem will be easier. But the easiest number to divide by is one,... so why not pick a number that changes the denominator of the fraction over a fraction to be a one... that is, multiply by its reciprocal. So for 2/3 divided by 5/7 we can write

And I hope that makes it clear.... questions and comments are gratefully received