Perigal's dissection proof of the Pythagorean theorem

This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Shown here is the famous dissection proof of the Pythagorean theorem by Henry Perigal.

It can be shown in animation form, and that is what we have done here. Please click on the link given below and follow the instructions.

Perigal's dynamic proof of the Pythagorean theorem

For some biographical details of Henry Perigal, it is best to look at the Wikipedia link: Henry Perigal

RMO 2016 – a few solutions

The Regional Mathematics Olympiad 2016 was held on October 9. The format followed was the usual one; it was a 3 hour paper, with 6 problems from the usual Olympiad topics: Algebra, Number Theory, Euclidean geometry, Combinatorics.

The problem in geometry was particularly challenging, though I think it was inaccurately framed.

Anyway, here are the solutions to four of the problems (problems 3, 4, 5 and 6). Do have a look!

Click on this link: solutions to some RMO 2016 problems

Problem concerning a rational number

Here's an interesting problem from number theory.

Which positive rational numbers have the property that the sum of the number and its reciprocal is an integer?

A little bit of experimentation will probably lead you to the answer. But it would be interesting to find one's own proof for that answer.

I have written up my analysis of the problem in the following PDF file: Problem about a rational number.

Do have a look at it. Your comments would be very welcome.

A Challenging Integral

Some weeks back, during a Math Teachers Workshop being conducted by the AMTI at the Institute of Mathematical Sciences (IMSc), Chennai, for a group of teachers from the Jawahar Navodaya Vidyalas, one of the teachers (Mr Sasi Kumar, of Palakkad, Kerala) posed a challenge to the group: the evaluation of a definite integral. It proved to be very challenging indeed. Indeed, no one got it.

On getting back and struggling with it for some days I finally cracked it. I have posted the solution here: ToughIntegral.

What is interesting to me is that it resisted all my attempts at an elementary solution. I wonder if there exists any such solution at all. My solution ultimately had to depend on Cauchy's Residue Theorem. If some reader finds an elementary solution, I would be most interested in seeing it.

Morley's Miracle

This GeoGebra applet demonstrates the famous theorem discovered by Frank Morley towards the end of the nineteenth century.

Triangle ABC is arbitrary. Each internal angle of the triangle is trisected, and the trisectors meet at points P, Q and R, as shown (the trisectors closest to BC meet at P, the trisectors closest to CA meet at Q, and the trisectors closest to AB meet at R).

And now the miracle: triangle PQR is equilateral, regardless of the shape of triangle ABC!


Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)


You can test it out by dragging the vertices using the mouse!

For more on Frank Morley, you can study the Wikipedia site MorleyWiki: or this site: MorleyBiography. I was impressed by the fact that he was a very good chess player and even managed to beat the great Emanuel Lasker while Lasker was the World Chess Champion.

NASA Finds Earth's First Trojan Asteroid

A Trojan asteroid has been found by NASA in the same orbit as the Earth! Here is a link that tells you about it:

NASA's WISE Finds Earth's First Trojan Asteroid - NASA Jet Propulsion Laboratory

And here is an animation that depicts the orbit of the asteroid. It is quite striking that the orbit takes it regularly above and below the plane of the Earth's orbit; enjoy! --

Euler, Madhava and pi

Euler and the Basel problem -

A pattern of signs

Signs

Prof R C Gupta and early Indian Mathematics

Yesterday I made an entry about the award of a prize to Prof R C Gupta for his huge opus of work in early Indian mathematics. I would like to say a bit more about this.

Prof Gupta has worked extensively on the work in mathematics and astronomy of the early mathematicians and astronomers like Aryabhata, Bhaskara I, Brahmagupta, Mahavira, Govindaswami, Bhaskara II and so on; also on early Jain mathematics in general. If you study the texts associated with all these mathematicians you do not see the subject developed as it is in today's books. Typically you see either extensive tables, or rules of some kind which are clearly the verbal equivalent of formulas, and these are generally given in the form of rhyming verse (in Sanskrit, of course). 

The first problem, obviously, is to interpret these verses correctly and meaningfully. But the really interesting question is: How did these mathematicians construct the formulas or the tables? 

For example, in Aryabhata's book Aryabhatiya, we find a table of first differences of the sine table, captured very compactly and conveniently in just two lines, in a form of verse. It seems amazing that in that distant era (5th century AD), Aryabhata was examining tables of differences of a function! And how did he even construct such a table? It compares very well with the modern table. Unfortunately, this question may never be fully answered, because the records are so scanty, but historians like Prof Gupta have looked very carefully at the question. (There are now many historians in the west too who have a great interest in such questions.) In the case of the later mathematicians (Bhaskara I, Mahavira, Bhaskara II) the records yield a little more. For example, it appears that some of these later mathematicians used second order interpolation methods to construct more accurate tables. Here is a paper of Prof Gupta's on just this topic, which you should have a look at: R C Gupta - 2nd order interpolation.

With regard to the Kerala school of mathematics - Madhava, Nilakantha and so on - the records are much clearer; in this school, the writers developed a tradition of describing how they got their results, in sharp contrast to (say) what Aryabhata or Bhaskara I did. So they describe clearly how they got the power series expansion for $\tan^{-1} x$. The famous series

\[ \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \cdots \]

which is nowadays called the Leibnitz-Gregory series for $\pi$ should more correctly be called the Madhava series, or the Madhava-Gregory-Leibnitz series, because it appears in the work of the Kerala school - the Yuktibhasa - and this work predates the work of Gregory and Leibnitz by over two centuries.

The proof of the series for $\tan^{-1} x$ shows quite clearly that the Kerala school had an understanding of the limit process. What is not clear is whether they were able to generalize this notion and apply it to the study of functions in general. If they had done this, then they would have had the differential calculus in their hands, with all the power and glory that it brings. But it is probable that they did not; so in that sense one can say that these mathematicians "almost" discovered the calculus for themselves, but they did not quite get there.

A divisibility problem

It is embarrassing to see how for long I have not posted anything on this blog! But the summer break broke my momentum, and too many events intervened subsequently. Thankfully I am able to make an entry today, after a break of close to ten weeks.

A reader - Mr Vinod Kumar, lecturer in mathematics, of Payyanur College, Kannur District, Kerala - has sent in the following problem:


Consider all four digit numbers with the property that the sum of the first two digits is equal to the sum of the last two digits. Examples of such numbers are $4123$, $6372$ and $0413$. (In this problem we shall include the numbers with one, two or three digits among the four digit numbers - we simply consider their leading digits to be $0$. So we shall write $311$ as $0311$, $12$ as $0012$, and so on.)


Here is the problem: Show that the sum of all such four digit numbers is divisible by $11$.


A second question is: If we consider instead all the six digit numbers with the corresponding property - that is, the sum of the first three digits equals the sum of the last three digits - what would be the number that divides the sum of all such numbers? Once again, we pad the number with enough zeros from the left side to ensure that it has six digits.


Readers are invited to solve these problems!

Prof R C Gupta to be honored

The renowned math historian Prof R C Gupta was to be honored last year at the ICHM (Budapest, July 2009),

ICHM Special Session Henk Bos Valedictory Symposium

but he was unable to be present for the occasion. So the prize - the Kenneth O. May Prize - will be given to him this year instead - at the ICM to be held in Hyderabad (19-27 August 2010). Here is a link to the ICM:

http://www.icm2010.in/about-icm-2010

Prof Gupta has done an amazing amount of work on early Indian mathematics, and is a highly accomplished historian of mathematics. I have had the pleasure and honor of interacting with him (in Warangal in December 2007, during the annual AMTI conference), and can testify to how simple and unassuming a person he is.

A curious occurrence of the Fibonacci numbers

Anyone who has some familiarity with the Fibonacci numbers,
\[0, \,\ 1, \,\ 1, \,\ 2, \,\ 3, \,\ 5, \,\ 8, \,\ 13, \,\ 21, \,\ 34, \,\ 55, \,\ 89, \,\ \ldots,\]
will know that they have a way of cropping up in the most unlikely places. In today's posting we document another such instance.

They arise when we study the compositions of positive integers. I prefer to call them "ordered partitions" but "compositions" is the accepted term.

As you know, a partition of a positive integer $n$ is a way of writing $n$ as a sum of (an unspecified number of) positive integers, order not being regarded as important. Thus, there are $2$ partitions of $2$ (namely, $2$ and $1+1$); $3$ partitions of $3$:
\[3, \quad 2+1, \quad 1+1+1\]
and $5$ partitions of $4$:
\[4, \quad 3+1, \quad 2+2, \quad 2+1+1 \quad 1+1+1+1.\]
Note that we do not distinguish between $3+1$ and $1+3$; that is, the sums are unordered. The above lists show that the partition numbers of $2$, $3$, $4$ are $2$, $3$, $5$ respectively; we write $P(2) = 2$, $P(3) = 3$ , $P(4) = 5$, where $P$ denotes the partition function. The $P$-function grows at terrific speed; for example,

• $P(100) = 190,569,292$;
• $P(200) = 3,972,999,029,388$;
• $P(300) = 9,253,082,936,723,602$.

Big numbers!

The partition function is the setting for many famous and beautiful (and almost unbelievable) results, proved by mathematicians like Ramanujan, Hardy, Littlewood and Rademacher. But we shall not dwell on these results in this post.

A variation - "ordered partitions"

Instead, we shall choose to regard order as important, in which case we get ordered partitions (or compositions, as noted above). Let $f(n)$ denote the number of ordered partitions of $n$. We get the following values of $f(n)$:

• $f(1) = 1$, as $1$ has just one ordered partition ($1$ itself);
• $f(2) = 2$, as $2$ has the following two ordered partitions: $2$ and $1+1$;
• $f(3) = 4$, as $4$ has the following four ordered partitions: $3$, $2+1$,$1+2$, $1+1+1$;
• $f(4) = 8$, as $4$ has the following eight ordered partitions: $4$, $3+1$, $1+3$, $2+2$, $2+1+1$, $1+2+1$, $1+1+2$, $1+1+1+1$.

Observe the sequence of values: $1$, $2$, $4$, $8$. Why, we seem to have a doubling sequence! Can we expect, then, that $f(5) = 16$? Let's check: the ordered partitions of $5$ are $5$, $4+1$, $1+4$, $3+2$, $2+3$, $3+1+1$, $1+3+1$, $1+1+3$, $2+2+1$, $2+1+2$,$1+2+2$, $2+1+1+1$, $1+2+1+1$, $1+1+2+1$, $1+1+1+2$, $1+1+1+1+1$. The count is indeed $16$.

It appears then that $f(n) = 2^{n-1}$ for all $n \ge 1$. But how may one prove this compact and elegant result? Try it out!

Another variation - we leave out the number 1

Now let us introduce a simple variation: in the partitions we shall not allow the number $1$; thus, all the summands must exceed $1$. In this case $5$ has only three such ordered partitions: $5$, $3+2$, $2+3$.

In general, let $g(n)$ denote the number of ordered partitions of $n$ in which every summand exceeds $1$. We get the following values of $g(n)$:

• $g(1) = 0$ (no partitions are possible if $1$ cannot be used);
• $g(2) = 1$ (there is just one such partition, $2$ itself);
• $g(3) = 1$ (there is just one such partition, $3$ itself);
• $g(4) = 2$ (the two partitions are $4$ and $2+2$);
• $g(5) = 3$ (obtained above);
• $g(6) = 5$ (the five partitions are $6$, $4+2$, $2+4$, $3+3$, $2+2+2$);
• $g(7) = 8$ (the eight partitions are $7$, $5+2$, $2+5$, $4+3$, $3+4$, $3+2+2$, $2+3+2$, $2+2+3$);
• $g(8) = 13$ (the thirteen partitions are $8$, $6+2$, $2+6$, $5+3$, $3+5$, $4+4$, $4+2+2$, $2+4+2$, $2+2+4$, $3+3+2$, $3+2+3$, $2+3+3$, $2+2+2+2$).

We get the following sequence of values of the $g$ function: $0$, $1$, $1$, $2$, $3$, $5$, $8$, $13$. Why, we seem to have obtained the Fibonacci sequence! Is this really so?

Let us verify whether $g(9) = 21$ (for the next Fibonacci number after $13$ is $21$); the ordered partitions of $9$ in which every summand exceeds $1$ are $9$, $7+2$, $2+7$, $6+3$, $3+6$, $5+4$, $4+5$, $5+2+2$, $2+5+2$, $2+2+5$, $4+3+2$, $4+2+3$, $3+4+2$, $3+2+4$, $2+3+4$, $2+4+3$, $3+3+3$, $3+2+2+2$, $2+3+2+2$, $2+2+3+2$, $2+2+2+3$. Indeed there are $21$ such partitions!

Is it not remarkable that the Fibonacci numbers have arisen in such a context? To check that we are not being fooled by "circumstantial evidence", we need to prove logically that $g(n)$ is a Fibonacci number; the proof cannot be numerical or circumstantial. This is not as difficult as may seem at first sight, and may be shown by proving that the $g$'s obey the basic recursive law governing the Fibonacci numbers, namely:
\[g(n) = g(n-1) + g(n-2)\]
for any integer $n \ge 3$. Since $g(1)$ and $g(2)$ are already known to be the first two consecutive Fibonacci numbers, the rest then follows.

Here are some references to Fibonacci numbers that point out some further aspects of the Fibonacci numbers.

• Wolfram Mathworld - Fibonacci 
• Wikipedia - Mathworld 

The Wolfram reference lists several contexts where the Fibonacci numbers arise in a natural yet unexpected manner.

Pretty decimals

Everyone is fond of recurring decimals (well, I certainly am ...). Who can resist being charmed by the fact that
\[\frac{1}{7} = 0.142857 \ 142857 \ 142857 \ \ldots,\]
with the portion $142857$ recurring indefinitely (it is called the repetend) and having these several striking properties: first,

• $142857 \times 2 = 285714$
• $142857 \times 3 = 428571$
• $142857 \times 4 = 571428$
• $142857 \times 5 = 714285$
• $142857 \times 6 = 857142$

with the circular arrangement of digits in each product being exactly the same as in $142857$; and next, if we "slice" the number in two halves, like this:
\[\fbox{$142$} \ \fbox{$857$}\]
and add the two "halves", we get
\[142 + 857 = 999,\]
a number wholly composed of $9$'s.

If we try this with the fraction $1/13$, we find something very similar. The repetend is now $076923$, since
\[\frac{1}{13} = 0.076923 \ 076923 \ 076923 \ \ldots,\]
and this number has nearly the same features as the number $142857$; but there are also some curious differences. Thus, its multiples with the numbers $1$, $3$, $4$, $9$, $10$, $12$ yield numbers that have the same circular pattern of digits as $076923$:

• $076923 \times 1 = 076923$
• $076923 \times 3 = 230769$
• $076923 \times 4 = 307692$
• $076923 \times 9 = 692307$
• $076923 \times 10 = 769230$
• $076923 \times 12 = 923076$

whereas its multiples with the numbers $2$, $5$, $6$, $7$, $8$, $11$ yield another set of numbers which too bear the very same relationship amongst themselves:

• $076923 \times 2 = 153846$
• $076923 \times 5 = 384615$
• $076923 \times 6 = 461538$
• $076923 \times 7 = 538461$
• $076923 \times 8 = 615384$
• $076923 \times 11 = 846153$

Note that the numbers
\[153846, \quad 384615, \quad 461538, \quad 538461, \quad 615384, \quad 846153\]
all have the same circular arrangement of digits. How very curious!

You may well ask, is there any other property that bonds the numbers $1$, $3$, $4$, $9$, $10$, $12$ together, and similarly bonds the numbers $2$, $5$, $6$, $7$, $8$, $11$ together? Yes, there is! - try to find it.

Finally, note that if we slice the number $076923$ in two halves,
\[\fbox{$076$} \ \fbox{$923$}\]
and add the two halves, we get
\[076 + 923 = 999,\]
exactly like earlier.

Is this a general feature that we find with all recurring decimals? Try it out - go and explore more such decimals! You will almost certainly unearth more patterns of interest.

Here is an associated problem which is of some interest: Find the smallest positive integer with the feature that if the units digit is taken to the "opposite end" of the number, the new number is exactly twice the original number.

Powers sequences strung on a line

Infinite decimals have still more morsels to offer. For example, consider the decimal arising from the fraction $1/99998$:
\[\frac{1}{99998} = 0.00001 \  00002 \ 00004 \ 00008 \ 00016 \ 00032 \ 00064 \ \ldots.\]
Well, we see the powers of $2$ on display, all strung out on a line!

If we change the fraction to $1/99997$ we get something just as nice:
\[\frac{1}{99997} = 0.00001 \ 00003 \ 00009 \ 00027 \ 00081 \ 00243 \ 00729 \ \ldots.\]
Now we see the powers of $3$. Isn't that a pretty sight?

Once we get the general idea, it isn't at all hard to generalize it. For example, here is $1/9995$:
\[\frac{1}{99995} = 0.00001 \ 00005 \ 00025 \ 00125 \ 00625 \ 03125 \ 15625 \ldots.\]
But I hear you jumping up and saying something urgently to me:

"Hold on, now! These are the decimal expansions of rational numbers, so they ought to yield recurring decimals, right? But where are the recurring sequences in the above decimals? I don't see any!" 

Well, I'll leave it to you to figure that one out!

One can even get the Fibonacci numbers strung out on a line:
\[\frac{1000}{998999} = 0.001 \ 001 \ 002 \ 003 \ 005 \ 008 \ 013 \ 021 \ 034 \ 055 \ 089 \  \ldots.\]
Try to figure out which fraction will yield the Lucas numbers.

(The Lucas sequence is less well known than the Fibonacci sequence, but is defined in a very similar way. Its starting numbers are $2$ and $1$, in that order, and following these, each subsequent number is the sum of the two that precede it
\[2, \,\ 1, \,\ 3, \,\ 4, \,\ 7, \,\ 11, \,\ 18, \,\ 29, \,\ 47, \,\ 76, \,\ \ldots.\]
These numbers have extremely close interconnections with the Fibonacci numbers.)

\[\clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit\]

Primes in arithmetic progression

How many primes can you find in arithmetic progression (prime AP for short)? Recall that in an AP, the differences between successive members are the same, so the AP has the form $a$, $a+d$, $a+2d$, $a+3d$, $a+4d$, .... Here $a$ is called the "first term" while $d$ is the "common difference". For example, $1$, $4$, $7$, $10$ is a four term AP with common difference $d = 3$.

Here are some examples of prime APs:

• {$3$, $5$, $7$}, with length $3$ and common difference $2$;
• {$5$, $11$, $17$, $23$, $29$}, with length $5$ and common difference $6$;
• {$7$, $37$, $67$, $97$, $127$, $157$}, with length $6$ and common difference $30$.
• {$7$, $157$, $307$, $457$, $607$, $757$, $907$}, with length $7$ and common difference $150$.

It is easy to check that the prime AP {$3$, $5$, $7$}, with length $3$ and common difference $2$, is the only such prime AP, and a prime AP with common difference $2$ cannot exceed a length of $3$. To see why, we only need to examine the remainders left by these numbers on division by $3$.

It is also easy to see the following:

• If a prime AP has length greater than $2$, then its common difference must be an even number, and it must feature only odd primes (because $2$ is the only even prime).
• If a prime AP has length greater than $3$, then its common difference must be a multiple of $3$, and therefore a multiple of $6$ as well (since it also must be an even number). To see why, we only need to examine the remainders left by these numbers on division by $3$.
• If a prime AP has length greater than $5$, then its common difference must be a multiple of $5$, and therefore a multiple of $30$ as well (since it also must be a multiple of $6$). To see why, we only need to examine the remainders left by these numbers on division by $5$.
• If a prime AP has length greater than $7$, then its common difference must be a multiple of $7$, and therefore a multiple of $210$ as well (since it also must be a multiple of $30$). To see why, we only need to examine the remainders left by these numbers on division by $7$.

This chain of results may clearly be extended indefinitely.

By these means we can list innumerable necessary conditions for the existence of long prime APs.

But finding such APs is quite a different matter. As of now, it is largely based on trial-and-error, but guided by clever heuristics. In other words: try, try, and try again! To get a sense of the difficulties involved in this, try finding a prime AP of length $10$ or more - on your own!

Recently (earlier this month, in fact), researchers found a prime AP of length $26$. Here is a link to the site where the discovery was announced: AP26 discovery. The first prime of this AP is
\[43142746595714191,\]
and its common difference is this number:
\[23681770 \times (2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23).\]
In other words, this AP consists of the numbers
\[43142746595714191 + 5283234035979900 \times n\]
for $n = 0$, $1$, $2$, ..., $25$.

Note that the number
\[2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 = 223092870,\]
which is a factor of the common difference, is simply the product of the first $9$ primes. (You will find numbers of this kind repeatedly featuring in such searches.) You can check that the last member of this prime AP is the number
\[175223597495211691.\]
To verify that all these $26$ numbers are really prime would require a good computer algebra system, like Derive or Mathematica or the free open source package, Maxima. Here is what Maxima tells you when you ask it whether the $27$-th term of the AP is prime or not - that is, the number
\[43142746595714191 + 5283234035979900 \times 26;\]
Maxima returns the factorization
\[29 \times 31 \times 41 \times 59 \times 85433250011.\]
So this particular AP cannot be extended beyond $26$ terms.

As of now, this is the "champion" - the holder of the "world record"! No prime AP of length greater than $26$ is currently known.

But just a few years back, Terence Tao and Ben Green proved that there exist prime APs of any desired length! (We had featured Terence Tao in a previous posting.) In other words, there must exist a prime AP of length exceeding one billion; only, we may never ever find it, for it may well involve stupendously large numbers ...

Here are two other sites which will tell you a lot about this topic: Prime AP records and Wikipedia-primes in AP.
\[\clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit\]