Chaitanya's Random Pages

January 30, 2016

Patterns early in the digits of pi

Filed under: mathematics — ckrao @ 8:59 pm

Recently when taking a look at the early decimal digits of \pi I made the following observations:


  • The first run of seven distinct digits (8327950, shown underlined) appears in the 26th to 32nd decimal place. Curiously the third such run (5923078, also underlined) in decimal places 61 to 67 contains the same seven digits. (There is also a run of seven distinct digits in places 51 to 57 with 5820974.)
  • Decimal digits 60 to 69 (shown in bold) are distinct (i.e. all digits are represented once in this streak). The same is true for digits 61 to 70 as both digits 60 and 70 are ‘4’.

Assume the digits of \pi are generated independently from a uniform distribution. Firstly, how often would we expect to see a run of 7 distinct digits? Places k to k+6 are distinct with probability

\displaystyle \frac{9}{10} \times \frac{8}{10} \times \frac{7}{10} \times \frac{6}{10} \times \frac{5}{10} \times \frac{4}{10} = \frac{9!}{3.10^6} = \frac{189}{3125}.

Hence we expect runs of 7 distinct digits to appear 3125/189 \approx 16.5 places apart. This includes the possibility of runs such as 12345678 which contain two runs of 7 distinct digits that are only 1 place apart.

How often would we expect to see the same 7 digits appearing in a run as we did in places 26-32 and 61-67? Furthermore let’s assume the two runs have no overlap, so we discount possibilities such as 12345678 which have a six-digit overlap. We expect a given sequence (e.g. 1234567, in that order) to appear 1/10^7 of the time. There are 7! = 5040 permutations of such a sequence, but of these 1 has overlap 6 (2345671), 2! has overlap 5 (3456712 or 3456721), 3! has overlap 4, …, 6! has overlap 1 with the original sequence. This leaves us with 5040 - (1 + 2 + 6 + 24 + 120 + 720) = 4167 possible choices of the next run to have the same 7 digits but non-overlapping (or to appear in precisely the same order – overlap 7). Hence we expect the same 7 digits to recur (no overlap with the original run) after approximately 10^7/4167 \approx 2400 places apart so what we saw in the first 100 places was remarkable.

Now let us turn to runs of all ten distinct digits. Repeating the argument above, such runs occur every 10^{10}/10! \approx 2756 places. According to [1] the next time we see ten distinct digits is in decimal places 5470 to 5479.

To answer the question of when we would expect to see the first occurrence of ten distinct digits, we adopt an argument from renewal-reward theory based on Sec 7.9.2 of [2] (there also exist approaches based on setting up recurrence relations, or martingale theory (modelling a fair casino), see [3]-[4]). Firstly we let T be the first time we get 10 consecutive distinct values – we wish to find E[T] where E denotes the expected value operator. Note that this will be more than the 2756 answer we obtained above since we make no assumption of starting with a run of ten distinct digits – there is no way T could be 1 for example, but we could have two runs of ten distinct digits that are 1 apart.

From a sequence of digits we first define a renewal process in which after we get 10 consecutive distinct values (at time T) we start over and wait for the next run of 10 consecutive distinct values without using any of the values  up to time T. Such a process will then have an expected length of cycle of E[T].

Next, suppose we earn a reward of $1 every time the last 10 digits of the sequence are distinct (so we would have obtained $1 at each of decimal places 69 and 70 in the \pi example). By an important result in renewal-reward theory, the long run average reward is equal to the expected reward in a cycle divided by the expected length of a cycle.

In a cycle we will obtain

  • $1 at the end
  • $1 at time 1 in the cycle with probability 1/10 (if that digit is the same as ten digits before it)
  • $1 at time 2 in the cycle with probability 2/100 (if the last two digits match those ten places before it)
  • $1 at time 9 in the cycle with probability 9!/10^9 (if the last nine digits match those ten places before it)

Hence the expected reward in a cycle is given by

\displaystyle 1 + \sum_{i=1}^9 \frac{i!}{10^i} = \sum_{i=0}^9 \frac{i!}{10^i}.

We have already seen that the long run average reward is 10!/10^{10} at each decimal place. Hence the expected length of a cycle E[T] (i.e. the expected number of digits before we expect the first run of ten consecutive digits) is given by

\frac{10^{10}}{10!}\sum_{i=0}^9 \frac{i!}{10^i} \approx 3118.

Hence it is pretty cool that we see it so early in the decimal digits of \pi. 🙂


[1] A258157 – Online Encyclopedia of Integer Sequences

[2] S. Ross, Introduction to Probability Models, Academic Press, 2014.

[3] Combinatorics Problem on Expected Value – Problem Solving: Dice Rolls – Daniel Wang | Brilliant

[4] A Collection of Dice

December 30, 2015

Novak Djokovic against younger players

Filed under: sport — ckrao @ 9:11 pm

Novak Djokovic had a stellar tennis record in 2015, winning three out of the four grand slams (runner-up in the other), the ATP World Tour finals for the fifth time, a record six Masters titles and reached 15 consecutive finals (winning 11 of them) in attaining an 82-6 record (31 wins against top 10 players!!) and US$21.6m for the year.

I noticed that all his losses for the year were to players older than him and I looked up his record against younger players. Here is a list of the players younger than Djokovic who have beaten him in his entire ATP career. The month refers to when the associated tournament started.

  • Ernests Gulbis (Jan 2009)
  • Filip Krajinovic (May 2010 – Djokovic retired after losing the first set)
  • Juan Martin del Potro (Sep 2011, Jul 2012 and Mar 2013)
  • Kei Nishikori (Oct 2011 and Aug 2014)
  • Sam Querry (Oct 2012)
  • Grigor Dimitrov (May 2013)

That’s right, only six players have had this honour for a total of nine losses. After his loss to Dimitrov at the Madrid Masters he has won 71 of his last 72 matches against younger players, losing only to Nishikori in the 2014 US Open semi-finals. He is currently on a 40-match winning streak and to the end of 2015 his record is 130-9 against younger players (career: 686-146). As a comparison here is the record of Murray, Nadal and Federer against younger players not named Novak Djokovic, Andy Murray or Rafael Nadal.

  • Murray: 118-19
  • Nadal: 177-17 (including 41 in a row from Nov 2009 to the end of 2011)
  • Federer: 517-61 (the higher numbers show the clear generation gap)



Tennis Abstract: ATP and WTA Match Results, Splits, and Analysis

December 23, 2015

Areas of sections of a triangle from distances to its sides

Filed under: mathematics — ckrao @ 12:35 pm

If a point P is in the interior of triangle ABC distance x, y and z from the sides, what is the ratio of the area of quadrilateral BXPZ to that of ABC?


One way of determining this is to draw parallels to the sides of the triangles through P. Let X_1 and X_2 be where these parallels meet side BC as shown below.


Let the sides of the triangles have lengths a, b, c with corresponding altitudes h_a, h_b, h_c.

Then as \triangle PX_1 X_2 and \triangle ACB are similar,

\begin{aligned}|PX_1X| &= |PX_1X_2| \frac{X_1X}{X_1X_2}\\ &= |PX_1X_2| \frac{b\cos C}{a}\\ &= |PX_1X_2| \frac{b (a^2 + b^2 - c^2)}{2a^2b} \quad \text{ (cosine rule)}\\ &= \left(\frac{x}{h_a} \right)^2|ABC|\frac{(a^2 + b^2 - c^2)}{2a^2}\\ &= |ABC|\left(\frac{ax}{ax+by+cz}\right)^2\frac{(a^2 + b^2 - c^2)}{2a^2}\\&= \frac{|ABC|x^2(a^2 + b^2 - c^2)}{2(ax+by+cz)^2},\quad\quad (1) \end{aligned}

where the second last line follows from twice the area of |ABC| being ah_a = ax + by + cz.


\displaystyle |PY_1Z| = \frac{|ABC|z^2(b^2 + c^2 - a^2)}{2(ax+by+cz)^2}.\quad \quad (2)


\begin{aligned}|X_1Y_1B| &= \left(\frac{h_b-y}{h_b}\right)^2|ABC|\\ &= \left(1-\frac{by}{bh_b}\right)^2 |ABC|\\ &= \left(1-\frac{by}{2|ABC|}\right)^2|ABC|\\ &= \left(1-\frac{by}{ax+by+cz}\right)^2|ABC|\\ &= \left(\frac{ax +cz}{ax+by+cz}\right)^2|ABC|. \quad\quad(3)\end{aligned}

Combining (1), (2) and (3), we obtain our desired answer as

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{|X_1Y_1B|-|PX_1X|-|PY_1Z|}{|ABC|}\\&= \left(\frac{ax +cz}{ax+by+cz}\right)^2-\frac{x^2(a^2 + b^2 - c^2)}{2(ax+by+cz)^2}-\frac{z^2(b^2 + c^2 - a^2)}{2(ax+by+cz)^2}\\&=\frac{(ax+cz)^2 - x^2(a^2 +b^2-c^2)/2 - z^2(b^2+c^2-a^2)/2}{(ax+by+cz)^2}\\ &= \frac{2axcz + x^2(a^2 - b^2 + c^2) + z^2(c^2 +a^2-b^2)}{(ax+by+cz)^2}\\&= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}.\quad\quad(4)\end{aligned}

Similar formulas can be found for quadrilaterals XPYC and YPZA by permuting variables. Note that if P is outside the triangle or if the triangle is obtuse-angled, care must be taken in the signs of the areas (the quadrilaterals may not be convex) and variables x, y, z.

Note that (4) may also be written as

\displaystyle \frac{|BXPZ|}{|ABC|} = \frac{ac(2xz + (x^2 + z^2)\cos B)}{(ax+by+cz)^2}.\quad\quad(5)

Special cases

1) If \triangle ABC is equilateral, a=b=c and from (4) we obtain

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2xz + (x^2 + z^2)(a^2)}{2a^2(x+y+z)^2}\\ &= \frac{4xz + x^2 + z^2}{2(x+y+z)^2}.\quad\quad(6)\end{aligned}

2) If P is at the incentre of \triangle ABC, then x = y = z = r (the inradius) and from (4) we have

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2xz + (x^2 + z^2)(a^2)}{2a^2(x+y+z)^2}\\ &= \frac{4xz + x^2 + z^2}{2(x+y+z)^2}.\quad\quad(7)\end{aligned}

3) If \triangle P is right-angled at B, then quadrilateral BXPZ is a rectangle with area xz and \triangle ABC has area ac/2 and from (5),

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{2acxz )}{(ax+by+cz)^2}\\ &= \frac{2acxz )}{(ac)^2}\\ &= \frac{2xz}{ac}.\quad \quad (8)\end{aligned}

as expected.

4) If a=c and x=z (symmetric isosceles triangle case) then from (4),

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2x^2 + 2x^2(2a^2-b^2)}{2(2ax+by)^2}\\ &= \frac{x^2(4a^2 -b^2)}{(2ax+by)^2}.\quad\quad(9)\end{aligned}

November 30, 2015

Lowest highest scores in test matches in recent times

Filed under: cricket,sport — ckrao @ 10:46 am

The past week saw two low scoring test cricket matches between India and South Africa in Nagpur and Australia vs New Zealand in Adelaide. According to this link, only 3 times since the 1930s has a completed test match (i.e. with a result) had no player make 50 in any of the four innings. Here are some other matches since 1980 where the batting average for the match was at most that of the Nagpur test (14.9 runs per wicket). Some of them may ring familiar for cricket fans.

Also only 14 times since 1980 has the top score been less than 66 which was the highest score in Adelaide (one other in Australia when McGrath had match figures of 10/27!).

Here are stats highlights from the two matches:

Lowest top score in an Adelaide Test Cricket – ESPN Cricinfo

Ashwin’s records and lowest top scores Cricket –  ESPN Cricinfo


November 25, 2015

An identity based on three numbers summing to zero

Filed under: mathematics — ckrao @ 11:06 am

Here is a nice identity which according to [1] appeared in a 1957 Chinese mathematics competition.

If x + y + z = 0 then

\displaystyle \left(\frac{x^2 + y^2 + z^2}{2} \right)\left(\frac{x^5 + y^5 + z^5}{5} \right) = \left(\frac{x^7 + y^7 + z^7}{7} \right).\quad \quad (1)

An elegant proof of this avoids any lengthy expansions. Let x, y and z be roots of the cubic polynomial

\begin{aligned} (X-x)(X-y)(X-z) &= X^3 - (x+y+z)X^2 + (xy + yz + zx)X - xyz\\ &:= X^3 + aX + b.\quad \quad (2)\end{aligned}


\begin{aligned} x^2 + y^2 + z^2 &= (x+y+z)^2 - 2(xy + yz + xz)\\ &= 0 -2a\\ &= -2a\quad\quad (3)\end{aligned}

and summing the relation X^3 = -aX - b for each of X=x, X=y and X=z, gives

\begin{aligned}x^3 + y^3 + z^3 &= -a(x + y + z) - 3b\\ &= -3b.\quad\quad (4)\end{aligned}

In a similar manner, X^4 = -aX^2 - bX and so

\begin{aligned} x^4 + y^4 + z^4 &= -a(x^2 + y^2 + z^2) - b(x + y + z)\\ &= -a(-2a)\\ &= 2a^2.\quad \quad (5)\end{aligned}

Next, X^5 = -aX^3 - bX^2 and so

\begin{aligned} x^5 + y^5 + z^5 &= -a(x^3 + y^3 + z^3) - b(x^2 + y^2 + z^2)\\ &= -a(-3b) -b(-2a)\\ &= 5ab.\quad \quad (6)\end{aligned}

Finally, X^7 = -aX^5 - bX^4 and so

\begin{aligned} x^7 + y^7 + z^7 &= -a(x^5 + y^5 + z^5) - b(x^4 + y^4 + z^4)\\ &= -a(5ab) -b(2a^2)\\ &= -7a^2b.\quad \quad (7)\end{aligned}

We then combine (3), (6) and (7) to obtain (1). It seems that x^n + y^n + z^n for higher values of n are more complicated expressions in a and latex $b$, so we don’t get as pretty a relation elsewhere.


[1] Răzvan Gelca and Titu Andreescu, Putnam and Beyond, Springer, 2007.

October 31, 2015

Highest ODI batting averages over 100 consecutive innings

Filed under: cricket,sport — ckrao @ 11:13 am

The 3 centuries by A B de Villiers in the most recent ODI series against India has given him a sizeable lead over the rest in the following list of highest batting averages in 100 consecutive one day international cricket innings. On 92 out of those 100 innings he came in at number 4 or lower (and never opened in that time span), and still he managed to score over 54 runs per innings. All of his ODI hundreds to date have been scored at a strike rate of at least 100. Some very impressive numbers have been posted for other batsmen in the list too.


Player Innings Not out Runs HS Average Balls faced Strike rate 100s 50s 0s Time period
de Villiers* 100 21 5454 162* 69.04 4935 110.52 20 27 1 Nov 09-Oct 15
Bevan 100 40 3726 108* 62.10 4844 76.92 3 26 2 Sep 94-Aug 99
Dhoni* 100 34 4027 139* 61.02 4587 87.79 5 28 2 Feb 09-Jan 14
Amla* 100 7 5388 159 57.94 5991 89.93 20 27 2 Nov 08-Mar 15
Richards 100 19 4611 189* 56.93 5103 90.36 8 34 4 Jun 75-Nov 86
Kohli* 100 16 4668 183 55.57 5016 93.06 17 21 7 Mar 11-Mar 15
Tendulkar 100 11 4789 186* 53.81 5273 90.82 18 16 5 Apr 98-Jun 02
Hussey 100 33 3574 109* 53.34 4114 86.87 2 27 2 Feb 05-Nov 09
Jones 100 19 4317 145 53.30 5667 76.18 7 32 1 Jan 85-Feb 91
Sangakkara 100 10 4736 169 52.62 5417 87.43 14 28 5 Aug 11-Mar 15
Chanderpaul 100 21 4144 149* 52.46 5627 73.64 8 30 2 Sep 04-Jun 10
Lara 100 11 4575 169 51.40 5738 79.73 11 30 3 Mar 92-Dec 97
Kallis 100 21 4044 139 51.19 5474 73.88 7 27 3 Mar 00-Feb 04
Ponting 100 11 4397 164 49.40 5068 86.76 11 29 4 Nov 03-Dec 07
Greenidge 100 11 4382 133* 49.24 6583 66.57 10 27 3 Dec 75-Dec 85

Statistics are from ESPN Cricinfo. I would be interested to know if there are others who averaged 50+ over 100 consecutive ODI innings.

October 28, 2015

The product of distances to a point from vertices of a regular polygon

Filed under: mathematics — ckrao @ 11:12 am

Here is a cool trigonometric identity I recently encountered:

\displaystyle \prod_{k=1}^n \sin \frac{(2k-1)\pi}{4n} = \prod_{k=1}^n \cos \frac{(2k-1)\pi}{4n} = \frac{\sqrt{2}}{2^n}.

For example, for n = 9:

\displaystyle \sin 5^{\circ} \sin 15^{\circ} \sin 25^{\circ} \ldots \sin 85^{\circ} = \frac{\sqrt{2}}{2^9}.

After thinking about it for some time I realised that the terms on the left side can each be seen as half the lengths of chords of a unit circle with 4n evenly spaced points that can then be rearranged to be distances from a point on the unit circle to half of the points of a regular (2n)-gon, as shown in the figure below.


With the insight of this figure we then write

\begin{aligned} \prod_{k=1}^n \sin \frac{(2k-1)\pi}{4n} &= \left(\prod_{k=1}^{2n} \left| \sin \frac{(2k-1)\pi}{4n} \right|\right)^{1/2} \quad \text{ (all terms are positive)}\\ &= \prod_{k=1}^{2n} \left| \frac{\exp(i\frac{(2k-1)\pi}{4n}) - \exp(-i\frac{(2k-1)\pi}{4n})}{2i} \right|^{1/2} \\ &= \frac{1}{2^n} \prod_{k=1}^{2n} \left| \exp\left(-i \frac{(2k+1)\pi}{4n}\right) \left(\exp\left(i\frac{4k\pi}{4n}\right) - \exp\left(i\frac{2\pi}{4n}\right)\right) \right|^{1/2}\\ &= \frac{1}{2^n} \prod_{k=1}^{2n} \left| \exp\left(-i \frac{(2k+1)\pi}{4n}\right) \right|^{1/2} \left| \left(\exp\left(i\frac{4k\pi}{4n}\right) - \exp\left(i\frac{2\pi}{4n}\right)\right) \right|^{1/2}\\ &= \frac{1}{2^n} \prod_{k=1}^{2n} \left| \left(\exp\left(i\frac{2k\pi}{2n}\right) - \exp\left(i\frac{\pi}{2n}\right)\right) \right|^{1/2}\\ &= \frac{1}{2^n} \left|\prod_{k=1}^{2n} \left(z-\exp\left(i\frac{2k\pi}{2n}\right)\right) \right|^{1/2} \quad \text{where }z = \exp\left(i\frac{\pi}{2n}\right)\\ &= \frac{1}{2^n} |(z^{2n}-1)|^{1/2}\\ &= \frac{1}{2^n} |-1-1|^{1/2}\\ &= \frac{\sqrt{2}}{2^n}. \end{aligned}

(The cosine formula can be derived in a similar manner.)

In general, the product of the distances of any point z in the complex plane to the n roots of unity \omega_n  is

\displaystyle \prod_{k=0}^{n-1} |z-\omega_n| = |z^n - 1|.

The above case was where z^n = -1. Two more cases are illustrated below, this time for n = 10. In the left example the product of distances is

\displaystyle \prod_{k=0}^9 |(1+i)-\exp(2\pi i k/10)| = |(1+i)^{10}-1| = 5\sqrt{41}

while for the right example it is

\displaystyle \prod_{k=0}^9 |1/2-\exp(2\pi i k/10)| = |(1/2)^{10}-1| = 1023/1024.


Note that earlier in the year I posted on the distances to a line from vertices of a regular polygon.

September 28, 2015

First space probes to visit bodies of the Solar System

Filed under: science — ckrao @ 11:11 am

The table below shows the first space probes to visit various bodies of the Solar System (and the year) by mission type (flyby, orbit, impact or soft landing). Notably 2015 had three events: MESSENGER ended its four year orbit of Mercury with the first impact on the planet, Dawn became the first spacecraft to orbit a dwarf planet (Ceres, in the asteroid belt), and New Horizons flew by Pluto.


flyby orbit impact soft landing
Sun Helios 2 1976 (within 43m km) Luna 1 1959
Mercury Mariner 10 1974 MESSENGER 2011 MESSENGER 2015
Venus Venera 1 1961 Venera 9 1975 Venera 3 1966 Venera 9 1975
Mars Mariner 4 1965 Mariner 9 1971 Mars 2 1971 Mars 3 1971
Jupiter Pioneer 10 1973 Galileo 1995 Galileo 1995
Saturn Pioneer 11 1979 Cassini 2004
Uranus Voyager 2 1986
Neptune Voyager 2 1989
Pluto New Horizons 2015
Ceres Dawn 2015
Moon Luna 1 1959 Luna 10 1966 Luna 2 1959 Luna 9 1966
Titan Huygens 2005
asteroid Galileo asteroid 951 Gaspra – 1991 NEAR Shoemaker asteroid 433 Eros – 2000 NEAR Shoemaker asteroid 433 Eros – 2000
comet ICE comet Giacobini-Zinner – 1985 Rosetta comet Churyumov-Gerasimenko – 2014 Deep Impact – Impactor comet Tempel -2005 Philae comet Churyumov-Gersimenko – 2014




September 20, 2015

The simplest Heronian triangles

Filed under: mathematics — ckrao @ 12:05 pm

Heronian triangles are those whose side lengths and area have integer value. Most of the basic ones are formed either by right-angled triangles of integer sides, or by two such triangles joined together. Following the proof in [1] it is not difficult to show that such triangles have side lengths proportional to (x,y,z) = (n(m^2 + h^2), m(n^2 + h^2), (m+n)(mn-h^2)) where m,n and h are integers with mn > h^2.  Firstly, if a triangle has integer side lengths and area, its altitudes must be rational, being twice the area divided by a side length. Also by the cosine rule, the cosine of its angles must be rational, so z_1 and z_2 in the diagram below are rational too (here assume z is the longest side, so that the altitude is inside the triangle).

heronian_setupThis gives us the equations

\displaystyle h^2 = x^2 - z_1^2 = y^2 - z_2^2, z_1 + z_2 = z,\quad \quad (1)

where h, z_1, z_2 \in \mathbb{Q}. Letting x + z_1 = m and y + z_2 = n it follows from the above equations that x - z_1 = h^2/m, y-z^2 = h^2/n from which

\displaystyle (x,y,z) = \left(\left(\frac{1}{2}(m + \frac{h^2}{m}\right), \frac{1}{2}\left(n + \frac{h^2}{n}\right), \frac{1}{2}\left( m - \frac{h^2}{m} + (n - \frac{h^2}{n}\right)\right). \quad\quad (2)

Scaling the sides up by a factor of 2mn, the sides are proportional to

(x',y',z') = (n(m^2 + h^2), m(n^2 + h^2), (m+n)(mn-h^2)).\quad\quad(3)

Next, letting d be the common denominator of the rational numbers h, z_1 and z_2, we multiply the rational solution (x', y', z') in (3) each by d^3 to obtain an integral solution. The altitude upon side length z is proportional to 2hmn and the area is hmn(m+n)(mn-h^2). Hence if we start with positive m,n,h with no common factor and with mn > h^2, then (3) gives the side lengths of a Heronian triangle that can then be made primitive by dividing by a common factor.

Below the 20 primitive Heronian triangles with area less than 100 are illustrated to scale, where the first row has been doubled in size for easier viewing (a larger list is here). Note that all but one of them is either an integer right-angled triangle or decomposable into two such triangles as indicated by the blue numbers and sides. Refer to [2] for more on triangles which are not decomposable into two integer right-angled triangles. Here are the primitive Pythagorean triples that feature in the triangles:

  • 3-4-5
  • 5-12-13
  • 8-15-17
  • 20-21-29
  • 7-24-25
  • 28-45-53



[1] Carmichael, R. D., 1914, “Diophantine Analysis”, pp.11-13; in R. D. Carmichael, 1959, The Theory of Numbers and Diophantine Analysis, Dover Publications, Inc.

[2] Yiu, Paul (2008), Heron triangles which cannot be decomposed into two integer right triangles (PDF), 41st Meeting of Florida Section of Mathematical Association of America.

August 31, 2015

Recap of the 2015 World Championships in Athletics

Filed under: sport — ckrao @ 1:18 pm

The 15th IAAF World Championships in Athletics recently concluded in Beijing, with Kenya and Jamaica each winning 7 gold medals. Below are some of the many highlights that are worth mentioning.

  • Usain Bolt became the most decorated World Championship athlete of all time taking his tally to 11 gold and 2 silver. In the 100m his best time this year out of just three races leading into the meet was 9.87s. After recovering from an early stumble in the semi final to qualify, he came over the top in the final by 0.01s to end Justin Gatlin‘s 28-race winning streak.
  • Ashton Eaton broke his own world record in the decathlon, the only world record to fall at the meet. His time of 45.00s for the 400m was the fastest ever in a decathlon.
  • Christian Taylor had the second longest triple jump of all time of 18.21m, just 8cm off the world record set in 1995. He and Pedro Pichardo were threatening something this special after both have cleared 18m this year.
  • In the women’s hammer throw Anita Włodarczyk continued her stellar 2015 with another second throw beyond 80m this month. She now has the 9 of the top 13 throws of all time.
  • The women’s 200m was possibly the best such race ever with Dafne Schippers and Elaine Thompson beating their personal best times by around 0.4s and becoming the 3rd and 5th fastest over the distance of all time.
  • Almaz Ayana won the women’s 5000m by over 100 metres, with her last 3000m covered in 8:20.
  • In the men’s marathon Ghirmay Ghebreslassie became the youngest ever world champion in the event at just 19 years of age. He also gave Eritrea its first ever gold medal at the championships.
  • Allyson Felix won her 9th world championship gold medal with a personal best of 49.26s in the 400m. She also ran the third fastest 400m split ever (47.72s) in the 4x400m relay.
  • Mo Farah repeated his 5000-10000m double from the World Championships two years ago. His last 800m of the 5000 was completed in 1:48.6.
  • Ezekiel Kemboi won the 3000m steeplechase for the fourth time in a row.
  • Jesús Ángel Garcia made his 12th appearance at the World Championships (50km walk) dating back to 1993. The event was won by Matej Tóth by almost 2 minutes despite a bathroom break in the middle.
  • Only 0.05s separated the top five in the women’s 100m sprint with Shelly-Ann Fraser-Pryce triumphant again (she now has 7 gold meals at the Worlds).
  • The men’s 4x100m had plenty of average baton handovers (including by winners Jamaica), but China’s were flawless and were rewarded with silver from lane 9.
  • Kenya triumphed in the men’s 400m hurdles and javelin events (Nicholas Bett and Julius Yego respectively, not only the distance events.
  • The women’s javelin was one of the most exciting events with multiple lead changes and Katharina Molitor won with a personal best and 2015-best in the very last throw of the competition.
  • Asbel Kiprop became three-time champion in the men’s 1500m with just 0.41s separating the top 5.
  • Recent world record holder Genzebe Dibaba won the women’s 1500m with 1:57.3s for the final 800m, faster than the winning time in the women’s 800m.
  • The men’s 400m was very fast with the top three (led by Wayde van Niekerk) finishing under 44 seconds for the first time ever. Also an Asian record of 43.93s was set by Yousef Masrahi from Saudi Arabia in the heats.
  • Jessica Ennis-Hill won the hepathlon, just over a year after giving birth to a second child.
  • World record holder Aries Merritt won bronze in the 110m hurdles with his kidneys functioning at just 20%, just two days before a scheduled kidney transplant.
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