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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.

August 28, 2015

The discriminant trick

Filed under: mathematics — ckrao @ 12:22 pm

Suppose you wish to find the maximum value of y =\frac{8}{2x+1} - \frac{1}{x} for x > 0. One way to do this without calculus is to massage the expression until one can apply an elementary inequality such as u + \frac{1}{u} \geq 2. To apply this particular result we aim to minimise 1/y and apply polynomial division.

\begin{aligned}  \frac{1}{y} &= \frac{1}{\frac{8}{2x+1} - \frac{1}{x}}\\  &= \frac{x(2x+1)}{8x - (2x+1)}\\  &=\frac{2x^2+x}{6x-1}\\  &= \frac{(6x-1)(x/3 + 2/9) + 2/9}{6x-1}\\  &= \frac{1}{9}\left( 3x+2 + \frac{2}{6x-1} \right)\\  &= \frac{1}{9}\left( \frac{5}{2} + \frac{6x-1}{2} + \frac{2}{6x-1} \right)\\  &\geq \frac{1}{9}\left(\frac{5}{2} + 2 \right)\\  &= \frac{1}{9}\left(\frac{9}{2}\right)\\  &= \frac{1}{2}\quad\quad \quad \quad \quad (1)  \end{aligned}

In the above steps we assume 6x-1 > 0, otherwise y is non-positive for x > 0. Hence y \leq 2 with equality when \frac{6x-1}{2} = 1 or x = 1/2.

Another elementary way that applies to quotients of quadratic polynomials is to re-write the expression as a quadratic in x:

\begin{aligned}  y &= \frac{8}{2x+1} - \frac{1}{x}\\  x(2x+1)y &= 8x - (2x+1)\\  2x^2y + (y-6)x + 1 &= 0.\quad \quad \quad \quad \quad \quad (2)  \end{aligned}

For fixed y this quadratic equation will have 0, 1 or 2 solutions in x depending on whether its discriminant is negative, zero, or positive respectively. At any maximum or minimum value y_0 of the function, the discriminant will be zero since on one side of y_0 the quadratic equation will have a solution (discriminant non-negative) while on the other it will not (discriminant negative). In the image below a maximum is reached at y_0 = 2 while it is of opposite signs either side of this.


Hence setting the discriminant of the left side of (2) to 0, (y-6)^2 - 4(2y) = 0 from which y^2 - 20y + 36 = (y-18)(y-2) = 0. Hence extrema are at y=2 and y = 18. We can solve (y-18)(y-2) \geq 0 to find that y \leq 2 or y \geq 18 is the range of the function y = \frac{8}{2x+1} - \frac{1}{x}. This tells us that y = 2 is a local maximum (illustrated above) and y = 18 is a local minimum (occurring when x < 0).


One advantage of this method is that unlike elementary calculus, one bypasses the step of finding the corresponding x value (i.e. by solving dy/dx = 0) before substituting this into the function to find the extremum value for y.

Another advantage is that the equation need not be polynomial in y. For example below is a plot of x^2 + 4x\sin y + 1 = 0. Using the above-mentioned discriminant trick we solve 16 \sin^2 y - 4 \geq 0 and find the range of the function is when \displaystyle \sin^2 y \geq 1/4, or y \in \bigcup_{k \in \mathbb{Z}} [k\pi + \pi/6, k\pi + 5\pi/6] . Below is a plot confirming this using WolframAlpha.


The reader is encouraged to try out other examples, for example this method should work for any equation of the form \displaystyle h(y) = \frac{ax^2 + bx + c}{dx^2 + ex + f} where h is a continuous function of y. Of course one should also take care in noting when the function is defined before cross-multiplying.

Further Reading

[1] Finding the range of rational functions – Mathematics Stack Exchange

[2] Find Range of Rational Functions –

[3] Critical point (mathematics): Use of the discriminant – Wikipedia, the free encyclopedia

July 30, 2015

Nineteenth century non-avian dinosaur discoveries

Filed under: nature,science — ckrao @ 11:05 am

Below is an attempted chronological list of non-avian dinosaur discoveries of the 19th century that today are considered valid genera. There may still be some where there are only scant remains of the fossil (e.g. a tooth or single bone remain). The list came from [1] with some help from [2] and Wikipedia to filter out doubtful names. Many of the best known dinosaurs are listed here and it looks like most of the major groups are covered. Good histories of dinosaur paleontology are in [3] and [4].


Genus Discoverer Year Dinosaur type
Megalosaurus Buckland 1824 tetanuran (stiff-tailed) theropod
Iguanodon Mantell 1825 beaked ornithopod
Streptospondylus von Meyer 1830 megalosaurid
Hylaeosaurus Mantell 1833 armoured
Thecodontosaurus Riley & Stutchbury 1836 prosauropod
Plateosaurus von Meyer 1837 prosauropod
Poekilopleuron Eudes-Deslongchamps 1838 megalosaurid
Cardiodon Owen 1841 sauropod
Cetiosaurus Owen 1841 sauropod
Pelorosaurus Mantell 1850 brachiosaur
Aepisaurus Gervais 1852 sauropod
Oplosaurus Gervais 1852 sauropod
Massospondylus Owen 1854 prosauropod
Nuthetes Owen 1854 maniraptoran
Troodon Leidy 1856 raptor
Stenopelix von Meyer 1857 pachycephalosaur
Astrodon Johnston 1858 sauropod
Hadrosaurus Leidy 1858 duckbilled ornithopod
Compsognathus J. A. Wagner 1859 coelurosaur (fuzzy theropod)
Scelidosaurus Owen 1859 armoured
Echinodon Owen 1861 heterodontosaurid (early bird-hipped dinosaur)
Polacanthus Owen vide [Anonymous] 1865 armoured
Calamospondylus Fox 1866 oviraptorosaur
Euskelosaurus Huxley 1866 prosauropod
Acanthopholis Huxley 1867 armoured
Hypselosaurus Matheron 1869 sauropod
Hypsilophodon Huxley 1869 beaked ornithopod
Rhabdodon Matheron 1869 beaked ornithopod
Ornithopsis Seeley 1870 sauropod
Struthiosaurus Bunzel 1870 armoured
Craterosaurus Seeley 1874 stegosaurian
Chondrosteosaurus Owen 1876 sauropod
Macrurosaurus Seeley 1876 sauropod
Allosaurus Marsh 1877 carnosaur
Apatosaurus Marsh 1877 sauropod
Camarasaurus Cope 1877 sauropod
Dryptosaurus Marsh 1877 tyrannosaur
Dystrophaeus Cope 1877 sauropod
Nanosaurus Marsh 1877 early bird-hipped dinosaur
Stegosaurus Marsh 1877 plated dinosaur
Diplodocus Marsh 1878 sauropod
Brontosaurus Marsh 1879 sauropod
Anoplosaurus Seeley 1879 armoured
Coelurus Marsh 1879 coelurosaur (fuzzy theropod)
Mochlodon Seeley 1881 beaked ornithopod
Craspedodon Dollo 1883 horned dinosaur
Ceratosaurus Marsh 1884 theropod
Anchisaurus Marsh 1885 prosauropod
Camptosaurus Marsh 1885 beaked ornithopod
Aristosuchus Seeley 1887 coelurosaur (fuzzy theropod)
Ornithodesmus Seeley 1887 raptor
Cumnoria Seeley 1888 beaked ornithopod
Priconodon Marsh 1888 armoured
Coelophysis Cope 1889 early theropod
Nodosaurus Marsh 1889 armoured
Triceratops Marsh 1889 horned dinosaur
Barosaurus Marsh 1890 sauropod
Claosaurus Marsh 1890 duckbilled ornithopod
Ornithomimus Marsh 1890 ostrich dinosaur
Ammosaurus Marsh 1891 prosauropod
Torosaurus Marsh 1891 horned dinosaur
Argyrosaurus Lydekker 1893 sauropod
Sarcolestes Lydekker 1893 armoured
Dryosaurus Marsh 1894 beaked ornithopod


[1] Dinosaur Genera List –

[2] Genus List for Holtz (2007) Dinosaurs

[3] Benton, M. J. 2000. A brief history of dinosaur paleontology. Pp. 10-44, in Paul, G. S. (ed.), The Scientific American book of dinosaurs. St Martin’s Press, New York. –

[4] Equatorial Minnesota The generic history of dinosaur paleontology 1699 to 1869 –

July 25, 2015

Compositions of positive integers where adjacent parts have different parity

Filed under: mathematics — ckrao @ 5:39 am

Here is an interesting problem based on one I found in the 2008 AIME (Q11).

How many compositions of the positive integer n have adjacent parts with different parity?

For example, a valid composition of n=14 is

14 = 1 + 2 + 5 + 2 + 1 + 2 + 1,

since the 1,2,5,2,1,2,1 alternate between being odd and even.

The first terms of the sequence are (OEIS A062200)

1, 1, 3, 2, 6, 6, 11, 16, 22, 37, 49, 80, …

Interestingly the sequence decreases between n=3 and 4 since the only valid compositions of 4 are 4 = 1 + 2 + 1 and 4, while 3 = 1 + 1 + 1 = 2 + 1 = 1 + 2.

Here is the first approach I came up with to solve the problem. Let a_n be the number of such compositions starting with an even number and let b_n be the number of compositions starting with an odd number. We wish to find t_n := a_n + b_n. We identify four cases:

  • If n is odd and the composition starts with an even number 2c, we remove that even number and have a composition of n-2c starting with an odd number. Summing over choices of c,
    a_{2k+1} = b_{2k-1} + b_{2k-3} + \ldots + b_1. \quad (1)
  • If n is odd and the composition starts with an odd number 2c+1, we either have the single term composition n, or we remove that odd number and have a composition of n-(2c+1) starting with an even number. Summing over choices of c,
    b_{2k+1} = 1 + a_{2k} + a_{2k-2} + \ldots + a_2. \quad (2)
  • If n is even  and the composition starts with an even number 2c, we either have the single term composition n, or we remove that even number and have a composition of n-2c starting with an odd number. Summing over choices of c,
    a_{2k} = 1 + b_{2k-2} + b_{2k-4} + \ldots + b_2. \quad (3)
  • If n is even and the composition starts with an odd number 2c+1, we remove that odd number and have a composition of n-(2c+1) starting with an even number. Summing over choices of c,
    b_{2k} = a_{2k-1} + a_{2k-3} + \ldots + a_1.\quad (4)

Recursions (1)-(4) with the initial conditions a_1 = 0, b_1 = 1, a_2 = 1, b_2 = 0 allows us to generate the following table.

n 1 2 3 4 5 6 7 8 9 10 11 12
a_n 0 1 1 1 3 2 6 6 11 16 22 37
b_n 1 0 2 1 3 4 5 10 11 21 27 43
t_n 1 1 3 2 6 6 11 16 22 37 49 80

After writing out some terms I noticed the pattern

\displaystyle a_n = a_{n-2} + b_{n-2} \text{ and } b_n = a_{n-1} + b_{n-2}.\quad \quad (5)

I later saw why this is so. Firstly, a composition of n that starts with an even number can be obtained from any composition of n-2 by adding 2 to the first term if even or inserting 2 at the start if odd. Hence a_n = t_{n-2} = a_{n-2} + b_{n-2}. For the second relation, if the composition of n starts with an odd number, it either is 1 plus an even-starting composition of n-1 or it can be obtained from an odd-starting composition of n-2 by adding 2 to the first term. Hence b_n = a_{n-1} + b_{n-2}.

 From (5) and using t_n = a_n + b_n we find that

\begin{aligned}  t_n &= a_n + b_n\\  &= t_{n-2} + (a_{n-1} + b_{n-2})\\  &= t_{n-2} + t_{n-3} + t_{n-2} - a{n-2}\\  &= 2t_{n-2} + t_{n-3} - t_{n-4}.\quad \quad (6)  \end{aligned}

Together with the initial terms t_1 = 1, t_2 = 1, t_3 = 3, t_4 = 2, (6) enables us to generate our desired terms without requiring a_n or b_n.

The OEIS page also gives the following expression for t_n:

\displaystyle t_n = \sum_{k=0}^{n+1} \binom{n-k+1}{3k-n+1}.\quad \quad (7)

Here is how to show this result, with help from the solution at AoPS here. For ease of exposition convert the problem to one of binary strings: we require the number of length-n 0-1 strings so that maximal continguous blocks of 1s have even length and maximal contiguous blocks of 0s have odd length.

Suppose there are b maximal blocks of 0s. All but the last such block must necessarily be followed by a pair of 1s. That is, the string has the form

\displaystyle [....][....0][11....][....0][11....] \ldots [....0][....],

where blocks enclosed by [] contain the same digit (0 or 1) and 0 or more pairs of digits are in place of the sets of four dots. We have shown 2b+1 blocks above but the first and last may be empty.

Let k = (n-b)/2, which is the number of pairs of digits to use after b 0s are used (noting that n-b must be even). We have b 0s used above along with b-1 pairs of 1s. Therefore (n-(b + 2(b-1)))/2 pairs of 0s or 1s need to inserted into the 2b+1 blocks above. The blocks that they end up determine whether they are 0 or 1.

Now x pairs can be inserted into y blocks in \binom{x+y-1}{x} ways, since we write out x+y-1 symbols in a row and choose x of them to be pairs, with the remaining symbols representing separators between blocks. In our case,

x = \frac{n-(b + 2(b-1))}{2}, y = 2b+1 and b = n-2k, so

\begin{aligned}  x &= \frac{n-3b+2}{2}\\  &= \frac{n-3n+6k+2}{2}\\  &= 3k-n+1,\end{aligned}

and x+y-1 = 3k-n+1 + 2(n-2k) + 1 = n-k+1.

Summing over choices of k gives (7) as desired. Note that some of the terms will be 0 as k ranges from 0 to n+1.

For example, when n=11 and 12, we have

\displaystyle t_{11} = \binom{8}{2} + \binom{7}{5} = 28 + 21 = 49,

\displaystyle t_{12} = \binom{9}{1} + \binom{8}{4} + \binom{7}{7} = 9 + 70 + 1= 80,

matching the values given in the table above.


June 30, 2015

The Pakistan heat wave of 2015

Filed under: climate and weather — ckrao @ 11:39 am

This month a heatwave has killed over 1500 people in Pakistan, most of them in Karachi. Here are the temperatures recorded at Karachi airport at that time: note that the average for this time of year is 28-35°C with high humidity. Rarely does the temperature reach 39.5°C three days in a row but here it did so six days in a row plus there seemed to be no relief at night.

Date 16-Jun 17-Jun 18-Jun 19-Jun 20-Jun 21-Jun 22-Jun 23-Jun 24-Jun
Min (°C) 30 29.3 29.5 31 31.6 33 33 33 30.5
Max (°C) 36 38.5 39.5 40.5 44.8 42.5 42.5 41.2 37

Most of the dead were homeless and it was also during the time of Ramadan where fasting is observed during daylight hours. This and the recent heatwave in India are two of the three most fatal on the Indian subcontinent in recent times.


[1] Temperatures: [2] Average Weather For Karachi, Pakistan – WeatherSpark [3] Dr. Jeff Masters’ WunderBlog: Pakistan’s Deadliest Heat Wave on Record Kills at Least 800: Weather Underground [4] News: Majority of Karachi heatwave victims were homeless minister – The Express Tribune

June 22, 2015

Places we see powers of 2

Filed under: mathematics — ckrao @ 11:44 am

Here are some obvious and less obvious places where we encounter the sequence f(n) = 1,2,4,8,16,.... Most of these come from the corresponding OEIS page.

  • as a solution to the recurrence f(n+1) = 2f(n), f(0) = 1.
  • the number of subsets of a n-element set: To each element there are two choices – to include it or not include it in the subset. Hence there are 2\times 2\times \ldots \times 2 = 2^n subsets.
  • the sum of the numbers in the n’th row of Pascal’s triangle This is a consequence of the binomial theorem applied to (1+1)^n = \sum_{i=0}^n \binom{n}{i}.
  • the sum of the coefficients of (x+1)^n, which is the sum of binomial coefficients
  • the number of even or odd-sized subsets of an (n+1)-element set, i.e. \sum_{i=0}^{\lfloor{(n+1)/2}\rfloor} \binom{n+1}{2i} = \sum_{i=0}^{\lfloor{n/2}\rfloor} \binom{n+1}{2i+1} = 2^n, a consequence of the binomial theorem (1-1)^n = \sum_{i=0}^n (-1)^i \binom{n}{i}.
  • each term in the sequence is the smallest natural number that is not the sum of any number of distinct earlier terms, an example of a sum-free sequence
  • the sequence forms the set of every natural number that is not the sum of 2 or more consecutive positive integers: Such a sum has the form a + (a+1) + \ldots + (a+ k) = (2a + k)(k+1)/2. If this were a power of 2 then (2a+k) and (k+1) must both be powers of 2 greater than or equal to 2. But their difference 2a-1 is odd, which is impossible for the difference of even powers of 2. Conversely if a natural number is not a power of two it is of the form 2^m d where d is an odd number at least 3. Then 2^m d may be written as the sum of either d consecutive positive integers with middle (mean) term 2^m (if (d+1)/2 \leq 2^m) or 2^{m+1} consecutive positive integers with mean d/2 (if 2^m < d/2). For example, 48 = 2^4\times 3 and 48 = 15 + 16 + 17 while 44 = 2^2\times 11 and 44 = 2+3+4+5+6+7+8+9.
  • the number of ordered partitions of n+1 is 2^n: for example, the eight ordered partitions of 4 are: 1+1+1+1, 2+1+1, 1+2+1, 1+1+2, 2+2, 3+1, 1+3, 4 To see why this result is true, imagine n+1 items in a row. Imagine separating them with dividers so that the number of items between each divider gives us the ordered partition. For example with 4 items we may place a divider between the third and fourth item to give the partition 3+1. We have n possible locations for dividers (between any adjacent items) and each location has a choice of being assigned to a divider or not. This leads to 2^n possible partitions. Alternatively, an ordered partition of n+2 may be derived inductively from an ordered partition of n+1 = a_1 + \ldots + a_k by either forming n+2 = 1 + a_1 + \ldots + a_k or n+2 = (a_1+1) + a_2 + \ldots + a_k. Also any ordered partition of n+2 has one of these forms as we may form from it an ordered partition of n+1 by either subtracting 1 from the first element of the partition (if it is greater than 1) or deleting the first element if it is 1. Hence the number of partitions doubles as n+1 is increased by 1 and for n = 1 we have the single partition.
  • the number of permutations of {1,2,...,n+1} with exactly one local maximum: for example, for n=3 we have the permutations 1234, 1243, 1342, 2341, 1432, 2431, 3421, 4321. Here we have the highest element n+1 somewhere and there are 2^n possible subsets of {1,2,...,n} that we may choose to precede it. These elements must be in ascending order, while the elements after n+1 are in descending order to ensure exactly one local maximum at n+1. Hence the order of terms is determined once the subset is chosen and we have 2^n permutations.
  • the number of permutations of n+1 elements where no element is more than one position to the right of its original place: for example, for n=3 the allowed permutations are 1234, 1243, 1324, 1423, 2134, 2143, 3124, 4123. This can be seen inductively: if a_1a_2\ldots a_{n+1} is a valid permutation of n+1 elements, then 1(1+a_1)(1+a_2)\ldots (1+a_{n+1}) and (1+a_1)1(1+a_2)\ldots (1+a_{n+1}) are valid permuations of n+2 elements. Also any valid (n+2) element permutation is of this form as it may be reduced to a valid (n+1) element permutation by deleting the 1 (which must occur in either the first or second position) and subtracting 1 from every other element. This describes a one to two mapping between valid permutations of n+1 and n+2 elements. Finally for 1 element there is 1 = 2^0 allowed permutation, so the result follows by induction.

May 31, 2015

Test pace bowlers to have conceded 100 runs in an innings the most

Filed under: cricket,sport — ckrao @ 10:16 am


Below is a list of pace bowlers who have conceded 100 or more runs in an innings at least 14 times (* indicates active players).

Name Tests Innings Bowled Wickets times conceded 100 runs innings per century
Botham 102 168 383 31 5.42
Kapil Dev 131 227 434 25 9.08
Ntini 101 190 390 23 8.26
Hadlee 86 150 431 21 7.14
Caddick 62 105 234 20 5.25
Imran Khan 88 142 362 20 7.1
M Johnson* 64 123 283 20 6.15
Vaas 111 194 355 20 9.7
Lillee 70 132 355 19 6.95
McDermott 71 124 291 19 6.53
I Sharma* 61 107 187 18 5.94
Anderson* 104 194 401 17 11.41
Gough 58 95 229 17 5.59
Srinath 67 121 236 17 7.12
F Edwards 55 97 165 16 6.06
Lawson 46 78 180 16 4.88
Mohammad Sami 36 66 85 16 4.13
Sarfraz Nawaz 55 95 177 16 5.94
Hoggard 67 122 248 15 8.13
Lee 76 150 310 15 10
Martin 71 126 233 15 8.4
Morrison 48 76 160 15 5.07
Steyn* 78 147 396 15 9.8
Bedser 51 92 236 14 6.57
Z Khan 92 165 311 14 11.79


Here are the numbers for remaining pace bowlers with at least 200 test wickets (Ray Lindwall a stand-out here):

Name Tests Innings Bowled Wickets times conceded 100 runs innings per century
Waqar Younis 87 154 373 13 11.85
Harmison 63 115 226 13 8.85
Wasim Akram 104 181 414 12 15.08
McKenzie 60 113 246 12 9.42
Cairns 62 104 218 12 8.67
Walsh 132 242 519 11 22
Sobers (not always pace) 93 159 235 11 14.45
Snow 49 93 202 11 8.45
Broad 79 143 285 11 13
Donald 72 129 330 10 12.9
Pollock 108 202 421 10 20.2
Willis 90 165 325 10 16.5
Hughes 53 97 212 10 9.7
Thomson 51 90 200 10 9
McGrath 124 243 563 9 27
Trueman 67 127 307 9 14.11
Statham 70 129 252 8 16.13
Flintoff 79 137 226 8 17.13
Streak 65 102 216 8 12.75
Roberts 47 90 202 8 11.25
Morkel 62 117 217 7 16.71
Gillespie 71 137 259 6 22.83
Marshall 81 151 376 5 30.2
Holding 60 113 249 5 22.6
Garner 58 111 259 4 27.75
Kallis 166 272 292 4 68
Lindwall 61 113 228 1 113

The numbers are generally higher for spin bowlers – the top six centurions overall are all spin bowlers:

Muralitharan (61), Kumble (57), Harbhajan Singh (43), Warne (40), Kaneria (39), Vettori (34).


[1] Bowling records:  Test matches:  Cricinfo Statsguru – ESPN Cricinfo

[2] Records:  Test matches:  Bowling records:  Most wickets in career – ESPN Cricinfo

May 24, 2015

A collection of proofs of Ptolemy’s Theorem

Filed under: mathematics — ckrao @ 11:39 am

Here we collect some proofs of the following nice geometric result.

If ABCD is a quadrilateral, then AB.CD + BC.DA \geq AC.BD with equality if ABCD is cyclic.


In words, the sum of the product of the lengths of the opposite sides of a quadrilateral is at least the product of the lengths of its diagonals.

The equality for a cyclic quadrilateral is known as Ptolemy’s theorem while the more general inequality applying to any quadrilateral is called Ptolemy’s inequality. Many of the proofs below that establish Ptolemy’s theorem can be modified slightly to prove the inequality.

Proofs of Ptolemy’s Theorem

Proof 1:

Choose point P on line CD extended beyond D as shown so that \angle DAP = \angle BAC.



a) \triangle ABC \sim \triangle ADP (\angle ABC = \angle ADP and \angle BAC = \angle DAP).

b) \triangle BAD \sim \triangle CAP (\angle BAD = \angle CAP and \angle ADB = \angle APC).

From a) BC.DA = BA.DP so AB.CD + BC.DA = AB(CD + DP) = AB.CP = AC.BD by b) and we are done.

Proof 2: (a slightly different choice of constructed point)

Let K be the point on AC so that \angle CBK = \angle ABD.



a) \triangle ABD \sim \triangle KBC (\angle ABD = \angle KBC and \angle BDA = \angle BCK) so KC = AD.BC/BD.

b) \triangle ABK \sim \triangle DBC (\angle ABK = \angle DBC and \angle KAB = \angle CDB) so AK = DC.AB/DB.

Adding the two gives AC = (AD.BC + DC.AB)/BD or AC.BD = AD.BC + DC.AB as required.

Proof 3: (ref:

Let the diagonals AC, BD intersect at P and construct E on the circumcircle so that CE is parallel to BD. A short angle chase shows that BCED is an isosceles trapezium.


Then \sin \angle ABE = \sin \angle ACE = \sin \angle BPC (angles on common arc AE and using BD \parallel CE). Also triangles BDE and BDC have the same base and height and hence the same area.

Hence the area of ABCD is the sum of the areas of triangles ABE and ADE, or

\begin{aligned}  & \frac{1}{2} \left( AB.BE\sin \angle ABE + AD.DE\sin \angle ADE \right)\\  &= \frac{1}{2} \left( AB.BE\sin \angle ABE + AD.DE\sin \angle ABE \right) \quad \text{as } \angle ABE \text{ and } \angle ADE \text{ are supplementary }\\  &= \frac{1}{2} \left( AB.BE+ AD.DE \right)\sin \angle ABE\\  &= \frac{1}{2} \left(AB.CD + AD.BC \right)\sin \angle ABE, \quad (1)  \end{aligned}

where the last step follows from BDEC being an isosceles trapezium.

But also this area is \frac{1}{2} AC.BD\sin \angle BPC and recall from above that \sin \angle BPC = \sin \angle ABE. Therefore equating this with (1) we end up with AB.CD + AD.BC = AC.BD.

Proof 4 (ref: [1])

Here three of the triangles of the cyclic quadrilateral are scaled to fit together to form a parallelogram. Namely,

  • \triangle ABD is scaled by AC,
  • \triangle ABC is scaled by AD,
  • \triangle ACD is scaled by AB.

A simple angle chase based on the coloured angles shown reveals that a parallelogram is formed when the triangles are joined together. Ptolemy’s theorem then follows from equating two of its opposite side lengths.


Proof 5: (from here)

Using the cosine rule in triangles ABC and ADC respectively gives

AC^2 = AB^2 + BC^2 - 2AB.BC\cos \angle ABC = AD^2 + CD^2 - 2AD.CD\cos \angle ADC.

Since \angle ABC + \angle ADC = 180^{\circ}, this becomes

\begin{aligned}  AB^2 + BC^2 - 2AB.BC\cos \angle ABC &= AD^2 + CD^2 + 2AD.CD\cos \angle ABC\\  \Rightarrow \cos \angle ABC &= \frac{AB^2 + BC^2 - AD^2 - CD^2}{2(AB.BC + AD.CD)}.  \end{aligned}


\begin{aligned}  AC^2 &= AB^2 + BC^2 - 2AB.BC\cos \angle ABC\\  &= AB^2 + BC^2 - 2AB.BC \frac{AB^2 + BC^2 - AD^2 - CD^2}{2(AB.BC + AD.CD)}\\  &= \frac{(AB^2+BC^2)(AB.BC + AD.CD) - AB.BC(AB^2 + BC^2 - AD^2 - CD^2)}{AB.BC + AD.CD}\\  &= \frac{(AB^2+BC^2)(AD.CD) + AB.BC(AD^2 + CD^2)}{AB.BC + AD.CD}\\  &= \frac{AB.AD(AB.CD+BC.AD) + BC.CD(BC.AD+AB.CD)}{AB.BC + AD.CD}\\  &= \frac{(AB.AD+BC.CD)(AB.CD+BC.AD)}{AB.BC + AD.CD}.\\  \end{aligned}

By a similar argument we may write

\displaystyle BD^2 = \frac{(AB.BC + AD.CD)(BC.AD+AB.CD)}{BC.CD + AB.AD}.

Multiplying these last two equations by each other and taking the square root gives

\displaystyle AC.BD = AB.CD + BC.AD,

which is Ptolemy’s theorem.

Proof 6 (via this):

Let the angles subtended by AB, BC, CD respectively be \alpha, \beta, \gamma and let the circumradius of the quadrilateral be R. By the sine rule AB = 2R \sin \alpha, BC = 2R\sin \beta, CD = 2R \sin \gamma, AD = 2R\sin (\alpha + \beta + \gamma ), AC = 2R \sin (\alpha + \beta), BD = 2R \sin (\beta + \gamma ). Then the equality to be proved is equivalent to

\displaystyle \sin (\alpha + \beta)\sin (\beta + \gamma) = \sin \alpha \sin \gamma + \sin \beta \sin (\alpha + \beta + \gamma).

But we have

\begin{aligned}  & \sin (\alpha + \beta)\sin (\beta + \gamma)\\ &= (\sin \alpha \cos \beta + \cos \alpha \sin \beta)(\sin \beta \cos \gamma + \cos \beta \sin \gamma)\\  &= \sin \alpha \sin \beta \cos \beta \cos \gamma + \sin \alpha \cos^2 \beta \sin \gamma + \cos \alpha \sin^2 \beta \cos \gamma + \cos \alpha \sin \beta \cos \beta \sin \gamma\\  &= \sin \alpha \sin \beta \cos \beta \cos \gamma + \sin \alpha (1- \sin^2 \beta ) \sin \gamma + \cos \alpha \sin^2 \beta \cos \gamma + \cos \alpha \sin \beta \cos \beta \sin \gamma\\  &= \sin \alpha \sin \gamma + \sin \beta (\sin \alpha \cos \beta \cos \gamma - \sin \alpha \sin \beta \sin \gamma + \cos \alpha \sin \beta \cos \gamma + \cos \alpha \cos \beta \sin \gamma )\\  &= \sin \alpha \sin \gamma + \sin \beta (\sin(\alpha + \beta ) \cos \gamma + \cos (\alpha + \beta ) \sin \gamma )\\  &= \sin \alpha \sin \gamma + \sin \beta \sin (\alpha + \beta + \gamma ),  \end{aligned}

as required.

Proofs of Ptolemy’s Inequality (all make use of the triangle inequality)

Proof 7:

Denote the points A, B, C, D by vectors or complex numbers a, b, c, d. Then we have the equality

\displaystyle (a-b).(c-d) + (a-d).(b-c) = (a-c).(b-d).

Applying the modulus (length) to both sides and then the triangle inequality leads to

\displaystyle |(a-b).(c-d)| + |(a-d).(b-c)| \geq |(a-c).(b-d)|,

which is Ptolemy’s inequality.

Proof 8: (ref: [2])

Place the origin at the point D so that A, B, C are represented by the vectors a, b, c respectively. Let a' = a/|a|^2, b' = b/|b|^2, c' = c/|c|^2. Then

\begin{aligned} |a'-b'|^2 &= \left| \frac{a}{|a|^2} - \frac{b}{|b|^2}\right|^2\\  &= \frac{|a|^2}{|a|^4} + \frac{|b|^2}{|b|^4} - 2\frac{\langle a, b \rangle}{|a|^2|b|^2}\\  &= \frac{|b|^2 + |a|^2 - 2\langle a, b \rangle}{|a|^2|b|^2}\\  &= \frac{|a-b|^2}{|a|^2|b|^2}.  \end{aligned}

Similarly, |b'-c'| = \frac{|b-c|}{|b||c|} and |c'-a'| = \frac{|c-a|}{|c||a|}. By the triangle inequality, \displaystyle |a'-b'| \leq |b'-c'| + |c'-a'|, or in other words,

\begin{aligned}  \frac{|a-b|}{|a||b|} &\leq \frac{|b-c|}{|b||c|} + \frac{|c-a|}{|c||a|}\\  \Rightarrow |a-b||c| \leq |b-c||a| + |c-a||b|,  \end{aligned}

which is another way of writing Ptolemy’s inequality.

Proof 9: (inversion)

Recall that under inversion under a point P a circle passing through P maps to a line. If X, Y are points on the circle mapping to X, Y' respectively under inversion in a circle of radius R centred at P, then

\displaystyle X'Y' = R^2.XY/(PX.PY).


To see this, since XP.X'P = YP.Y'P = R^2 by the definition of the inverse, XP/YP = Y'P/X'P. This together with the fact that angle P is common implies that triangles XPY and Y'PX' are similar. This gives us the relationship X'Y' = YX.PY'/PX = XY.R^2/(PX.PY) as desired.


For our quadrilateral ABCD apply an inversion centred at D with radius R. Then A, B, C map to points A', B', C' which are collinear if ABCD is cyclic. Using the result above A'B' = AB.R^2/(DA.DB) and similarly A'C' = AC.R^2/(DA.DC) and B'C' = BC.R^2/(DB.DC). By the triangle inequality A'C' \leq AB' + BC' and so this becomes

\begin{aligned}  \frac{AC.R^2}{DA.DC} &\leq \frac{AB.R^2}{DA.DB} + \frac{BC.R^2}{DB.DC}\\  \Rightarrow AC.DB &\leq AB.DC + BC.DA,  \end{aligned}

which is Ptolemy’s inequality.

Proof 10: (ref: [3])

Construct E on AC so that EC = BC, then draw F on CD with EF \parallel AD. Finally rotate \triangle FEC about C to map to \triangle F'BC.


Then F'B = FE and by similarity of triangles ACD and ECF, FE/DA = CE/CA = CB/CA so that

(a) F'B = FE = AD.BC/CA

Secondly, \triangle DF'C \sim ABC as F'C/BC = FC/EC = DC/AC and \angle F'CD = \angle BCA. This gives

(b) DF' = AB.CD/AC

Adding (a) and (b) and applying the triangle inequality,

\displaystyle BD \leq BF' + F'D = (AD.BC + AB.CD)/AC,

from which

\displaystyle AC.BD \leq AD.BC + AB.CD

which is Ptolemy’s inequality.


[1] A. Bogomolny, Ptolemy Theorem – Proof Without Words from Interactive Mathematics Miscellany and Puzzles, Accessed 30 May 2015

[2] W.H. Greub, Linear Algebra, Springer Science & Business Media, 2012 (p 190).

[3] C. Alsina, R. B. Nelsen, When Less is More: Visualizing Basic Inequalities, MAA, 2009.

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