Kinetic energy is that associated with motion and is defined as for a particle with mass , velocity and momentum . If the mass is a fluid in motion (e.g. wind) with density and volume through cross-sectional area , then .

Work is the result of a force applied over a displacement and is given by the line integral

This has the simple form when force is constant and displacement is linear where is the angle between the force and displacement vectors.

Using Newton’s 2nd law and the relation this can be written as

This is the work-energy theorem which says that work is the change in kinetic energy by a net force. It can also be written as where is momentum.

The above has the rotational analogue where is moment of inertia and is angular velocity and the equation for work becomes

,

where is a torque vector.

This has the simple form in the special case of a constant magnitude tangential force where is the torque resulting from force applied at distance from the centre of rotation.

Note that the time derivative of work is defined as power, so work can also be expressed as the time integral of power:

If the work done by a force field depends only on a particle’s end points and not on its trajectory (i.e. conservative forces), one may define a potential function of position, known as potential energy satisfying . By convention positive work is a reduction in potential, hence the minus sign. It then follows that in such force fields the sum of kinetic and potential energy is conserved.

Some types of potential energy:

- due to a gravitational field: , where are the masses of two bodies, the distance between their centre of masses and is Newton’s gravitation constant.
- due to earth’s gravity at the surface: where and is the object’s height above ground (small compared with the size of the earth).
- due to a spring obeying Hooke’s law: where is the spring constant and the displacement from an equilibrium position.
- due to an electrostatic field: where is Coulomb’s constant and are charges. This can be written as where is a potential function measured in volts.
- for a system of point charges: .
- for a system of conductors: where the charge on conductor is and its potential is .
- for a charged dielectric: the above may be generalised to the volume integral where is charge density and is the potential corresponding to the electric field.
- for an electric dipole in an electric field: where is directed from the negative to positive charge and has magnitude equal to the product of the positive charge and charge separation distance.
- for a current loop in a magnetic field: where is directed normal to the loop and has magnitude equal to the product of the current through the loop and its area.

In electric circuits the voltage drop across an inductance is and the current though a capacitance is . These inserted into the relationship lead to the formulas and for the energy stored in a capacitor and inductor respectively.

Also in electromagnetism the energy flux (flow per unit area per unit time) is the Poynting vector , the cross product of the electric and magnetising field vectors. The electromagnetic energy in a volume is given by ([1])

,

where is the electric displacement field and is the magnetic field. This is more commonly written as when the relationships hold.

In special relativity energy is the time component of the momentum 4-vector. That is, energy and momentum are mixed in a similar way to how space and time are mixed at high velocities. Computing the norm of the momentum four-vector gives the energy-momentum relation

.

This leads to for massless particles (such as photons) and more generally , the mass-energy equivalence relation (here and is rest mass).

In quantum mechanics the energy of a photon is also written as (Planck-Einstein relation) where is Planck’s constant and are frequency and wavelength respectively. Energies of quantum systems are based on the eigenstates of the Hamiltonian operator, an example of which is .

Force is also equal to pressure times area, so another formula for work (e.g. done by an expanding gas) is the volume integral . In thermodynamics heat is energy transferred through the random motion of particles. The fundamental equation of thermodynamics quantifies the internal energy which disregards kinetic or potential energy of a system as a whole (only considering microscopic kinetic and potential energy):

where is temperature, is entropy, is the number of particles and the chemical potential of species . Similar formulas exist for other thermodynamic potentials such as Gibbs energy, enthalpy and Helmholtz energy.

The mean translational kinetic energy of a bulk substance is related to its temperature by where is Boltzmann’s constant.

In thermal transfer the change in internal energy is given by where is mass and is the heat capacity which may apply to constant volume or constant pressure.

The power per unit area emitted by a body is given by the Stefan-Boltzmann law where is the emissivity (=1 for black body radiation) and is the Stefan–Boltzmann constant. This equation may be used to determine the energy emitted by stars using their emission spectrum.

The latent heat (thermal energy change during a phase transition) of mass of a substance with specific latent heat constant is given by .

Finally, the energy of a single wavelength of a mechanical wave is where the mass of a wavelength, the amplitude and the angular frequency [2]. This can be applied to finding the energy density of ocean waves for example [3].

[1] Poynting Vector. *Brilliant.org*. Retrieved 22:24, December 28, 2018, from https://brilliant.org/wiki/poynting-vector/

[2] Power of a Wave. *Brilliant.org*. Retrieved 21:23, December 30, 2018, from https://brilliant.org/wiki/power-of-a-wave/

[3] Wikipedia contributors, “Wave power,” *Wikipedia, The Free Encyclopedia,* https://en.wikipedia.org/w/index.php?title=Wave_power&oldid=875183814 (accessed December 30, 2018).

[4] Wikipedia contributors, “Work (physics),” *Wikipedia, The Free Encyclopedia,* https://en.wikipedia.org/w/index.php?title=Work_(physics)&oldid=874162163 (accessed December 30, 2018).

[5] Wikipedia contributors, “Potential energy,” *Wikipedia, The Free Encyclopedia,* https://en.wikipedia.org/w/index.php?title=Potential_energy&oldid=873393028 (accessed December 30, 2018).

[6] Wikipedia contributors, “Electric potential energy,” *Wikipedia, The Free Encyclopedia,* https://en.wikipedia.org/w/index.php?title=Electric_potential_energy&oldid=868852409 (accessed December 30, 2018).

[7] Wikipedia contributors, “Thermodynamic equations,” *Wikipedia, The Free Encyclopedia,* https://en.wikipedia.org/w/index.php?title=Thermodynamic_equations&oldid=865388931 (accessed December 30, 2018).

[8] H. Ohanian, *Physics*, 2nd edition, Norton & Company, 1989.

Win-loss breakdown by tournament (Masters tournament finals changed to best of 3 sets from 2007):

- French Open: 86-2
- Davis Cup: 18-0
- Barcelona Open: 2-0
- Monte Carlo Masters: 2-0
- Rome Masters: 2-0
- Stuttgart: 1-0

Win-loss breakdown by number of sets (overall he has won 331 and lost 36 completed sets so even winning a set against him is a big deal!):

- 5 sets: 4-0 (Coria, Federer, Isner, Djokovic – 5th set scores 7-6 (6), 7-6 (5), 6-4, 9-7 respectively)
- 4 sets: 22-1 (loss to Söderling)
- 3 sets: 83-1

Most common opponents (2 or more matches):

- Djokovic: 7-1 (lost 7 sets)
- Federer: 7-0 (lost 7 sets)
- Ferrer: 4-0 (lost 1 set)
- Almagro: 4-0
- Hewitt: 4-0 (lost 1 set)
- Söderling: 3-1 (lost 3 sets)
- Thiem: 3-0
- del Potro: 3-0 (lost 1 set)
- Gasquet: 3-0
- Seppi: 2-0 (lost 1 set)
- Murray: 2-0
- Roddick: 2-0 (lost 1 set)
- Coria: 2-0 (lost 3 sets)
- Ljubicic: 2-0
- Monaco: 2-0
- Bolelli: 2-0
- Wawrinka: 2-0 (same score of 6-2 6-3 6-1 both times)
- Bellucci: 2-0

Breakdown by set score (almost the same likelihood of winning a set 6-2, 6-3 or 6-4):

- 6-0: 26
- 6-1: 61
- 6-2: 68
- 6-3: 66
- 6-4: 67
- 7-5: 18
- 7-6: 24
- 9-7: 1
- 6-7: 9
- 5-7: 6
- 4-6: 8
- 3-6: 6
- 2-6: 3 (Federer in 2006 Rome, Söderling in 2009 French Open, Djokovic in 2012 French Open)
- 1-6: 3 (Federer in 2006 French Open, del Potro in 2011 Davis Cup, Djokovic in 2015 French Open)
- 0-6: 1 (Coria in 2005 Monte Carlo Masters)

(only one incomplete set 2-0 after which Pablo Carreno Busta retired)

(1) Tennis Abstract: Rafael Nadal ATP Match Results, Splits, and Analysis

]]>**Example 1**: The number of ways of colouring an m by n grid black or white so that there is an even number of 1s in each row and column is

**Proof**: The first rows and columns may be coloured arbitarily. This then uniquely determines how the bottom row and rightmost column are coloured (restoring even parity). The bottom right square will be black if and only if the number of black squares in the remainder of the grid is odd, hence this is also uniquely determined by the first rows and columns. Details are also given here.

**Example 2**: The number of ways of colouring an m by n grid black or white so that every 2 by 2 square has an odd number (1 or 3) of black squares is

**Proof**: First colour the first row and first column arbitarily (there are such squares each with 2 possibilities). This uniquely determines how the rest of the grid must be coloured by considering the colouring of adjacent squares above and to the left.

By the same argument, the above is the same as the number of colouring an m by n grid black or white so that every 2 by 2 square has an even number (0, 2 or 4) of black squares.

**Example 3**: The number of ways of colouring an m by n grid black or white so that every 2 by 2 square has two of each type is

**Proof**: If there are two adjacent squares of the same colour with one above the other, the remaining squares of the corresponding two rows are uniquely determined as being the same alternating between black and white. The remainder of the grid is then determined by the colouring of first column ( possibilities where we omit the two cases of alternating colours down the first column). Such a grid cannot have two horizontally adjacent squares of the same colour. By a similar argument a colouring that has two adjacent colours with one left of the other can be done in ways. Finally we have the two additional configurations where there are no adjacent squares of the same colour, which is uniquely determined by the colour of the top left square. Hence in total we have possible colourings.

This question for was in the 2017 Australian Mathematics Competition and the general solution is also discussed here.

**Example 4**: The number of ways of colouring an m by n grid black or white so that each row and each column contain at least one black square is (OEIS A183109)

Proof: First we count the number of colourings where a fixed subset of columns is entirely white and each row has at least one black square. The remaining columns and rows can be coloured in ways. To count colourings where each column has at least one black square we apply the principle of inclusion-exclusion and arrive at the above result.

Another inclusion-exclusion example shown here counts the number of 3 by 3 black/white grids in which there is no 2 by 2 black square. The answer is 417 with more terms for n by n grids in OEIS A139810.

**Example 5**: Suppose we wish to count the number of colourings of an m by n grid in which row i has black squares and column j has black squares (, ). Following [1], the number of ways this can be done is the coefficient of in the polynomial

To see this note that expanding the product gives products of terms of the form where such a term included corresponds to the ‘th row and th column being coloured black. Hence the coefficient of is the number of ways in which the system has a solution (, ) for equal to 1 if and only if row and column are coloured black and 0 otherwise.

Let us evaluate this in the special case of 2 black squares in every row and every column for an n by n square grid (i.e. and ). Picking two squares in each column to colour black means viewing the expansion as a polynomial in the coefficient of has sums of products of terms of the form . Then using notation to denote the coefficient of an expression, we have

Here the second last line follows from considering the number of ways that products of terms of the form arise in the product (which is ) and products of terms of the form can be formed in the product (which is ).

For example, when this is equivalent to finding the coefficient of in . Products are either paired up in complementary ways such as in ( ways) or we have the three products ( ways). This gives us a total of 90 (this question appeared in the 1992 Australian Mathematics Competition). More terms of the sequence are found in OEIS A001499 and the 6 by 4 case (colouring two shaded squares in each row and three in each column in 1860 ways) appeared in the 2007 AIME I (see Solution 7 here).

**Example 6**: If we wish to count the number of grid configurations in which reflections or rotations are considered equivalent, we may make use of Burnside’s lemma that the number of orbits of a group is the average number of points fixed by an element of the group. For example, to find the number of configurations of 2 by 2 grids up to rotational symmetry, we consider the cyclic group . For quarter turns there are configurations fixed (a quadrant determines the colouring of the remainder of the grid) while for half turns there are configurations as one half determines the colouring of the other half. This gives us an answer of

which is part of OEIS A047937. If reflections are also considered equivalent we need to consider the dihedral group and we arrive at the sequence in OEIS A054247.

If we want to count the number of 3 by 3 grids with four black squares up to equivalence, this is equivalent to the number of full noughts and crosses configurations. A nice video by James Grime explaining this is here (the answer is 23).

**Example 7**: The number of ways of colouring an m by n grid black or white so that the regions form 2 by 1 dominoes has the amazing form

For example, the 36 ways of tiling a 4 by 4 grid are given here. A proof of the above formula using the Pfaffian of the adjacency matrix of the corresponding grid graph is given in chapter 10 of [2].

[1] L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions (pp 235-6), D. Reidel Publishing Company, 1974.

[2] M. Aigner, A Course in Enumeration, Springer, 2007.

]]>Note that the first data point in 1974 corresponds to a game that was reduced to 50 overs per side after originally intended to be a 55 over game.

If we slice the data into eras marked by calendar years of roughly equal numbers of games, the mean score had a slight slow-down in the rate of increase from 2008-2012, then accelerated again in the past five years.

Era | Number of matches | Mean score batting first |

1974-1994 | 427 | 229 |

1995-1999 | 383 | 247 |

2000-2003 | 368 | 257 |

2004-2007 | 380 | 267 |

2008-2012 | 393 | 272 |

2013-2017 | 398 | 288 |

1974-2017 |
2349 |
260 |

The histograms below show how rarely teams score less than 200 runs in recent times when using the full quota of 50 overs. In fact these days a team is more likely to score over 400 than below 200 if using the full quota of 50 overs!

Comparing the distribution of first innings winning versus losing scores we find that the mean scores are 275 vs 236 respectively with sample sizes 1392 vs 901 (34 games had no result and 22 were tied). Restricting to the past five years, the median score batting first for the full 50 overs in winning matches is exactly 300.

Interestingly if we break down the runs scatter plot by team, the trends are not the same across the board. In particular England and South Africa have had more dramatic increases in recent times than the other teams, especially compared with India, Pakistan, Sri Lanka and West Indies.

Restricting to the last five years (2013-2017), here are the mean first innings scores for each team based on the match result (assuming they bat the full 50 overs).

Team | Result | mean score | # matches |

Afghanistan | lost | 249 | 6 |

Afghanistan | won | 260 | 12 |

Australia | lost | 295 | 13 |

Australia | n/r | 253 | 3 |

Australia | won | 310 | 31 |

Bangladesh | lost | 263 | 16 |

Bangladesh | won | 275 | 15 |

Canada | lost | 230 | 3 |

England | lost | 282 | 13 |

England | won | 329 | 22 |

Hong Kong | won | 283 | 4 |

India | lost | 282 | 12 |

India | won | 310 | 27 |

Ireland | lost | 244 | 6 |

Ireland | tied | 268 | 1 |

Ireland | won | 289 | 3 |

Kenya | lost | 260 | 1 |

Netherlands | lost | 265 | 1 |

New Zealand | lost | 277 | 12 |

New Zealand | tied | 314 | 1 |

New Zealand | won | 308 | 27 |

P.N.G. | lost | 218 | 2 |

P.N.G. | won | 232 | 1 |

Pakistan | lost | 266 | 9 |

Pakistan | n/r | 296 | 1 |

Pakistan | tied | 229 | 1 |

Pakistan | won | 290 | 20 |

Scotland | lost | 238 | 6 |

Scotland | won | 284 | 8 |

South Africa | lost | 258 | 7 |

South Africa | n/r | 301 | 1 |

South Africa | won | 321 | 36 |

Sri Lanka | lost | 249 | 15 |

Sri Lanka | n/r | 268 | 2 |

Sri Lanka | tied | 286 | 1 |

Sri Lanka | won | 305 | 22 |

U.A.E. | lost | 279 | 3 |

U.A.E. | won | 267 | 3 |

West Indies | lost | 265 | 10 |

West Indies | won | 298 | 10 |

Zimbabwe | lost | 247 | 9 |

Zimbabwe | tied | 257 | 1 |

Zimbabwe | won | 276 | 1 |

The England and South Africa numbers stand out the most here in winning causes. Also Australia has a particularly high average score of 294 in losing causes. Sri Lanka has the largest difference (56 runs) between average winning and losing scores.

**Edit**: The following shows the mean scores in the 100 matches prior to and after key rule changes (still focusing on first innings 50-over scores). Note that in two of the three cases, the average scores reduced.

- Restriction of 2 outside the 30-yard circle in the first 15 overs (’92 World Cup)

03 Jan 88 to 20 Jan 92:**231**

12 Feb 92 to 16 Feb 94:**222** - Introduction of Powerplay overs

13 Mar 04 to 30 Jun 05:**267**

07 Jul 05 to 08 Sep 06:**267** - Removal of powerplay, fifth fielder allowed outside the circle in the last ten overs

17Aug 14 to 24 Jun 15:**301**

10 Jul 15 to 19 Jan 17:**289**

The important difference between classical and quantum mechanics is that in the latter, the order in which measurements are taken sometimes matters. This is because obtaining the value of one measurement can disturb the system of interest to the extent that a consistently precise value of the other cannot be found. A famous example is position and momentum of a quantum particle – the Heisenberg uncertainty relation states that the product of their uncertainties (variances) in measurement is strictly greater than zero.

If measurements are treated as real-valued functions of the state space of system, we will not be able to capture the fact that the measurements do not commute. Since linear operators (e.g. matrices) do not commute in general, we use algebras of operators instead. We make use of the spectral theory leading from a special class of algebras with norm and adjoint known as von Neumann algebras which in turn are a special case of C*-algebras. The **spectrum** of an operator A is the set of numbers for which does not have an inverse. Self-adjoint operators have a real spectrum and will represent the set of values that an observable (a physical variable that can be measured) can take. Hence we have this correspondence between self-adjoint operators and observables.

By the Gelfand-Naimark theorem C*-algebras can be represented as bounded operators on a Hilbert space . See Section II.6.4 of [**3**] for proof details. If the C*-algebra is commutative the representation is as continuous functions on a locally compact Hausdorff space that vanish at infinity. Furthermore we assume the C*-algebra and corresponding Hilbert space are separable, meaning the space contains a countable dense subset (analogous to how the subset of rationals are dense in the set of real numbers). This ensures that the Stone-von Neumann theorem holds which was used to show that the Heisenberg and Schrödinger pictures of quantum physics are equivalent [see pp7-8 here].

The link between C*-algebras and Hilbert spaces is made via the notion of a **state** which is a positive linear functional on the algebra of norm 1. A state evaluated on a self-adjoint operator outputs a real number that will represent the expected value of the observable corresponding to that operator. Note that it is impossible to have two different states that have the same expected values across over observables. A state is called **pure** if it is an extreme point on the boundary of the (convex) space of states. In other words, we cannot write a pure state as where are states and ). A state that is not pure is called **mixed**.

Now referring to a Hilbert space , for any mapping of bounded operators to expectation values such that

- (it makes sense that the identity should have expectation value 1),
- self-adjoint operators are mapped to real numbers with positive operators (those with positive spectrum) mapped to positive numbers and
- is continuous with respect to the strong convergence in – i.e. if for all , then ,

then there is a is a unique self-adjoint non-negative trace-one operator (known as a **density matrix**) such that for all (see [1] Proposition 19.9). (The trace of an operator is defined as where is an orthonormal basis in the separable Hilbert space – in the finite dimensional case it is the sum of the operator’s eigenvalues.) Hence states are represented by positive self-adjoint operators with trace 1. Such operators are compact and so have a countable orthonormal basis of eigenvectors.

When corresponds to a projection operator onto a one-dimensional subspace it has the form where and . In this case we can show , which recovers the alternative view that unit vectors of correspond to states (known as **vector states**) so that the expected value of an observable corresponding to the operator is . This is done by choosing the orthonormal basis where and computing

Trace-one operators can be written as a convex combination of rank one projection operators: . From this it can be shown that those density operators which cannot be written as a convex combination of other states (called pure states) are precisely those of the form . Hence vector states and pure states are equivalent notions. Mixed states can be interpreted as a probabilistic mixture (convex combination) of pure states.

Let us now look at the similarity with probability theory. A measure space is a triple where is a set, is a collection of measurable subsets of called a -algebra and is a additive measure. If is a non-negative integrable function with it is called a **density function** and then we can define a **probability measure** by

.

A **random variable** maps elements of a set to real numbers in such a way that for any Borel subset of . This enables us to compute their expectation with respect to the density function as

.

This is like the quantum formula with our density operator playing the role of and operator playing the role of random variable . Hence a probability density function is the commutative probability analogue of a quantum state (density operator).

While Borel sets are the **events** from which we define simple functions and then random variables, in the non-commutative case we define operators in terms of **projections** (equivalently closed subspaces) of a Hilbert space . A projection operator is self-adjoint, satisfies and has the discrete spectrum . Hence they are analogous to 0-1 indicator random variables, the answers to yes/no events. For any unit vector the expected value

is interpreted as the probability the observable corresponding to will have value 1 when measured in the state corresponding to . In particular this probability will be 1 if and only if is in the invariant subspace of . We define meet and join operations on these closed subspaces to create a **Hilbert lattice** :

Borel sets form a algebra in which the distributive law holds for any elements of . However in the Hilbert lattice the corresponding rule (where are projection operators) only holds some of the time (see here for an example). This failure of the distributive law is equivalent to the general non-commutativity of projections.

A **quantum probability measure** can be defined by combining projections in a -additive way, namely and where are mutually orthogonal projections (). **Gleason’s theorem** says that for Hilbert space dimension at least 3 a state is uniquely determined by the values it takes on the orthogonal projections – a quantum probability measure can be extended from projections to bounded operators to obtain , similar to how characteristic functions are extended to integrable functions. Hence this is a key result for non-commutative integration (note: the continuity conditions defining in 1-3 above are stronger). We choose von Neumann algebras over C*-algebras since the former contain all spectral projections of their self-adjoint elements while the latter may not [ref].

So far we have seen that expected values of observables are derived via the formula . To derive the distribution itself, we make of the **spectral theorem** and for self-adjoint operators with continuous spectrum this requires **projection valued measures**. A self-adjoint operator has a corresponding function mapping Borel sets to projections so that represents the event that the outcome of measuring observable is in the set : we require that and is a complex additive function (measure) for all . We use as shorthand for . Similar to the way a finite dimensional self-adjoint matrix may be eigen-decomposed in terms of its eigenvalues and normalised eigenvectors as

the spectral theorem for more general self-adjoint operators allows us to write

which means that for every ,

.

Here, the integrals are over the spectrum of . Through this formula we can work with functions of operators and in particular the distribution of the random variable corresponding to operator in state will be

.

The similarities we have seen here between classical probability and quantum mechanics are summarised in the table below, largely taken from [2] which greatly aided my understanding. Note how the pairing between trace class and bounded operators is analogous to the duality of and functions.

Classical Probability |
Quantum Mechanics(non-commutative probability) |

– measure space | – Hilbert space model of QM |

– set | – Hilbert space |

– Boolean algebra of Borel subsets of called events |
– orthomodular lattice of projections (equivalently closed subspaces) of |

disjoint events | orthogonal projections |

– additive positive measure | – functional |

– integrable functions (probability density functions) | – trace class operators (density operators) |

– probability measure mapping Borel sets to numbers in [0,1] in a sigma-additive way |
– quantum state mapping projections to numbers in [0,1] in a sigma-additive way |

– essentially bounded measurable functions (bounded random variables) |
– von Neumann algebra of bounded operators (bounded observables) |

– expectation value of with respect to |
– expectation value of in state |

In summary, the fact that measurements don’t always commute lead us to consider non-commutative operator algebras. This leads us to the Hilbert space representation of quantum mechanics where a quantum state is a trace-one density operator and an observable is a bounded linear operator. We also saw that projections can be viewed as 0-1 events. The spectral theorem is used to decompose operators into a sum or integral of projections.

The richer mathematical setting for quantum mechanics allows us to model non-classical phenomena such as quantum interference and entanglement. We have not mentioned the time evolution of states, but in short, state vectors evolve unitarily according to the Schrödinger equation, generated by an operator known as the Hamiltonian.

[1] Hall, B.C., *Quantum Theory for Mathematicians*, Springer, Graduate Texts in Mathematics #267, June 2013 (relevant section)

[2] Redei, M., Von Neumann’s work on Hilbert space quantum mechanics

[3] Blackadar, B., Operator Algebras: Theory of C*-Algebras and von Neumann Algebras

[4] Wilce, Alexander, “Quantum Logic and Probability Theory“, *The Stanford Encyclopedia of Philosophy * (Spring 2017 Edition), Edward N. Zalta (ed.).

[6] Planetmath.org – Lattice of Projections

[7] Planetmath.org – Spectral Measure

[8] quantum mechanics – Intuitive meaning of Hilbert Space formalism – Physics Stack Exchange

[9] This answer to: mathematical physics – Quantum mechanics in a metric space rather than in a vector space, possible? – Physics Stack Exchange

[10] functional analysis – Resolution of the identity (basic questions) – Mathematics Stack Exchange

]]>As expected the main freeways carry the most traffic with the West Gate Freeway near the Western Link (CityLink) carrying the maximum average of 196,000 vehicles per day. The busiest segment of non-freeway is a stretch of Kings Way between Albert Road and Queens Road (99,000 vehicles per day).

]]>If there are no restrictions, the number of ways the votes are tallied is the binomial coefficient . The number of favourable outcomes (the numerator of the desired probability) in which A remains ahead can counted recursively in a similar way to Pascal’s triangle (each number the sum of the two neighbours above it) except no number may appear to the left of the vertical midline, as illustrated below. For example, the second element of the fifth row (3) corresponds to the case (AAAB, AABA, ABAA). More generally, dividing into the cases where the final vote is A or B, the number of ways in which A remains ahead of B is equal to where if . This sequence appears as A008313 in the OEIS and is the reversed form of Catalan’s triangle.

A way of generating the general term is making use of a beautiful reflection principle that gives a 1-1 correspondence between the number of tallies leading to a tie at some point and the number of tallies in which the first vote goes to candidate B: simply interchange A with B for all votes up to and including that tie. This amounts to reflecting the random walk about the midline, as illustrated below with the blue path corresponding to ABAA and the the red path BAAA.

Since , the probability candidate A always leads is 1 minus the probability the sequence ties at some point. But the bijection above shows an equal number of these start with A and with B, so our desired probability is

The numbers in the triangle are also formed by differences of adjacent entries of Pascal’s triangle, namely row p+q has terms of the form

This can be interpreted as the number of unrestricted sequences with p As and q Bs of length (p+q) that start with A minus the corresponding number that start with B, again following from the reflection principle.

As an aside, looking at the bottom row above we see , or equivalently

Finally we note that the Catalan numbers arise from the following parts of the triangle above:

- as entries in the first column (counting Dyck paths)
- as the sum of squares of each row
- as the sum of entries in NE-SW diagonals

Catalan’s triangle can be generalised to a trapezium in which we count the number of strings consisting of n As and k Bs such that in every initial segment of the string the number of Bs does not exceed the number of As by m or more.

]]>54 tests | 5251 SPD Smith (AUS) | 5210 SM Gavaskar (INDIA) | 4991 L Hutton (ENG) | 4840 JB Hobbs (ENG) |

54 tests | 61.06 SPD Smith (AUS) | 60.73 H Sutcliffe (ENG) | 59.51 GS Sobers (WI) | 59.02 JB Hobbs (ENG) |

100 innings | 5354 JB Hobbs (ENG) | 5345 GS Sobers (WI) | 5279 WR Hammond (ENG) | 5251 SPD Smith (AUS) |

100 innings | 61.06 SPD Smith (AUS) | 60.74 GS Sobers (WI) | 60.68 WR Hammond (ENG) | 58.48 L Hutton (ENG) |

A more complete list of the top 10 scorers in these categories after n tests/innings is below. Statistics are from ESPN Cricinfo and are current to 9 January 2018. Corrections are welcome.

The following players have been ranked first for some n (as of 9 January 2018):

- highest aggregate after n innings: Foster, Rowe, Javed Miandad, Kambli, Weekes, Bradman, Hobbs, Sobers, Smith, Hammond, Hutton, Sehwag, Tendulkar, Sangakkara, Lara
- highest aggregate after n tests: Rowe, Foster, Javed Miandad, Gavaskar, Bradman, Smith, Sobers, Hutton, Sehwag, Sangakkara, Younis Khan, Lara, Ponting, Kallis, Dravid, Tendulkar
- highest average after n innings: Foster, Rowe, Bell, Trott, Kambli, Gavaskar, Harvey, Bradman, Voges, Sutcliffe, Hobbs, Smith, Barrington, Sobers, Hammond, Hutton, Dravid, Tendulkar, Ponting, Sangakkara, Kallis
- highest average after n tests: Rowe, Rudolph, Bell, Gavaskar, Kambli, Samaraweera, Harvey, Bradman, Voges, Sutcliffe, Smith, Hobbs, Hammond, Sobers, Dravid, Tendulkar, Ponting, Sangakkara, Kallis

The first problem I found via the Romantics of Geometry Facebook group: let be the point of tangency of the incircle of with and let be the foot of the perpendicular from the incentre of the to . Then show bisects .

The second problem is motivated by the above and problem 2 of the 2008 USAMO: this time let be a symmedian of and be the foot of the perpendicular from the circumcentre of to . Then show that bisects .

Here is a solution to the first problem inspired bythat of Vaggelis Stamatiadis. Let the line through the other two points of tangency of the incircle with intersect line at the point as shown below. Note that since and are tangents to the circle, line is the polar of with respect to the incircle.

Since is on the polar of , by La Hire’s theorem, is on the polar of . The polar of also passes through (as is a tangent to the circle at ). We conclude that the polar of is the line through and .

Next, let intersect at . By theorem 5(a) at this link, the points form a harmonic range. Since the cross ratio of collinear points does not change under central projection, considering the projection from , also form a harmonic range. (Alternatively, this follows from the theorems of Ceva and Menelaus using the Cevians intersecting at the Gergonne point and transveral ). Also, as both and are perpendicular to polar of .

Considering a central projection from of line to a line parallel to through , we see that form a harmonic range. Since is a point at infinity, this implies is the midpoint of and so triangles and are congruent (equality of two pairs of sides and included angle is ). Hence bisects as was to be shown.

For the second problem, we use the following characterisation of a symmedian: extended concurs with the lines of tangency of the circumcircle at and . (For three proofs of this see here.)

Define as the intersection of with and as the intersection of with the tangents at . Note that line is the polar of with respect to the circumcircle. By La Hire’s theorem, must be on the polar of . This polar is perpendicular to (the line joining to the centre of the circle) and as by construction of , it follows that line is the polar of . Again by theorem 5(a) in reference (2), form a harmonic range. Following the same argument as the previous proof, this together with imply bisects as required.

By similar arguments, one can prove the following, left to the interested reader. If is the -excentre of , the ex-circle’s point of tangency of , and the foot of the perpendicular from to line , then bisects .

(1) Alexander Bogomolny, __Poles and Polars__* from Interactive Mathematics Miscellany and Puzzles* http://www.cut-the-knot.org/Curriculum/Geometry/PolePolar.shtml, Accessed 19 March 2017

(2) Poles and Polars – Another Useful Tool! | The Problem Solver’s Paradise

(3) Yufei Zhao, Lemmas in Euclidean Geometry

]]>Test |
ODI |
T20I |
IPL |

200 (283) | 91 (97) | 90* (55) | 75 (51) |

44 (90) | 59 (67) | 59* (33) | 79 (48) |

3 (8) | 117 (117) | 50 (36) | 33 (30) |

4 (17) | 106 (92) | 7 (12) | 80 (63) |

9 (10) | 8 (11) | 49 (51) | 100* (63) |

18 (40) | 85* (81) | 56* (47) | 14 (17) |

9 (28) | 9 (13) | 41* (28) | 52 (44) |

45 (65) | 154* (134) | 23 (27) | 108* (58) |

211 (366) | 45 (51) | 55* (37) | 20 (21) |

17 (28) | 65 (76) | 24 (24) | 7 (7) |

40 (95) | 82* (51) | 109 (55) | |

49* (98) | 89* (47) | 75* (51) | |

167 (267) | 16 (9) | 113 (50) | |

81 (109) | 54* (45) | ||

62 (127) | 0 (2) | ||

6* (11) | 54 (35) | ||

235 (340) | |||

15 (29) | |||

1215 @ 75.9 |
739 @ 92.37, SR 100 |
641 @ 106.8, SR 140.3 |
973 @ 81.2, SR 152.0 |