# Chaitanya's Random Pages

## January 26, 2019

### 49+ °C temperatures in Australia

Filed under: climate and weather — ckrao @ 12:32 pm

Below is a list of recorded instances of maximum temperatures of 49 degrees Celsius or more in Australia, based on [1] and [2] from Australia’s Bureau of Meteorology. Out of the 48 occasions, 22 have occurred in this decade including 8 (so far) during this summer alone! I believe all the stations have been recording temperatures for at least 20 years except Port Augusta (which started in 2001).

 Temperature (°C) Date Station Name State 50.7 2-Jan-60 Oodnadatta Airport SA 50.5 19-Feb-98 Mardie WA 50.3 3-Jan-60 Oodnadatta Airport SA 49.8 21-Feb-98 Emu Creek Station WA 49.8 13-Jan-79 Forrest Aero WA 49.8 3-Jan-79 Mundrabilla Station WA 49.7 10-Jan-39 Menindee Post Office NSW 49.6 12-Jan-13 Moomba Airport SA 49.5 24-Jan-19 Port Augusta Aero SA 49.5 24-Dec-72 Birdsville Police Station QLD 49.4 21-Dec-11 Roebourne WA 49.4 16-Feb-98 Emu Creek Station WA 49.4 7-Jan-71 Madura Station WA 49.4 2-Jan-60 Marree Comparison SA 49.4 2-Jan-60 Whyalla (Norrie) SA 49.3 27-Dec-18 Marble Bar WA 49.3 2-Jan-14 Moomba Airport SA 49.3 9-Jan-39 Kyancutta SA 49.2 24-Jan-19 Kyancutta SA 49.2 21-Feb-15 Roebourne Aero WA 49.2 10-Jan-14 Emu Creek Station WA 49.2 22-Dec-11 Onslow Airport WA 49.2 1-Jan-10 Onslow WA 49.2 11-Jan-08 Onslow WA 49.2 9-Feb-77 Mardie WA 49.2 1-Jan-60 Oodnadatta Airport SA 49.2 3-Jan-22 Marble Bar Comparison WA 49.2 11-Jan-05 Marble Bar Comparison WA 49.1 24-Jan-19 Tarcoola Aero SA 49.1 23-Jan-19 Red Rocks Point WA 49.1 13-Jan-19 Marble Bar WA 49.1 27-Dec-18 Onslow Airport WA 49.1 3-Jan-14 Walgett Airport AWS NSW 49.1 2-Jan-10 Emu Creek Station WA 49.1 18-Feb-98 Roebourne WA 49.1 23-Dec-72 Moomba SA 49 15-Jan-19 Tarcoola Aero SA 49 23-Jan-15 Marble Bar WA 49 13-Jan-13 Birdsville Airport QLD 49 9-Jan-13 Leonora WA 49 21-Dec-11 Roebourne Aero WA 49 1-Jan-10 Mardie WA 49 10-Jan-09 Emu Creek Station WA 49 11-Jan-08 Port Hedland Airport WA 49 11-Jan-08 Roebourne WA 49 12-Jan-88 Marla Police Station SA 49 6-Dec-81 Birdsville Police Station QLD 49 22-Dec-72 Marree SA

## December 30, 2018

### A collection of energy formulas

Filed under: science — ckrao @ 10:58 am

Energy is a quantity that is conserved as a consequence of the time translation invariance of the laws of physics. Below are some formulas calculating energy of different forms.

Kinetic energy is that associated with motion and is defined as $K = \frac{1}{2} mv^2 = \frac{p^2}{2m}$ for a particle with mass $m$, velocity $v$ and momentum $p$. If the mass is a fluid in motion (e.g. wind) with density $\rho$ and volume $A v t$ through cross-sectional area $A$, then $K = \frac{1}{2} At\rho v^3$.

Work is the result of a force $F$ applied over a displacement $\mathbf{s}$ and is given by the line integral

$\displaystyle W = \int_C \mathbf{F} . \mathrm{d}\mathbf{s} = \int_{t_1}^{t_2} \mathbf{F} . \frac{\mathrm{d}\mathbf{s}}{\mathrm{d}t} \ \mathrm{d}t= \int_{t_1}^{t_2} \mathbf{F}.\mathbf{v}\ \mathrm{d}t .$

This has the simple form $W = Fs \cos \theta$ when force is constant and displacement is linear where $\theta$ is the angle between the force and displacement vectors.

Using Newton’s 2nd law and the relation $\frac{\mathrm{d}}{\mathrm{d}t} (\mathbf{v}^2) = 2\mathbf{v}.\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}$ this can be written as

$\displaystyle W = m\int_{t_1}^{t_2} \frac{d\mathbf{v}}{dt} . \mathbf{v} \mathrm{d}t = \frac{1}{2}m\int_{t_1}^{t_2} \frac{\mathrm{d}}{\mathrm{d}t} (\mathbf{v}^2) \mathrm{d}t = \frac{1}{2}m\int_{v_1^2}^{v_2^2} \mathrm{d}(\mathbf{v}^2) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2.$

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 $W = \int_{v_1}^{v_2} m \mathbf{v}.\mathrm{d}\mathbf{v} = \int_{p_1}^{p_2} \mathbf{v}.d\mathbf{p}$ where $\mathbf{p} = m\mathbf{v}$ is momentum.

The above has the rotational analogue $K = \frac{1}{2} I \omega^2$ where $I$ is moment of inertia and $\omega$ is angular velocity and the equation for work becomes

$\displaystyle W = \int_{t_1}^{t_2} \mathbf{T} . \mathbf{\omega}\ \mathrm{d}t$,

where $\mathbf{T}$ is a torque vector.

This has the simple form $W = Fr \omega = \tau \omega$ in the special case of a constant magnitude tangential force where $\tau = Fr$ is the torque resulting from force $F$ applied at distance $r$ 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:

$W = \int P(t)\ \mathrm{d}t = \int_{t_1}^{t_2} \mathbf{F}.\mathbf{v}\ \mathrm{d}t.$

If the work done by a force field $\mathbf{F}$ 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 $U$ satisfying $\mathbf{F} = -\nabla U$. 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: $\mathbf{F} = -(GMm/r^2) \hat{r}, U = -GMm/r$, where $M, m$ are the masses of two bodies, $r$ the distance between their centre of masses and $G$ is Newton’s gravitation constant.
• due to earth’s gravity at the surface: $\mathbf{F} = -mg, U = mgh$ where $g \approx 9.8 ms^{-2}$ and $h$ is the object’s height above ground (small compared with the size of the earth).
• due to a spring obeying Hooke’s law: $\mathbf{F} = -kx, U = kx^2/2$ where $k$ is the spring constant and $x$ the displacement from an equilibrium position.
• due to an electrostatic field: $\mathbf{F} = q\mathbf{E} = (k_e qQ/r^2) \hat{r}, U = k_e qQ/r$ where $k_e$ is Coulomb’s constant $1/(4\pi \epsilon_0)$ and $q, Q$ are charges. This can be written as $U = qV$ where $V$ is a potential function measured in volts.
• for a system of point charges: $\displaystyle U =k_e \sum_{1 \leq i < j \leq n} \frac{q_i q_j}{r_{ij}}$.
• for a system of conductors: $U = \frac{1}{2} \sum_{i=1}^n Q_i V_i$ where the charge on conductor $i$ is $Q_i$ and its potential is $V_i$.
• for a charged dielectric: the above may be generalised to the volume integral $U = \frac{1}{2} \int_V \rho \Phi \ \mathrm{d}v$ where $\rho$ is charge density and $\Phi$ is the potential corresponding to the electric field.
• for an electric dipole in an electric field: $U = -\mathbf{p}.\mathbf{E}$ where $\mathbf{p}$ 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: $U = -\mathbf{\mu}.\mathbf{B}$ where $\mathbf{\mu}$ 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 $L$ is $v = L di/dt$ and the current though a capacitance $C$ is $i = C dv/dt$. These inserted into the relationship $E = \int i(t)v(t) \ \mathrm{d}t$ lead to the formulas $E = \frac{1}{2}L(\Delta I)^2$ and $E = \frac{1}{2}C(\Delta V)^2$ 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 $\mathbf{S} = \mathbf{E} \times \mathbf{H}$, the cross product of the electric and magnetising field vectors. The electromagnetic energy in a volume $V$ is given by ([1])

$\displaystyle \frac{1}{2}\int_V \mathbf{B}.\mathbf{H} + \mathbf{E}.\mathbf{D} \ \mathrm{d}v$,

where $\mathbf{D}$ is the electric displacement field and $\mathbf{B}$ is the magnetic field. This is more commonly written as $\displaystyle \frac{1}{2} \int_V \epsilon_0 |E|^2 + |B^2|/\mu_0 \ \mathrm{d}v$ when the relationships $\mathbf{D} = \epsilon_0\mathbf{E}, \mathbf{B} = \mu_0 \mathbf{H}$ 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

$E^2 = (pc)^2 + (m_0c^2)^2$.

This leads to $E = pc$ for massless particles (such as photons) and more generally $E = \gamma m_0 c^2$ , the mass-energy equivalence relation (here $\gamma = (1 - (v/c)^2)^{-1/2}$ and $m_0$ is rest mass).

In quantum mechanics the energy of a photon is also written as $E = hf = hc/\lambda$ (Planck-Einstein relation) where $h$ is Planck’s constant and $f, \lambda$ are frequency and wavelength respectively. Energies of quantum systems are based on the eigenstates of the Hamiltonian operator, an example of which is $\displaystyle {\hat {H}}=-{\frac {\hbar ^{2}}{2m}}\nabla^2+V(x)$.

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

$\displaystyle U = \int \left(T \text{d}S - p \mathrm{d}V + \sum_i \mu_i \mathrm{d}N_i \right)$

where $T$ is temperature, $S$ is entropy, $N_i$ is the number of particles and $\mu_i$ the chemical potential of species $i$. 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 $\bar{E} = \frac{3}{2}k_B T$ where $k_B$ is Boltzmann’s constant.

In thermal transfer the change in internal energy is given by $\Delta U = m C \Delta T$ where $m$ is mass and $C$ 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 $P = A \epsilon \sigma T^4$ where $\epsilon$ is the emissivity (=1 for black body radiation) and $\sigma$ 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 $m$ of a substance with specific latent heat constant $L$ is given by $Q = mL$.

Finally, the energy of a single wavelength of a mechanical wave is $\displaystyle \frac{1}{2} m\omega^2 A^2$ where $m$ the mass of a wavelength, $A$ the amplitude and $\omega$ the angular frequency [2]. This can be applied to finding the energy density of ocean waves for example [3].

#### References

[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.

## June 12, 2018

### Rafael Nadal in best of five set matches on clay

Filed under: sport — ckrao @ 1:28 pm

Following Rafael Nadal‘s 11th French Open win, it’s worth looking at just how amazing his best-of-five set record on clay is. He now has a 111-2 win-loss record, with his only two losses against Söderling and Djokovic.

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)

## April 1, 2018

### A collection of binary grid counting problems

Filed under: mathematics — ckrao @ 3:52 am

The number of ways of colouring an m by n grid one of two colours without restriction is $2^{mn}$. The following examples show what happens when varying restrictions are placed on the colouring.

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

$\displaystyle 2^{(m-1)(n-1)}.$

Proof: The first $m-1$ rows and $n-1$ 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 $m-1$ rows and $n-1$ 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

$\displaystyle 2^{m+n-1}.$

Proof: First colour the first row and first column arbitarily (there are $m+n-1$ 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

$\displaystyle 2^m + 2^n - 2.$

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 ($2^m - 2$ 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 $2^n-2$ 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 $(2^m-2) + (2^n-2) + 2 = 2^m + 2^n - 2$ possible colourings.

This question for $m = n = 8$ 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)

$\displaystyle \sum_{j=0}^m (-1)^j \binom{m}{j} (2^{m-j}-1)^n.$

Proof: First we count the number of colourings where a fixed subset of $j$ columns is entirely white and each row has at least one black square. The remaining $m-j$ columns and $n$ rows can be coloured in $(2^{m-j}-1)^n$ 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 $k_i$ black squares and column j has $l_j$ black squares ($i = 1, 2, \ldots m$, $j = 1, 2, \ldots, n$). Following [1], the number of ways this can be done is the coefficient of $x_1^{k_1}x_2^{k_2} \ldots x_m^{k_m}y_1^{l_1}y_2^{l_2}\ldots y_n^{l_n}$ in the polynomial

$\displaystyle \prod_{i=1}^m \prod_{j=1}^n (1 + x_i y_j).$

To see this note that expanding the product gives products of terms of the form $(x_i y_j)$ where such a term included corresponds to the $i$‘th row and $j$th column being coloured black. Hence the coefficient of $x_1^{k_1}x_2^{k_2} \ldots x_m^{k_m}y_1^{l_1}y_2^{l_2}\ldots y_n^{l_n}$ is the number of ways in which the system $\sum_{j=1}^n a_{ij} = k_i, \sum_{i=1}^m a_{ij} = l_j$ has a solution ($i = 1, 2, \ldots m$, $j = 1, 2, \ldots, n$) for $a_{ij}$ equal to 1 if and only if row $i$ and column $j$ 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. $k_i = l_j = 2$ and $m = n$). Picking two squares in each column to colour black means viewing the expansion as a polynomial in $y_1, \ldots, y_n$ the coefficient of $y_1^2y_2^2\ldots y_n^2$ has sums of products of $n$ terms of the form $x_ix_j$. Then using $[]$ notation to denote the coefficient of an expression, we have

\begin{aligned} \left[x_1^2x_2^2 \ldots x_n^2y_1^2y_2^2\ldots y_n^2 \right] \prod_{i=1}^n \prod_{j=1}^n (1 + x_i y_j) &= \left[x_1^2x_2^2 \ldots x_n^2 \right] \left( \sum_{i=1}^n\sum_{j=i+1}^n x_i x_j \right)^n\\&= \left[x_1^2x_2^2 \ldots x_n^2 \right] 2^{-n} \left( \left( \sum_{i=1}^n x_i\right)^2 - \sum_{i=1}^n x_i^2 \right)^n\\ &= \left[x_1^2x_2^2 \ldots x_n^2 \right] 2^{-n} \sum_{k=0}^n (-1)^k \binom{n}{k} \left( \sum_{i=1}^n x_i^2 \right)^k\left(\sum_{i=1}^n x_i\right)^{2(n-k)}\\ &= 2^{-n} \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{n!}{(n-k)!} \frac{(2n-2k)!}{2^{n-k}}\\ &= 4^{-n} \sum_{k=0}^n (-1)^k \binom{n}{k}^2 2^k (2n-2k)!. \end{aligned}

Here the second last line follows from considering the number of ways that products of $k$ terms of the form $x_i^2$ arise in the product $\left( \sum_{i=1}^n x_i^2 \right)^k$ (which is $\frac{n!}{(n-k)!}$) and products of $(n-k)$ terms of the form $x_i^2$ can be formed in the product $\left(\sum_{i=1}^n x_i\right)^{2(n-k)}$ (which is $\frac{(2n-2k)!}{2^{n-k}}$).

For example, when $n=4$ this is equivalent to finding the coefficient of $a^2b^2c^2d^2$ in $(ab + bc + ac + bc + bd + cd)^4$. Products are either paired up in complementary ways such as in $ab.ab.cd.cd$ ($3 \times \binom{4}{2} = 18$ ways) or we have the three products $ab.bc.cd.ad, ab.bd.cd.ad, ac.bc.bd.ad$ ($3 \times 4! = 72$ 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 $C_4$. For quarter turns there are $2^4$ configurations fixed (a quadrant determines the colouring of the remainder of the grid) while for half turns there are $2^8$ configurations as one half determines the colouring of the other half. This gives us an answer of

$\displaystyle \frac{2^{16} + 2.2^4 + 2^8}{4} = 16456,$

which is part of OEIS A047937. If reflections are also considered equivalent we need to consider the dihedral group $D_4$ 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

$\displaystyle 2^{mn/2} \prod_{j=1}^{\lceil m/2 \rceil} \prod_{k=1}^{\lceil n/2 \rceil} \left(4 \cos^2 \frac{\pi j}{m+1} + 4 \cos^2 \frac{\pi k}{n+1}\right).$

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].

#### References

[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.

## December 26, 2017

### The evolution of ODI team totals

Filed under: cricket,sport — ckrao @ 11:46 am

Over the years one day international cricket scores have been on the rise and this post intends to look into this in some detail. We shall restrict ourselves to first innings scores where the team batting first lasted exactly 50 overs. Hence games greater than 50 overs per team long or where a team was bowled out prematurely are omitted. There are 2349 (out of 3945) such matches according to this query on Cricinfo Statsguru and on average  7 wickets fall over the 50 overs. The plot below shows a scatter plot of the scores over time. The red curve shows that mean scores were steady around 225 during the 1980s and have been on the rise since 1990 so that now the mean score is approaching 300.

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.

1. 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
2. Introduction of Powerplay overs
13 Mar 04 to 30 Jun 05: 267
07 Jul 05 to 08 Sep 06: 267
3. 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

## September 8, 2017

### Notes on von Neumann’s algebra formulation of Quantum Mechanics

Filed under: mathematics,science — ckrao @ 9:49 pm

The Hilbert space formulation of (non-relativistic) quantum mechanics is one of the great achievements of mathematical physics. Typically in undergraduate physics courses it is introduced as a set of postulates (e.g. the Dirac-von Neumann axioms) and hard to motivate without some knowledge of functional analysis or at least probability theory.  Some of that motivation and the connection with probability theory is summarised in the notes here – in fact it can be said that quantum mechanics is essentially non-commutative probability theory [2]. Furthermore having an algebraic point of view seems to provide a unified picture of classical and quantum mechanics.

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 $\lambda$ for which $(A-\lambda I)$ 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 ${\cal H}$. 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 $\omega$ 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 $\omega$ as $\omega = \lambda \omega_1 + (1-\lambda) \omega_2$ where $\omega_1 \neq \omega_2$ are states and $0 < \lambda < 1$). A state that is not pure is called mixed.

Now referring to a Hilbert space ${\cal H}$, for any mapping $\Phi$ of bounded operators $B({\cal H})$ to expectation values such that

1. $\Phi(I) = 1$ (it makes sense that the identity should have expectation value 1),
2. self-adjoint operators are mapped to real numbers with positive operators (those with positive spectrum) mapped to positive numbers and
3. $\Phi$ is continuous with respect to the strong convergence in $B({\cal H})$ – i.e. if $\lVert A_n \psi - A \psi \rVert \rightarrow 0$ for all $\psi \in H$, then $\Phi (A_n) \rightarrow \Phi (A)$,

then there is a is a unique self-adjoint non-negative trace-one operator $\rho$ (known as a density matrix) such that $\Phi (A) = \text{trace}(\rho A)$ for all $A \in B(H)$ (see [1] Proposition 19.9). (The trace of an operator $A$ is defined as $\sum_k \langle e_k, Ae_k \rangle$ where $\{e_k \}$ 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 $\rho$ corresponds to a projection operator onto a one-dimensional subspace it has the form $\rho = vv^*$ where $v \in {\cal H}$ and $\lVert v \rVert = 1$. In this case we can show $\text{trace}(\rho A) = \langle v, Av \rangle = v^*Av$, which recovers the alternative view that unit vectors of ${\cal H}$ correspond to states (known as vector states) so that the expected value of an observable corresponding to the operator $A$ is $\langle v, Av \rangle$. This is done by choosing the orthonormal basis $\{e_k \}$ where $e_1 = v$ and computing

\begin{aligned} \text{trace}(\rho A) &= \sum_k \langle e_k, vv^*Ae_k \rangle\\ &= \sum_k e_k^* v v^* Ae_k\\ &= e_1^* e_1 e_1^*Ae_1 \quad \text{ (as }e_k^*v = \langle e_k, v \rangle = 0\text{ for } k > 1\text{)}\\ &= e_1^*Ae_1\\ &= \langle v, Av \rangle. \end{aligned}

Trace-one operators $\rho$ can be written as a convex combination of rank one projection operators: $\rho = \sum \lambda_k v_k v_k^*$. 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 $\rho = vv^*$. 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 $(X, {\cal S}, \mu)$ where $X$ is a set, ${\cal S}$ is a collection of measurable subsets of $X$ called a $\sigma$-algebra and $\mu:{\cal S} \rightarrow \mathbb{R} \cup \infty$ is a $\sigma-$additive measure. If $g$ is a non-negative integrable function with $\int g \ d\mu = 1$ it is called a density function and then we can define a probability measure $p_g:{\cal S} \rightarrow [0,1]$ by

$\displaystyle p_g(S) = \int_S g\ d\mu \in [0,1], S \in {\cal S}$.

A random variable $f:X\rightarrow \mathbb{R}$ maps elements of a set to real numbers in such a way that $f^{-1}(B) \in {\cal S}$ for any Borel subset of $\mathbb{R}$. This enables us to compute their expectation with respect to the density function $g$ as

$\displaystyle \int_X f \ dp_g = \int_X fg\ d\mu$.

This is like the quantum formula $\text{Tr}(\rho A)$ with our density operator $\rho$ playing the role of $g$ and operator $A$ playing the role of random variable $f$. 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 ${\cal H}$. A projection operator $P$ is self-adjoint, satisfies $P^2 = P$ and has the discrete spectrum $\{0,1\}$. Hence they are analogous to 0-1 indicator random variables, the answers to yes/no events. For any unit vector $v \in {\cal H}$ the expected value

$\displaystyle \langle v, Pv \rangle = \langle v, P^2v \rangle = \langle Pv, Pv \rangle = \lVert Pv \rVert^2$

is interpreted as the probability the observable corresponding to $P$ will have value 1 when measured in the state corresponding to $v$. In particular this probability will be 1 if and only if $v$ is in the invariant subspace of $P$. We define meet and join operations $\vee, \wedge$ on these closed subspaces to create a Hilbert lattice $({\cal P}({\cal H}), \vee, \wedge, \perp)$:

• $A \wedge B = A \cap B$
• $A \vee B = \text{closure of } A + B$
• $A^{\perp} = \{u: \langle u,v \rangle = 0\ \forall v \in A\}$

Borel sets form a $\sigma-$algebra in which the distributive law $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ holds for any elements of ${\cal S}$. However in the Hilbert lattice the corresponding rule $A \wedge (B \vee C) = (A \wedge B) \vee (A \wedge C)$ (where $A, B, C$ 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 $\phi:{\cal P} \rightarrow [0,1]$ can be defined by combining projections in a $\sigma$-additive way, namely $\phi(0) = 0, \phi(I) = 1$ and $\phi(\vee_i P_i) = \sum_i \phi(P_i)$ where $P_i$ are mutually orthogonal projections ($P_i \leq P_j^{\perp}, i \neq j$). 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 $\phi(A) = \text{Tr}(\rho_{\phi} A)$, 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 $\Phi$ 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 $A$ are derived via the formula $\text{Tr}(\rho A)$. 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 $A$ has a corresponding function $E_A:{\cal S} \rightarrow {\cal P}({\cal H})$ mapping Borel sets to projections so that $E_A(S)$ represents the event that the outcome of measuring observable $A$ is in the set $S$: we require that $E_A(X) = I$ and $S \mapsto \langle u,E_A(S)v \rangle$ is a complex additive function (measure) for all $u, v \in {\cal H}$. We use $E_A(\lambda)$ as shorthand for $E_A(\{x:x\leq \lambda\})$. Similar to the way a finite dimensional self-adjoint matrix $M$ may be eigen-decomposed in terms of its eigenvalues $\lambda_i$ and normalised eigenvectors $u_i$ as

\begin{aligned} M &= \sum_i \lambda_i u_i u_i^T \\ &= \sum_i \lambda_i P_i \quad \text{(where }P_i := u_i u_i^T \text{ is a projection)}\\ &= \sum_i \lambda_i (E_i - E_{i-1}), \quad \text{(where } E_i := \sum{k \leq i} P_k\text{ ),} \end{aligned}

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

$A = \int_{\sigma(A)} \lambda dE_A(\lambda)$

which means that for every $u, v \in {\cal H}$,

$\langle u, Av \rangle = \int_{\sigma(A)} \lambda d\langle u,E_A v \rangle$.

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

$\text{Pr}(X \leq x) = E\left[ 1_{\{X \leq x\} }\right] = \text{Tr} \left( \rho\int_{-\infty}^x dE_A(\lambda) \right) = \text{Tr} \left( \rho E_A(x) \right)$.

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 $L^1$ and $L^{\infty}$ functions.

 Classical Probability Quantum Mechanics (non-commutative probability) $(X,{\cal S}, \mu)$ – measure space $({\cal H}, {\cal P}({\cal H}), \text{Tr})$ – Hilbert space model of QM $X$ – set ${\cal H}$ – Hilbert space ${\cal S}$ – Boolean algebra of Borel subsets of $X$ called events ${\cal P}({\cal H})$ – orthomodular lattice of projections (equivalently closed subspaces) of ${\cal H}$ disjoint events orthogonal projections $\mu:{\cal S} \rightarrow {\mathbb R}^{+} \cup \infty$ – $\sigma-$additive positive measure $\text{Tr}$ – functional $g \in L^1(X,\mu), g \geq 0, \int g \ d\mu = 1$ – integrable functions (probability density functions) $\rho \in {\cal T}({\cal H}), \rho \geq 0, \text{Tr}(\rho) = 1$ – trace class operators (density operators) $p_g(S) = \int \chi_S g\ d\mu \in [0,1], S \in {\cal S}$ – probability measure mapping Borel sets to numbers in [0,1] in a sigma-additive way $\phi(S) = \text{Tr}(\rho_{\phi } S) \in [0,1], \rho_{\phi } \in {\cal T}({\cal H}), S \in {\cal P}({\cal H})$ – quantum state mapping projections to numbers in [0,1] in a sigma-additive way $f \in L^{\infty}(X,\mu)$ – essentially bounded measurable functions (bounded random variables) $A \in {\cal B}({\cal H})$ – von Neumann algebra of bounded operators (bounded observables) $\int fg\ d\mu, g \in L^1(X,\mu)$ – expectation value of $f \in L^{\infty}(X,\mu)$ with respect to $p_g$ $\text{Tr}(\rho A), \rho \in {\cal T}({\cal H})$ – expectation value of $A \in {\cal B}({\cal H})$ in state $\rho$

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.).

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

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

## August 27, 2017

Filed under: geography — ckrao @ 11:16 am

VicRoads has a large collection of open data, which include traffic count estimates for main roads (excluding toll roads). I have taken the kml file from that link, colour coded the roads by two-way counts and shown only those with at least 35,000 vehicles per day (two-way traffic, 2017 estimates, averaged over a year) so that 1,465 road segments are shown. The results are mapped below (click on a road for count information).

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).

## June 30, 2017

### The ballot problem and Catalan’s triangle

Filed under: Uncategorized — ckrao @ 10:15 pm

The ballot problem asks for the probability that candidate A is always ahead of candidate B during a tallying process if they respectively end up with $p$ and $q$ votes where $p > q$. For example if $p = 2, q = 1$ there are 3 ways in which the three votes are counted (AAB, ABA, BAA) but the only favourable outcome in which A remains ahead throughout is occurs if the tally appears as AAB. Hence the probability A remains ahead is 1/3.

If there are no restrictions, the number of ways the votes are tallied is the binomial coefficient $\binom{p+q}{p}$. 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 $p = 3, q = 1$ (AAAB, AABA, ABAA). More generally, dividing into the cases where the final vote is A or B, the number of ways $N_{p,q}$ in which A remains ahead of B is equal to $N_{p-1,q} + N_{p,q-1}$ where $N_{p,q} = 0$ if $q \geq p$. 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 $p > q$, 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

$\displaystyle 1 - 2 \text{Pr(sequence starts with B)} = 1 - 2\frac{q}{p+q} = \frac{p-q}{p+q}.$

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

\displaystyle \begin{aligned} N_{p,q} &= \binom{p+q}{p}\frac{p-q}{p+q}\\ &= \frac{(p+q-1)!(p-q)}{p!q!}\\&= \binom{p+q-1}{q}-\binom{p+q-1}{p}.\end{aligned}

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 $N_{8,6} = N_{10,4} = 429$, or equivalently

$\displaystyle \binom{13}{4} - \binom{13}{3} = \binom{13}{6} - \binom{13}{5} = 429.$

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.

## April 9, 2017

### Highest aggregates and averages after n test matches/innings

Filed under: cricket,sport — ckrao @ 9:14 am

Soon after the 2017 India-Australia test series I noticed that among players who have played 54 tests, nobody has scored more than Steven Smith’s 5251 runs. Here is a list of the top three aggregates and averages after 54 tests and 100 innings.

 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

## March 19, 2017

### Two similar geometry problems based on perpendiculars to cevians

Filed under: mathematics — ckrao @ 7:18 am

In this post I wanted to show a couple of similar problems that can be proved using some ideas from projective geometry.

The first problem I found via the Romantics of Geometry Facebook group: let $M$ be the point of tangency of the incircle of $\triangle ABC$ with $BC$ and let $E$ be the foot of the perpendicular from the incentre $X$ of the $\triangle ABC$ to $AM$. Then show $EM$ bisects $\angle BEC$.

The second problem is motivated by the above and problem 2 of the 2008 USAMO: this time let $AM$ be a symmedian of $ABC$ and $E$ be the foot of the perpendicular from the circumcentre $X$ of $\triangle ABC$ to $AM$. Then show that $EM$ bisects $\angle BEC$.

Here is a solution to the first problem inspired bythat of Vaggelis Stamatiadis. Let the line through the other two points of tangency $P, Q$ of the incircle with $ABC$ intersect line $BC$ at the point $N$ as shown below. Note that since $AP$ and $AQ$ are tangents to the circle, line $NPQ$ is the polar of $A$ with respect to the incircle.

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

Next, let $MN$ intersect $PQ$ at $R$. By theorem 5(a) at this link, the points $(N, R, P, Q)$ form a harmonic range. Since the cross ratio of collinear points does not change under central projection,  considering the projection from $A$, $(N,M,B,C)$ 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 $NPQ$). Also, $NE \perp EM$ as both $NI$ and $IE$ are perpendicular to polar $AM$ of $N$.

Considering a central projection from $E$ of line $NMBC$ to a line $N', M, P', C'$ parallel to $NE$ through $M$, we see that $(N', M, P', C')$ form a harmonic range. Since $N'$ is a point at infinity, this implies $M$ is the midpoint of $B'C'$ and so triangles $B'EM$ and $C'EM$ are congruent (equality of two pairs of sides and included angle is $90^{\circ}$). Hence $EM$ bisects $\angle BEC$ as was to be shown.

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

Define $N$ as the intersection of $XE$ with $BC$ and $D$ as the intersection of $AM$ with the tangents at $B, C$. Note that line $NBMC$ is the polar of $D$ with respect to the circumcircle. By La Hire’s theorem, $D$ must be on the polar of $N$. This polar is perpendicular to $NX$ (the line joining $N$ to the centre of the circle) and as $ED \perp EX$ by construction of $E$, it follows that line $AEMD$ is the polar of $N$. Again by theorem 5(a) in reference (2), $(N, M, B, C)$ form a harmonic range. Following the same argument as the previous proof, this together with $NE \perp EM$ imply $EM$ bisects $\angle BEC$ as required.

By similar arguments, one can prove the following, left to the interested reader. If $X$ is the $A$-excentre of $\triangle ABC$, $M$ the ex-circle’s point of tangency of $BC$, and $E$ the foot of the perpendicular from $X$ to line $AM$, then $EM$ bisects $\angle BEC$.

#### References

(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

(3) Yufei Zhao, Lemmas in Euclidean Geometry

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