# Chaitanya's Random Pages

## February 15, 2022

### Some cricket stats by test series

Filed under: cricket,sport — ckrao @ 10:31 am

Below is a list of some statistics related to performances across test series. They are current up to 15 February 2022. Data is from ESPNcricinfo.

#### Most series scoring at least r runs:

(Click on the image to change r in Tableau.)

#### Most series taking at least w wickets:

(Click on the image to change w in Tableau.)

#### Most series scoring at least r runs and taking at least w wickets:

(Click on the image to change r and w in Tableau.)

#### Never batting having played in 3+ tests in a series:

All the above played 3 tests in the series – note that 3 players did not bat in the 2015/16 series.

## December 31, 2021

### Calculating the locations of stars in the sky

Filed under: geography,mathematics,science — ckrao @ 10:09 am

We present here a calculation of the location of a star as a function of its declination, the observer’s latitude and time after it is at its highest position in the sky. We use the fact that stars trace out circular paths about a fixed point in the sky.

The image below shows a celestial sphere centred on an observer at the location $O$. Assume the observer is in the northern hemisphere and the point $A$ represents the north celestial pole about which the stars appear to rotate anti-clockwise as the earth rotates on its axis (in the southern hemisphere stars rotate clockwise about the south celestial pole). The red ellipse illustrates the circular path of a star over a 24-hour period – point $B$ is its highest location while point $C$ is where it will be after a quarter of a day. Also shown in the diagram are axes $OA$ (pointing east), $OY$ (pointing north), $OZ$ (pointing directly overhead) and corresponding unit basis vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$.

Our aim is to find the location of the star (in 3-dimensional coordinates) on the red circle given:

• $t$ – the time (as a proportion of a day) after the star has reached its highest point
• $\delta$ – the angle of declination of the star (-90° to 90°) – the angle between the star and the celestial equator (which is the earth’s equator projected skyward)
• $\phi$ – the latitude of the observer (-90° to 90°)

The point $A$ is fixed as the earth rotates and in the north direction with an angle of elevation equal to the location’s latitude. Hence we may write

$\displaystyle \vec{OA} = 0\ \mathbf{i} + \cos \phi\ \mathbf{j} + \sin \phi\ \mathbf{k}.\quad\quad ...(1)$

Next consider the angle between $OA$ and $OB$. If $\delta = 90^{\circ}$, we would have a northern pole star and $\angle AOB = 0^{\circ}$. If $\delta = 0^{\circ}$ we would have a star on the celestial equator and $\angle AOB = 90^{\circ}$. More generally,

$\displaystyle \angle AOB = 90^{\circ}-\delta. \quad\quad ...(2)$

The point $B$ has an angle of elevation of $\phi + 90^{\circ}-\delta$, hence we have

\displaystyle \begin{aligned}\vec{OB} &= 0\ \mathbf{i} + \cos (\phi + 90^{\circ}-\delta ) \mathbf{j} + \sin (\phi + 90^{\circ}-\delta ) \mathbf{k}\\ &= \sin (\delta - \phi ) \mathbf{j} + \cos (\delta - \phi ) \mathbf{k}. \quad \quad ...(3)\end{aligned}

The circular path of the star can be regarded as a point traced around by a line having fixed angle $(90^{\circ}-\delta)$ from the fixed line $OA$. This circle has its centre at the point $A^{'}$ which is the projection of $OB$ onto the segment $OA$. Note that $A^{'}$ does not lie on the sphere but $A$, $B$ and $C$ do. We have

$\displaystyle \vec{OA^{'}} = \cos (90^{\circ}-\delta) \vec{OA} = \sin \delta \quad\quad ...(4)$

and

$\displaystyle |\vec{A^{'}B}| = \cos \delta.\quad\quad ...(5)$

We parameterise any point on this circular path by expressing it in terms of the basis vectors $\vec{A^{'}B}$ and $\vec{A^{'}C}$. For any point $P$ on the path we can write

$\displaystyle \vec{A^{'}P} = \cos 2\pi t\ \vec{A^{'}B} + \sin 2 \pi t\ \vec{A^{'}C}, \quad t \in [0, 1].\quad\quad ...(6)$

For example,

• if $t = 0$, $\vec{A^{'}P} = \vec{A^{'}B}$
• if $t = 0.25$, $\vec{A^{'}P} = \vec{A^{'}C}$
• if $t = 0.5$, $\vec{A^{'}P} = -\vec{A^{'}B}$
• if $t = 0.75$, $\vec{A^{'}P} = -\vec{A^{'}C}$

Since $A^{'}C$ has the same length as $|\vec{A^{'}B}|$ and is pointing directly west (i.e. perpendicular to both $\vec{OA}$ and $\vec{OB}$), we have from (5):

$\vec{A^{'}C} = -\cos \delta\ \mathbf{i}.\quad\quad ...(7)$

We now have all the ingredients we need to find $\vec{OP}$:

\displaystyle\begin{aligned}\vec{OP} &= \vec{OA^{'}} + \vec{A^{'}P}\\&= \vec{OA^{'}} + \cos 2\pi t\ \vec{A^{'}B} + \sin 2 \pi t\ \vec{A^{'}C}\quad \text{(by (6))}\\&=\vec{OA^{'}} + \cos 2\pi t\ (\vec{A^{'}O} + \vec{OB}) + \sin 2 \pi t\ \vec{A^{'}C}\\&= \vec{OA^{'}} - \cos 2\pi t\ \vec{OA^{'}} + \cos 2\pi t\ \vec{OB} + \sin 2 \pi t\ \vec{A^{'}C}\\&= (1- \cos 2\pi t) \vec{OA^{'}} + \cos 2\pi t\ \vec{OB} + \sin 2 \pi t\ \vec{A^{'}C}\\&=\left(1 - \cos 2\pi t \right)\sin \delta\ \vec{OA} + \cos 2\pi t\ \vec{OB} + \sin 2 \pi t\ \vec{A^{'}C}\quad\text{(by (4))}\\&=\left(1 - \cos 2\pi t \right)\sin \delta \left(\cos \phi\ \mathbf{j} + \sin \phi\ \mathbf{k}\right) + \cos 2\pi t \left(\sin (\delta - \phi) \mathbf{j} + \cos (\delta - \phi) \mathbf{k} \right)\\&\quad\quad + \sin 2 \pi t \left(-\cos \delta\ \mathbf{i} \right)\quad\text{(by (1), (3) and (7))}\\&=-\sin 2\pi t \cos \delta\ \mathbf{i} + \left(\sin \delta \cos \phi + \cos 2\pi t \left( \sin (\delta - \phi ) - \sin \delta \cos \phi \right) \right) \mathbf{j}\\&\quad\quad + \left( \sin \delta \sin \phi + \cos 2 \pi t \left( \cos (\delta - \phi ) - \sin \delta \sin \phi \right) \right) \mathbf{k}\\&=-\sin 2\pi t \cos \delta\ \mathbf{i} + \left( \sin \delta \cos \phi - \cos 2\pi t \cos \delta \sin \phi \right) \mathbf{j}\\&\quad\quad + \left(\sin \delta \sin \phi + \cos 2 \pi t \cos \delta \cos \phi \right) \mathbf{k}\quad\text{(using trigonometric identities).}\quad\quad ...(8)\end{aligned}

In this formula, the angle of elevation $\theta$ is found by setting $\sin \theta$ equal to the $\mathbf{k}$-component, or

$\displaystyle \theta = \arcsin \left(\sin \delta \sin \phi + \cos 2 \pi t \cos \delta \cos \phi \right).\quad \quad ...(9)$

Let us test out formula (8) in some special cases:

• If $\delta = 90^{\circ}$ we recover the formula $\vec{OP} = \cos \phi\ \mathbf{j} + \sin \phi\ \mathbf{k} = \vec{OA}$, the fixed location of the pole star.
• If $\delta = 0^{\circ}$ we recover the formula $\vec{OP} = -\sin 2 \pi t\ \mathbf{i} - \cos 2\pi t \left(\sin \phi\ \mathbf{j} - \cos \phi\ \mathbf{k} \right)$. This traces a circle with diameter joining $-\sin \phi \ \mathbf{j} + \cos \phi\ \mathbf{k}$ and $\sin \phi \ \mathbf{j} - \cos \phi\ \mathbf{k}$ (two antipodal points), hence it is a great circle on the celestial sphere.
• If $\phi = 0^{\circ}$ (observer on the equator) we obtain $\vec{OP} = -\sin 2\pi t \cos \delta\ \mathbf{i} + \sin \delta \mathbf{j} + \cos 2 \pi t \cos \delta\ \mathbf{k}$, hence the $\mathbf{j}$-component remains fixed and a circular path is traced.
• If $\phi = 90^{\circ}$ (observer at the north pole) we obtain $\vec{OP} = -\sin 2\pi t \cos \delta\ \mathbf{i} - \cos 2\pi t \cos \delta\ \mathbf{j} + \sin \delta\ \mathbf{k}$, hence the star traces out a circle with constant angle of elevation.

We can also determine the times at which a star rises or sets by setting the $\mathbf{k}$-component of position in (8) to 0:

\displaystyle \begin{aligned} \sin \delta \sin \phi + \cos 2 \pi t \cos \delta \cos \phi &= 0\\ \Rightarrow \quad \cos 2 \pi t &= -\frac{\sin \delta \sin \phi }{\cos \delta \cos \phi }\\&= -\tan \delta \tan \phi.\quad \quad ...(9)\end{aligned}

This is the so-called sunrise equation and was derived differently in an earlier blog post in [1].

Next let us look at the case when the star is our sun. The declination changes during the year between -23.4° and +23.4° (earth’s axial tilt) between the winter and summer soltice (for the northern hemisphere) as the earth revolves around the sun. Assume that the centre of the sun is at the origin and that earth’s orbit is the x-y plane having the form $r = R(\cos 2 \pi s\ \mathbf{i} + \sin 2\pi s\ \mathbf{j})$ (assuming a circular orbit with radius $R$ starting at $R\ \mathbf{i}$ at $s=0$). The parameter $s$ here represents the fraction of year after the winter solstice. Earth’s axis of rotation points in the direction $a = \sin 22.4^{\circ}\ \mathbf{i} + \cos 22.4^{\circ}\ \mathbf{k}$, being tilted away from the sun in the north when $s = 0$.

From the earth’s point of view the sun is in the direction $-\hat{r} = -(\cos 2 \pi s\ \mathbf{i} + \sin 2\pi s\ \mathbf{j})$. Then the angle $(90 - \delta)$ between the sun’s position and the earth’s axis satisfies

\displaystyle \begin{aligned}\sin \delta = \cos (90 - \delta) &= -\hat{r} . a\\&= -(\cos 2 \pi s\ \mathbf{i} + \sin 2\pi s\ \mathbf{j}).(\sin 22.4^{\circ}\ \mathbf{i} + \cos 22.4^{\circ}\ \mathbf{k})\\&= -\sin 22.4^{\circ}\cos 2 \pi s\\ \Rightarrow \delta &= \arcsin (-\sin 22.4^{\circ}\cos 2 \pi s)\\ &\approx \arcsin(-0.398 \cos 2 \pi s).\quad\quad ...(10) \end{aligned}

For example if $s = 0$, we have $\delta = -22.4^{\circ}$. If $s = 0.25, 0.75$, we have the equinoxes and $\delta = 0^{\circ}$. Finally if $s = 0.5$ then $\delta = 22.4^{\circ}$.

Substituting (10) into (8) gives the location of the sun given the time of year and time relative to when it’s at its highest point in the sky (solar noon). Note that the equation is not perfect since it does not take into account atmospheric refraction or the fact that the earth’s orbit is not perfectly circular and therefore the earth is not uniform in speed. For more exercises on the length of days based on latitude and the sun’s angle of declination refer to [2].

Finally, we can determine the bearing of a star when it rises or sets in the following manner. The position $P$ of the sun is in the x-y plane and so has the form

$\displaystyle \vec{OP} = x\ \mathbf{i} + y\ \mathbf{j}.\quad\quad ...(11)$

Secondly $\angle POA = 90^{\circ} - \delta$ and hence

\displaystyle \begin{aligned}\sin \delta &= \cos (90^{\circ} - \delta)\\ &= \vec{OP}.\vec{OA}\quad \text{(the dot product of two unit vectors is cosine of the angle between them)}\\ &= (x\ \mathbf{i} + y\ \mathbf{j}).(\cos \phi\ \mathbf{j} + \sin \phi\ \mathbf{k})\quad\quad\text{(from (1) and (11))}\\&=y \cos \phi.\quad\quad ...(12)\end{aligned}

Hence $y = \sin \delta/\cos \phi$. For example taking the sun on the summer solstice at Melbourne, Australia we have $\delta = -22.4^{\circ}$, $\phi = -38^{\circ}$ and so $y = \sin (-22.4^{\circ})/\cos(-38^{\circ}) \approx -0.484$. This corresponds to a bearing of $\arcsin(0.484) + 90^{\circ} \approx 119^{\circ}$, around 20 degrees south of east. The answer given in timeanddate.com is $121^{\circ}$, slightly more perhaps because it takes into account refraction and the fact that the sun is not pointlike.

### Reference

[1] The Shortest Day of the Year, Chaitanya’s Random Pages – https://ckrao.wordpress.com/2012/06/23/the-shortest-day-of-the-year/

[2] Alan Champneys, The Length of Dayshttps://www.bristol.ac.uk/media-library/sites/engineering/engineering-mathematics/documents/modelling/teacher/daylength_t.pdf

## September 24, 2021

### International men’s cricket match streaks

Filed under: Uncategorized — ckrao @ 8:58 am
Tags: ,

I have created a few interactive charts with Tableau on winning/losing/drawing international men’s cricket streaks using data from ESPNcricinfo.com’s Statsguru. You can click on any of the links or graphs to interact with it in a new page.

1. Most consecutive tests played with N matches of a given result (won/lost/drawn)

2. Fewest consecutive tests played with N matches of a given result (won/lost/drawn)

3. Most consecutive one-day internationals played with N matches of a given result (won/lost/drawn)

4. Fewest consecutive one-day internationals played with N matches of a given result (won/lost/drawn)

5. Most consecutive T20 internationals played with N matches of a given result (won/lost/drawn)

6. Fewest consecutive T20 internationals played with N matches of a given result (won/lost/drawn)

Here are a few observations.

Tests:

• Largely due to timeless tests in Australia up to World War II, Australia had 0 drawn tests at home Australia for 87 consecutive tests until 1946. For similar reasons England had 0 drawn tests away from home for 66 consecutive tests up to 1914 when they drew in South Africa.
• The top streaks without a loss for any team anywhere are 27, 26 and 25 by West Indies (1982-84), England (1968-71) and Australia (1946-51) respectively.
• England amazingly lost only 1 test away from home in 40 matches between 1963 and 1972.
• At home Pakistan lost only 1 test out of 40 from 1969 to 1986 while India only lost once in a span of 35 tests between 2012 and 2019.
• West Indies lost only 2 out of 53 tests between 1980 and 1986 and just 10 out of 100 from 1976 to 1988.
• Australia won 12 consecutive tests at home between 1999 and 2001 and 49 out of 61 home tests between 1998 and 2008. India is not far behind with 40 out of 53 home test wins from 2010 to 2021.
• In any ground Australia won 76 out of 100 tests between 1999 and 2008.
• South Africa has played 32 consecutive matches (anywhere) without a draw from 2017, an active streak. The next best is 26 by Zimbabwe from 2005 to 2017.
• New Zealand has an impressive active streak of 17 matches without a loss dating back to 2017 with 8 of their 13 wins by an innings.

One-day Internationals:

• Australia lost just 2 matches out of 32 anywhere between 2002 and 2003.
• India lost only 1 match out of 16 away from home between 2016 and 2018.
• West Indies lost only 7 matches out of 50 in away or neutral grounds between 1981 and 1985.
• Four teams have won 16 consecutive ODIs at home – Australia (2014-16), South Africa (2016-17), Sri Lanka (1996-98) and West Indies (1986-90). The best is 18 by Australia and Sri Lanka.
• Australia (1999-2004), Sri Lanka (1994-2004) and South Africa (1994-2001) all won 53/65 matches at home.

T20 Internationals:

• Afghanistan and Romania share the record of consecutive wins (12).
• Zimbabwe had 16 consecutive losses (2010-13).
• Afghanistan (2016-19) and Pakistan (2017-18) have both won 23 out of 26 consecutive games.
• Afghanistan won 40 out of 50 matches (2014-19).

## December 30, 2020

### An identity involving the product of reciprocals

Filed under: mathematics — ckrao @ 10:18 am

In this post I would like to prove the following identity, motivated by this tweet.

$\displaystyle n! \prod_{k=0}^n \frac{1}{x+k} = \frac{1}{x\binom{x+n}{n}} = \sum_{k=0}^n \frac{(-1)^k \binom{n}{k}}{x+k}$

The first of these equalities is straightforward by the definition of binomial coefficients. To prove the second, we make use of partial fractions. We write the expansion

$\displaystyle \prod_{k=0}^n \frac{1}{x+k} = \sum_{k=0}^n \frac{A_k}{x+k},$

where the coefficients $A_k$ are to be determined. Multiplying both sides by $(x+k')$ for some $k' \in \{0, 1, \ldots, n\}$,

\displaystyle \begin{aligned} (x+k') \prod_{k=0}^n \frac{1}{x+k} &= \sum_{k=0}^n \frac{A_k(x+k')}{x+k}\\\hbox{i.e.}\quad \prod_{\substack{k=0\\k \neq k'}}^n \frac{1}{x+k} &= A_{k'} + \sum_{\substack{k=0\\k \neq k'}}^n \frac{A_k(x+k')}{x+k}.\end{aligned}

Then setting $x = -k$ in both sides gives

\displaystyle \begin{aligned} \prod_{\substack{k=0\\k \neq k'}}^n \frac{1}{k-k'} &= A_{k'} + \sum_{\substack{k=0\\k \neq k'}}^n \frac{A_k(0)}{x+k}\\&= A_{k'}\\\hbox{i.e.}\quad A_{k'} &= \frac{1}{-k'} \cdot \frac{1}{-k'+1}\cdot \ldots \cdot \frac{1}{-1} \cdot \frac{1}{1}\cdot \frac{1}{2}\cdot \ldots \cdot \frac{1}{n-k'}\\&= \frac{(-1)^{k'}}{k'! (n-k')!}\\&= \frac{(-1)^{k'}}{n!} \binom{n}{k'},\end{aligned}

As a few examples of this identity we have:

\displaystyle \begin{aligned}n=1&:& \frac{1}{x(x+1)} &= \frac{1}{x} - \frac{1}{x+1}\\n = 2&:& \frac{1}{x(x+1)(x+2)} &= \frac{1}{2} \left[ \frac{1}{x} - \frac{2}{x+1} + \frac{1}{x+2} \right]\\n=3&:& \frac{1}{x(x+1)(x+2)(x+3)} &= \frac{1}{6} \left[ \frac{1}{x} - \frac{3}{x+1} + \frac{3}{x+2} - \frac{1}{x+3}\right] \end{aligned}

I had previously posted an identity with similar, but trickier derivation here.

## July 12, 2020

### Charting Test and ODI cricket performances over consecutive innings

Filed under: cricket,sport — ckrao @ 2:23 am

Adding to my earlier blog post about the highest proportion of test scores above m after n innings, I have created some new interactive charts for best streaks early and mid-career in both bowling and batting in tests and ODIs. The data is mostly from https://data.world/cclayford/cricinfo-statsguru-data.

#### 4. ODI bowling

Below are shown a few sample charts. Clicking on the chart will take you to a new page where you can interact further.

Test batting:

Test bowling:

ODI batting:

ODI bowling:

## July 8, 2020

### The largest parallelogram in a triangle

Filed under: mathematics — ckrao @ 11:59 pm

In this post we find the largest parallelogram, rhombus, rectangle and square that can be contained in a given triangle. We will see that in the first three cases we can achieve half the area of the triangle but no more, while it is generally less than this for a square.

#### 1. The largest parallelogram inside a triangle

It can be readily seen that one can obtain a parallelogram having half the area of a triangle by connecting a vertex with the three midpoints of the sides. (This has half the base length of the triangle and half its height.)

Is it possible to obtain a larger parallelogram? As outlined in [1], if two or fewer vertices of the parallelogram are on sides of the triangle, a smaller similar triangle can be created by drawing a line parallel to the triangle’s side through a vertex of the parallelogram that is interior triangle. (This is done three times in the figure below.) This reduces the problem to the next case.

We are left to consider the case where three or more vertices of the parallelogram on the triangle. We can draw a line from a vertex to the opposite side parallel to a pair of sides (in the figure below $AH$ is drawn parallel to $DG$), thus dissecting the triangle into two. Each of the smaller triangles then has an inscribed parallelogram where two of the vertices are on a side of each triangle. Then by drawing lines parallel to sides if required, we create two sub-problems each having four vertices of the parallelogram on the sides of the triangle.

Finally, if all four vertices of the parallelogram are on sides of the triangle, by the pigeonhole principle, two of them are on a side (say $BC$ as shown in the figure below). In this figure, if we let $AD/AB = k$ and the height of ABC from $BC$ be $h_a$, then by the similarity of triangles $ADE$ and $ABC$, $DE = k.BC$ and $\triangle ADE$ has height $kh_a$. Then the area of the parallelogram $DEFG$ is $DE \times (1-k)h_a = k(1-k) DE.h_a$ which is $2k(1-k)$ times the area of $\triangle ABC$. This quantity has maximum value 1/2 when $k=1/2$ so we conclude that the parallelogram does not exceed half the triangle’s area.

#### 2. The largest rhombus in a triangle

Constraining sides of the parallelogram to be equal (forming a rhombus), we claim that the largest rhombus that can be inscribed in a triangle is also half its area. This can be formed with two of the vertices on the second longest side of the triangle. Suppose $a \geq b \geq c$ are the sides of the triangle with $b=AC$ the second longest side length. Then let the segment $MN$ joining the midpoints of $AB$ and $BC$ form one side of the rhombus of length $b/2$. It remains to be shown that there exist parallel segments of this same length from $M, N$ to $AC$. The longest possible such segment has length $NC = a/2$ and the shortest has length $h_b/2$, half the length of the altitude of $\triangle ABC$ from $B$. We wish to show that $h_b/2 \leq b/2 \leq a/2$. This follows from

$\displaystyle h_b = c \sin A \leq c \leq b \leq a.$

The area of this rhombus is clearly half the area of the triangle as it has half the length of its base and half the height.

#### 3. The largest rectangle in a triangle

Here if $BC$ is the longest side of the triangle we form the rectangle from midpoints $D, E$ of $AB$ and $AC$ respectively, dropping perpendiculars onto $BC$ forming rectangle $DEFG$:

The area of this rectangle is half the area of the triangle as it has half the length of its base and half the height.

Interestingly the reflections of the vertices of the triangle in the sides of the rectangle coincide, showing a paper folding interpretation of this result [2].

Since rhombuses and rectangles are special cases of parallelograms and we found that inscribed parallelograms in a triangle occupy no more than half its area, the rhombus and rectangle constructions here are optimal.

#### 4. The largest square in a triangle

Here we shall see that the best we can do may not be half the area of the triangle. As before, if two or fewer vertices of the square are not on the sides of the triangle it is possible to scale up the square (or scale down the triangle) so that three of the square’s vertices are on the sides. We claim that the largest square must have two of its vertices on a side of the triangle. Suppose this is not the case and we have the figure below.

Consider squares $LMNO$ inscribed in $\triangle ABC$ so that one vertex $L$ is on $AB$, $M$ is on $BC$ and $O$ is on $CA$. We claim that the largest such square is either $DEFG$ (two vertices on $BC$) or $HIJK$ (two vertices on $AC$). Suppose on the contrary that neither of these squares is the largest. Then we make use of the fact that all 90-45-45 triangles $LMO$ inscribed in $\triangle ABC$ have a common pivot point $P$. This is the point at the intersection of the circumcircles of triangles $OAL$, $LBM$ and $MCO$. To show these circles intersect at a single point, we can prove that if the circumcircles of triangles $OAL$ and $LBM$ intersect at $P$ then the points $O,P,M,C$ are cyclic by the following equality:

\displaystyle \begin{aligned} \angle MPO &= 360^{\circ} - \angle LPM - \angle OPL\\ &= (180^{\circ} - \angle LPM) + (180^{\circ} - \angle OPL)\\ &= \angle B + \angle A \\ &= 180^{\circ} - \angle C,\end{aligned}

where the second last equality makes use of quadrilaterals $BMPL$, $ALPO$ being cyclic.

\displaystyle \begin{aligned}\angle BPC &= \angle BPM + \angle MPC\\ &= \angle BLM + \angle MOC\\ &= (\angle LAM + \angle AML) + (\angle MAO + \angle OMA)\\&=(\angle LAM + \angle MAO ) + (\angle AML + \angle OMA )\\&= \angle BAC + \angle OML\\&=\angle A + 45^{\circ}, \end{aligned}

with similar relations for $\angle APB$ and $\angle CPA$. Hence $P$ is the unique point satisfying

\displaystyle \begin{aligned}\angle APB &= \angle ACB + 90^{\circ}\\\angle BPC &= \angle BAC + 45^{\circ}\\\angle CPA &= \angle CBA +45^{\circ}.\end{aligned}

(Each equation defines a circular arc, they intersect at a single point. Note that $P$ may be outside triangle $ABC$.) This point is the centre of spiral similarity of 90-45-45 triangles $LMO$ with $L, M, O$ respectively on the sides $AB, BC, CA$ of the triangle. Consider the locus of the points of the square as $L, M, O$ vary on straight line segments pivoting about $P$. It follows that the fourth point of the square $N$ also traces a line segment, between the points $F$ and $J$ so as to be contained within the triangle.

As the side length of the square is proportional to the distance of a vertex to its pivot point, the largest square will be where $NP$ is maximised. We have seen that the point $N$ varies along a line segment, so $NP$ will be maximised at one of the extreme points – either when $N=F$ or $N=J$. We therefore conclude that the largest square inside a triangle will have two points on a side.

If the triangle is acute-angled, by calculating double the area of the triangle in two ways, the side length $x$ of a square on the side of length $a$ with altitude $h_a$ is derived as

\displaystyle \begin{aligned}ah_a &= 2x^2 + (a-x)x + x(h_a-x)\\&=2x^2 + ax - x^2 + xh_a - x^2\\&= ax + xh_a\\Rightarrow \quad x &= \frac{ah_a}{a + h_a}.\end{aligned}

If the triangle is obtuse-angled, the square erected on a side may not touch both of the other two sides. In the figure below the side length of square $BDEF$ is the same as if $A$ were moved to $G$, where $\triangle GBC$ is right-angled. In this case the square’s side length is $BC.BG/(BC + BG)$.

The largest square erected on a side may be constructed using the following beautiful construction [2]: simply erect a square CBDE external to the side $BC$ and find the intersection points $F = AD \cap BC, G = AE \cap BC$.

These points define the base of the square to be inscribed since by similar triangles

$\displaystyle \frac{IF}{BD} = \frac{AF}{AD} = \frac{FG}{DE} = h_a/(a + h_a)$

so that

$\displaystyle IF = FG = \frac{ah_a}{a + h_a}.$

One can use this interactive demo to view the largest square in any given triangle. One needs to find the largest of the three possibilities of the largest square erected on each side. In the acute-angled-triangle case, the largest square is on the side that minimises the sum of that side length and its corresponding perpendicular height – as their product is fixed as twice the triangle’s area, this will occur when the side and height have minimal difference. For a right-angled triangle with legs $a, b$ and hypotenuse $c$, we wish to compare the quantities $(a+b)$ and $(c+h)$, the two possible sums of the base and height of the triangle. We always have $a + b - c = 2(s-c) = 2r < h$ because the diameter of the incircle of the triangle is shorter than the altitude from the hypotenuse (i.e. the incircle is inside the triangle). We conclude that the largest square in a right-angled triangle is constructed on its two legs rather than its hypotenuse.

#### References

[1] I. Niven, Maxima and Minima without Calculus, The Mathematical Association of America, 1981.

[2] M. Gardner, Some Surprising Theorems About Rectangles in Triangles, Math Horizons, Vol. 5, No. 1 (September 1997), pp. 18-22.

[3] Jaime Rangel-Mondragon “Largest Square inside a Triangle” http://demonstrations.wolfram.com/LargestSquareInsideATriangle/ Wolfram Demonstrations Project Published: March 7 2011

## April 16, 2020

### Highest proportion of test scores above m after n innings

Filed under: cricket,sport — ckrao @ 6:02 am

I created an interactive workbook with Tableau to determine the test batsmen who have had the highest proportion of innings scoring at least m runs after having played n test innings. Below are some screenshots for particular choices of m runs. The data is available from [1]. As expected Bradman comes up on top in many scenarios but it is interesting to see other names that appear up there.

(Click on the above image to go to the Tableau page if you wish to change the parameters. You can also select the “Innings by innings” tab to look up a player’s list of innings.)

Below we some examples for different m (full-career stats among players who have played at least 20 innings).

m=1: A total of 22 players have had an entire career of 20+ innings getting off the mark each time, with RA Duff (Australia, 1902-1905) the only to play 40 innings (note that JW Burke played 44 innings without a duck, but made 0 not out in one innings).

Angelo Mathews (SL) has managed just 2 ducks in his 154 innings to date.

m=10: Hobbs, Hutton, Kanhai and Sobers stand out here, having played over 100 innings and reaching double figures at least 80% of the time (Hobbs over 86%). Labuschagne, Hetmyer, Handscomb and Head are recent players to feature highly here.

m=25: Bradman starts to distance himself from the rest. Hammond, Smith, Sobers and de Villiers also impress here.

m=50: Smith has matched Barrington’s career figures of 50+ starts. Sutcliffe had 33 50+ scores in his first 64 innings, the same as Bradman.

m=75: Barrington’s numbers are amazing here and Katich is ahead of Kohli, Tendulkar and Lara.

m=100: Smith and Kohli are currently higher than Sangakkara, the highest among recent retirees.

m=125: Only Bradman (6) had more 140+ scores than Labuschagne after playing 23 test innings, equal with Graeme Smith (who had 4 150+ scores in his first 17 innings!).

m=150: Bradman is so far ahead of the rest here. Lara and Sangakkara are just one behind Tendulkar with the most 150+ scores despite almost 100 fewer innings.

m=175: Again Lara and Sangakkara have the same number of 175+ scores (15).

m=200: Kohli is similar to Hammond’s career figures at this stage, with 6 of his 7 double centuries coming within 33 innings between July 2016 and December 2017.

## December 7, 2019

### Coordinates of special points of the 3-4-5 triangle

Filed under: mathematics — ckrao @ 3:40 am

One thing I observed is that the 3-4-5 triangle is rather attractive in solving problems using coordinates. If the vertices are placed at (0,3), (0,0) and (4,0) the following are the coordinates of points and equations of some lines of interest.

Line AC: $x/4 + y/3 = 1$

Incentre: $(1, 1)$

Centroid: $(4/3, 1)$

Circumcentre: $(2, 1.5)$

Orthocentre: $(0, 0)$

Nine-point centre: $(1, 3/4)$ (midpoint of the midpoints of AB and BC)

Angle bisectors: $y = x, y=-2x +3, y=4/3-x/3$

Ex-centres (intersection of internal and external bisectors): $(3,- 3), (6, 6), (-2, 2)$

Lines joining the excentres (in red above): $y=-x, y=x/2 +3, y = 3(x-4)$

Altitude to the hypotenuse: $y = 4x/3$

Euler line: $y=3x/4$

Foot of altitude to the hypotenuse: $(36/25, 48/25)$ (where $x/4 + y/3 = 1$ intersects $y=4x/3$)

Symmedian point (midpoint of the altitude to the hypotenuse [1]): $(18/25, 24/25)$

Contact points of incircle and triangle: $(1,0), (0,1), (8/5, 9/5)$

Gergonne point (intersection of Cevians that pass through the contact points of the incircle and triangle = the intersection of $y=3-3x$ and $y=1-x/4$): $(8/11, 9/11)$

Nagel point (intersection of Cevians that pass through the contact points of the ex-circles and triangle = the intersection of $y=3-x$ and $y=2-x/2$: $(2,1)$

#### Reference

[1] Weisstein, Eric W. “Symmedian Point.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/SymmedianPoint.html

## 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 52 occasions, 22 26 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 and Keith West (which both started in 2001). Edited: 20 Dec 2019

 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 49.9 19-Dec-19 Nullarbor SA 49.8 19-Dec-19 Eucla WA 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 19-Dec-19 Forrest WA 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 20-Dec-19 Keith West 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 an inductor and capacitor 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.

Next Page »

Blog at WordPress.com.