# Chaitanya's Random Pages

## April 30, 2015

### Number of countries that have abolished the death penalty by year

Filed under: Uncategorized — ckrao @ 11:36 am

By 1978 only 20 or so countries had abolished the death penalty but over 80 countries have followed suit since. Around 36 countries still retain capital punishment in law and practice with the remaining 50+ countries inactive in its use.

## April 19, 2015

### Distances to a line from vertices of a regular polygon

Filed under: mathematics — ckrao @ 11:01 am

If we take a regular polygon and any line through its centre, then the sum of the squares of the distances from the vertices of the polygon to the line is independent of the orientation of the line or polygon. For example, in the following two diagrams, the sum of the squares of the lengths of the 5 blue line segments is the same.

Such a result is amenable to a proof via complex numbers. Without losing generality the real number line may be our line of interest and the points of the n-sided polygon may be described by the complex numbers $z_k := R\exp(2\pi i k / n + i \phi)$ for $k = 0, 1, ..., n-1$, where $R$ is the circumradius of the polygone and $\phi$ is an arbitrary real-numbered phase. Then the squared distance from a point to the real line is $R^2\cos^2 (2\pi k / n + \phi) = \left(z_k + \overline{z_k})\right)^2/4$ where $\overline{z_k} = R^2/z_k$. Summing this over $k$ gives our sum of squared distances as

\begin{aligned} \frac{1}{4} \sum_{k=0}^{n-1} \left(z_k + \overline{z_k}\right)^2 &= \frac{1}{4} \sum_{k=0}^{n-1} \left(z_k^2 + 2z_k\overline{z_k} + \overline{z_k}^2\right)\\ &= \frac{1}{4} \sum_{k=0}^{n-1} \left(z_k^2 + 2R^2 + \overline{z_k}^2\right)\\ &= \frac{nR^2}{2} + \frac{1}{4} \left(\sum_{k=0}^{n-1} z_k^2 + \sum_{k=0}^{n-1} \overline{z_k}^2\right).\quad\quad(1)\end{aligned}

Each of these sums is geometric in nature so can be simplified. For example,

\begin{aligned}\sum_{k=0}^{n-1} z_k^2 &= \sum_{k=0}^{n-1} R^2\exp(4\pi i k / n + 2i\phi)\\&= R^2 \exp(2i\phi)\sum_{k=0}^{n-1} \exp(4\pi i k / n)\\ &= \begin{cases}R^2 \exp(2i\phi)\frac{1-\exp(4\pi i n / n)}{1- \exp(4\pi i / n)}, & \text{if }\exp(4\pi i/n)\neq 1 \\ R^2 \exp(2i\phi)n, & \text{if }\exp(4\pi i/n)= 1 \end{cases}\\ &= R^2 \exp(2i\phi)\frac{1-\exp(4\pi i)}{1- \exp(4\pi i / n)}, \quad \text{as }n \geq 3\text{ means }\exp(4\pi i/n)\neq 1\\ & = 0.\quad\quad(2)\end{aligned}

Similarly, $\sum_{k=0}^{n-1} \overline{z_k}^2 = 0$, being the conjugate of (2), and so from (1) our required sum of squared distances is $nR^2/2$, which is independent of $\phi$ proving the orientation independence.

More generally,

\begin{aligned}\sum_{k=0}^{n-1} z_k^m &= R^m\exp(i m \phi) \sum_{k=0}^{n-1} \exp(2\pi i k m / n) \\ &= \begin{cases} R^m\exp(i m \phi) n, & \text{if }m/n \text{ is an integer}\\ 0, & \text{otherwise}\end{cases}.\quad\quad(3)\end{aligned}

This enables us to generalise the above result to the following.

Given a regular n-sided polygon and line through its circumcentre, the sum of the mth power signed distances from the vertices of the polygon to the line is independent of the orientation when n > m.

By signed distances, we mean that points on different sides of the line will have distances of opposite sign.

To prove this, we define $z_k := R\exp(2\pi i k / n + i \phi)$ as before and this time our desired sum is

\begin{aligned}\sum_{k=0}^{n-1} \left(\frac{z_k + \overline{z_k}}{2}\right)^m &= \frac{1}{2^m}\sum_{k=0}^{n-1} \sum_{j=0}^m z_k^j \overline{z_k}^{m-j} \binom{m}{j}\\ &= \frac{1}{2^m}\sum_{k=0}^{n-1} \sum_{j=0}^m z_k^j (R^2/z_k)^{m-j}\binom{m}{j}\\ &= \frac{1}{2^m}\sum_{k=0}^{n-1} \sum_{j=0}^m z_k^{2j-m} (R^2)^{m-j}\binom{m}{j}\\ &= \frac{1}{2^m} \sum_{j=0}^m (R^2)^{m-j}\binom{m}{j}\sum_{k=0}^{n-1}z_k^{2j-m}. \quad\quad(4)\\ \end{aligned}

By (3), $\sum_{k=0}^{n-1}z_k^{2j-m} = 0$ unless $(2j-m)/n$ is an integer. For $m < n$ this can only occur in the case $j = m/2$ (if $m$ is even). Hence (4) becomes

\begin{aligned}\sum_{k=0}^{n-1} \left(\frac{z_k + \overline{z_k}}{2}\right)^m &= \begin{cases} \left(\frac{R}{2}\right)^m\binom{m}{m/2}n, & \text{if }m\text{ is even,}\\ 0, & \text{otherwise.}\end{cases}\end{aligned} \quad\quad(5)

Finally, what if the line does not pass through the centre of the polygon but is instead at distance $d$ from the centre?

This corresponds to replacing $\left(\frac{z_k + \overline{z_k}}{2}\right)$ with $\left(\frac{z_k + \overline{z_k}}{2}-d\right)$ in the above calculations and we find

\begin{aligned} \sum_{k=0}^{n-1} \left(\frac{z_k + \overline{z_k}}{2}-d\right)^m &= \sum_{k=0}^{n-1} \sum_{j = 0}^m\left(\frac{z_k + \overline{z_k}}{2}\right)^j (-d)^{m-j} \binom{m}{j}\\ &= \sum_{j = 0}^m (-d)^{m-j} \binom{m}{j}\sum_{k=0}^{n-1} \left(\frac{z_k + \overline{z_k}}{2}\right)^j\\ &= \sum_{i = 0}^{\lceil{m/2}\rceil} (-d)^{m-2i} \binom{m}{2i} \left(\frac{R}{2}\right)^{2i}\binom{2i}{i} n\\ &= (-1)^m n \sum_{i = 0}^{\lceil{m/2}\rceil} \frac{m!}{(m-2i)!i!i!}d^{m-2i}\left(\frac{R}{2}\right)^{2i}.\quad\quad(6)\end{aligned}

Once again we find the sum is independent of the orientation of the line or polygon. In the particular case of $m=2$ this sum is $nR^2/2 + nd^2$, which also may be obtained by an application of the parallel axis theorem.

## March 31, 2015

### 2015 ICC Cricket World Cup Attendances

Filed under: cricket,sport — ckrao @ 10:26 am

The 2015 ICC Cricket World Cup held in Australia and New Zealand saw over one million people attend its 49 matches, with attendances shown below, taken from here.

If we compare these numbers with the ground capacities shown below (largely taken from ground pages at ESPNcricinfo), we can make a graph of ground occupancy for each game.

 Ground Capacity Melbourne Cricket Ground 95,000 Adelaide Oval 50,000 Sydney Cricket Ground 44,000 Eden Park, Auckland 41,000 Gabba, Brisbane 37,000 Westpac Stadium, Wellington 33,500 WACA Ground, Perth 24,500 Hagley Oval, Christchurch 18,000 Blundstone Arena, Hobart 16,200 Manuka Oval, Canberra 12,000 Seddon Park, Hamilton 12,000 McLean Park, Napier 10,500 Saxton Oval, Nelson 6,000 University Oval, Dunedin 5,000

We see that most of the first 10 matches (up to the wash-out between Australia and Bangladesh) were close to capacity, while some of the later matches leading up the quarter final had attendances well under 50% of capacity.

Finally, the table below shows the total attendances for games involving each country, as well as average ground occupancy percentages for their matches.

 Team total attendance average % occupancy #matches attendance per match New Zealand 277,024 94.2 9 30,780 Australia 360,086 83.7 8 45,011 India 288,888 73.8 8 36,111 South Africa 223,803 65.1 8 27,975 Scotland 39,518 63.7 6 6,586 Afghanistan 48,847 63.1 6 8,141 West Indies 96,788 60.4 7 13,827 Sri Lanka 138,893 58.6 7 19,842 England 166,221 57.8 6 27,704 Bangladesh 118,337 57.6 6 19,723 Pakistan 136,419 51.3 7 19,488 Ireland 42,352 47.8 6 7,059 Zimbabwe 60,490 47.4 6 10,082 UAE 25,138 23.8 6 4,190

Australia had a lowish attendance for its game against Afghanistan while New Zealand had every match at least 86% full.

## March 28, 2015

### A triangle centre arising from central projections

Filed under: mathematics — ckrao @ 10:23 pm

In this recent post, the following figure was formed by considering two triangular regions in three-dimensional space, where $a,b,c > 0$:

$\displaystyle ax + by + cz = 1, 0 \leq x, 0 \leq y, 0 \leq z,$

$\displaystyle ax + by + cz = 1, x \leq 1, y \leq 1, z \leq 1.$

The intersection of the two triangles is the cross-section of a cube, but in this post we wish to explore further the centre of similarity of the two triangles.

The line joining $(1/a, 0, 0)$ and $((1-b-c)/a, 1, 1)$ satisfies

\begin{aligned} (x,y,z) &= (1/a, 0, 0) + t\left[ ((1-b-c)/a, 1, 1)-(1/a, 0, 0) \right]\\ &= (t,t,(1-ta-tb)/c), t \in \mathbb{R}. \end{aligned}

Similarly, the line joining $(0,1/b,0)$ and $(1,(1-a-c)/b,1)$ satisfies

$(x,y,z) = (u, (1-ua-uc)/b, u), u \in \mathbb{R}.$

Equating the two expressions gives $t = u$ and $t = (1-ta-tc)/b$ from which $t = u = 1/(a+b+c)$. The point of intersection is therefore at $(t,t,(1-ta-tb)/c) = (1/(a+b+c), 1/(a+b+c), 1/(a+b+c))$. By symmetry of this expression the line joining $(0,0,1/c)$ and $(1,1,(1-a-b)/c)$ also passes through this point. This is the point on the plane $ax + by + cz = 1$ that is equi-distant from the xy-, yz- and xz- coordinate planes. It is also the central projection of the origin onto the plane along the vector parallel to $(1,1,1)$.

In terms of the original two triangles this point is neither the centroid, incentre, orthocentre, circumcentre nor other commonly encountered triangle centre. Let us find the barycentric coordinates of this point (call it $S$) in terms of the triangle with vertices at $A(1/a,0,0), B(0,1/b,0), C(0,0,1/c)$.

The first barycentric coordinate will be the ratio of the area of $\triangle SBC$ to the area of $\triangle ABC$. Since $B$ and $C$ have the same x-coordinate, this will be the ratio of the x-coordinates of $S$ to $A$, which is $1/(a+b+c) / (1/a) = a/(a+b+c)$. By symmetry it follows that the barycentric coordinates have the attractive form

$\displaystyle \frac{a}{a+b+c} : \frac{b}{a+b+c} : \frac{c}{a+b+c}.$

Let the side lengths of $\triangle ABC$ be $BC = x, CA = y, AB = z$. Then by Pythagoras’ theorem, $x^2 = 1/b^2 + 1/c^2, y^2 = 1/a^2 + 1/c^2, z^2 = 1/a^2 + 1/b^2$. Hence

$x^2 + y^2 - z^2 = 2/c^2$.

By the cosine rule, $x^2 + y^2 - z^2 = 2xy \cos C$ (where $C = \angle ACB$) which equals $2/c^2$ from the above expression. Therefore $c^2 = \sec C/xy$ and similarly we obtain $a^2 = \sec A/yz, b^2 = \sec B/xz$. Then

\begin{aligned} \frac{a}{a+b+c} &= \frac{\sqrt{\sec A}/\sqrt{yz}}{\sqrt{\sec B}/\sqrt{xz} + \sqrt{\sec C}/\sqrt{xy} + \sqrt{\sec A}/\sqrt{yz}}\\ &= \frac{\sqrt{x\sec A}}{\sqrt{x\sec A} + \sqrt{y\sec B} + \sqrt{z\sec C}}.\end{aligned}

By the sine rule, $x = 2R\sin A$ ($R$ being the circumradius of $\triangle ABC$) from which $x \sec A = 2R\sin A/\cos A = 2R\tan A$. Hence the barycentric coordinates of $S$ may be written in non-normalised form as

$\displaystyle {\sqrt{\tan A}} : {\sqrt{\tan B}} : {\sqrt{\tan C}}.$

Comparing this with the coordinates of the orthocentre $\tan A : \tan B : \tan C$, the point $S$ is known as the square root of the orthocentre (see Theorem 1 of [1]). Note that the real existence of the point requires $\triangle ABC$ to be acute, which it is when $a,b,c > 0$. A geometric construction of the square root of a point is given in Section 8.1.2 of [2].

#### References

[1] Miklós Hoffmann, Paul Yiu. Moving Central Axonometric Reference Systems, Journal for Geometry and Graphics, Volume 9 (2005), No. 2, 127–134.

[2] Paul Yiu. “Introduction to the Geometry of the Triangle”. http://math.fau.edu/Yiu/GeometryNotes020402.pdf

## February 28, 2015

### Large US+Canada box office openings

Filed under: movies and TV — ckrao @ 1:38 pm

The site boxofficemojo.com lists the opening weekend grosses of movies in the US and Canada dating back to the early 1980s. Via this page on top opening weekends, I worked out movies that at their time of release attained the n’th highest grossing opening weekend where n ranges from 1 to 10 (all dollar amounts in $US). It gives a perspective on how big some movies were at the time. It also shows how movie grosses have grown through inflation and more frontloading over the years. Note that only opening weekends are shown here – for example Superman’s third weekend was once the largest grossing weekend at the time, but is not listed here. n=1 (i.e. current and previous record-breaking openings):  Title Opening Date (mm/dd/yyyy) Marvel’s The Avengers$207,438,708 5/4/2012 Harry Potter and the Deathly Hallows Part 2 $169,189,427 7/15/2011 The Dark Knight$158,411,483 7/18/2008 Spider-Man 3 $151,116,516 5/4/2007 Pirates of the Caribbean: Dead Man’s Chest$135,634,554 7/7/2006 Spider-Man $114,844,116 5/3/2002 Harry Potter and the Sorcerer’s Stone$90,294,621 11/16/2001 The Lost World: Jurassic Park $72,132,785 5/23/1997 Batman Forever$52,784,433 6/16/1995 Jurassic Park $47,026,828 6/11/1993 Batman Returns$45,687,711 6/19/1992 Batman $40,489,746 6/23/1989 Ghostbusters II$29,472,894 6/16/1989 Indiana Jones and the Last Crusade $29,355,021 5/24/1989 Beverly Hills Cop II$26,348,555 5/20/1987 Indiana Jones and the Temple of Doom $25,337,110 5/23/1984 Return of the Jedi$23,019,618 5/25/1983 Star Trek II: The Wrath of Khan $14,347,221 6/4/1982 Superman II$14,100,523 6/19/1981 Star Trek: The Motion Picture $11,926,421 12/7/1979 Every Which Way But Loose$10,272,294 12/20/1978

n=2 (i.e. the second largest opening at the time)

 Title Opening Date Iron Man 3 $174,144,585 5/3/2013 Star Wars: Episode III – Revenge of the Sith$108,435,841 5/19/2005 Shrek 2 $108,037,878 5/19/2004 The Matrix Reloaded$91,774,413 5/15/2003 Planet of the Apes (2001) $68,532,960 7/27/2001 The Mummy Returns$68,139,035 5/4/2001 Star Wars: Episode I – The Phantom Menace $64,820,970 5/19/1999 Independence Day$50,228,264 7/3/1996 Lethal Weapon 3 $33,243,086 5/15/1992 Terminator 2: Judgment Day$31,765,506 7/3/1991 Rocky III $12,431,486 5/28/1982 The Cannonball Run$11,765,654 6/19/1981 Smokey and the Bandit II $10,883,835 8/15/1980 The Empire Strikes Back$10,840,307 6/20/1980

n=3:

 Title Opening Date The Dark Knight Rises $160,887,295 7/20/2012 The Hunger Games$152,535,747 3/23/2012 The Twilight Saga: New Moon $142,839,137 11/20/2009 Shrek the Third$121,629,270 5/18/2007 Harry Potter and the Prisoner of Azkaban $93,687,367 6/4/2004 Harry Potter and the Chamber of Secrets$88,357,488 11/15/2002 Star Wars: Episode II – Attack of the Clones $80,027,814 5/16/2002 Hannibal$58,003,121 2/9/2001 Mission: Impossible II $57,845,297 5/24/2000 Toy Story 2$57,388,839 11/24/1999 Austin Powers: The Spy Who Shagged Me $54,917,604 6/11/1999 Men in Black$51,068,455 7/2/1997 The Lion King $40,888,194 6/24/1994 Rambo: First Blood Part II$20,176,217 5/22/1985 Star Trek III: The Search for Spock $16,673,295 6/1/1984 n=4:  Title Opening Date X-Men: The Last Stand$102,750,665 5/26/2006 Harry Potter and the Goblet of Fire $102,685,961 11/18/2005 X2: X-Men United$85,558,731 5/2/2003 Austin Powers in Goldmember $73,071,188 7/26/2002 Rush Hour 2$67,408,222 8/3/2001 Pearl Harbor $59,078,912 5/25/2001 Mission: Impossible$45,436,830 5/22/1996 Twister $41,059,405 5/10/1996 Back to the Future Part II$27,835,125 11/22/1989 Rocky IV $19,991,537 11/27/1985 Beverly Hills Cop$15,214,805 12/5/1984 Jaws 3-D $13,422,500 7/22/1983 Superman III$13,352,357 6/17/1983

n=5:

 Title Opening Date The Twilight Saga: Breaking Dawn Part 1 $138,122,261 11/18/2011 Iron Man 2$128,122,480 5/7/2010 Pirates of the Caribbean: At World’s End $114,732,820 5/25/2007 How the Grinch Stole Christmas$55,082,330 11/17/2000 Interview with the Vampire $36,389,705 11/11/1994 Home Alone 2: Lost in New York$31,126,882 11/20/1992 Bram Stoker’s Dracula $30,521,679 11/13/1992 Star Trek IV: The Voyage Home$16,881,888 11/26/1986 The Best Little Whorehouse in Texas $11,874,268 7/23/1982 E.T.: The Extra-Terrestrial$11,835,389 6/11/1982

n=6:

 Title Opening Date The Hunger Games: Catching Fire $158,074,286 11/22/2013 Harry Potter and the Deathly Hallows Part 1$125,017,372 11/19/2010 Alice in Wonderland (2010) $116,101,023 3/5/2010 The Passion of the Christ$83,848,082 2/25/2004 Monsters, Inc. $62,577,067 11/2/2001 X-Men$54,471,475 7/14/2000 Ace Ventura: When Nature Calls $37,804,076 11/10/1995 Robin Hood: Prince of Thieves$25,625,602 6/14/1991 Total Recall $25,533,700 6/1/1990 Teenage Mutant Ninja Turtles$25,398,367 3/30/1990 Ghostbusters $13,578,151 6/8/1984 Staying Alive$12,146,143 7/15/1983

n=7:

 Title Opening Date Transformers: Revenge of the Fallen $108,966,307 6/24/2009 Spider-Man 2$88,156,227 6/30/2004 Batman and Robin $42,872,605 6/20/1997 Lethal Weapon 2$20,388,800 7/7/1989 Star Trek V: The Final Frontier $17,375,648 6/9/1989 n=8:  Title Opening Date The Twilight Saga: Breaking Dawn Part 2$141,067,634 11/16/2012 The Lord of the Rings: The Return of the King $72,629,713 12/17/2003 Godzilla$44,047,541 5/20/1998 The Flintstones $29,688,730 5/27/1994 Gremlins$12,511,634 6/8/1984

n=9:

 Title Opening Date Finding Nemo $70,251,710 5/30/2003 The Mummy$43,369,635 5/7/1999 Deep Impact $41,152,375 5/8/1998 n=10:  Title Opening Date Toy Story 3$110,307,189 6/18/2010 Indiana Jones and the Kingdom of the Crystal Skull $100,137,835 5/22/2008 Iron Man$98,618,668 5/2/2008 Dick Tracy \$22,543,911 6/15/1990

(To create the above lists the movie lists in decreasing order of gross were pasted into Excel and the opening weekend date was converted to a number by creating a new column with formula =–TEXT(,”mm/dd/yyyy”). This was then converted to a rank by a countif formula to count the number of occurrences with higher gross that predated each movie. Finally a filter was applied to select ranks 1 to 10.)

## February 27, 2015

### Cross sections of a cube

Filed under: mathematics — ckrao @ 9:59 pm
When a plane intersects a cube there is a variety of shapes of the resulting cross section.
• a single point (a vertex of the cube)
• a line segment (an edge of the cube)
• a triangle (if three adjacent faces of the cube are intersected)
• a parallelogram (if two pairs of opposite faces are intersected – this includes a rhombus or rectangle)
• a trapezium (if two pairs of
• a pentagon (if the plane meets all but one face of the cube)
• a hexagon (if the plane meets all faces of the cube)

The last five of these (the non-degenerate cases) are illustrated below and at http://cococubed.asu.edu/images/raybox/five_shapes_800.png . Some are demonstrated in the video below too.

One can use [1] to experiment interactively with cross sections given points on the edges or faces, while [2] shows how to complete the cross section geometrically if one is given three points on the edges.

Let us be systematic in determining properties of the cross sections above. Firstly, if the plane is parallel to an edge (any of four parallel edges), the cross section can be seen to be a line or rectangle with the longer dimension of length at most $\sqrt{2}$ times the other. That rectangle becomes a square if the plane is parallel to a face.

If the plane is not parallel to a face, we may set up a coordinate system where a unit cube is placed in the first octant aligned with the coordinate axes and the normal to the plane has positive x, y and z coordinates. In other words, we may assume the plane has equation $ax + by + cz = 1$ and intercepts at $(1/a,0,0), (0,1/b,0)$ and $(0,0,1/c)$, where $a, b, c$ are positive.

The cross section satisfies $ax + by + cz = 1$ and the inequalities $0 \leq x \leq 1$, $0 \leq y \leq 1$ and $0 \leq z \leq 1$. This can be considered the intersection of the two regions

$\displaystyle ax + by + cz = 1, 0 \leq x, 0 \leq y, 0 \leq z,$

$\displaystyle ax + by + cz = 1, x \leq 1, y \leq 1, z \leq 1,$

each of which is an acute-angled triangle in the same plane (acute because one can show that the sum of the squares of any two sides is strictly greater than the square of the third side). Note that the triangles have parallel corresponding sides, being bounded by the pairs of parallel faces of the cube $x = 0, x=1, y = 0, y= 1, z=0, z=1$. Hence the two triangles are oppositely similar with a centre of similarity.

The following diagram shows the coordinates of the vertices of the two triangles, which in this case intersect in a hexagon.

The centre of similarity of the two triangles is the intersection of two lines joining corresponding sides – this can be found to be the point $(1/(a+b+c), 1/(a+b+c), 1/(a+b+c))$, which is the intersection of the unit cube’s diagonal from the origin (to $(1,1,1)$) and the plane $ax + by + cz = 1$.

Side lengths of the triangles and distances between corresponding parallel sides may be found by Pythagoras’ theorem and are shown below for one pair of corresponding sides (the remaining lengths can be found by cyclically permuting $a,b,c$).

To sum up, all of the possible cross sections of a cube where the plane is not parallel to an edge can be described by the intersection of two oppositely similar triangles with corresponding sides parallel.

The type of polygon obtained depends on which vertices of the figure below are selected, as determined by the values of $a,b,c$.

In this figure a vertex for the cross-sectional polygon is chosen if the constraint associated with it is satisfied. A red vertex has a conflicting constraint with its neighbouring two blue vertices, so either a red point or one or more blue points in this area can be chosen. Note that for the plane to intersect the cube at all we require $(1,1,1)$ to be on the different side of the plane from the origin, or in other words, $a + b + c \geq 1$.

Let us look at a few examples. Firstly, if $a, b, c$ are all greater than 1 we choose the following triangle.

Similarly if $a+b, b+c, c+a$ all are less than 1, the oppositely similar triangle on the red vertices would be chosen.

Next, if $c > 1, a < 1, a+b < 1$ we obtain the following parallelogram.

If $c > 1, a < 1, a+b > 1$ we obtain either a pentagon (parallelogram truncated at a vertex) or a trapezium depending on whether $b < 1$ or $b \geq 1$ respectively.

$b < 1$:

$b\geq 1$:

Finally, if $a, b, c$ are less than 1 and $a+b, b+c, c+a$ are greater than 1, we obtain a hexagon.

For details on calculating the areas of such polygons refer to [3], especially the method applying the area cosine principle that relates an area of a figure to its projection. For calculating volumes related to regions obtained by the cross section refer to [4].

#### References

[1] Cross Sections of a Cube: http://www.wou.edu/~burtonl/flash/sandbox.html

[2] Episode 16 – Cross sections of a cube: http://sectioneurosens.free.fr/docs/premiere/s02e16s.pdf

## January 30, 2015

### AB de Villiers’ fastest ODI century

Filed under: cricket,sport — ckrao @ 12:51 pm

Recently we witnessed the fastest century in one-day cricket history, with AB de Villiers coming in during the 39th over at the fall of South Africa’s first wicket (1/247) and proceeding to blast an incredible 149 off 44 balls in just 59 minutes propelling the team to 2/439 (Amla also made 153* and Rossouw 128). The match produced all types of records including many instances of most runs in n consecutive overs, documented in other blog post here.

Here is the ball-by-ball breakdown of his innings followed by a graph of runs versus balls.

4 2 1 | 1 4 6 4 6 | 6 . 2 2 LB | 1 6 6 6 4 | 6 1 4 1 | 1 . 4 4 | . 6 | 4 6 6 6 | LB 6 | 1 | 6 6 4 6 6 2 | 2 . out

We see that three times he hit 28 runs in 5 balls and 26 runs in another space of 5 balls – that’s 110 runs in just 20 balls with 15 6s and 5 4s right there! He was already 82 off 27 balls (7 6s, 7 4s) and then made 63 off his next 13 balls including 9 6s and 2 4s to surge to 145 off 40 balls!! I have never seen such a concentration of 6 hitting. More analysis of his strike rate is at this @dualnoise post.

Here is what he scored (with balls faced) against the four bowlers who were up against him:

Taylor: 30 (13)
Holder: 45 (9)
Russell: 35 (12)
Smith: 39 (10)

Some other amazingly fast centuries in limited over cricket are in this blog post. Here is the breakdown of the previous fastest ODI century (36 balls) by Corey Anderson in this match:

1 | . 4 1 | . 1 | 6 1 . 2 1 | 4 6 1 | 6 6 . 6 . 6 | 1 . . 6 | 6 6 6 6 1 | . 1 | 4 4 1 1 | 6 6 4 1 | 2 4 1 | 6 1 | 2 2 1

## January 26, 2015

### Affine Transformations of Cartesian Coordinates

Filed under: mathematics — ckrao @ 11:43 am

A very common exercise in high school mathematics is to plot transformations of some standard functions. For example, to plot $y = 2\sin (5x + \frac{2\pi}{3}) - 1$ we may start with a standard sine curve and apply the following transformations in turn:

• squeeze it by a factor of 5 in the $x$-direction
• shift it left by $\frac{2\pi}{3}$
• stretch it by a factor of 2 in the $y$-direction
• shift it down by $1$

This leads to the plot shown.

For sine and cosine graphs an alternative is to plot successive peaks/troughs of the curve and interpolate accordingly. For example, to plot $y = 2\sin (5x + \frac{2\pi}{3}) - 1$ we may proceed as follows.

• Since $\sin(x)$ has a peak at $\frac{\pi}{2}$, solve $5x + \frac{2\pi}{3} = \frac{\pi}{2}$ to find $x = -\frac{\pi}{30}$ as a point where there is a peak at $y = 2\times 1 - 1 = 1$. Hence plot the point $(-\frac{\pi}{30},1)$.
• Since the angular frequency is 5, the period is $\frac{2\pi}{5}$ and we may plot successive peaks spaced $\frac{2\pi}{5}$ apart from the point $(-\frac{\pi}{30},1)$.
• Troughs will be equally spaced halfway between the peaks at $y = 2\times (-1) - 1 = -3$ (at $x = -\frac{\pi}{30} + \frac{\pi}{5} + k\frac{2\pi}{5}$). Then join the dots with a sinusoidal curve.
• Additionally $x-$ and $y-$ intercepts may be found by setting $y = 0$ and $x = 0$ respectively. We find that the $x-$intercepts are at $x = \frac{2\pi k}{5} - \frac{\pi}{10}, \frac{2\pi k}{5} + \frac{\pi}{30}\ (k \in \mathbb{Z})$ and $y-$intercept is $y = 2\sin (\frac{2\pi}{3}) - 1 = \sqrt{3}-1$.

The first approach is more generalisable to plotting other functions. Instead of thinking of the graph transforming, we also may consider it as a change of coordinates. For example, if we translate the parabola $y = x^2$ so that its turning point is at $(2,1)$, this is equivalent to keeping the parabola fixed and shifting axes so that the new origin is at $(-2,-1)$ with respect to the old coordinates. This is illustrated below where the black coordinates are modified to the red ones. The parabola has equation $y = x^2$ under the black coordinates and $y - 1 = (x-2)^2$ or $y = (x-2)^2 + 1$ under the red coordinates.

As another example suppose we take a unit circle and stretch it by a factor of $2$ in the $x$ direction and a factor of $3$ in the $y$ direction. This is the equivalent of changing scale so that the $x$-axis is squeezed by $2$ and the $y$-axis is squeezed by $3$.

Under this stretching of the circle or squeezing of axes, the unit circle equation $x^2 + y^2 = 1$ transforms to that of the ellipse $\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = 1$.

More generally, by stretching a Cartesian graph $f(x,y) = 0$ by $a$ in the $x$-direction and $b$ in the $y$-direction, then shifting it along the vector $(h,k)$, we obtain the equation

$\displaystyle f\left(\frac{x-h}{a}, \frac{y-k}{b}\right) = 0.$

This uses the fact that $\frac{x-h}{a}$ and $\frac{y-k}{b}$ are the inverses of $ax+h$ and $by+k$ respectively. Note that if $|a|$ or $|b|$ are less than 1, the stretch becomes a squeezing of the graph, while $a < 0$ or $b < 0$ correspond to a reflection in the $x$ or $y$ axes.

We can extend this idea to the rotation of a graph. Suppose for example we wish to rotate the hyperbola $y = 1/x$ by 45 degrees anti-clockwise. This is equivalent to a rotation of the axes by 45 degrees clockwise and the matrix corresponding to this linear transform is

$\displaystyle \left[ \begin{array} {c} x' \\ y' \end{array} \right] = \frac{1}{\sqrt{2}}\left[ \begin{array}{rr} 1 & 1\\ -1 & 1 \end{array} \right] \left[ \begin{array} {c} x \\ y \end{array} \right] = \left[ \begin{array} {c} \frac{x+y}{\sqrt{2}} \\ \frac{-x+y}{\sqrt{2}}\end{array} \right].$

(Here the columns of the change of basis matrix correspond to where the basis vectors (1,0) and (0,1) map to under a 45 degree clockwise rotation.)

In other words we replace $x'$ with $(x+y)/\sqrt{2}$ and $y'$ with $(-x+y)/\sqrt{2}$ in the equation $y' = 1/x'$ and obtain $(-x+y)/\sqrt{2} = 1/( (x+y)/\sqrt{2})$ or $y^2 - x^2 = 2$.

Here is the same transformation applied to the parabola $y = x^2$ to obtain $\frac{-x+y}{\sqrt{2}} = \frac{(x+y)^2}{2}$ or $x^2 + y^2 + 2xy + \sqrt{2}(x-y) = 0$:

If a graph is affinely transformed (by an invertible map) so that $(1,0)$ maps to $(a,b)$ and $(0,1)$ maps to $(c,d)$ followed by a shift along the vector $(h,k)$, then this is equivalent to the coordinates shifting by $(-h,-k)$ and then transforming under the inverse mapping $\displaystyle \left[ \begin{array}{cc} a & c\\ b & d\end{array} \right]^{-1} = \frac{1}{ad-bc} \left[ \begin{array}{cc} d & -c\\ -b & a\end{array} \right]$:

\displaystyle \boxed{\begin{aligned} x' &= (x-h) \frac{d}{ad-bc} - (y-k) \frac{c}{ad-bd}\\ y' &= -(x-h) \frac{b}{ad-bc} + (y-k) \frac{a}{ad-bd}\end{aligned}}

Here are some special cases of this formula:

• rotation of the graph by $\theta$ anti-clockwise: $\displaystyle x' = x\cos \theta + y \sin \theta, y' = - x \sin \theta + y\cos \theta$ (the example $\theta = \pi/4$ was done above)
• reflection of the graph in $y = x$: $x' = y, y' = x$
• reflection of the graph in $y = mx$ where $m = \tan \theta$: verify that $a = \cos 2\theta, b = \sin 2\theta, c = \sin 2\theta, d = -\cos 2\theta$ so $\displaystyle x' = x \cos 2\theta + y \sin 2\theta, y' = x \sin 2\theta - y \cos 2\theta$
• reflection of the graph in $y = mx + c$ where $m = \tan \theta$: this is equivalent to a reflection in the line $y = mx$ followed by a shift along the vector $(h,k) = (-c \sin 2\theta, c (1+\cos 2 \theta) )$ so
\displaystyle \begin{aligned} x' &= (x+c \sin 2\theta)\cos 2\theta + (y-c(1+\cos 2\theta))\sin 2\theta\\ &= x \cos 2\theta + (y-c)\sin 2\theta,\\ y' &= (x+c \sin 2\theta)\sin 2\theta - (y-c(1+\cos 2\theta))\cos 2\theta\\ &= x \sin 2\theta - (y-c) \cos 2\theta + c \end{aligned}
• reflection of the graph in $y = x + c$ (special instance of the previous case with $\sin 2\theta = 1, \cos 2\theta = 0$): $x'= y - c, y' = x+c$

## December 15, 2014

### Types of -saurs that are not dinosaurs

Filed under: nature — ckrao @ 12:06 pm

Below is a reference for myself of types of (mostly) prehistoric animals that are not dinosaurs but have names ending in -saur (sauria means lizard but most of these are not that closely related to lizards).

 Group Prefix meaning When it lived Brief description Aetosaur eagle late Triassic heavily armoured Anteosaur Antaeus (son of Poseidon and Gaia) 272-260 Ma large Dinocephalians (therapsid) Cotylosaur cup late Carboniferous-Permian basal reptile (also known as Captorhinids) Ichthyosaur fish 245-90 Ma dolphin-like marine reptile Ictidosaur ferret late Triassic – mid Jurassic mammal-like cynodonts, also known as tritheledontids Mesosaur middle 299-280 Ma like a small aligator Mosasaur Meuse River late Cretaceous marine reptile similar to monitor lizards Nothosaur false/hybrid Triassic slender marine reptile Pachypleurosaur thick-ribbed Triassic like an aquatic lizard Pareiasaur shield 270-250 Ma large anapsid Pelycosaur axe or bowl 320-251 Ma non-therapsid synapsids (e.g. Dimetrodon) Phytosaur plant 228-200 Ma long-snouted archosauriforms Plesiosaur close to/near 210-65 Ma marine reptile with broad flat body and short tail Pliosaur closely 200-89 Ma short-necked plesiosaur Poposaur discovered on Popo Agie River (ref) late Triassic carnivorous paracrocodylomorphs Protorosaur early Permian-Triassic long-necked archosauromorphs Pterosaur winged 228-65 Ma closest relatives to dinosauromorphs Rhynchosaur beaked Triassic beaked archosauromorphs Teleosaur end/last early Jurassic – early Cretaceous marine crocodyliforms Thalattosaur ocean Triassic marine reptile with long flat tail Trilophosaur three ridged late Triassic lizard-like archosauromorphs Xenosaur strange present (Cenozoic) knob-scaled lizards

## December 14, 2014

### Basic combinatorics results

Filed under: mathematics — ckrao @ 8:11 pm

The following lists most of the introductory combinatorics formulas one might see in a first course expressed in terms of the number of arrangements of letters in which repetition or order matters.

 number of letters alphabet size letters repeated? order matters? formula comments $k$ $n$ yes yes $n^k$ samples with replacement $k$ $n$ no yes \begin{aligned} & n(n-1)\ldots (n-k+1)\\ &= \frac{n!}{(n-k)!} = P(n,k) = (n)_k\end{aligned} samples without replacement (permutations) $n$ $n$ no yes $n!$ if letters in a line $(n-1)!$ if letters in a ring and rotations are considered equivalent $(n-1)!/2$ if letters in a ring and rotations & reflections are considered equivalent $k$ $n$ no no $\frac{n!}{k!(n-k)!} = \binom{n}{k} = C(n,k)$ binomial coefficient (combinations) $n$ 2 yes: $k$ of type 1, $n-k$ of type 2 yes $\binom{n}{k}$ $n$ $m$ yes: $k_1$ of type $i$ for $i = 1,\ldots, m$ yes $\frac{n!}{k_1! k_2! \ldots k_m!} = \frac{(k_1 + k_2 + \ldots + k_m)!}{k_1! k_2! \ldots k_m!} = \binom{n}{k_1, k_2, \ldots, k_m}$ multinomial coefficient (multiset permutations) e.g. number of arrangements of “BANANA” is $\frac{6!}{3!2!1!}$ $n$ $k$ yes no $\binom{n+k-1}{k-1} = \binom{n+k-1}{n} = \left(\!\!{n \choose k}\!\!\right)$ multiset coefficient (combinations with repetition): number of non-negative integer solutions to $n_1 + n_2 + \ldots + n_k = n$ $n$ $2$ no two consecutive letters of type 1 yes $F_{n+2}$ Fibonacci number where $F_{n+2} = F_{n+1} + F_n$ and $F_1 = F_2 = 1$ e.g. number of sequences of 6 coin tosses with no two consecutive heads is $F_8 = 21$
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