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.

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 for , where is the circumradius of the polygone and is an arbitrary real-numbered phase. Then the squared distance from a point to the real line is where . Summing this over gives our sum of squared distances as

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

Similarly, , being the conjugate of (2), and so from (1) our required sum of squared distances is , which is independent of proving the orientation independence.

More generally,

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 as before and this time our desired sum is

By (3), unless is an integer. For this can only occur in the case (if is even). Hence (4) becomes

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

This corresponds to replacing with in the above calculations and we find

Once again we find the sum is independent of the orientation of the line or polygon. In the particular case of this sum is , which also may be obtained by an application of the parallel axis theorem.

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.

Match #

Date

Match

Venue

Attendance

1

14/02/15

New Zealand d Sri Lanka

Hagley Oval, Christchurch

17,228

2

14/02/15

Australia d England

Melbourne Cricket Ground

84,336

3

15/02/15

South Africa d Zimbabwe

Seddon Park, Hamilton

8,332

4

15/02/15

India d Pakistan

Adelaide Oval

41,587

5

16/02/15

Ireland d West Indies

Saxton Oval, Nelson

4,143

6

17/02/15

New Zealand d Scotland

University Oval, Dunedin

4,684

7

18/02/15

Bangladesh d Afghanistan

Manuka Oval, Canberra

10,972

8

19/02/15

Zimbabwe d UAE

Saxton Oval, Nelson

2,643

9

20/02/15

New Zealand d England

Westpac Stadium, Wellington

30,148

10

21/02/15

West Indies d Pakistan

Hagley Oval, Christchurch

14,461

11

21/02/15

Australia v Bangladesh

Gabba, Brisbane

washed out

12

22/02/15

Sri Lanka d Afghanistan

University Oval, Dunedin

2,711

13

22/02/15

India d South Africa

Melbourne Cricket Ground

86,876

14

23/02/15

England d Scotland

Hagley Oval, Christchurch

12,388

15

24/02/15

West Indies d Zimbabwe

Manuka Oval, Canberra

5,544

16

25/02/15

Ireland d UAE

Gabba, Brisbane

5,249

17

26/02/15

Afghanistan d Scotland

University Oval, Dunedin

3,229

18

26/02/15

Sri Lanka d Bangladesh

Melbourne Cricket Ground

30,012

19

27/02/15

South Africa d West Indies

Sydney Cricket Ground

23,612

20

28/02/15

New Zealand d Australia

Eden Park, Auckland

40,053

21

28/02/15

India d UAE

WACA Ground, Perth

8,718

22

1/3/2015

Sri Lanka d England

Westpac Stadium, Wellington

18,183

23

1/3/2015

Pakistan d Zimbabwe

Gabba, Brisbane

9,847

24

3/3/2015

South Africa d Ireland

Manuka Oval, Canberra

8,831

25

4/3/2015

Pakistan d UAE

McLean Park, Napier

2,406

26

4/3/2015

Australia d Afghanistan

WACA Ground, Perth

12,710

27

5/3/2015

Bangladesh d Scotland

Saxton Oval, Nelson

3,491

28

6/3/2015

India d West Indies

WACA Ground, Perth

17,557

29

7/3/2015

Pakistan d South Africa

Eden Park, Auckland

22,713

30

7/3/2015

Ireland d Zimbabwe

Blundstone Arena, Hobart

4,048

31

8/3/2015

New Zealand d Afghanistan

McLean Park, Napier

10,022

32

8/3/2015

Australia d Sri Lanka

Sydney Cricket Ground

39,951

33

9/3/2015

Bangladesh d England

Adelaide Oval

11,963

34

10/3/2015

India d Ireland

Seddon Park, Hamilton

10,192

35

11/3/2015

Sri Lanka d Scotland

Blundstone Arena, Hobart

3,549

36

12/3/2015

South Africa d UAE

Westpac Stadium, Wellington

4,901

37

13/03/15

New Zealand d Bangladesh

Seddon Park, Hamilton

10,347

38

13/03/15

England d Afghanistan

Sydney Cricket Ground

9,203

39

14/03/15

India d Zimbabwe

Eden Park, Auckland

30,076

40

14/03/15

Australia d Scotland

Blundstone Arena, Hobart

12,177

41

15/03/15

West Indies d UAE

McLean Park, Napier

1,221

42

15/03/15

Pakistan d Ireland

Adelaide Oval

9,889

43

18/03/15

QF1: South Africa d Sri Lanka

Sydney Cricket Ground

27,259

44

19/03/15

QF2: India d Bangladesh

Melbourne Cricket Ground

51,552

45

20/03/15

QF3: Australia d Pakistan

Adelaide Oval

35,516

46

21/03/15

QF4: New Zealand d West Indies

Westpac Stadium, Wellington

30,250

47

24/03/15

SF1: New Zealand d South Africa

Eden Park, Auckland

41,279

48

26/03/15

SF2: Australia d India

Sydney Cricket Ground

42,330

49

29/03/15

Final: Australia d New Zealand

Melbourne Cricket Ground

93,013

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.

In this recent post, the following figure was formed by considering two triangular regions in three-dimensional space, where :

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 and satisfies

Similarly, the line joining and satisfies

Equating the two expressions gives and from which . The point of intersection is therefore at . By symmetry of this expression the line joining and also passes through this point. This is the point on the plane 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 .

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 ) in terms of the triangle with vertices at .

The first barycentric coordinate will be the ratio of the area of to the area of . Since and have the same x-coordinate, this will be the ratio of the x-coordinates of to , which is . By symmetry it follows that the barycentric coordinates have the attractive form

Let the side lengths of be . Then by Pythagoras’ theorem, . Hence

.

By the cosine rule, (where ) which equals from the above expression. Therefore and similarly we obtain . Then

By the sine rule, ( being the circumradius of ) from which . Hence the barycentric coordinates of may be written in non-normalised form as

Comparing this with the coordinates of the orthocentre , the point is known as the square root of the orthocentre (see Theorem 1 of [1]). Note that the real existence of the point requires to be acute, which it is when . 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.

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.)

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 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 and intercepts at and , where are positive.

The cross section satisfies and the inequalities , and . This can be considered the intersection of the two regions

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 . 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 , which is the intersection of the unit cube’s diagonal from the origin (to ) and the plane .

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 ).

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 .

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 to be on the different side of the plane from the origin, or in other words, .

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

Similarly if all are less than 1, the oppositely similar triangle on the red vertices would be chosen.

Next, if we obtain the following parallelogram.

If we obtain either a pentagon (parallelogram truncated at a vertex) or a trapezium depending on whether or respectively.

:

:

Finally, if are less than 1 and 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].

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.

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:

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:

A very common exercise in high school mathematics is to plot transformations of some standard functions. For example, to plot we may start with a standard sine curve and apply the following transformations in turn:

squeeze it by a factor of 5 in the -direction

shift it left by

stretch it by a factor of 2 in the -direction

shift it down by

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 we may proceed as follows.

Since has a peak at , solve to find as a point where there is a peak at . Hence plot the point .

Since the angular frequency is 5, the period is and we may plot successive peaks spaced apart from the point .

Troughs will be equally spaced halfway between the peaks at (at ). Then join the dots with a sinusoidal curve.

Additionally and intercepts may be found by setting and respectively. We find that the intercepts are at and intercept is .

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 so that its turning point is at , this is equivalent to keeping the parabola fixed and shifting axes so that the new origin is at with respect to the old coordinates. This is illustrated below where the black coordinates are modified to the red ones. The parabola has equation under the black coordinates and or under the red coordinates.

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

Under this stretching of the circle or squeezing of axes, the unit circle equation transforms to that of the ellipse .

More generally, by stretching a Cartesian graph by in the -direction and in the -direction, then shifting it along the vector , we obtain the equation

This uses the fact that and are the inverses of and respectively. Note that if or are less than 1, the stretch becomes a squeezing of the graph, while or correspond to a reflection in the or axes.

We can extend this idea to the rotation of a graph. Suppose for example we wish to rotate the hyperbola 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

(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 with and with in the equation and obtain or .

Here is the same transformation applied to the parabola to obtain or :

If a graph is affinely transformed (by an invertible map) so that maps to and maps to followed by a shift along the vector , then this is equivalent to the coordinates shifting by and then transforming under the inverse mapping :

Here are some special cases of this formula:

rotation of the graph by anti-clockwise: (the example was done above)

reflection of the graph in :

reflection of the graph in where : verify that so

reflection of the graph in where : this is equivalent to a reflection in the line followed by a shift along the vector so

reflection of the graph in (special instance of the previous case with ):

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).

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

yes

yes

samples with replacement

no

yes

samples without replacement (permutations)

no

yes

if letters in a line

if letters in a ring and rotations are considered equivalent

if letters in a ring and rotations & reflections are considered equivalent