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June 30, 2015

The Pakistan heat wave of 2015

Filed under: Uncategorized — ckrao @ 11:39 am

This month a heatwave has killed over 1500 people in Pakistan, most of them in Karachi. Here are the temperatures recorded at Karachi airport at that time: note that the average for this time of year is 28-35°C with high humidity. Rarely does the temperature reach 39.5°C three days in a row but here it did so six days in a row plus there seemed to be no relief at night.

Date 16-Jun 17-Jun 18-Jun 19-Jun 20-Jun 21-Jun 22-Jun 23-Jun 24-Jun
Min (°C) 30 29.3 29.5 31 31.6 33 33 33 30.5
Max (°C) 36 38.5 39.5 40.5 44.8 42.5 42.5 41.2 37

Most of the dead were homeless and it was also during the time of Ramadan where fasting is observed during daylight hours. This and the recent heatwave in India are two of the three most fatal on the Indian subcontinent in recent times.

References

[1] Temperatures: http://www.ogimet.com/cgi-bin/gsynres?lang=en&ind=41780&ano=2015&mes=6&day=30&hora=6&min=0&ndays=30
[2] Average Weather For Karachi, Pakistan – WeatherSpark

[3] Dr. Jeff Masters’ WunderBlog: Pakistan’s Deadliest Heat Wave on Record Kills at Least 800: Weather Underground

[4] News: Majority of Karachi heatwave victims were homeless minister – The Express Tribune

June 22, 2015

Places we see powers of 2

Filed under: mathematics — ckrao @ 11:44 am

Here are some obvious and less obvious places where we encounter the sequence f(n) = 1,2,4,8,16,.... Most of these come from the corresponding OEIS page.

  • as a solution to the recurrence f(n+1) = 2f(n), f(0) = 1.
  • the number of subsets of a n-element set: To each element there are two choices – to include it or not include it in the subset. Hence there are 2\times 2\times \ldots \times 2 = 2^n subsets.
  • the sum of the numbers in the n’th row of Pascal’s triangle This is a consequence of the binomial theorem applied to (1+1)^n = \sum_{i=0}^n \binom{n}{i}.
  • the sum of the coefficients of (x+1)^n, which is the sum of binomial coefficients
  • the number of even or odd-sized subsets of an (n+1)-element set, i.e. \sum_{i=0}^{\lfloor{(n+1)/2}\rfloor} \binom{n+1}{2i} = \sum_{i=0}^{\lfloor{n/2}\rfloor} \binom{n+1}{2i+1} = 2^n, a consequence of the binomial theorem (1-1)^n = \sum_{i=0}^n (-1)^i \binom{n}{i}.
  • each term in the sequence is the smallest natural number that is not the sum of any number of distinct earlier terms, an example of a sum-free sequence
  • the sequence forms the set of every natural number that is not the sum of 2 or more consecutive positive integers: Such a sum has the form a + (a+1) + \ldots + (a+ k) = (2a + k)(k+1)/2. If this were a power of 2 then (2a+k) and (k+1) must both be powers of 2 greater than or equal to 2. But their difference 2a-1 is odd, which is impossible for the difference of even powers of 2. Conversely if a natural number is not a power of two it is of the form 2^m d where d is an odd number at least 3. Then 2^m d may be written as the sum of either d consecutive positive integers with middle (mean) term 2^m (if (d+1)/2 \leq 2^m) or 2^{m+1} consecutive positive integers with mean d/2 (if 2^m < d/2). For example, 48 = 2^4\times 3 and 48 = 15 + 16 + 17 while 44 = 2^2\times 11 and 44 = 2+3+4+5+6+7+8+9.
  • the number of ordered partitions of n+1 is 2^n: for example, the eight ordered partitions of 4 are: 1+1+1+1, 2+1+1, 1+2+1, 1+1+2, 2+2, 3+1, 1+3, 4 To see why this result is true, imagine n+1 items in a row. Imagine separating them with dividers so that the number of items between each divider gives us the ordered partition. For example with 4 items we may place a divider between the third and fourth item to give the partition 3+1. We have n possible locations for dividers (between any adjacent items) and each location has a choice of being assigned to a divider or not. This leads to 2^n possible partitions. Alternatively, an ordered partition of n+2 may be derived inductively from an ordered partition of n+1 = a_1 + \ldots + a_k by either forming n+2 = 1 + a_1 + \ldots + a_k or n+2 = (a_1+1) + a_2 + \ldots + a_k. Also any ordered partition of n+2 has one of these forms as we may form from it an ordered partition of n+1 by either subtracting 1 from the first element of the partition (if it is greater than 1) or deleting the first element if it is 1. Hence the number of partitions doubles as n+1 is increased by 1 and for n = 1 we have the single partition.
  • the number of permutations of {1,2,...,n+1} with exactly one local maximum: for example, for n=3 we have the permutations 1234, 1243, 1342, 2341, 1432, 2431, 3421, 4321. Here we have the highest element n+1 somewhere and there are 2^n possible subsets of {1,2,...,n} that we may choose to precede it. These elements must be in ascending order, while the elements after n+1 are in descending order to ensure exactly one local maximum at n+1. Hence the order of terms is determined once the subset is chosen and we have $2^n$ permutations.
  • the number of permutations of n+1 elements where no element is more than one position to the right of its original place: for example, for n=3 the allowed permutations are 1234, 1243, 1324, 1423, 2134, 2143, 3124, 4123. This can be seen inductively: if a_1a_2\ldots a_{n+1} is a valid permutation of n+1 elements, then 1(1+a_1)(1+a_2)\ldots (1+a_{n+1}) and (1+a_1)1(1+a_2)\ldots (1+a_{n+1}) are valid permuations of n+2 elements. Also any valid (n+2) element permutation is of this form as it may be reduced to a valid (n+1) element permutation by deleting the 1 (which must occur in either the first or second position) and subtracting 1 from every other element. This describes a one to two mapping between valid permutations of n+1 and n+2 elements. Finally for 1 element there is 1 = 2^0 allowed permutation, so the result follows by induction.

May 31, 2015

Test pace bowlers to have conceded 100 runs in an innings the most

Filed under: cricket,sport — ckrao @ 10:16 am

 

Below is a list of pace bowlers who have conceded 100 or more runs in an innings at least 14 times (* indicates active players).

Name Tests Innings Bowled Wickets times conceded 100 runs innings per century
Botham 102 168 383 31 5.42
Kapil Dev 131 227 434 25 9.08
Ntini 101 190 390 23 8.26
Hadlee 86 150 431 21 7.14
Caddick 62 105 234 20 5.25
Imran Khan 88 142 362 20 7.1
M Johnson* 64 123 283 20 6.15
Vaas 111 194 355 20 9.7
Lillee 70 132 355 19 6.95
McDermott 71 124 291 19 6.53
I Sharma* 61 107 187 18 5.94
Anderson* 104 194 401 17 11.41
Gough 58 95 229 17 5.59
Srinath 67 121 236 17 7.12
F Edwards 55 97 165 16 6.06
Lawson 46 78 180 16 4.88
Mohammad Sami 36 66 85 16 4.13
Sarfraz Nawaz 55 95 177 16 5.94
Hoggard 67 122 248 15 8.13
Lee 76 150 310 15 10
Martin 71 126 233 15 8.4
Morrison 48 76 160 15 5.07
Steyn* 78 147 396 15 9.8
Bedser 51 92 236 14 6.57
Z Khan 92 165 311 14 11.79

 

Here are the numbers for remaining pace bowlers with at least 200 test wickets (Ray Lindwall a stand-out here):

Name Tests Innings Bowled Wickets times conceded 100 runs innings per century
Waqar Younis 87 154 373 13 11.85
Harmison 63 115 226 13 8.85
Wasim Akram 104 181 414 12 15.08
McKenzie 60 113 246 12 9.42
Cairns 62 104 218 12 8.67
Walsh 132 242 519 11 22
Sobers (not always pace) 93 159 235 11 14.45
Snow 49 93 202 11 8.45
Broad 79 143 285 11 13
Donald 72 129 330 10 12.9
Pollock 108 202 421 10 20.2
Willis 90 165 325 10 16.5
Hughes 53 97 212 10 9.7
Thomson 51 90 200 10 9
McGrath 124 243 563 9 27
Trueman 67 127 307 9 14.11
Statham 70 129 252 8 16.13
Flintoff 79 137 226 8 17.13
Streak 65 102 216 8 12.75
Roberts 47 90 202 8 11.25
Morkel 62 117 217 7 16.71
Gillespie 71 137 259 6 22.83
Marshall 81 151 376 5 30.2
Holding 60 113 249 5 22.6
Garner 58 111 259 4 27.75
Kallis 166 272 292 4 68
Lindwall 61 113 228 1 113

The numbers are generally higher for spin bowlers – the top six centurions overall are all spin bowlers:

Muralitharan (61), Kumble (57), Harbhajan Singh (43), Warne (40), Kaneria (39), Vettori (34).

References

[1] Bowling records:  Test matches:  Cricinfo Statsguru – ESPN Cricinfo

[2] Records:  Test matches:  Bowling records:  Most wickets in career – ESPN Cricinfo

May 24, 2015

A collection of proofs of Ptolemy’s Theorem

Filed under: mathematics — ckrao @ 11:39 am

Here we collect some proofs of the following nice geometric result.

If ABCD is a quadrilateral, then AB.CD + BC.DA \geq AC.BD with equality if ABCD is cyclic.

Ptolemy

In words, the sum of the product of the lengths of the opposite sides of a quadrilateral is at least the product of the lengths of its diagonals.

The equality for a cyclic quadrilateral is known as Ptolemy’s theorem while the more general inequality applying to any quadrilateral is called Ptolemy’s inequality. Many of the proofs below that establish Ptolemy’s theorem can be modified slightly to prove the inequality.

Proofs of Ptolemy’s Theorem

Proof 1:

Choose point P on line CD extended beyond D as shown so that \angle DAP = \angle BAC.

Ptolemy2

Then

a) \triangle ABC \sim \triangle ADP (\angle ABC = \angle ADP and \angle BAC = \angle DAP).

b) \triangle BAD \sim \triangle CAP (\angle BAD = \angle CAP and \angle ADB = \angle APC).

From a) BC.DA = BA.DP so AB.CD + BC.DA = AB(CD + DP) = AB.CP = AC.BD by b) and we are done.

Proof 2: (a slightly different choice of constructed point)

Let K be the point on AC so that \angle CBK = \angle ABD.

Ptolemy3

Then

a) \triangle ABD \sim \triangle KBC (\angle ABD = \angle KBC and \angle BDA = \angle BCK) so KC = AD.BC/BD.

b) \triangle ABK \sim \triangle DBC (\angle ABK = \angle DBC and \angle KAB = \angle CDB) so AK = DC.AB/DB.

Adding the two gives AC = (AD.BC + DC.AB)/BD or AC.BD = AD.BC + DC.AB as required.

Proof 3: (ref: http://mathafou.free.fr/themes_en/kptol.html)

Let the diagonals AC, BD intersect at P and construct E on the circumcircle so that CE is parallel to BD. A short angle chase shows that BCED is an isosceles trapezium.

Ptolemy5

Then \sin \angle ABE = \sin \angle ACE = \sin \angle BPC (angles on common arc AE and using BD \parallel CE). Also triangles BDE and BDC have the same base and height and hence the same area.

Hence the area of ABCD is the sum of the areas of triangles ABE and ADE, or

\begin{aligned}  & \frac{1}{2} \left( AB.BE\sin \angle ABE + AD.DE\sin \angle ADE \right)\\  &= \frac{1}{2} \left( AB.BE\sin \angle ABE + AD.DE\sin \angle ABE \right) \quad \text{as } \angle ABE \text{ and } \angle ADE \text{ are supplementary }\\  &= \frac{1}{2} \left( AB.BE+ AD.DE \right)\sin \angle ABE\\  &= \frac{1}{2} \left(AB.CD + AD.BC \right)\sin \angle ABE, \quad (1)  \end{aligned}

where the last step follows from BDEC being an isosceles trapezium.

But also this area is \frac{1}{2} AC.BD\sin \angle BPC and recall from above that \sin \angle BPC = \sin \angle ABE. Therefore equating this with (1) we end up with AB.CD + AD.BC = AC.BD.

Proof 4 (ref: [1])

Here three of the triangles of the cyclic quadrilateral are scaled to fit together to form a parallelogram. Namely,

  • \triangle ABD is scaled by AC,
  • \triangle ABC is scaled by AD,
  • \triangle ACD is scaled by AB.

A simple angle chase based on the coloured angles shown reveals that a parallelogram is formed when the triangles are joined together. Ptolemy’s theorem then follows from equating two of its opposite side lengths.

Ptolemy7

Proof 5: (from here)

Using the cosine rule in triangles ABC and ADC respectively gives

AC^2 = AB^2 + BC^2 - 2AB.BC\cos \angle ABC = AD^2 + CD^2 - 2AD.CD\cos \angle ADC.

Since \angle ABC + \angle ADC = 180^{\circ}, this becomes

\begin{aligned}  AB^2 + BC^2 - 2AB.BC\cos \angle ABC &= AD^2 + CD^2 + 2AD.CD\cos \angle ABC\\  \Rightarrow \cos \angle ABC &= \frac{AB^2 + BC^2 - AD^2 - CD^2}{2(AB.BC + AD.CD)}.  \end{aligned}

Hence

\begin{aligned}  AC^2 &= AB^2 + BC^2 - 2AB.BC\cos \angle ABC\\  &= AB^2 + BC^2 - 2AB.BC \frac{AB^2 + BC^2 - AD^2 - CD^2}{2(AB.BC + AD.CD)}\\  &= \frac{(AB^2+BC^2)(AB.BC + AD.CD) - AB.BC(AB^2 + BC^2 - AD^2 - CD^2)}{AB.BC + AD.CD}\\  &= \frac{(AB^2+BC^2)(AD.CD) + AB.BC(AD^2 + CD^2)}{AB.BC + AD.CD}\\  &= \frac{AB.AD(AB.CD+BC.AD) + BC.CD(BC.AD+AB.CD)}{AB.BC + AD.CD}\\  &= \frac{(AB.AD+BC.CD)(AB.CD+BC.AD)}{AB.BC + AD.CD}.\\  \end{aligned}

By a similar argument we may write

\displaystyle BD^2 = \frac{(AB.BC + AD.CD)(BC.AD+AB.CD)}{BC.CD + AB.AD}.

Multiplying these last two equations by each other and taking the square root gives

\displaystyle AC.BD = AB.CD + BC.AD,

which is Ptolemy’s theorem.

Proof 6 (via this):

Let the angles subtended by AB, BC, CD respectively be \alpha, \beta, \gamma and let the circumradius of the quadrilateral be R. By the sine rule AB = 2R \sin \alpha, BC = 2R\sin \beta, CD = 2R \sin \gamma, AD = 2R\sin (\alpha + \beta + \gamma ), AC = 2R \sin (\alpha + \beta), BD = 2R \sin (\beta + \gamma ). Then the equality to be proved is equivalent to

\displaystyle \sin (\alpha + \beta)\sin (\beta + \gamma) = \sin \alpha \sin \gamma + \sin \beta \sin (\alpha + \beta + \gamma).

But we have

\begin{aligned}  & \sin (\alpha + \beta)\sin (\beta + \gamma)\\ &= (\sin \alpha \cos \beta + \cos \alpha \sin \beta)(\sin \beta \cos \gamma + \cos \beta \sin \gamma)\\  &= \sin \alpha \sin \beta \cos \beta \cos \gamma + \sin \alpha \cos^2 \beta \sin \gamma + \cos \alpha \sin^2 \beta \cos \gamma + \cos \alpha \sin \beta \cos \beta \sin \gamma\\  &= \sin \alpha \sin \beta \cos \beta \cos \gamma + \sin \alpha (1- \sin^2 \beta ) \sin \gamma + \cos \alpha \sin^2 \beta \cos \gamma + \cos \alpha \sin \beta \cos \beta \sin \gamma\\  &= \sin \alpha \sin \gamma + \sin \beta (\sin \alpha \cos \beta \cos \gamma - \sin \alpha \sin \beta \sin \gamma + \cos \alpha \sin \beta \cos \gamma + \cos \alpha \cos \beta \sin \gamma )\\  &= \sin \alpha \sin \gamma + \sin \beta (\sin(\alpha + \beta ) \cos \gamma + \cos (\alpha + \beta ) \sin \gamma )\\  &= \sin \alpha \sin \gamma + \sin \beta \sin (\alpha + \beta + \gamma ),  \end{aligned}

as required.

Proofs of Ptolemy’s Inequality (all make use of the triangle inequality)

Proof 7:

Denote the points A, B, C, D by vectors or complex numbers a, b, c, d. Then we have the equality

\displaystyle (a-b).(c-d) + (a-d).(b-c) = (a-c).(b-d).

Applying the modulus (length) to both sides and then the triangle inequality leads to

\displaystyle |(a-b).(c-d)| + |(a-d).(b-c)| \geq |(a-c).(b-d)|,

which is Ptolemy’s inequality.

Proof 8: (ref: [2])

Place the origin at the point D so that A, B, C are represented by the vectors a, b, c respectively. Let a' = a/|a|^2, b' = b/|b|^2, c' = c/|c|^2. Then

\begin{aligned} |a'-b'|^2 &= \left| \frac{a}{|a|^2} - \frac{b}{|b|^2}\right|^2\\  &= \frac{|a|^2}{|a|^4} + \frac{|b|^2}{|b|^4} - 2\frac{\langle a, b \rangle}{|a|^2|b|^2}\\  &= \frac{|b|^2 + |a|^2 - 2\langle a, b \rangle}{|a|^2|b|^2}\\  &= \frac{|a-b|^2}{|a|^2|b|^2}.  \end{aligned}

Similarly, |b'-c'| = \frac{|b-c|}{|b||c|} and |c'-a'| = \frac{|c-a|}{|c||a|}. By the triangle inequality, \displaystyle |a'-b'| \leq |b'-c'| + |c'-a'|, or in other words,

\begin{aligned}  \frac{|a-b|}{|a||b|} &\leq \frac{|b-c|}{|b||c|} + \frac{|c-a|}{|c||a|}\\  \Rightarrow |a-b||c| \leq |b-c||a| + |c-a||b|,  \end{aligned}

which is another way of writing Ptolemy’s inequality.

Proof 9: (inversion)

Recall that under inversion under a point P a circle passing through P maps to a line. If X, Y are points on the circle mapping to X, Y' respectively under inversion in a circle of radius R centred at P, then

\displaystyle X'Y' = R^2.XY/(PX.PY).

inversion

To see this, since XP.X'P = YP.Y'P = R^2 by the definition of the inverse, XP/YP = Y'P/X'P. This together with the fact that angle P is common implies that triangles XPY and Y'PX' are similar. This gives us the relationship X'Y' = YX.PY'/PX = XY.R^2/(PX.PY) as desired.

Ptolemy4

For our quadrilateral ABCD apply an inversion centred at D with radius R. Then A, B, C map to points A', B', C' which are collinear if ABCD is cyclic. Using the result above A'B' = AB.R^2/(DA.DB) and similarly A'C' = AC.R^2/(DA.DC) and B'C' = BC.R^2/(DB.DC). By the triangle inequality A'C' \leq AB' + BC' and so this becomes

\begin{aligned}  \frac{AC.R^2}{DA.DC} &\leq \frac{AB.R^2}{DA.DB} + \frac{BC.R^2}{DB.DC}\\  \Rightarrow AC.DB &\leq AB.DC + BC.DA,  \end{aligned}

which is Ptolemy’s inequality.

Proof 10: (ref: [3])

Construct E on AC so that EC = BC, then draw F on CD with EF \parallel AD. Finally rotate \triangle FEC about C to map to \triangle F'BC.

Ptolemy6

Then F'B = FE and by similarity of triangles ACD and ECF, FE/DA = CE/CA = CB/CA so that

(a) F'B = FE = AD.BC/CA

Secondly, \triangle DF'C \sim ABC as F'C/BC = FC/EC = DC/AC and \angle F'CD = \angle BCA. This gives

(b) DF' = AB.CD/AC

Adding (a) and (b) and applying the triangle inequality,

\displaystyle BD \leq BF' + F'D = (AD.BC + AB.CD)/AC,

from which

\displaystyle AC.BD \leq AD.BC + AB.CD

which is Ptolemy’s inequality.

References

[1] A. Bogomolny, Ptolemy Theorem – Proof Without Words from Interactive Mathematics Miscellany and Puzzles
http://www.cut-the-knot.org/proofs/PtolemyTheoremPWW.shtml, Accessed 30 May 2015

[2] W.H. Greub, Linear Algebra, Springer Science & Business Media, 2012 (p 190).

[3] C. Alsina, R. B. Nelsen, When Less is More: Visualizing Basic Inequalities, MAA, 2009.

April 30, 2015

Number of countries that have abolished the death penalty by year

Filed under: Uncategorized — ckrao @ 11:36 am

The graph below is generated from data at this Wikipedia page.

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.

deathpenalty

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.

regularpolywithline

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?

regularpolywithline2

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.

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

 

attendancesbymatch

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.

2trianglesThe 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).

barycentric

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.

cs1 cs2 cs3 cs4 cs5

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.

coordsystem

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.

2triangles

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).

lengthsofsides

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.constraints

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.

tricase

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.

parallelogramcase

 

 

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:

pentagoncaseb\geq 1:
trapeziumcase

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

hexagoncase

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

[3] calculus – Area of the polygon formed by cutting a cube with a plane – Mathematics Stack Exchange

[4] integration – Volume of cube section above intersection with plane – Mathematics Stack Exchange

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