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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.

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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

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