# Chaitanya's Random Pages

## 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

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