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April 28, 2013

A few fast centuries

Filed under: cricket,sport — ckrao @ 8:58 am

Last week Chris Gayle scored a phenomenal 175* off 66 balls in an IPL cricket game for Bangalore featuring 17 6s. It was the best display of sustained hitting by a batsman I have seen. He didn’t give the fielders any chance of catching him and he barely played a false shot. Here is his ball-by-ball scoring off legal deliveries (not including leg byes).

0 0 1 | 4 4 0 4 4 0 4 | | 1 | 6 6 4 0 6 6 | | 4 6 1 6 0 | 6 6 4 6 6 | 0 1 6 1 | 0 1 1 0 | 1 1 | 0 0 6 1 0 | 1 0 1 6 | 4 1 | 4 6 6 4 6 | 1 1 | 0 1 | 4 1 4 1 |  1 0 | 6 1 1

Notice that he scored 28, 28, 26 and 20 off four of the overs… over a hundred runs just there! Interestingly he never hit 3 sixes in a row, though he managed 10 in the space of 16 balls (73 runs total)! The graph below shows how he took his time during the middle of his innings (102 in his first 30 balls, just 14 off his next 15) before surging again to 150.

Gayle batting chart

Apparently, while playing for a Jamaican XI in 2006 he scored 196 off 68 balls and was dismissed in the 15th over!!

Below are a few more fast centuries of the past. More can be seen at the following ESPN CricInfo links.

Here are those centuries from the above lists scored off fewer than 40 balls.

Name Balls Faced Teams Venue Season/scorecard Game type
CH Gayle 30 Royal Challengers Bangalore v Pune Warriors Bangalore 2013 T20
A Symonds 34 Kent v Middlesex Maidstone 2004 T20
D Hookes 34 South Australia v Victoria Adelaide 1982/3 FC
LP van der Westhuizen 35 Namibia v Kenya Windhoek 2011/12 T20
GD Rose 36 Somerset v Devon Torquay 1990 List A
YK Pathan 37 Rajasthan Royals v Mumbai Indians Mumbai 2009/10 T20
SB Styris 37 Sussex v Gloucestershire Hove 2012 T20
Shahid Afridi 37 Pakistan v Sri Lanka Nairobi 1996 ODI

There is also a list here of the fastest centuries of each English county season of recent years. More on fast innings through declaration bowling here.

A few centuries worth mentioning: I still have a soft spot for Shahid Afridi’s 102 off 40 balls for Pakistan vs Sri Lanka with his century off 37 balls – amazingly it was his first ODI innings.

0 6 1 0 4 0 0 6 0 0 6 6 1 1 6 6 2 6 4 4 0 0 6 6 1 4 1 1 0 4 1 6 0 6 0 2 4 1 0 X

Yusuf Pathan’s IPL century off 37 balls featured 11 consecutive legal deliveries sent for 4 or 6.

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

Scott Styris’s century off 37 balls featured 38 in one over:

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

Finally, Don Bradman scored a century off just 22 balls in three eight-ball overs (not even one dot ball!) in a game for Blackheath XI vs Lithgow:

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

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April 22, 2013

A few geometric inequalities proved using complex numbers

Filed under: mathematics — ckrao @ 11:33 am

The following inequalities all follow from the elementary triangle inequality for complex numbers:

\displaystyle \left|z_1 \right| + \left| z_2 \right| \geq \left| z_1 + z_2 \right|.

Equality holds if and only if 0, z_1, z_2 are collinear or in other words z_2/z_1 is a real number when z_1 \neq 0.

1. Ptolemy’s inequality: we start with the equality

\begin{aligned} (p-q)(r-s) + (p-s)(q-r) &= pr - qr - ps + qs + pq - qs - pr + rs\\ &= pq - qr - ps + rs\\ &= (p-r)(q-s) \end{aligned}

Hence

\begin{aligned}\left| p-r \right| \left| q-s\right| &= \left| (p-r)(q-s) \right| \\ &= \left| (p-q)(r-s) + (p-s)(q-r)\right| \\&\leq \left| (p-q)(r-s)\right| + \left| (p-s)(q-r)\right|\end{aligned}

Hence if P, Q, R, S are four points in the plane,

\displaystyle PR \times QS \leq PQ \times RS + PS \times QR, \quad \quad \quad (1)

which is Ptolemy’s inequality. Equality holds if and only if

\displaystyle \frac{(p-s)(q-r)}{(p-q)(r-s)} \in \mathbb{R}. The ratio (p-s)/(p-q) is a complex number with argument \angle QPS while the ratio (q-r)/(r-s) has argument \pi - \angle SRQ. Hence for the product of these ratios to be real means \angle QPS + \angle SRQ is a multiple of \pi. In other words, the points P, Q, R, S lie on a circle.

2. We start with the equality

\begin{aligned}\frac{bc}{(a-b)(a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)} &= \frac{bc(c-b) + ca(a-c) + ab(b-a)}{(a-b)(b-c)(c-a)} \\ &= 1 \end{aligned}

From this,

\begin{aligned}\frac{ |b| |c|}{|a-b| |a-c|} + \frac{|c| |a|}{|b-c| |b-a|} + \frac{|a| |b|}{|c-a| |c-b|} &\geq \left| \frac{bc}{(a-b)(a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)}\right|\\ &= 1. \end{aligned}

In other words, if P, A, B, C are four points in the plane, (PB\times PC)/(AB \times AC) + (PC \times PA)/(BC \times AB) + (PA \times PB)/(AC \times BC) \geq 1, or

\displaystyle (BC \times PB \times PC) + (CA \times PC \times PA) + (AB \times PA \times PB) \geq AB \times BC \times CA.\quad (2)

3. Similarly we have

\begin{aligned} a^2(b-c) + b^2(c-a) + c^2(a-b) &= -(a-b)(b-c)(c-a), \end{aligned}

from which

\displaystyle PA^2 \times BC + PB^2 \times CA + PC^2 \times AB \geq AB \times BC \times CA.\quad \quad \quad (3)

4. Finally, we have the equality

\begin{aligned} & a^3(b-c) + b^3(c-a) + c^3(a-b) \\&= (a + b + c)(a^2(b-c) + b^2(c-a) + c^2(a-b)) - \left[ (b+c) a^2(b-c) + (c + a) b^2 (c-a) + (a+b)c^2(a-b) \right]\\ &= -(a+b+c)(a-b)(b-c)(c-a) + \left[ a^2(b^2-c^2) + b^2(c^2 - a^2) + c^2(a^2-b^2) \right] \quad \text{(from the equality in 3.)}\\ &= -(a-b)(b-c)(c-a)(a + b + c)\end{aligned}

from which

\displaystyle PA^3 \times BC + PB^3 \times CA + PC^3 \times AB \geq 3 AB \times BC \times CA \times PG, \quad \quad \quad (4)

where G is the centroid of \triangle ABC (i.e. the vector sum PA + PB + PC is 3PG).

See [2] for more details of special cases of these inequalities.

References

[1] A. Bogomolny, Complex Numbers and Geometry from Interactive Mathematics Miscellany and Puzzles  http://www.cut-the-knot.org/arithmetic/algebra/ComplexNumbersGeometry.shtml#cycl, Accessed 22 April 2013.

[2] T. Andreescu and D. Andrica, Proving some geometric inequalities by using complex numbers, Educatia Matematica Vol. 1, Nr. 2 (2005), pp. 19–26.

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