# Chaitanya's Random Pages

## 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).

## 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.

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 $\angle DAP = \angle BAC$.

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 $\angle CBK = \angle ABD$.

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.

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.

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

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

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.

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 $\triangle F'BC$.

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.

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