# Chaitanya's Random Pages

## October 29, 2011

### The orthocentre of an isosceles triangle of base length 2

Filed under: mathematics — ckrao @ 9:48 pm

During one of my recent mathematical explorations, I wanted to calculate the distance of the points of an isosceles triangle to its orthocentre (points where the altitudes meet).

🙂

In other words, I firstly wanted to find $x = PH$ in terms of $h$ in the following figure.

Once $x$ is found, the required distances will be $\sqrt{1+x^2}$ and $|h - x|$. In this post I will show two ways of finding $x$.

Here is the approach I initially came up with. Since the angles at $P$ and $Q$ are each $90$ degrees, quadrilateral $QAPH$ is cyclic. Then by the intersecting chords theorem (or by the similarity of triangles $BAQ$ and $BHP$),

$\displaystyle 1.2 = \sqrt{1+x^2} \sqrt{4-y^2}$

from which $\displaystyle 1 + x^2 = \frac{4}{4-y^2}$       …(1)

Also, since triangles $BAQ$ and $CAP$ are similar,

$\displaystyle \frac{y}{2} = \frac{1}{\sqrt{1+h^2}}$

from which $\displaystyle \frac{4}{y^2} = 1+h^2$       …(2)

Combining (1) and (2),

$\displaystyle 1 + x^2 = \frac{4}{4-y^2} = \frac{4/y^2}{4/y^2 - 1} = \frac{1 + h^2}{h^2} = 1 + \frac{1}{h^2}$.

It follows that $\displaystyle x = 1/h$. Who would have thought that such a simple answer would be found! Hence, we have rediscovered a way of constructing the reciprocal of a positive number $h$: draw an isosceles triangle with base length $2$ and height $h$, then the height of its orthocentre above the base is $1/h$.

Naturally, the question comes up of whether there is an easier way of calculating this answer and surely enough there is. If we extend $CH$ until it meets the circumcircle of $ABC$ at $C'$, we find that $\angle HBA = ACC'$ are equal (each being the complement of $\angle BAC$). Then we also have $\angle ACC' = \angle ABC'$, both angles lying on the arc $AC'$. Both acute and obtuse angled cases are illustrated below.

It follows by Angle-Angle-Side that triangles HPB and C’PB are congruent, from which $x = HP = PC'$. In other words, the reflection of $H$ in $AB$ lies on the circumcircle! This fact is true for the reflection of $H$ in any side, and $ABC$ need not be isosceles.

By the intersecting chords theorem for $CC'$ and $AB$ intersecting at $P$ (or equivalently, by the similarity of triangles $CAP$ and $BC'P$),

$\displaystyle 1.1 = x.h$

or $x = 1/h$ as required.

I later found that the second proof is also given in pp18-19 of Honsberger’s book entitled “Episodes in 19th and 20th century Euclidean Geometry”.

## October 26, 2011

### Country of origin of prominent West Indian cricketers

Filed under: cricket,sport — ckrao @ 11:17 am

After watching the movie Fire in Babylon (I have now seen it twice – well worth seeing if you are a cricket fan!) I was curious to find out the country of origin of prominent players of the West Indies. There are some 15 countries that make up the West Indies (English speaking part of the Caribbean, note that it does not include Bermuda, Bahamas, Turks and Caicos Islands or the Cayman Islands). Some of which are dependent territories – these are shown in italics below. The six first class teams of the West Indies are:

Of these nations, the only ones never to have a test cricket representative are the British and US Virgin Islands and Sint Maarten. See here for a full list of West Indian test players by origin.

From the table see how many great players the tiny nations of Barbados and Antigua and Barbuda produced!

 Nation Land Area (sq km) Population Players Tests/ODIs Barbados 431 273,000 Ian Bradshaw 5/62 Sherwin Campbell 52/90 George Challenor 3 Corey Collymore 30/84 Pedro Collins 32/30 Anderson Cummins+ 5/76 Fidel Edwards 47/50* Joel Garner 58/98 John Goddard 27 Gordon Greenidge 108/128 Charlie Griffith 28 Desmond Haynes 116/238 Wes Hall 48 Vanburn Holder 40 Sir Conrad Hunte 44 Malcolm Marshall 81/136 Seymour Nurse 29 Dwayne Smith 10/77 Sir Garfield Sobers 93/1 Sir Clyde Walcott 44 Sir Everton Weekes 48 Sir Frank Worrell 51 Trinidad and Tobago 5,128 1,318,000 Ian Bishop 43/84 Dwayne Bravo 40/117* Baron Constantine 18 Mervyn Dillon 38/108 Daren Ganga 48/35 Larry Gomes 60/83 Gerry Gomez 29 Brian Lara 131/299 Gus Logie 52/158 Deryck Murray 62/26 Sonny Ramadhin 43 Denesh Ramdin 42/84 Ravi Rampaul 11/62* Phil Simmons 26/143 Jeff Stollmeyer 32 Jamaica 10,831 2,706,000 Jimmy Adams 54/127 Jeff Dujon 81/169 Chris Gayle 91/228* George Headley+ 22 Wavell Hinds 45/119 Michael Holding 60/102 Patrick Patterson 28/59 Daren Powell 37/55 Ricardo Powell+ 2/109 Lawrence Rowe 30/11 Marlon Samuels 33/120* Jerome Taylor 29/66* Alf Valentine 36 Courtney Walsh 132/205 Guyana 214,969 785,000 Basil Butcher 44 Shivnarine Chanderpaul 134/268* Roy Fredericks 59/12 Lance Gibbs 79/3 Roger Harper 25/105 Carl Hooper 102/227 Alvin Kallicharran 66/31 Rohan Kanhai 79/7 Clive Lloyd 110/87 Ramnaresh Sarwan 87/173* Antigua and Barbuda 443 89,000 Curtley Ambrose 98/176 Keith Arthurton 33/105 Winston Benjamin 21/85 Ridley Jacobs 65/147 Sir Viv Richards 121/187 Richie Richardson 86/224 Andy Roberts 47/56 Grenada 344 104,000 Junior Murray 33/55 Devon Smith 33/42* Saint Kitts and Nevis 261 52,000 Stuart Williams 31/57 Saint Lucia 606 167,000 Darren Sammy 17/65* Saint Vincent and the Grenadines 389 109,000 Dominica 751 68,000 United States Virgin Islands 346 106,000 Sint Maarten 34 37,000 British Virgin Islands 151 28,000 Anguilla 91 15,000 Montserrat 102 6,000 Totals 234,877 5,863,000

+Anderson Cummins played some ODIs for Canada.
+George Headley was born in Panama.
+Ricardo Powell later played for Trinidad and Tobago.

## October 16, 2011

### Optimisation problems given a point inside a given angle

Filed under: mathematics — ckrao @ 10:59 pm

Given a point inside an angle, it easy to come up with many minimisation problems to solve. Here are some mostly taken from [1], with the answers given below, but proofs are left as exercises to the interested reader. 🙂

Let $XOY$ be the angle, $P$ the point inside the angle and $MN$ be a variable straight line through $P$ with $M$ on $OX$ and $N$ on $OY$. Let $A, B$ be variable points on $OX, OY$ respectively.

Problems:

1. Find $M, N$ so that the area of triangle $MON$ is minimised.
2. Find $M, N$ so that the perimeter of triangle $MON$ is minimised.
3. Find $M, N$ so that the length $MN$ is minimised.
4. Find $M, N$ so that $MP.PN$ is minimised.
5. Find $M, N$ so that $1/MP + 1/PN$ is maximised.
6. Find $M, N$ so that $OM^p.ON^q$ is minimised (p, q > 0).
7. Find $M, N$ so that $OA = OB$ and $AP + PB$ is minimised.
8. Find $M, N$ so that the perimeter of triangle $PAB$ is minimised.

Do try at least one of these before reading below for the answers!

1. [to minimise the area of $MON$] Choose $MN$ so that $MP = PN$ (extend $OP$ to twice its length, then complete the parallelogram with $P$ as centre and $MN$ as diagonal).
2. [to minimise the perimeter of $MON$] Construct a circle tangent to the angle and through $P$, then $MN$ is its tangent through $P$. The circle is formed by first choosing any point on the angle bisector and drawing any initial circle tangent to $XOY$ (its radius can be found by dropping a perpendicular to $OX$). This initial circle can be scaled up or down by drawing parallel lines so that the final circle passes through $P$.
3. [to minimise the length of $MN$] The solution is Philo’s line, mentioned in a previous blog entry. The points $M$ and $N$ cannot be found by straight edge and compass, but should be chosen so that they are equidistant to the midpoint of $OP$.
4. [to minimise $MP.PN$] Choose $MN$ so that $OMN$ is isosceles (construct the angle bisector of angle $XOY$, then draw the perpendicular to this line through $P$).
5. [to maximise $1/MP + 1/PN$] Choose $MN$ perpendicular to $OP$.
6. [to minimise $OM^p.ON^q$] Choose $M, N$ so that $MP/PN = q/p$. Given the lengths $p, q$ drawn from $P$ this can be done via parallel lines and similar triangles as shown:
7. [to minimise $AP + PB$ so that $OA = OB$] Rotate $P$ by $\angle XOY$ about $O$ to $P'$. Then $A$ is found by the intersection of $MN$ and $PP'$ and $B$ is constructed so that $OA=OB$.
8. [to minimise the perimeter of triangle $PAB$] Reflect $P$ in $OX$ and $OY$. Then $A$ and $B$ are where the line joining these reflected images meets $OX$ and $OY$. If $\angle XOY \geq 90^{\circ}$ then $A=B=O$.

So given the same setup we have at least 8 different problems with 8 different constructions and answers!

Reference

[1] T. Andreescu, O. Mushkarov, L. Stoyanov, Geometric Problems on Maxima and Minima, Birkhäuser, 2006.

## October 7, 2011

### 2011 Normalised AFL ladder and successful sports teams over five year periods

Filed under: sport — ckrao @ 1:59 pm

Firstly, I thought I would see what this year’s AFL ladder would look like if each team played each other once (in reality with 17 teams playing 22 games each, a team plays six others twice and the other ten once). To do this I simply halved the results (scores and premiership points) of those games between teams that played each other twice during the home and away season. Here is that “normalised” ladder in the final two columns compared with the actual one. It is as though each team played every other once, so that the maximum possible number of points is 16*4 = 64.

 # TEAM Played Wins Losses Draws % PTS Norm Pts Norm % 1 Collingwood 22 20 2 0 167.66 80 60 168.82 2 Geelong 22 19 3 0 157.38 76 54 158.85 3 Hawthorn 22 18 4 0 144.12 72 52 139.46 4 West Coast 22 17 5 0 130.32 68 46 121.83 5 Carlton 22 14 7 1 130.88 58 41 130.13 6 St Kilda 22 12 9 1 112.76 50 38 110.94 7 Sydney 22 12 9 1 109.34 50 36 110.12 8 Essendon 22 11 10 1 100 46 35 103.02 9 North Melbourne 22 10 12 0 101.15 40 30 103.94 10 Western Bulldogs 22 9 13 0 95.59 36 30 101.13 11 Fremantle 22 9 13 0 83.11 36 28 85.66 12 Richmond 22 8 12 1 86.35 34 22 83.84 13 Melbourne 22 8 13 1 85.27 34 26 84.84 14 Adelaide 22 7 15 0 79.43 28 18 77.97 15 Brisbane Lions 22 4 18 0 80.98 16 10 78.30 16 Port Adelaide 22 3 19 0 64.51 12 8 65.41 17 Gold Coast 22 3 19 0 56.27 12 10 56.25

We see that the top 8 would be unchanged, and only Richmond-Melbourne and Gold Coast-Port Adelaide would have been swapped. This year, no team can really complain about having an unfortunate draw.

Secondly, after reading that Geelong has compiled a 105-20 (84%) win-loss record in their past 5 years (in fact winning 103 of their last 120 games including streaks of 15, 15, 13, 13, 12!), I thought I would compare it with a few other top team sports results over a five year period. We are not necessarily looking at teams that won championships in every year of that period, more winning percentage. Games with many draws such as test cricket, ice hockey and soccer had to be excluded since they give lower winning percentages. Note that a draw or no result counts as half a win in the percentages below. It can be seen that Geelong’s winning percentage compares favourably with those of many great sporting teams of the past.

• The New England patriots compiled a 77-17 (81.9%) record from 2003-2007 [ref].
• The Australian one-day cricket team had a 107-26 win-loss record plus 2 ties and 5 no results (78.9%) between 2001 and 2005 [ref].
• Between 1981 and 1986 the Boston Celtics had a 363-128 record (73.9%) including playoff games. As far as I can tell this exceeds any five-year period during their dynasty of the 1950s and 60s, as well as any best 5-year period of the LA Lakers or Chicago Bulls.
• The St George Dragons had a 87-1-13 win-draw-loss record (86.6%) in Australian Rugby League between 1957 and 1961 (they won 11 consecutive premierships from 1956 to 1966!)
• Worth a mention, tennis doubles “team” Martina Navratilova and Pam Shriver had a mind-boggling 217-5 (97.7%) record from 1983 to 1987. This included 109 consecutive wins between April 1983 and Wimbledon 1985 in which they only lost 14 sets! (Data taken manually from the WTA website, do correct me if I am wrong!)