Chaitanya's Random Pages

May 31, 2014

The world’s fastest growing countries by population

Filed under: geography — ckrao @ 7:24 am

I was amazed to learn recently how rapidly the population of some African countries is increasing. The following table shows those countries that grew by at least half a million people in the mid 2012 to mid 2013 year (see references below). Note that Nigeria is at least 50% ahead of every country other than India and China. Also note the complete absence of European countries (Italy had the largest growth there of 416,000 which is only 47th in the world). Note that the figures are estimates only.

Country Continent Annual Population Growth (mid 2012 – mid 2013)
India Asia 20,290,000
China Asia 6,688,000
Nigeria Africa 5,551,000
Pakistan Asia 3,696,000
Indonesia Asia 3,553,000
Democratic Republic of Congo Africa 2,334,000
United States North America 2,281,000
Ethiopia Africa 2,253,000
Bangladesh Asia 2,081,000
Mexico North America 2,026,000
Egypt Africa 1,893,000
Philippines Asia 1,825,000
Brazil South America 1,685,000
Kenya Africa 1,266,000
Uganda Africa 1,232,000
Tanzania Africa 1,204,000
Myanmar Asia 1,160,000
Iraq Asia 1,051,000
Saudi Arabia Asia 997,000
Iran Asia 976,000
Vietnam Asia 922,000
Sudan Africa 863,000
Algeria Africa 792,000
Mozambique Africa 790,000
Malaysia Asia 734,000
Yemen Asia 725,000
Ivory Coast Africa 717,000
South Africa Africa 704,000
Ghana Africa 659,000
Niger Africa 649,000
Angola Africa 647,000
Madagascar Africa 585,000
Burkina Faso Africa 550,000
Colombia South America 544,000
Cameroon Africa 543,000
Mali Africa 532,000
Syria Asia 531,000
Afghanistan Asia 517,000
Thailand Asia 508,000

 

The next table shows that more than 6/7 of the world’s current population growth is from Asia or Africa.

Continent Annual population growth in millions from mid 2012 to mid 2013 (% of world total)
Asia 51.3 (55.2%)
Africa 29.3 (31.5%)
North America 6.2 (6.6%)
South America 4.4 (4.7%)
Europe 1.1 (1.2%)
Oceania 0.66 (0.7%)
World 92.9

References from Wikipedia

[1] List of Asian countries by population

[2] List of African countries by population

[3] List of North American countries by population

[4] List of South American countries by population

[5] List of European countries by population

[6] List of Oceanian countries by population

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May 17, 2014

Forms of Stewart’s theorem

Filed under: mathematics — ckrao @ 10:41 am

Stewart’s theorem finds the length of a cevian d=AD in terms of the side lengths of the triangle ABC and the lengths m, n into which point D on BC divides that side. Here are some forms of the same formula.

Stewart

1. The most common form we see is

\displaystyle b^2 m + c^2 n = a(d^2 + mn) \quad \Rightarrow \quad d^2 = \frac{b^2m + c^2n}{a} - mn.\quad\quad(1)

An easy-to remember form of this is rewriting the above as man + dad = bmb + cnc (a man and his dad hid a bomb in the sink!). This can be proved by applying the cosine rule to triangles ACD and then ABC:

\begin{aligned}  d^2 &= AD^2\\  &= AC^2 + CD^2 - 2AC.CD\cos \angle DCA\\  &= AC^2 + CD^2 - 2AC.CD \cos \angle BCA\\  &= AC^2 + CD^2 - 2AC.CD \frac{CA^2 + CB^2 - AB^2}{2CA.CB}\\  &= b^2 + n^2 - 2bn \frac{b^2 + a^2 - c^2}{2ba}\\  &= b^2 + n^2 - n \frac{b^2 + a^2 - c^2}{a}\\  &= \frac{b^2(m+n) - n(b^2 + a^2 - c^2)}{a} + n^2\\  &= \frac{b^2 m + c^2n}{a} + n^2 - \frac{na^2}{a}\\  &= \frac{b^2 m + c^2n}{a} + n(n-a)\\  &= \frac{b^2 m + c^2n}{a} - mn.\\  \end{aligned}

2. If D divides the side BC in the ratio BD:DC = r:s,

\displaystyle d^2 = \frac{rb^2 + sc^2}{r+s} - \frac{rsa^2}{(r+s)^2} = \frac{r^2b^2 + s^2c^2 + rs(b^2 + c^2 - a^2)}{(r+s)^2} .\quad\quad(2)

3. Similar to (2) but substituting \displaystyle \lambda = r/(r+s) = BD:BC,

\displaystyle d^2 = \lambda b^2 + (1-\lambda)c^2 - \lambda(1-\lambda)a^2.\quad\quad(3)

This and the previous form are conveniently proved using vectors. Writing the vector \mathbf{AD} = \lambda \mathbf{AC} + (1-\lambda) \mathbf{AB},

\begin{aligned}  d^2 &= \mathbf{AD}. \mathbf{AD}\\  &= \left(\lambda \mathbf{AC} + (1-\lambda) \mathbf{AB}\right).\left(\lambda \mathbf{AC} + (1-\lambda) \mathbf{AB}\right)\\  &= \lambda^2 \mathbf{AC}.\mathbf{AC} + (1-\lambda)^2 \mathbf{AB}.\mathbf{AB} + 2\lambda(1-\lambda)\mathbf{AB}.\mathbf{AC}\\  &= \lambda^2 b^2 + (1-\lambda)^2 c^2 + \lambda(1-\lambda)\left(b^2 + c^2 - a^2\right)\\  &= b^2(\lambda^2 + \lambda(1-\lambda)) + c^2((1-\lambda)^2 + \lambda(1-\lambda)) - a^2\lambda(1-\lambda)\\  &= \lambda b^2 + (1-\lambda)c^2 - \lambda(1-\lambda)a^2.  \end{aligned}

Note that this is valid for any real \lambda, so D may lie beyond segment BC.

4. Writing (3) as a quadratic in \lambda:

\displaystyle d^2 = \lambda^2 a^2 + \lambda(c^2-b^2 -a^2) + b^2.\quad\quad(4)

5. A symmetric form [1], where the following distances are taken as directed segments (CD = -DC etc.)

\displaystyle \frac{BA^2}{BC.BD} + \frac{CA^2}{CB.CD} + \frac{DA^2}{DB.DC} = 1\quad\quad(5)

 Note that this is equivalent to BA^2 . CD + CA^2 . DB = DA^2.CB + CD.DB.CB = CB(DA^2 + CD.DB) which is (1).

 

Here are a few special cases of this formula applying form (3).

  • D = C (i.e. \lambda = 1): d^2 = b^2
  • D is a third of the way along CB (closer to C) (\lambda = 2/3): \displaystyle d^2 = 2b^2/3 + c^2/3 - 2a^2/9
  • AD is the internal angle bisector of \angle BAC (\lambda = c/(b+c)):

\begin{aligned}  d^2 &= cb^2/(b+c) + bc^2/(b+c) - bca^2/(b+c)^2\\  &= bc\left[\frac{b}{b+c} + \frac{c}{b+c}- \left(\frac{a}{b+c}\right)^2\right]\\  &= bc\left[1 - \left(\frac{a}{b+c}\right)^2\right]\\  &= bc - mn.  \end{aligned}

  • AD is the external angle bisector of \angle BAC (assume b > c so \lambda = -c/(b-c)):

Stewart external bisector

\begin{aligned}  d^2 &= \frac{-c}{b-c} b^2 + \frac{b}{b-c}c^2 + \frac{bc}{(b-c)^2}a^2\\  &= bc\left[\frac{-b}{b-c} + \frac{c}{b-c}+ \left(\frac{a}{b-c}\right)^2\right]\\  &= bc\left[ \left(\frac{a}{b-c}\right)^2 - 1\right]\\  &= mn - bc.  \end{aligned}

Reference

[1] Darij Grinberg – Unpublished Notes – Geometry

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