# 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

## 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.

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 the midpoint of $BC$ ($\lambda = 1/2$): $\displaystyle d^2 = b^2/2 + c^2/2 - a^2/4$ or $\displaystyle b^2 + c^2 = 2(d^2 + (a/2)^2)$ (Apollonius’ theorem)
• $D$ 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 $\lambda = -c/(b-c)$):

\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

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