Chaitanya's Random Pages

April 29, 2014

Historical world population distribution

Filed under: geography — ckrao @ 2:31 pm

The following plots show the percentage distribution of the world’s population over a 2400 year period. To my surprise China had over a third of the world population during much of the 1800s, while the Indian subcontinent has historically had a higher share than now. It’s interesting to see the growth of the New World in the bottom right plot too. One expects Asia minus China and Africa to have significant proportional growth at least in the first half of this century.

population percentage distributions

Reference

http://www.worldhistorysite.com/population.html – using Colin McEvedy and Richard Jones, Atlas of World Population History (Penguin, 1978)

April 25, 2014

Three and four tangent circles

Filed under: mathematics — ckrao @ 12:14 pm

If we are given two tangent circles how do we find a third circle tangent to both of them? If the two circles are internally tangent, the third circle is also internally tangent to the larger circle.

internal

If the two circles are externally tangent, the third circle is either internally or externally tangent to both.

external

Note that we disregard cases where the third circle has the same point of tangency as the other two such as shown below.

disallowed

Locus of centre of third circle

If we let the centres of the three circles C_1, C_2, C_3 be O_1, O_2, O_3 and with respective radii R_1, R_2, R_3 (assume R_1 > R_2) then what is the locus of O_3 as O_1, R_1, O_2, R_2 are fixed while R_3 is allowed to vary?

We note in the internally tangent case,

O_2 O_3 + O_3 O_1 = (R_2 + R_3) + (R_1 - R_3) = R_1 + R_2.\quad\quad (1)

Since this distance is independent of R_3, O_3 is on an ellipse with foci at O_1 and O_2 shown in green below.

threecircles2

In the externally tangent case, there are two subcases depending on whether the third circle is internally or externally tangent to the other two:

O_3 O_2 - O_3 O_1 = (R_3 - R_2) - (R_3 - R_1) = R_1 - R_2.\quad\quad (2)

O_3 O_1 - O_3 O_2 = (R_3 + R_1) - (R_3 + R_2) = R_1 - R_2.\quad\quad (3)

This corresponds to the two branches of a hyperbola with foci at O_1 and O_2 shown in green below.

external2

Notice that similar results hold when the original two circles are not necessarily tangent to each other.

 

Distances

Denoting the points of tangency between C_2, C_3 by T_1 and similarly defining T_2, T_3, we shall find some distances in the figure below.

setup2

By the cosine rule in \triangle O_1O_2O_3,

\begin{aligned} \cos \angle O_1 O_2 O_3 &= \frac{O_1 O_2^2 + O_2O_3^2 - O_1O_3^2}{2O_1 O_2. O_2O_3}\\  &= \frac{(R_1 - R_2)^2 + (R_2 + R_3)^2- (R_1 - R_3)^2 }{2(R_1 - R_2)(R_2 + R_3)}\\  &=\frac{2R_2^2 - 2R_1 R_2 + 2R_2 R_3 + 2R_1 R_3}{2(R_1 - R_2)(R_2 + R_3)}\\  &= \frac{-(R_1 - R_2)(R_2+R_3) + 2R_1R_3}{(R_1 - R_2)(R_2 + R_3)}\\  &= -1 + \frac{2R_1R_3}{(R_1 - R_2)(R_2 + R_3)}.\quad\quad (4)  \end{aligned}

The perimeter of \triangle O_1O_2O_3 is (R_1 - R_2) + (R_1 - R_3) + (R_2 + R_3) = 2R_1, so by Heron’s formula its area is

\displaystyle \sqrt{R_1(R_1 - (R_1 - R_2)) (R_1 - (R_1 - R_3)) (R_1 - (R_2 + R_3))} = \sqrt{R_1R_2 R_3(R_1 - R_2 - R_3)}.\quad\quad (5)

If the points T_3, O_1, O_2 have coordinates (0,0), (R_1,0), (R_2,0) respectively, then O_3 has coordinates (R_2 + (R_2 + R_3)\cos \angle O_1 O_2 O_3, (R_2 + R_3)\sin \angle O_1 O_2 O_3). We have \cos \angle O_1 O_2 O_3 found above and \sin \angle O_1 O_2 O_3 can be found via

\displaystyle \frac{1}{2}O_1O_2.O_2O_3 \sin \angle O_1 O_2 O_3 = |O_1O_2O_3| = \sqrt{R_1R_2 R_3(R_1 - R_2 - R_3)}.\quad\quad (6)

Hence the coordinates of O_3 are

\begin{aligned} & (R_2 + (R_2 + R_3)\cos \angle O_1 O_2 O_3, (R_2 + R_3)\sin \angle O_1 O_2 O_3)\\ &= \left(R_2 + (R_2 + R_3)\left(-1 + \frac{2R_1R_3}{(R_1 - R_2)(R_2 + R_3)}\right), (R_2 + R_3)2\frac{\sqrt{R_1R_2 R_3(R_1 - R_2 - R_3)}}{O_1 O_2.O_2O_3}\right)\\  &= \left(R_2 - R_2 - R_3+ \frac{2R_1R_3}{(R_1 - R_2)}, (R_2 + R_3)2\frac{\sqrt{R_1R_2 R_3(R_1 - R_2 - R_3)}}{(R_1-R_2)(R_2+R_3)}\right)\\  &= \left(\frac{R_3(R_1 + R_2)}{R_1 - R_2}, \frac{2\sqrt{R_1R_2 R_3(R_1 - R_2 - R_3)}}{R_1-R_2}\right).\quad\quad (7)  \end{aligned}

Note here that the first coordinate varies with R_3 linearly for R_1, R_2 fixed. The distance T_3O_3 is found from

\begin{aligned} T_3O_3^2 &= \frac{1}{(R_1 - R_2)^2}\left([R_3(R_1 + R_2)]^2 + 4R_1R_2R_3(R_1 - R_2 - R_3)\right)\\  &= \frac{1}{(R_1 - R_2)^2}\left(R_3^2 ((R_1 + R_2)^2 - 4R_1 R_2) + 4R_1 R_2 R_3(R_1 - R_2)\right)\\  &= \frac{1}{(R_1 - R_2)^2}\left(R_3^2 (R_1 - R_2)^2 + 4R_1 R_2 R_3(R_1 - R_2)\right)\\  &= R_3^2 + \frac{4R_1 R_2 R_3}{R_1 - R_2}.\quad\quad (8)  \end{aligned}

Hence we have found that interestingly the power of the tangency point T_3 with respect to circle C_3 is the relatively simple form \frac{4R_1R_2R_3}{R_1 - R_2}.

We may also find the distance between points of tangency T_1T_3 as 2R_2 \cos (\angle O_1 O_2 O_3/2) using the identity \cos^2 (A/2) = (1 + \cos A)/2. From (4),

\begin{aligned} T_1 T_3 &= 2R_2 \cos (\angle O_1 O_2 O_3/2)\\  &= 2R_2 \sqrt{\frac{R_1R_3}{(R_1 - R_2)(R_2 + R_3)}}\\  &= \frac{2\sqrt{R_1R_2R_3} \sqrt{R_2}}{\sqrt{(R_1 - R_2)(R_2 + R_3)}}.\quad\quad(9)  \end{aligned}

Next we perform the same calculations (4)-(9) with the configuration of three circles tangent externally.

setup

By the cosine rule in \triangle O_1O_2O_3,

\begin{aligned} \cos \angle O_1 O_2 O_3 &= \frac{O_1 O_2^2 + O_2O_3^2 - O_1O_3^2}{2O_1 O_2. O_2O_3}\\  &= \frac{(R_1 + R_2)^2 + (R_2 + R_3)^2- (R_1 + R_3)^2 }{2(R_1 + R_2)(R_2 + R_3)}\\  &=\frac{2R_2^2 + 2R_1 R_2 + 2R_2 R_3 - 2R_1 R_3}{2(R_1 + R_2)(R_2 + R_3)}\\  &= \frac{(R_1 + R_2)(R_2+R_3) - 2R_1R_3}{(R_1 + R_2)(R_2 + R_3)}\\  &= 1 - \frac{2R_1R_3}{(R_1 + R_2)(R_2 + R_3)}. \quad\quad (10)  \end{aligned}

The perimeter of \triangle O_1O_2O_3 is (R_1 + R_2) + (R_1 + R_3) + (R_2 + R_3) = 2(R_1+R_2+R_3), so by Heron’s formula its area is the attractive form

\displaystyle \sqrt{R_1R_2 R_3(R_1 + R_2 + R_3)}.\quad\quad (11)

If the points T_3, O_1, O_2 have coordinates (0,0), (-R_1,0), (R_2,0) respectively, then O_3 has coordinates (R_2 - (R_2 + R_3)\cos \angle O_1 O_2 O_3, (R_2 + R_3)\sin \angle O_1 O_2 O_3). We have \cos \angle O_1 O_2 O_3 found above and \sin \angle O_1 O_2 O_3 can be found via

\displaystyle \frac{1}{2}O_1O_2.O_2O_3 \sin \angle O_1 O_2 O_3 = |O_1O_2O_3| = \sqrt{R_1R_2 R_3(R_1 + R_2 + R_3)}.\quad\quad (12)

Hence the coordinates of O_3 are

\begin{aligned} & (R_2 - (R_2 + R_3)\cos \angle O_1 O_2 O_3, (R_2 + R_3)\sin \angle O_1 O_2 O_3)\\ &= \left(R_2 - (R_2 + R_3)\left(1 - \frac{2R_1R_3}{(R_1 + R_2)(R_2 + R_3)}\right), (R_2 + R_3)2\frac{\sqrt{R_1R_2 R_3(R_1 + R_2 + R_3)}}{O_1 O_2.O_2O_3}\right)\\  &= \left(R_2 - R_2 - R_3+ \frac{2R_1R_3}{(R_1 + R_2)}, (R_2 + R_3)2\frac{\sqrt{R_1R_2 R_3(R_1 +R_2 + R_3)}}{(R_1+R_2)(R_2+R_3)}\right)\\  &= \left(\frac{R_3(R_1 - R_2)}{R_1 + R_2}, \frac{2\sqrt{R_1R_2 R_3(R_1 + R_2 + R_3)}}{R_1+R_2}\right).\quad\quad (13)  \end{aligned}

Note again that the first coordinate varies with R_3 linearly for R_1, R_2 fixed. Formulas (7) and (13) were used to generate the diagrams in this post.

The distance T_3O_3 is found from

\begin{aligned} T_3O_3^2 &= \frac{1}{(R_1 + R_2)^2}\left([R_3(R_1 - R_2)]^2 + 4R_1R_2R_3(R_1 + R_2 + R_3)\right)\\  &= \frac{1}{(R_1 + R_2)^2}\left(R_3^2 ((R_1 - R_2)^2 + 4R_1 R_2) + 4R_1 R_2 R_3(R_1 + R_2)\right)\\  &= \frac{1}{(R_1 + R_2)^2}\left(R_3^2 (R_1 + R_2)^2 + 4R_1 R_2 R_3(R_1 + R_2)\right)\\  &= R_3^2 + \frac{4R_1 R_2 R_3}{R_1 + R_2}.\quad\quad (14)  \end{aligned}

Hence we have found that the power of the tangency point T_3 with respect to circle C_3 is \frac{4R_1R_2R_3}{R_1 + R_2}.

The distance between points of tangency T_1T_3 is 2R_2 \sin (\angle O_1 O_2 O_3/2) and using the identity \sin^2 (A/2) = (1 - \cos A)/2 with (10),

\begin{aligned} T_1 T_3 &= 2R_2 \sin (\angle O_1 O_2 O_3/2)\\  &= 2R_2 \sqrt{\frac{R_1R_3}{(R_1 + R_2)(R_2 + R_3)}}\\  &= \frac{2\sqrt{R_1R_2R_3} \sqrt{R_2}}{\sqrt{(R_1 + R_2)(R_2 + R_3)}}.\quad\quad(15)  \end{aligned}

Many of the formulas here are found from the first case by replacing R_1 with -R_1. Other distances between points in the figure can be found through similar means.

Finally, we remark that the radical axes of each pair of circles intersect at the radical centre – that point is the circumcentre of the triangle passing through the three points of tangency. With this circle being orthogonal to the three existing circles, inversion in this circle preserves the figure.

radical

Four circles

For the case of four mutually tangential circles (where there is more than one point of tangency), there is an amazing relationship involving the complex number coordinates z_1, z_2, z_3, z_4 of the centres of the circles and their curvatures (reciprocals of radii) k_i= 1/R_i:

\displaystyle 2\left((k_1z_1)^2 + (k_2z_2)^2 + (k_3z_3)^2 + (k_4z_4)^2\right) = \left(k_1z_1 + k_2z_2 + k_3z_3 + k_4z_4\right)^2.\quad\quad (16)

This is known as the complex Descartes theorem. Note that for this formula to be valid for internally tangent circles, we make the sign of the curvature of the larger circle negative.

Interestingly this was only proven in 2001 in [1] with a nice proof shown in [2]. That proof assigns a sphere to each of the six points of tangency with curvature equal to the sums of the curvatures of the two circles meeting there. It then turns out that three spheres corresponding to the tangency points on the same circle are mutually tangent. If one does the same construction of spheres for the dual configuration of circles shown in blue below (the original four circles are in black), the same set of six spheres is obtained!

dualconfig

Note that we have more results from (16). Since the relationships remain when the points are translated by the same complex number z we have

\displaystyle 2\left((k_1(z_1-z))^2 + (k_2(z_2-z))^2 + (k_3(z_3-z))^2 + (k_4(z_4-z))^2\right) = \left(k_1(z_1-z) + k_2(z_2-z) + k_3(z_3-z) + k_4(z_4-z)\right)^2.\quad\quad (17)

Expanding this as a quadratic in z and comparing coefficients of z^2, z [1] gives further two relationships:

\begin{aligned} 2\left(k_1^2 + k_2^2 + k_3^2 + k_4^2\right) &= \left(k_1 + k_2 + k_3 + k_4\right)^2, \quad\quad & (18)\\  2\left(k_1^2 z_1 + k_2^2 z_2 + k_3^2 z_3 + k_4^2 z_4\right) &= \left(k_1 + k_2 + k_3 + k_4\right)\left(k_1 z_1 + k_2 z_2 + k_3 z_3 + k_4 z_4 \right).\quad\quad& (19)  \end{aligned}

 Equation (18) is known as Descartes’ Theorem, discovered in 1643.

To see (16) and (18) in action, let us set for example z_1 = 3, R_1 = 3, z_2 = 1, R_2 = 1, R_3 = 1. Then applying (7) we find

\begin{aligned} z_3 &= \frac{R_3(R_1 + R_2)}{R_1 - R_2} + \frac{2\sqrt{R_1R_2R_3(R_1 - R_2 - R_3)}}{R_1 - R_2}i\\  &= \frac{1(3+1)}{3-1} + \frac{2\sqrt{3.1.1(3-1-1)}}{3-1}i\\  &= 2 + i\sqrt{3}.\quad\quad(20)  \end{aligned}

From (18) we may write

k_4 = k_1 + k_2 + k_3 \pm 2\sqrt{k_1 k_2 + k_2 k_3 + k_1 k_3},\quad\quad(21)

which for k_1 = -1/3 (negative as the larger circle is internally tangent), k_2 = k_3 = 1 leads to k_4 = 5/3 \pm 2\sqrt{1/3}[ or R_4 = 1/k_4 = 3(7 \mp \sqrt{15})/34.]

From (16) we may write

z_4 = (k_1z_1 + k_2z_2 + k_3z_3 \pm 2\sqrt{k_1 k_2z_1z_2 + k_2 k_3z_2z_3 + k_1 k_3z_1z_3})/k_4,\quad\quad(22)

which for k_1 z_1 = -1, k_2 z_2 = 1, k_3 z_3 = 2 + i\sqrt{3}, k_4 = 5/3 \pm 2\sqrt{1/3} leads to

\begin{aligned} z_4 &= \left(-1 + 1 + 2 + \sqrt{3}i \pm 2\sqrt{-1}\right)/(5/3 \pm 2\sqrt{1/3})\\  &=\frac{2+(\sqrt{3}+ 2)i}{5/3 + 2/\sqrt{3}}\quad \text{or}\quad\frac{2+(\sqrt{3}- 2)i}{5/3 - 2/\sqrt{3}}.\quad\quad(23)\end{aligned}

This is used to generate the sketch below. As can be seen there are two possible circles shown in red.

example

Here is how one may prove (19) without the insight of the dual configuration, alluded to in [3] and [4]. Let us consider the externally tangent case with two possible fourth circles shown in red.

example2

If we perform an inversion of the configuration in a circle of radius R centred at the tangency point T_3 of C_1 and C_2, the circles C_1 and C_2 invert to parallel lines while the circles C_3, C_4 invert to equal-radii circles C_3', C_4' tangent to the parallel lines as shown below. Note that in this new configuration it is easier to see that there are two possible choices for C_4'.

example3

What distance separates the two parallel lines? We need only look at the image of the points diametrically opposite T_3 in C_1 and C_2. This leads to the distance d = R^2/(2R_1) + R^2/(2R_2). Hence the radii of the images of C_3 and C_4 is half this, or R^2(1/R_1 + 1/R_2)/4.

If we know the distance of the centre of the image of C_4 from T_3, the following lemma will enable us to determine R_4. We shall use the following useful lemma which tells us how to find the radius of a circle from quantities involving its inverse.

Lemma

If a circle has radius r and distance x from the origin, its inverse in a circle of radius R centred at the origin has radius r' = rR^2/|x^2-r^2| and its centre has distance x' = xR^2/|x^2 - r^2| from the origin. (Hence x'/r ' = x/r.)

Proof of lemma

lemmaproofIn the above figure the dashed circle is the circle of inversion while the separate cases x > r and x < r are shown in blue and red respectively.

The points on the original circle collinear with the origin are at distance x+r and |x-r| from the origin. The inverses of these points have distance R^2/(x+r) and R^2/(x-r) from the origin (this quantity is negative if x < r). Hence the centre of the inverted circle has distance x' = |R^2(1/(x+r) + 1/(x-r))/2| = xR^2/|x^2 - r^2| from the origin and radius r'= |R^2(1/(x-r) - 1/(x+r))/2| = rR^2/|x^2-r^2| as required. Note that if x=r the circle inverts to a line which can be thought of as a circle with centre at infinity with infinite radius.

From (12), if the origin is at T_3, O_3 has coordinates

\displaystyle (x,y):= \left(\frac{R_3(R_1 - R_2)}{R_1 + R_2}, \frac{2\sqrt{R_1R_2 R_3(R_1 + R_2 + R_3)}}{R_1+R_2}\right).\quad\quad (24)

Suppose O_3' has coordinates (x',y') (we note that while C_3' is the image of C_3 under the inversion, the centre O_3' is generally not the image of O_3). Then O_4' is at distance

\displaystyle d:= \frac{R^2}{2}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)\quad\quad (25)

above or below O_3' (the distance between the parallel lines), so has coordinates (x',y'\pm d). By the above lemma with r = d/2, the radius of C_4 with centre O_4 is dR^2/2(x'^2 + (y'\pm d)^2 - (d/2)^2). Hence its curvature is

\displaystyle \frac{1}{R_4} = \frac{2(x'^2 + y'^2 + 3d^2/4 \pm 2y'd)}{dR^2}.\quad\quad(26)

Firstly from (13)

x^2 + y^2 = T_3O_3^2 = R_3^2 + \frac{4R_1 R_2 R_3}{R_1 + R_2}, so by the lemma, we find x'^2 + y'^2 via the ratio of radii (d/2) to R_3:

\displaystyle x'^2 + y'^2 = (d/2R_3)^2(x^2 + y^2),\quad\quad (27)

so

\begin{aligned}\frac{2(x'^2 + y'^2 + 3d^2/4)}{dR^2} &= \frac{2(d/2R_3)^2(x^2 + y^2) + 3d^2/2}{dR^2}\\  &=\frac{d[(x^2 + y^2)/2R_3^2 + 3/2]}{R^2}\\  &=\frac{d}{R^2}\left[ \frac{R_3^2}{2R_3^2} + \frac{4R_1 R_2 R_3}{(R_1 + R_2)2R_3^2} + \frac{3}{2}\right]\\  &= \frac{d}{R^2}\left[ 2 + \frac{2R_1R_2}{(R_1 + R_2)R_3} \right]\\  &= \frac{2d}{R^2}\left[ \frac{R_3(R_1 + R_2) + R_1 R_2}{(R_1 + R_2)R_3} \right]\\  &= \left(\frac{1}{R_1} + \frac{1}{R_2}\right) \left[ \frac{R_3(R_1 + R_2) + R_1 R_2}{(R_1 + R_2)R_3} \right]\quad \text{(by (25))}\\  &= \frac{R_1 R_2 + R_2 R_3 + R_1 R_3}{R_1 R_2 R_3}\\  &= \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}.\quad\quad(28)  \end{aligned}

Next,

\begin{aligned} \frac{4y'}{R^2} &= \frac{4 y (d/2R_3)}{R^2}\\  &= \frac{4d\sqrt{R_1R_2 R_3(R_1 + R_2 + R_3)}}{R^2(R_1+R_2)R_3}\quad \text{(by (24))}\\  &= \frac{4\frac{(R^2)(R_1 + R_2)}{2R_1R_2}\sqrt{R_1R_2 R_3(R_1 + R_2 + R_3)}}{R^2(R_1+R_2)R_3}\quad \text{(by (25))}\\  &= \frac{2\sqrt{R_1R_2 R_3(R_1 + R_2 + R_3)}}{R_1R_2R_3}\\  &= 2\sqrt{\frac{1}{R_1R_2} + \frac{1}{R_2R_3} + \frac{1}{R_1R_3}}.\quad\quad(29)  \end{aligned}

Using (28) and (29) in (26) gives us

\begin{aligned} \frac{1}{R_4} &= \frac{2(x'^2 + y'^2 + 3d^2/4 \pm 2y'd)}{dR^2}\\  &= \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \pm 2\sqrt{\frac{1}{R_1R_2} + \frac{1}{R_2R_3} + \frac{1}{R_1R_3}}.\quad\quad(30)  \end{aligned}

With k_i = 1/R_i, this becomes k_4 = k_1 + k_2 + k_3 \pm 2\sqrt{k_1 k_2 + k_2 k_3 + k_1 k_3}, establishing (21).

For the case of C_2 and C_3 internally tangent to C_1 a similar computation to the above can be carried out and we obtain

\displaystyle \frac{1}{R_4} = -\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \pm 2\sqrt{-\frac{1}{R_1R_2} + \frac{1}{R_2R_3} - \frac{1}{R_1R_3}},\quad\quad(31)

which is the same as (28) with R_1 replaced with -R_1.

Note that Descartes’ theorem also holds in the special cases of two parallel lines tangent to two equal circles or a single line tangent to three mutually tangent circles. In the latter case we may set k_4 = 0 and assuming k_1 \geq k_2 we obtain from (18) the nice formula

\sqrt{k_3} = \sqrt{k_1} \pm \sqrt{k_2}.\quad\quad(32)

We see these two solutions below in the right figure.

degenerate_cases

Further generalisations of Descartes’ theorem to further dimensions are in [1], and more can be read about tangent circle packings in [5].

References

[1] J.C. Lagarias, C.L. Mallows, and A. Wilks, Beyond the Descartes Circle Theorem, American Mathematical Monthly, 109 (2002), 338-361. Available at http://www.arxiv.org/abs/math?papernum=0101066.

[2] S. Northshield, Complex Descartes Circle Theorem, to appear, American Mathematical Monthly.

[3] D. Pedoe, On a theorem in geometry, American Mathematical Monthly, 74 (1967), 627-640.

[4] P. Sarnak, Integral Apollonian Packings, American Mathematical Monthly, 118 (2011), 291-306.

[5] D. Austin, When Kissing Involves Trigonometry, Feature Column from the AMS, March 2006.

 

Create a free website or blog at WordPress.com.

%d bloggers like this: