The Games that Populations Play (Tuklas Vol. 13, No. 7 - Jan. 28, 2012)

THE GAMES THAT POPULATIONS PLAY

The outcome of making decisions can surprisingly (for some), actually be investigated within a mathematical framework. This branch of mathematics is called classical game theory; here, situations that call for decision-making are called games. In this framework, the game is being played by an individual; the prize for winning the game is quantified by the payoff. A special type of game known as the population game represents a situation wherein decision making is applied in the context of a collection of individual decision-makers who have the same set of strategies and choices.
Suppose that game theory is applied in explaining the phenomenon of gender ratio in the population. Why is the ratio of males to females in human (or animal) populations 50:50? Game theory phrases the answer to this question as follows: the ratio 50:50 is an evolutionarily stable strategy or ESS. An ESS is related to an outcome which best serves the decision maker in the midst of uncertainty. Consider a game described by the following:

• Male-to-female ratio is $\rho:1-\rho$, where $0\leq\rho \leq 1$
• All females only mate once in their lifetime; consequently engender $n$ offspring
• Males are polygamous and mate, on average, with $\frac{1-\rho}{\rho}$ distinct females in their lifetime
• Females are the ``decision makers;" either $D_1$: all offpsring are males; or $D_2$: all offspring are females

The female does not actually make a conscious decision about what type of offspring it produces. So in general, of the $n$ offsprings it produces, a proportion $p$ would be male, while $1-p$ would be female. This can be summarized as a vector $\sigma=\langle p, 1-p\rangle$, where $0\leq p\leq 1$. At a certain point in time, if one looks at the entire population and conducts a census the proportion of males is $x$, and that of females is $1-x$ so that the sex ratio is such that $\rho=x$, and can be summarized as the vector $\mathbf{x}=\langle x, 1-x\rangle$ wherein $0\leq x\leq 1$. The payoff for the pure decisions $D_1$ and $D_2$ is equivalent to the number of grandchildren that each female is expected to have: $$\begin{align*} \pi(D_1,\mathbf{x}) &= n^2\frac{1-\rho}{\rho}\\ \pi(D_2,\mathbf{x}) &= n^2 \end{align*}$$ The payoff $\pi(D_1)$ arises because all of the $n$ male offspring will mate with $(1-\rho)/\rho$ distinct females (on average) and engendering $n$ offpsrings in each mating. On the other hand, $\pi(D_2)$ arises because all $n$ female offpsring would each give birth to $n$ offspring in their lifetime. Finally, game theory prescribes that the payoff of the general strategy $\sigma$ can be written as follows: $$\begin{align*} \pi(\sigma,\mathbf{x}) &= p\cdot\pi(D_1,\mathbf{x}) + (1-p)\cdot \pi(D_2,\mathbf{x}) \\ &= pn^2 \frac{1-\rho}{\rho} + (1-p)n^2 \\ &= n^2\left[\left(\frac{1-2\rho}{\rho}\right)p+1\right] \end{align*}$$ The ESS, by its very definition, should correspond to the value of $p$ that results to the highest possible $\pi(\sigma,\mathbf{x})$ given $\rho$. From the above result, three cases can be enumerated:

• If $\rho < \frac{1}{2}$, then $(1-2\rho)/\rho > 0$ so that $p=1$ yields the highest possible value for $\pi(\sigma,\mathbf{x})$
• If $\rho > \frac{1}{2}$, then $(1-2\rho)/\rho < 0$ so that $p=0$ yields the highest possible value for $\pi(\sigma,\mathbf{x})$
• If $\rho = \frac{1}{2}$, then $(1-2\rho)/\rho = 0$ so that any value of $p$ yields the highest possible value for $\pi(\sigma,\mathbf{x})$

Since the female cannot really control the value of $p$ (i.e., it does not have the will to choose the gender of offspring that it produces) then the game should be played without any regard of the specific value of $p$. In other words, in finding the ESS of the population game, one should choose the case wherein any value of $p$ will do. This case corresponds only when $\rho = \frac{1}{2}$. Therefore, the ultimate consequence is that the ratio between males and females in the population is 50:50.

ABOUT THE AUTHOR:
Dranreb Earl Juanico is an Assistant Professor at the Ateneo de Manila University. He obtained his B.S., M.S., and Ph.D. in Physics at the University of the Philippines in 2002, 2004 and 2007, respectively.

OLYMPIAD CORNER

from the 57th Belarusian Mathematical Olympiad

Problem: Let $O$ be the intersection point of the diagonals of the convex quadrilateral $ABCD$, $AO = CO$. Points $P$ and $Q$ are marked on the segments $AO$ and $CO$, respectively, so that $PO = OQ$. Let $N$ and $K$ be the intersection points of the sides $AB$, $CD$ and lines $DP$, $BQ$ respectively. Prove that the points $N$, $O$, $K$ are collinear.

Solution: 

Draw $NM || KL || AC$. Let $a = BO$, $b = DO$, $c = PO = OQ$, $l = AO = OC$, $x = KL$, $u = OL$, $y = NM$, $v = OM$.

Since $\Delta BOQ \sim \Delta BLK$, it follows that $$\frac{x}{c} = \frac{a+u}{a} = 1 + \frac{u}{a}$$ Furthermore, since $\Delta DOC \sim \Delta DLK$, then $$\frac{x}{l} = \frac{b-u}{b} = 1 - \frac{u}{b}$$ Therefore $$x\left( \frac{1}{c} - \frac{1}{l} \right) = u\left( \frac{1}{a} + \frac{1}{b} \right) \Rightarrow \frac{x}{u} = \frac{a+b}{ab} \cdot \frac{cl}{l-c}$$ By a similar argument, we also obtain $$\frac{y}{v} = \frac{a+b}{ab} \cdot \frac{cl}{l-c}$$ Since $\frac{x}{u} = \frac{y}{v}$, and $\angle NMO = \angle KLO$, then $\Delta ONM \sim \Delta OKL$. This implies that $\angle NOM=\angle KOL$, which makes them vertical angles. Therefore, the points $N$, $O$ and $K$ must be collinear.

PROBLEMS

19. If $a_1 = 1$ and $$a_{n+1} = \frac{1}{1+\frac{1}{a_n}}, n = 1, 2, \ldots, 2011,$$ find the value of $$a_1 a_2 + a_2 a_3 + a_3 a_4 + \ldots + a_{2011} a_{2012}.$$
    
20. One commercially available ten-button lock may be opened by depressing - in any order - the correct five buttons. Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations should this allow?
21. Let $a$, $b$ and $c$ be three distinct positive integers. Show that among the numbers $$a^5 b - ab^5, b^5 c - bc^5, c^5 a - ca^5,$$ there must be one that is divisible by 8.

We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is February 4, 2012. 

SOLUTIONS

(for January 14 and 21, 2012)

13. $AB$ is a chord in a circle with center $O$ and radius 52 cm. The point $M$ divides the chord $AB$ such that $AM = 63$ cm and $MB = 33$ cm. Find the length of $OM$ in cm. (Singapore Secondary Schools Mathematical Olympiads for Junior Section, 2003)
    (solved by Emman Joshua B. Busto [PSHS - Main] and Dianne S. Garcia [AA])
    
    
    SOLUTION: 
    

    
    Let $D$ be a point on $AB$ such that $OD \perp AB$. We now have $$AD = BD = \frac{1}{2}(63+33) = 48.$$ Thus $$\begin{align*} MD = 48 - 33 &= 15 \\ {OD}^2 = {OB}^2 - {BD}^2 &= {52}^2 - {48}^2 = 400 \end{align*}$$ and $OM = \sqrt{{OD}^2 + {MD}^2} = 25$ cm.
14. If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that $$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}.$$ (Taken from A Primer for Mathematics Competitions by Hitchcock and Zawaira)
    
    
    SOLUTION: 
    Note that $$\begin{align*} \left( a + \frac{1}{a} \right)^2 + \left( b + \frac{1}{b} \right)^2 &= a^2 + \frac{1}{a^2} + b^2 + \frac{1}{b^2} + 4 \\ &=(a+b)^2 - 2ab + \left( \frac{1}{a} + \frac{1}{b} \right)^2 - \frac{2}{ab} + 4 \\ &= 1 - 2ab + \frac{1-2ab}{(ab)^2} + 4. \end{align*}$$ Furthermore, by the AM-GM inequality we have $$ab \leq \left( \frac{a+b}{2} \right)^2 = \frac{1}{4}.$$ Since $a$ and $b$ are positive real numbers, the second inequality yields the following: $$\begin{align*} ab \leq \frac{1}{4} \Rightarrow -2ab \geq -\frac{1}{2}, \\ ab \leq \frac{1}{4} \Rightarrow \frac{1}{(ab)^2} \geq 16. \end{align*}$$ Thus, we have, $$\left( a + \frac{1}{a} \right)^2 + \left( b + \frac{1}{b} \right)^2 \geq 1 - \frac{1}{2} + \left(1 - \frac{1}{2}\right)16 + 4 = \frac{25}{2}.$$
15. For any non-negative integers $m, n, p,$ prove that the polynomial $x^{3m} + x^{3n+1}+x^{3p+2}$ has the factor $x^2 + x + 1$. (Taken from Lecture Notes on Mathematical Olympiad Courses by Xu Jiagu)
    
    
    SOLUTION: 
    Consider $f(y) = y^m - 1$. Since $f(1) = 0$, we have $y^m - 1 = (y-1)q(y)$. Taking $y = x^3$, we have $$x^{3m}-1 = (x^3)^m - 1 = (x^3 - 1)q(x^3) = (x-1)(x^2+x+1)q(x^3).$$ Also, $$\begin{align*} x^{3n+1} - x &= x(x^{3n}-1), \\ x^{3p+2} - x^2 &= x^2(x^{3p}-1), \end{align*}$$ implying that all of these terms have a factor of $x^2+x+1$. Taking them together, we now have $$x^{3m} + x^{3n+1} + x^{3p+2} = (x^{3m}-1) + (x^{3n+1} - x) + (x^{3p+2} - x^2) + (x^2+x+1),$$ showing that the polynomial does have a factor of $x^2+x+1$.
16. Let $T$ be a triangle of perimeter 2, and $a$, $b$, $c$ be the lengths of its three sides. Prove that $$abc + \frac{28}{27} \geq ab + bc + ca \text{ and } abc + 1 \leq ab +bc +ca.$$ (Ireland Mathematical Olympiad, 2003)
    
    
    SOLUTION: 
    Since $a + b + c = 2$ then $0 < a, b, c < 1$ hence $$\begin{align*} 0 < (1-a)(1-b)(1-c) \leq \left( \frac{1-a+1-b+1-c}{3} \right)^3 = \frac{1}{27} \\ 0 < 1 - (a + b + c) + (ab +bc +ca) - abc \leq \frac{1}{27} \\ 0 < (ab + bc + ca) - 1 - abc \leq \frac{1}{27}, \end{align*}$$ and thus both inequalities are shown to be true.
17. If $a$, $b$ are two different positive integers, and the two quadratic equations $$(a-1)x^2 - (a^2+2)x +(a^2+2a) = 0, (b-1)x^2 - (b^2+2)x +(b^2+2b) = 0$$ have one common root, find the value of $$\frac{a^b + b^a}{a^{-b} + b^{-a}}.$$ (China Mathematical Competition for Primary Schools, 2003)
    
    
    SOLUTION: 
    We can deduce that $a, b > 1$ where $a \neq b$. Let $r$ be the common root. We have $$\begin{align*} (a-1)r^2 - (a^2+2)r+(a^2+2a) &= 0, \\ (b-1)r^2 - (b^2+2)r +(b^2+2b) &= 0, \end{align*}$$ where $r \neq 1$, since otherwise $a = b = 1$, a contradiction.
    Using the two equations, we eliminate $x^2$ and simplify to obtain $$(a-b)(ab-a-b-2)(r-1) = 0,$$ where $a - b \neq 0$ and $r \neq 1$, thus implying $ab - a - b - 2 = 0$. Since this equation (and the final equation we are considering) is symmetric, we can assume $a > b > 1$ without loss of generality. Then $b = 1 + \frac{b}{a} + \frac{2}{a} \leq 3$, so $b = 2, a = 4$.
    From there, $$\frac{a^b + b^a}{a^{-b}+b^{-a}} = (a^b + b^a)\frac{a^bb^a}{a^b+b^a} = a^bb^a = 256.$$
18. In the convex quadrilateral $ABCD$, the midpoints of $BC$ and $AD$ are $E$ and $F$ respectively. Prove that $[EDA] + [FBC] = [ABCD]$, where [] denotes the area of the region enclosed by the specified points. (IMO Shortlist, 1989)
    
    
    SOLUTION: 
    

    
    Suppose that $AE$ and $BF$ intersect at $Q$ and $CF$ and $DE$ intersect at $P$. Then it is sufficient to show that $[FQEP] = [ABQ] + [DPC]$. Let $h_1$, $h_2$ and $h_3$ denote the heights of triangles $ABE$, $FBC$ and $DEC$ respectively, giving us $h_2 = \frac{1}{2}(h_1+h_3)$. Thus $$\begin{align*} [FQEP] + [BEQ] + [ECP] &= [FBC] = \frac{1}{2}h_2BC \\ &= \frac{1}{4}(h_1+h_3)BC \\ &= \frac{1}{4}h_1(2BE) + \frac{1}{4}h_3(2EC) \\ &=([ABQ] + [BEQ]) + ([DPC] + [ECP]), \end{align*}$$ which gives us the desired equality.