The Will Rogers Phenomenon (Tuklas Vol. 13, No. 4 - Dec. 10, 2011)

THE WILL ROGERS PHENOMENON

In studying phenomenon that involve numerical data, it is nice to have number(s) that summarize the data. One commonly used number is the average or arithmetic mean. Given numbers $x_1,x_2,\ldots,x_n$, the arithmetic mean is computed
$$\bar{x} = \frac{x_1 + x_2 + \ldots + x_n}{n} = \frac{1}{n}\sum_{i=1}^n x_i.$$
It takes an equal contribution of $\frac{1}{n}$ from each data point and adds them all together.

Sometimes, however, this average may not be a good measure to summarize data. First, the arithmetic mean is sensitive to outliers. For example, in the data set
$$\{1,1,1,1,1,1,1,1,1,100\}$$
the average is $\frac{1}{10}(109) = 10.9$. However, this is clearly not representative of the data, as most of the numbers are 1 and the 100 appears to be an anomaly.

Our second example involves two groups. Suppose that in a certain math class the passing grade is $50$, and that the 10 students in class have grades $25,30,35,35,45,55,55,60,75,80$. There are five students who passed, and their average is $\frac{1}{5}(55+55+60+75+80) = 65$. Those who weren't as fortunate, however, have average $\frac{1}{5}(25+30+35+35+45) = 34$. 

What happens if the passing grade is moved to $60$? Obviously, there will only be three of them who would pass. The averages, however, tell an interesting story. The average of those who will pass becomes $\frac{1}{3}(60+75+80) \approx 71.66$ and of those who won't becomes $40$. We summarize this in the table

Averages	Pass	Didn't Pass
Passing Grade 50	65	34
Passing Grade 60	71.66	40

Just by changing the passing mark, the averages of both groups actually increased! Even though we did not actually help either group, the averages seem to tell us that both groups improved.By increasing the passing grade, we moved the two students who got $55$ to the group that did not pass. Of course, $55$ is already greater than the grades those who did not pass, and by including them in that group, the average increases. On the other hand, the two who got $55$ have grades lower than the others who did pass, and by moving them to the other group, the average of the remaining also rises.

While this may look like a trivial example, there are real-world concerns where this could fool people. For example, instead of grades, imagine that the numbers above represents the income of $10$ people. Just by redefining what it means to be "rich" (the "passing" grade), one could falsely say that both the "rich" and "poor" experienced a rise in average incomes, even though no change actually happened. 

Another application is in medicine [1], [2], specifically, stage migration. In stage migration, improved detection of diseases moves people from one group into another (healthy and sick). Just like in the above example, the movement of people from one category to another can cause averages to rise (say, for example, average lifespan). The principle is still the same. Of course, we are not arguing against improved detection methods, but rather that we be wary in looking at just averages and not understanding what took place.

This phenomenon is called the Will Rogers phenomenon. The origin of its name is quite interesting. The comedian Will Rogers is said to have joked: "When the Okies left Oklahoma and moved to California, they raised the average intelligence level in both states" [3].

ABOUT THE AUTHOR:
Emerson Escolar is a Lecturer at the Ateneo de Manila University. He is currently taking his M.S. in Mathematics at the same university.

REFERENCES:

[1] Feinstein AR, Sosin DM, Wells CK (June 1985). "The Will Rogers phenomenon. Stage migration and new diagnostic techniques as a source of misleading statistics for survival in cancer". The New England Journal of Medicine 312 (25): 1604-8.
[2] Sormani, M. P.; Tintorè, M.; Rovaris, M.; Rovira, A.; Vidal, X.; Bruzzi, P.; Filippi, M.; Montalban, X. (2008). "Will Rogers phenomenon in multiple sclerosis". Annals of Neurology 64 (4): 428-433.
[3] Wikipedia Contributors. "Will Rogers phenomenon". Wikipedia, The Free Encyclopedia. http://en.wikipedia.org/wiki/Will_Rogers_phenomenon. (Accessed 6 Dec. 2011).

OLYMPIAD CORNER

from the 57th Belarusian Mathematical Olympiad

Problem: Prove that the inequality  $$\frac{1}{x_{1}}+\frac{x_{1}}{x_{2}}+\frac{x_{1}x_{2}}{x_{3}}+\frac{x_{1}x_{2}x_{3}}{x_{4}}+\cdots +\frac{x_{1}x_{2}\ldots x_{n}}{x_{n+1}}\geq 4\left( 1-x_{1}\ldots x_{n+1}\right)$$  holds for all positive numbers $x_{1},\ldots ,x_{n+1}$.

Solution: 

We note that for all positive numbers $a$ and $b$, we have 

$$\begin{align*} \left( 2b-1\right)^{2} &\geq &0 \\ 1 &\geq &4b(1-b) \\ \frac{1}{b} &\geq &4(1-b)  \\ \frac{a}{b} &\geq &4\left( a-ab\right)  \end{align*}$$

Hence the inequalities $$\frac{1}{x_{1}}\leq 4(1-x_{1})$$ and $$\frac{x_{1}x_{2}\ldots x_{k}}{x_{k+1}}\geq 4\left(x_{1}x_{2}\cdots x_{k}-x_{1}x_{2}\ldots x_{k}x_{k+1}\right)$$ holds for any $k=1,...,n$. Therefore, 

$$\begin{align*} \frac{1}{x_1}+\frac{x_1}{x_2}+\frac{x_1 x_2}{x_3}+\cdots + \frac{x_1 x_2 \ldots x_n}{x_{n+1}} &\geq &4\left(1-x_1\right) +4\left( x_1 - x_1 x_2 \right) +\cdots  \\ +4\left( x_1 \ldots x_n -x_1 \ldots x_n x_{n+1}\right)  &=&4\left(1-x_1 \ldots x_{n+1}\right) \end{align*}$$

as required.

PROBLEMS

10. Let $a, b, c$ be positive real numbers with $abc = 8$. Prove that $$\frac{a-2}{a+1} + \frac{b-2}{b+1} + \frac{c-2}{c+1} \leq 0.$$
    
11. Let $\{x\} = x - \left\lfloor x \right\rfloor$ denote the fractional part of $x$. Prove that for every natural number $n$, $$\sum_{j=1}^{n^2} \left\{ \sqrt{j} \right\} \leq \frac{n^2 - 1}{2}.$$
12. Show that there are infinitely many ordered integral pairs $(a,b)$ such that $\frac{a^3+b^3}{a-b}$ is a perfect square.

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 January 14, 2012. 

SOLUTIONS

(for November 26, 2011)

1. Evaluate $$1+\frac{1}{2}+\frac{2}{2}+\frac{1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{3}{3}+\frac{2}{3}+\frac{1}{3} +\ldots+\frac{1}{n}+\frac{2}{n}+\ldots+\frac{n}{n}+\frac{n-1}{n}+\ldots+\frac{1}{n}.$$ (Taken from Principles and Techniques in Combinatorics by Chong and Meng) 
   (solved by Ferdinand John S. Briones [MSHS], Emman Joshua B. Busto [PSHS - Main], Dianne Garcia [AA], Albert Jason Z. Olaya [PSHS - Main] and Gerald M. Pascua [PSHS - Main])
   
   
   SOLUTION: 
   Note that for every positive integer $k$ we have $$\frac{1}{k} + \frac{2}{k} + \ldots + \frac{k}{k} + \frac{k-1}{k} + \ldots \frac{1}{k} = \frac{2\frac{(k-1)k}{2}+k}{k} = k,$$ Thus the sum reduces to $$1 + 2 + \ldots + n = \frac{n(n-1)}{2}.$$
2. $ABCD$ is a square, $E$ and $F$ are the midpoints of the sides $AB$ and $BC$ respectively. If $M$ is the point of intersection of $CE$ and $DF$, prove that $AM$ = $AD$. (Taken from Lecture Notes on Mathematical Olympiad Courses by Xu Jiagu)
   (solved by Ferdinand John S. Briones [MSHS] and Emman Joshua B. Busto [PSHS - Main])
   
   
   SOLUTION: 
   We have the following figure
   
   Since $BE = CF$ and $BC = CD$, we have $\Delta DFC \cong \Delta CEB$, meaning $\angle CDF = \angle BCE$ and thus $CE \perp DF$. 
   
   
   Extend $CE$ to intersect the extension of $DA$ at some point $N$.
   
   We now have $AE = BE$ and $\angle NEA = \angle BEC$, which means $\Delta AEN \cong \Delta BEC$ and hence $AN = BC = AD$, implying that $AM$ is the median of $ND$, which is the hypotenuse of $\Delta NMD$, giving us $AM = AN = AD$.
3. If $A, B$ and $C$ are the measurements of the angles in each of the vertices of the triangle $ABC$, show that $$\cos A + \cos B + \cos C \leq \frac{3}{2}.$$ (Taken from Inequalities by Manfrino, Ortega, and Delgado)
   
   

   (solved by Ferdinand John S. Briones [MSHS])
   
   
   SOLUTION: 
   Note that $$a(b^2+c^2-a^2)+b(c^2+a^2-b^2)+c(a^2+b^2-c^2)=(b+c-a)(c+a-b)(a+b-c)+2abc$$ thus we have $$\begin{align*} \cos A + \cos B + \cos C &=& \frac{b^2+c^2-a^2}{2bc} + \frac{c^2+a^2-b^2}{2ac} + \frac{a^2+b^2-c^2}{2ab} \\ &=& \frac{(b+c-a)(c+a-b)(a+b-c)}{2abc} + 1 \end{align*}$$ where $(b+c-a)(c+a-b)(a+b-c) \leq abc$, hence the result.

Additional Material and Announcements

The module handout for those taking Methods of Proof can be found by clicking here.

People who wish to submit answers to the problems in the November 26 issue (Angry Birds) can still do so until December 10, 2011. The 3rd issue of Tuklas has also been released, so check that out also! :)

Sequences in Resistors (Tuklas Vol. 13, No. 3 - Dec. 3, 2011)

SEQUENCES IN RESISTORS

Many of us are very familiar with the concept of series and parallel circuits in physics - quite a bold opening statement for a mathematics article, don’t you agree? In this issue, we will attempt to make a connection between concepts of mathematics as a result of some physical phenomenon. Allow me to refresh your memory regarding resistors arranged in series. The total, or in very fancy terms, the effective resistance, is simply the sum of the individual resistances present in the system. For simplicity, all the resistances we will deal with here have a value of 1. As an example, if resistors are arranged in a purely series circuit as in Fig(1), then the effective resistance is $R_{tot} = \sum_{i=1}^{3} R_i = 3$.

If resistors are arranged in parallel as in Fig(2), then the effective resistance is the reciprocal of the sum of the reciprocals of each resistance. Confusing? It is a lot simpler to write down the equation than to describe it: $\frac{1}{R_{tot}} = \sum_{i=1}^{3} \frac{1}{R_i} =\frac{1}{3}$. Now most of these concepts sound elementary if a circuit is entirely series or parallel. The challenge begins when both series and parallel are mixed and appear simultaneously in a system. A piece of unsolicited advice here (and I'm sure this always works): start from the outermost resistors and comb your way going in, combining series (the easiest) and parallel resistors (the easier) along the way as you see them. Don't forget the formulas, especially the reciprocals! And so Fig(1) is child’s play: the effective resistance is 3. If one loop is added as in Fig(3), then the effective resistance is $\frac{11}{4}$. With two loops, the effective resistance decreased to 2.75. So what happens if more and more loops are added? Will the effective resistance decrease all the way? The table below summarizes the effective resistances side-by-side the number of loops in a circuit:

No. of loops	Resistance
1	3
2	2.75
3	2.7333333
4	2.732142857
5	2.732057416
6	2.732051282
7	2.732050842
8	2.73205081
9	2.732050808
10	2.732050808
11	2.732050808

Well this may seem a random set of numbers, but the values obviously approach some number. Because the process of solving for the effective resistance of many loops becomes repetitive, the numbers on the right column must be generated by a certain function. In fact, we can derive the generating function by observing the first few terms of the sequence. It can be seen that if only 1 loop is present, the effective resistance is 3. We can assign that to be $x_1=3$. Following our equations for series and parallel resistances, $x_2=\frac{1}{\frac{1}{x_1}+1}+2$. In general, $x_n=\frac{1}{\frac{1}{x_{n-1}}+1}+2$. With a little courage, clean sheets of paper, and tons and tons of free time, you will see that the limiting value of the sequence can be written as $\lim_{n \to +\infty} x_n = 1+\sqrt{3}$. In fact, $1+\sqrt{3}$ is approximately 2.732050808. The sequence converges to the exact value rather quickly; and with the fifth loop, you have five correct decimal figures ready for a good approximation.

Finally, as a means of verification, a well known solution, which may require some leap of faith, is provided below. Fig(4) is an illustration of what this infinite blanket of resistors may look like. Since the problem desires the effective resistance of this infinity system be known, let us call this total resistance as $R$. Notice the boxed area extends all the way to infinity as well, so subtracting one loop from an infinite number loops will surely excite the great mathematician Hilbert - and he would most likely say: it still contains an infinite number of loops and the total resistance is still $R$. And so, with a leap of faith on the wings of prayer, we can reduce the system into bits that is easier to work on as in Fig(5). By applying the formulas again, I have the final steps of solving for $R$: $$\begin{align*} \left(\frac{1}{R}+1 \right)^{-1} + 2 &= R \\ \frac{R}{R+1} + 2 &= R \\ \frac{R}{R+1} &= R-2 \\ R &= R^2-R-2 \\ R^2-2R-2 &= 0 \end{align*}$$
Finally the quadratic equation will yield $R=1+\sqrt{3}$, which is exactly what we wanted.

ABOUT THE AUTHOR:
Clark Kendrick Go is an Instructor at the Ateneo de Manila University. He is currently taking his M.S. in Physics at the same university.

OLYMPIAD CORNER

from the Estonian Math Competitions, 2006-2007

Problem: Let $x,y,z$ be positive real numbers such that $x^{n},y^{n}$ and $z^{n}$ are side lengths of some triangle for all positive integers $n$. Prove that at least two of $x,y$ and $z$ are equal.

Solution: 

We first assume that $x,y,z$ are all different, and in particular, $x<y<z$ (without loss of generality). For any $n$, the triangle inequality implies $x^{n}+y^{n}>z^{n}$, or $$\left( \dfrac{x}{y} \right)^n+1 >\left( \dfrac{z}{y}\right)^n. \hspace{20px} (1)$$
Since $\dfrac{x}{y}<1$, then $\left( \dfrac{x}{y}\right)^n+1<2$ for any integer $n$. On the other hand, $\dfrac{z}{y}$ is larger than 1, hence there exists an integer $N$ large enough so that $\left( \dfrac{z}{y}\right)^{N}>2$. This means that equation $(1)$ cannot hold for all $n,$ leading to a contradiction.

As a consequence, $x,y,z$ cannot be distinct numbers; at least two of them must be equal.

PROBLEMS

7. Evaluate $$\frac{1}{(1+1)!} + \frac{2}{(2+1)!} + \ldots + \frac{n}{(n+1)!},$$ where $n$ is any natural number.
   
8. Suppose that the incircle of $ABC$ is tangent to the sides $BC, CA, AB$ at $D, E, F$ respectively. Prove that $$EF^2 + FD^2 + DE^2 \leq \frac{s^2}{3},$$ where $s$ is the semiperimeter of $ABC$.
9. On the blackboard some student has written 17 natural numbers, and their units digits are inside the set $\{0, 1, 2, 3, 4\}$. Prove that one can always select out 5 numbers from them such that their sum is divisible by 5.

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 December 10, 2011. 

SOLUTIONS

(for November 19, 2011)

1. From 2011 subtract half of it at first, then subtract $\frac{1}{3}$of the remaining number, next subtract $\frac{1}{4}$ of the remaining number, and so on, until $\frac{1}{2011}$ of the remaining number is subtracted. What is the final remaining number?  
   (solved by Daniel Berba [Ateneo HS], Ferdinand Briones [MSHS], Jungwon Hong [St. Paul Pasig], Ezekiel Ong [Uno HS], Samuel Ong [Uno HS], Andrea Onglao [St. Paul Pasig], Gerald M. Pascua [PSHS - Main], Lorenzo Quiogue [Ateneo HS], Simon San Pedro [MSHS] and Marisse T. Sonido [AA])
   
   SOLUTION:
   Note that $$\begin{align*} 2011 - \left(\frac{1}{2}\right)2011 &= 2011\left( 1 - \frac{1}{2} \right) \\ \left(\frac{1}{2}\right)2011 - \frac{1}{3}\left[\left(\frac{1}{2}\right)2011\right] &= \left(\frac{1}{2}\right)2011\left( 1 - \frac{1}{3} \right) \\ &= 2011\left( 1 - \frac{1}{2} \right)\left( 1 - \frac{1}{3} \right) \\ \end{align*}$$ and so on implies that the answer is $$2011\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\ldots\left(1-\frac{1}{2011}\right) = 2011\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\ldots\frac{2009}{2010}\cdot\frac{2010}{2011} = 1.$$
2. In triangle $ABC$, $\angle A = 96^{\circ}$. Extend $BC$ to an arbitrary point $D$. The angle bisectors of $\angle ABC$ and $\angle ACD$ intersect at $A_1$, and the angle bisectors of $\angle A_1BC$ and $\angle A_1CD$ intersect at $A_2$, and so on. The angle bisectors of $\angle A_4BC$ and $\angle A_4CD$ intersect at $A_5$. Find the size of $\angle A_5$ in degrees. (China Mathematical Competitions for Secondary Schools, 1998)  
   (solved by Daniel Berba [Ateneo HS], Ferdinand Briones [MSHS], Jungwon Hong [St. Paul Pasig], Samuel Ong [Uno HS], Andrea Onglao [St. Paul Pasig], Gerald M. Pascua [PSHS - Main] and Lorenzo Quiogue [Ateneo HS])
   
   SOLUTION:
   Constructing $A_1$, we have the following figure
   
   and upon repeating the process four other times, we obtain the resulting figure
   
   Note that since $A_1 B$ and $A_1 C$ bisect $\angle ABC$ and $\angle ACD$ respectively, $$\angle A = \angle ACD - \angle ABC = 2(\angle A_1 CD - \angle A_1 BC) = 2\angle A_1$$ giving us $\angle A_1 = \frac{1}{2} \angle A$. Similarly, we obtain $A_{k+1} = \frac{1}{2}A_k$ for $k = 1,2,3,4,$ thus $\angle A_5 = \frac{1}{2^5}\angle A = 3^{\circ}.$
3. Let $a_1 , \ldots , a_n, b_1, \ldots , b_n$ be positive numbers, prove that $$\sum_{i=1}^{n}{\frac{1}{a_{i}b_{i}}}\sum_{i=1}^{n}(a_{i}+b_{i})^{2} \geq 4n^{2}.$$ (Taken from Inequalities by Manfrino, Ortega, and Delgado)  
   (solved by Daniel Berba [Ateneo HS], Samuel Ong [Uno HS] and Lorenzo Quiogue [Ateneo HS])
   
   SOLUTION:
   By AM-GM, we have $$(a_i + b_i)^2 \geq 4a_i b_i.$$
   Thus $$\sum_{i=1}^{n} \frac{1}{a_i b_i} \sum_{i=1}^{n} (a_i + b_i)^2 \geq \sum_{i=1}^{n} \frac{1}{a_i b_i} \sum_{i=1}^{n} 4a_i b_i = 4 \sum_{i=1}^{n} \frac{1}{a_i b_i} \sum_{i=1}^{n} a_i b_i.$$
   Furthermore, by the Cauchy-Schwartz inequality, $$\begin{align*} 4 \sum_{i=1}^{n} \frac{1}{a_i b_i} \sum_{i=1}^{n} a_i b_i &= 4 \sum_{i=1}^{n} \left(\frac{1}{\sqrt{a_i b_i}}\right)^2 \sum_{i=1}^{n} \left(\sqrt{a_i b_i}\right)^2 \\ &\geq 4 \sum_{i=1}^{n} \left(\frac{1}{\sqrt{a_i b_i}} \sqrt{a_i b_i}\right)^2 \\ &= 4n^2. \end{align*}$$

Basic Mathematics of "Angry Birds" (Tuklas Vol. 13, No. 2 - Nov. 26, 2011)

BASIC MATHEMATICS OF "ANGRY BIRDS"

"Angry Birds" has been a popular touchscreen game that quenched the boredom and leisure of many since its inception in 2010. The mathematics behind the "Angry Birds" is found in describing the path that the bird takes to get to its target. Figure 1 provides a schematic diagram of the game. The goal is to launch the bird, by drawing the slingshot, to target green pigs situated on the other side of the screen. There is a story line behind the animosity the birds have over the pigs, but it suffices to say that they are motivated enough to sacrifice themselves as avian cannonballs. 

Figure 1: Schematic diagram of the "Angry Birds" game. The inset is
a zoomed-in version of the bird on the slingshot.

With a bird riding on it, the slingshot is drawn tight such that it makes a certain angle $\theta$ with respect to the horizontal within the reasonable interval $\left[0,\frac{\pi}{2}\right]$. This is referred to as the launching angle. It is also the angle between the horizontal and an arrow that represents the take-off velocity of the bird (see zoomed-in inset of Fig. 1). When the slingshot is released, the bird takes off with initial speed $v_0$ in the direction denoted by the arrow. 

Since the game is laid in 2d (i.e., on a plane), then position can be described by coordinates $(x,y)$, where $x$ denotes position along the horizontal axis and $y$ position along the vertical axis. Let us take the origin of the axis as the launching point for the angry bird, i.e., $(0,0)$ as depicted in Fig. 1. The position of the target (e.g., the green pig) is at $(p,q)$. For the sake of simplicity, let us take $q \approx 0$ (i.e., the target is situated at a vertical level close to the level from which the bird is launched). 

During its flight, the bird changes its position $(x,y)$ so that $x = x(t)$ and $y = y(t)$ are functions of time $t$. This motion, known as projectile motion, is a combination of both the forward horizontal motion and falling vertical motion. Indeed, not all falling objects fall straight down to the ground. It all depends on how the object is launched. Thanks to Galileo Galilei and Isaac Newton, who laid the foundations of our current understanding about objects in motion in the 17th century, we can write down the mathematical equations to quantify motion. 

The displacement is generally a product of the speed of motion and time. According to the findings of Galileo, the bird's horizontal speed should not change throughout its flight since nothing is pushing the bird to move forward horizontally (Note: the game always takes place on a windless day so that one can conveniently neglect the effect of air). That means $x(t) = (v_0 \cos \theta) t$ where $v_0 \cos \theta$ is the component of the bird's initial motion along the horizontal axis. On the other hand, some invisible agent is pulling the bird down so that it is bound to fall back to the ground. According to Newton, this agent is known as gravity which makes falling objects close enough to the ground (i.e., anything lower than Mt. Everest) gain speed at a constant rate, $g$, on its way down. Consequently, $y(t) = (v_0 \sin \theta) t - g\left(\frac{t^2}{2}\right)$, where $v_0 \sin \theta$ is the component along the vertical axis of the bird's initial motion. In summary, the equations of motion in terms of time t are: $$\begin{align} x(t) &= (v_0 \cos \theta) t \\ y(t) &= (v_0 \sin \theta) t - g\left(\frac{t^2}{2}\right) \end{align}$$

By eliminating $t$, an equation for $y = y(x)$ can be written as follows,  $$y(x) = (\tan \theta)x - \left(g\frac{{v_0}^2}{2}{\cos}^2 \theta\right)x^2$$ which is an equation for a parabola (concave downward) on the Cartesian plane. Indeed if you observe the paths traced out in the game, they do resemble the parabola.

What is more interesting is that one could maximize the range (i.e.,how far the bird projectile would land away from the slingshot for any given initial speed $v_0$) by simply adjusting the launching angle. In order to determine the optimal angle $\theta^{\ast}$, one solves for $t$ for $y = q \approx 0$. This leads to a quadratic equation whose roots are: $$t^{\ast} = 0 \text{ and } t^{\ast} = \frac{1}{g}(2 v_0 sin \theta)$$  

The only relevant solution is the latter one, which is now used to determine the final position $x(t^{\ast})$. The range $R$ is simply the difference between the horizontal position of the angry bird, i.e., $x(0) = 0$ and the final position; hence,  $$R= (v_0 \cos \theta)t^{\ast} = (v_0 \cos \theta)\left(\frac{1}{g}\right)(2v_0 \sin \theta) = \left(\frac{1}{g}\right)({v_0}^2 \sin 2\theta) = R(\theta).$$

By using the basic concepts of differential calculus, $R(\theta)$ is maximized such that $\frac{dR}{d\theta} = 0$ and $\frac{d^2R}{d^2\theta} < 0$. Hence, $$\begin{eqnarray*} \left.\frac{dR}{d\theta}\right|_{\theta=\theta^*} &= & \left.\frac{v_0^2}{g}\left(2\cos2\theta\right)\right|_{\theta=\theta^*} =0\\ & &\Rightarrow \cos2\theta^* = 0 \Rightarrow 2\theta^* = \frac{\pi}{2} \Rightarrow \theta^* = \frac{\pi}{4} \equiv 45^{\circ}\, . \end{eqnarray*}$$

Consequently, $\frac{d^2R}{d^2\theta} = -\sin 2\theta < 0$ for $\theta = \theta^{\ast}$. Therefore, $\theta^{\ast} = 45^{\circ}$ does give the maximum range. This can be verified through the game. One would see that if $\theta > 45^{\circ}$, the angry bird can go up so high but does not really travel forward by as much. On the other hand, if $\theta$ is too small, the bird lands on the ground right away before being able to move appreciably forward.

So next time you play "Angry Birds", find a protractor and mark your screens with the special angle $45^{\circ} = \frac{\pi}{4}$. It might be quite useful in aiming for those nasty swines.

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 14th Turkish Mathematical Olympiad, 2006

Problem: Points $E$ and $F$ on side $CD$ of a convex quadrilateral $ABCD$ are given such that $DE = FC$. The circumcircles of triangles $ADE$ and $ACF$ meet again at $K$ and those of $BDE$ and $BCF$ meet again at $L$. Show that the points $A, B, K, L$ lie on a circle.

Solution: 

Let $M$ be intersection of $DC$ and $AK$, and $N$ be theintersection of $DC$ and $BL$. Based on  the powers of the point $M$ with respect to the circles $ADE$ and $AFC$, we have $$ME\cdot MD=MK\cdot MA=MF\cdot MC \hspace{20px} (1)$$

Using the derived equality $ME\cdot MD=MF\cdot MC$ and the given $DE=FC$, $$\begin{align*} ME\cdot MD &= MF\cdot MC \\ ME\cdot ( ME+DE) &= MF\cdot ( MF+FC) \\ (ME)^2+ME\cdot DE &= ( MF)^2+MF\cdot FC \\ (ME)^2+ME\cdot DE &= (MF)^2+MF\cdot DE \\ (ME)^2-(MF)^2+ME\cdot DE-MF\cdot DE &= 0 \\ (ME-MF)(ME+MF)+DE(ME-MF) &= 0 \\ (ME-MF) (ME+MF+DE) &= 0 \end{align*}$$ Noting that $ME+MF+DE>0$, then $ME=MF$ and hence $M$ is the midpoint of segment $FE$. Since $DE=FC$, then $MD=MC$ and therefore $M$ is the midpoint of segment $CD$.

Likewise, based on the powers of point $N$, we have $$NE\cdot ND=NL\cdot NB=NF\cdot NC \hspace{20px} (2)$$ where can it also be shown, using the same arguments as above, that $N$ is the midpoint of segment $CD$. Hence $M$ and $N$ represent the same point.

From $(1)$ and $(2)$, we have $$MK\cdot MA\underset{(1)}{=}ME\cdot MD\underset{M=N}{=}NE\cdot ND\underset{(2)}{=}NL\cdot NB\underset{M=N}{=}ML\cdot MB$$ which tells us that $A,B,K,L$ all lie on a circle. 

PROBLEMS

4. Evaluate $$1+\frac{1}{2}+\frac{2}{2}+\frac{1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{3}{3}+\frac{2}{3}+\frac{1}{3} +\ldots+\frac{1}{n}+\frac{2}{n}+\ldots+\frac{n}{n}+\frac{n-1}{n}+\ldots+\frac{1}{n}.$$
   
5. $ABCD$ is a square, $E$ and $F$ are the midpoints of the sides $AB$ and $BC$ respectively. If $M$ is the point of intersection of $CE$ and $DF$, prove that $AM$ = $AD$.
6. If $A, B$ and $C$ are the measurements of the angles in each of the vertices of the triangle $ABC$, show that $$\cos A + \cos B + \cos C \leq \frac{3}{2}.$$

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 December 3, 2011 (extended to December 10, 2011).

Invariance (Tuklas Vol. 13, No. 1 - Nov. 19, 2011)

INVARIANCE

Consider this scenario: At first, a room is empty. Each minute, either one person enters or two persons leave. After exactly $3^{1999}$ minutes, could the room contain $3^{1000}+2$ people?

Problems such as this present a significant level of shock factor to those who would encounter them for the first time. Of course, we can always try to use brute force in solving problems like these, but it is highly impractical. So instead of trying to deal with the mess and complexity of the problems head on, it would be wise for us to "reduce" them by only considering the essential properties or entities that such problems possess. There are many ways in which we can do this, but for this article, we are mainly focusing on determining invariants.

As the term implies, an invariant is an aspect of a problem that does not change, even if the conditions and other properties of the problem do change. It can be numeric or non-numeric in nature, though in most cases, it is found to be a numerical quantity. 

An example of an invariant can be seen in Euler's Formula. This formula states that there is a relationship between the number of edges ($e$), vertices ($v$) and faces ($f$) of a polyhedron without "holes", in particular $$v - e + f = 2$$ Meaning whether the polyhedron is a cube, a tetrahedron or a "buckyball", the quantity $v-e+f$ is always equal to 2, and as such, it is an invariant. 

Now let us return to the problem posed at the beginning of the article. We can start by making a general case, wherein we let $n$ be the number of people in the room. After one minute, there will be either $n+1$ or $n-2$ people. Note that the difference between the two possible outcomes is 3. The next minute, the possible number people in the room can be $n+2, n-1$ or $n-4$. Observe that the difference between any two outcomes is either a 3 or 6. 

In the long run, we can actually conclude that, at any fixed time $t$, the possible values for the population of the room differ from one another by a multiple of 3. 

So after $3^{1999}$ minutes, it would be possible for us to have $3^{1999}$ people in the room (this means that 1 person comes in after every minute.). Based on our conclusion above, the other possible populations of the room should differ $3^{1999}$ by a multiple of 3. And since $3^{1999}$ is a multiple of 3, hence the other population outcomes should also be a multiple of 3. This means therefore that $3^{1000}+2$ will not be a valid population. 

The problem above gave us an example of a divisibility invariant (divisible by 3). The simplest divisibility invariant is parity, which merely refers to the oddness or evenness of a number. This concept will be explored in greater detail through the following problems. 

Let $a_1, a_2, \ldots, a_n$ represent an arbitrary arrangement of the numbers $1, 2, 3, \ldots, n$. Prove that if $n$ is odd, the product $$(a_1 - 1)(a_2 - 2)(a_3 - 3)\ldots(a_n - n)$$ is an even number. 

What we can do here is consider the sum of the terms.   $$\begin{align*} (a_1 - 1)(a_2 - 2)(a_3 - 3)\ldots(a_n - n) & = (a_1 - 1)(a_2 - 2)(a_3 - 3)\ldots(a_n - n)-(1+2+...+10) \\ & = (1+2+...+10) - (1+2+...+10) \\ & = 0 \end{align*}$$

The sum therefore is an invariant, since its always equal to zero (an even number), regardless of the arrangement of the numbers. Since the number of terms is odd, the only possible way for them to have an even sum is for the $n$ terms to have at least one even number. This basically solves the problem. 

In the figure below you may switch the signs of all numbers of a row, column, or a parallel to one of the diagonals. In particular, you may switch the sign of each square corner square. Prove that at least one -1 will remain in the table.

1	1	1	1
1	1	1	1
1	1	1	1
1	-1	1	1

Let $S$ represent the eight boundary squares (except the four corners). We note that the product of those squares is equal to -1. After every transformation, it is either none get switched (this is in the event that the sign of the corner square is altered), or two of the squares in $S$ get switched: 

1 and 1 becomes -1 and -1,

-1 and -1 becomes 1 and 1,

1 and -1 becomes -1 and 1,

-1 and 1 becomes 1 and -1.

Since the product of the resulting pair is the same as the original pair, the product of the squares in $S$ will always remain at -1. This means that there will always be at least one -1 in the table. 

The vertices of a cube are labeled $a, b, c, d, w, x, y, z$ as seen in the figure. Vertices $a$ and $c$ are initially given a value of 1, while the rest of the vertices are given a value of 0. If you are allowed to add 1 to each of any pair of adjacent vertices, can you eventually get all the vertices to have the same value? 

You can actually get your hands dirty for this problem, and no matter what you do, you will not be able to get the desired result of making all values equal. But how can you prove this? 

We can let $s_1 = a + c + w + y$ and $s_2 = b + d + x + z$. Hence at the start, $s_1 -s_2 = 2$. The end result we want is such that all the vertices will have the same value; this implies therefore that $s_1 - s_2 = 0$. 

What happens when we add 1 to each of any pair of adjacent vertices? Notice that every time we pick such a pair, one of them belongs to $s_1$, and the other belongs to $s_2$. Therefore each time we perform the process, 1 is added to both $s_1$ and $s_2$. This makes the difference $s_1 - s_2$ remain equal to 2, which means it is actually an invariant. 

Since the difference can never be equal to 0, thus we can conclude that we cannot get all the vertices to have the same value.

ABOUT THE AUTHOR:

Timothy Robin Teng is an Instructor at the Ateneo de Manila University. He is currently finishing his Ph.D. in Mathematics at the same university.

OLYMPIAD CORNER  

from the 17th Asian Pacific Mathematics Olympiad, March 2005

Problem: Prove that for every irrational real number $a$, there are irrational real numbers $b$ and $b^{\prime}$ such that $a+b$ and $ab^{\prime}$ are both rational while $ab$ and $a+b^{\prime}$ are both irrational.

Solution: 

Let $a$ be an irrational number. We first show that there exists an irrational number $b$ such that $a+b$ is rational and $ab$ is irrational.

For this, we consider the number $a^2$. If $a^2$ is irrational, let $b=-a$. Then $a+b=a-a=0$ is rational, and $ab=a(-a) =-a^2$ is irrational. If, on the other hand, $a^2$ is rational, let $b=a^2-a$ (difference between a rational and irrational number). Then $a+b=a+(a^2-a)=a^2$ is rational, and $ab=a(a^2-a)=a^2(a-1)$ is irrational (product of a rational number $a^2$ and irrational number $a-1$).

We next show that there exists an irrational number $b^{\prime}$ such that $ab^{\prime}$ is rational and $a+b^{\prime}$ is irrational. Let $$b^{\prime}=\frac{1}{a} \text{ or }b^{\prime }=\frac{2}{a}.$$
Then $ab^{\prime}$ is a rational number, which has a value of either 1 or 2. Now note that
$$a+b^{\prime}=\frac{a^2+1}{a}\text{ or  }a+b^{\prime}=\frac{a^2+2}{a}.$$
Next we take the difference, $$\frac{a^2+2}{a}-\frac{a^2+1}{a}=\frac{1}{a}$$ where the result is an irrational number. This implies that at least one of $\frac{a^2+1}{a}$ and $\frac{a^2+2}{a}$ is irrational (both of them cannot be rational since the difference of two rational numbers is a rational number). This shows that there exists an irrational $b$ such that $a+b^{\prime}$ is irrational.

 

PROBLEMS

1. From 2011 subtract half of it at first, then subtract $\frac{1}{3}$of the remaining number, next subtract $\frac{1}{4}$ of the remaining number, and so on, until $\frac{1}{2011}$ of the remaining number is subtracted. What is the final remaining number?
2. In triangle $ABC$, $\angle A = 96^{\circ}$. Extend $BC$ to an arbitrary point $D$. The angle bisectors of $\angle ABC$ and $\angle ACD$ intersect at $A_1$, and the angle bisectors of $\angle A_1BC$ and $\angle A_1CD$ intersect at $A_2$, and so on. The angle bisectors of $\angle A_4BC$ and $\angle A_4CD$ intersect at $A_5$. Find the size of $\angle A_5$ in degrees.
3. Let $a_1 , \ldots , a_n, b_1, \ldots , b_n$ be positive numbers, prove that $$\sum_{i=1}^{n}{\frac{1}{a_{i}b_{i}}}\sum_{i=1}^{n}(a_{i}+b_{i})^{2} \geq 4n^{2}.$$

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 November 26, 2011.

PEM Class List (Teachers)

PROGRAM FOR EXCELLENCE IN MATHEMATICS 2011-2012 

Teachers Part I: November 26 and December 10, 2011

Module 1
Problem Solving Strategies
Schedule: Sat, 9:30-11:30AM
Teacher: Dr. Ma. Alva Aberin

Room: CTC 102

#	School	Last Name	First Name	M.I.	Attendance
					OC	1	2
1	Caloocan High School	Bataller	Liza	D.A.	Y
2	Caloocan High School	Corona	Generieve	B.	Y
3	Caloocan High School	Rubino	Erlan	P.	Y
4	Colegio San Agustin - Makati	Gacutan	Rolando	B.	Y
5	Colegio San Agustin - Makati	Llanto	Paolo	A.	Y
6	Colegio San Agustin - Makati	Maristela	Jasmin	D.	Y
7	Immaculate Conception Academy	Baranda	Agnes	D.	Y
8	Lourdes School of Mandaluyong	Arpia	Dionelyn	S.	Y
9	Lourdes School of Mandaluyong	Duavit	James	V.	Y
10	Lourdes School of Mandaluyong	Sebastian	Modesto Jr.	C.	Y
11	Lourdes School Quezon City	Dela Cruz	Carlo	R.	Y
12	Lourdes School Quezon City	Lim	Carmela	D.	Y
13	Lourdes School Quezon City	Lumbre	Angelina	P.	Y
14	Manila Science High School	Aniban	Diana Grace	B.	Y
15	Manila Science High School	Pollo	Anancita	L.	Y
16	Manila Science High School	Ramirez	Julie Anne	O.	Y
17	Notre Dame of Greater Manila	Geraldoy	Rodel	V.	Y
18	Notre Dame of Greater Manila	Narciso	Ma. Rizaliana	D.C.	Y
19	PAREF Southridge School	Alcaraz	Rex		Y
20	PAREF Southridge School	Tadeo	Jigs	T.	Y
21	PAREF Southridge School	Tamayao	Nancisco	S.	Y
22	Philippine Pasay Chung Hua Academy	Espina	Esterlita	D.	Y
23	Philippine Pasay Chung Hua Academy	Franco	Helen	L.	Y
24	Philippine Science High School - Main	Granada	Kiel	F.	Y
25	San Sebastian College Manila	Fernandez	Marilyn		Y
26	San Sebastian College Manila	Noblejas	Marissa		Y
27	St. Matthew College	dela Cruz	James Robin	R.	Y
28	St. Scholastica's College, Manila	Muli	Eduardo, Jr.	D.	Y
29	St. Scholastica's College, Manila	Soto	Antonio		Y
30	St. Theresa's College of Quezon City	Abriol	Dave		Y
31	St. Theresa's College of Quezon City	Narvaez	Marvin		Y
32	St. Theresa's College of Quezon City	Viloria	Khriztoffer		Y
33	Uno High School	Ang	Kurt Byron	C.	Y
34	Uno High School	Rosario	Jay-Ar	S.	Y
35	Xavier School	Bague	Adelaida	H.	Y
36	Xavier School	Hernandez	Linda May	S.	Y