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, $$\begin{align*} 4 \sum_{i=1}^{n} \frac{1}{a_i b_i} \sum_{i=1}^{n} a_i b_i &= 4 \sum_{i=1}^{n} \left(\frac{1}{\sqrt{a_i b_i}}\right)^2 \sum_{i=1}^{n} \left(\sqrt{a_i b_i}\right)^2 \\ &\geq 4 \sum_{i=1}^{n} \left(\frac{1}{\sqrt{a_i b_i}} \sqrt{a_i b_i}\right)^2 \\ &= 4n^2. \end{align*}$$