**THE WILL ROGERS PHENOMENON**

\[ \bar{x} = \frac{x_1 + x_2 + \ldots + x_n}{n} = \frac{1}{n}\sum_{i=1}^n x_i. \]

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).

[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**

- Let \(a, b, c\) be positive real numbers with \(abc = 8\). Prove that \[\frac{a-2}{a+1} + \frac{b-2}{b+1} + \frac{c-2}{c+1} \leq 0.\]

- Let \(\{x\} = x - \left\lfloor x \right\rfloor\) denote the fractional part of \(x\). Prove that for every natural number \(n\), \[ \sum_{j=1}^{n^2} \left\{ \sqrt{j} \right\} \leq \frac{n^2 - 1}{2}.\]
- 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)**

*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 \[1 + 2 + \ldots + n = \frac{n(n-1)}{2}.\]*\(ABCD\) 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 figureSince \(BE = CF\) 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 \(AM = AN = AD\).*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.

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