A Divine Proportion (Tuklas Vol. 14, No. 5 - Jan. 12, 2013)

A DIVINE PROPORTION

Consider a string of finite length. Let us divide it into two parts. We do our division such that the ratio of the longer and shorter parts is equal to the ratio of the whole length and the longer part. It turns out that no matter how long the string is, the common value is a fixed positive number. Throughout history it has been called by many names such as the divine proportion, golden ratio, golden section, medial section and extreme and mean ratio. As a symbol, we denote it by $\varphi$(phi), the first letter in the name of the Greek sculptor Phidias.

We now use our string analogy to try and see what this number looks like. Two numbers \(a\) and \(b\), with \(a>b\), are said to be in the golden ratio if \(\frac{a+b}{a} = \frac{a}{b} = \varphi\). By rewriting the left-most quantity into \(1+\frac{b}{a}\), we get the equation \(\varphi = 1 + \frac{1}{\varphi}\). After some algebraic manipulations, we arrive at \[ \varphi ^2 - \varphi - 1 =0. \] Using the quadratic formula, we have \(\frac{1+\sqrt{5}}{2}\) and \(\frac{1-\sqrt{5}}{2}\) as solutions to the equation. Since \(\frac{1-\sqrt{5}}{2} < 0 < \frac{1+\sqrt{5}}{2}\), then \(\varphi = \frac{1+\sqrt{5}}{2}\). Using ten decimal places, the value of \(\varphi\) is approximately equal 1.6180339887. We can also write \(\varphi\) in terms of a continued fraction. We start with \(\varphi = 1 + \frac{1}{\varphi}\). By replacing \(\varphi\) with \(1 + \frac{1}{\varphi}\), we get \(\varphi = 1 + \frac{1}{1 + \frac{1}{\varphi}}\). Repeat the process infinitely many times and we get the equation shown below [1].

Let's move on to two dimensional objects. A golden rectangle is a rectangle whose side lengths are in the golden ratio. We can construct one such rectangle by starting with a right triangle with a base of half a unit and a height of one unit. From the Pythagorean theorem, the length of the hypotenuse is \(\frac{\sqrt{5}}{2}\) units. Rotate the hypotenuse in such a way that it extends the base of the triangle. Then we complete the rectangle. The golden rectangle that we have constructed has dimensions 1 unit by \(\varphi\) units. If we adjoin a square whose side is of length \(\varphi\) units, the resulting rectangle formed is also a golden rectangle. In fact, if we start with an arbitrary golden rectangle and adjoin a square whose side is of the same length as the longer side of the rectangle then we create another golden rectangle.

Remember Phidias, the guy from whom the symbol for the golden ratio originated? It is said that he incorporated golden rectangles into the design of Parthenon. The length and height of the building, the spacing between the columns and the pitch of the roof are all controlled by the golden section. Architects and artists believed that structures and paintings whose design is governed by the medial section are more pleasing to the eye.

Figure 1: The Greek Parthenon, and the superimposed golden rectangle. [2]

We can also observe the extreme and mean ratio in the natural sciences. For example, a biologist sees it in the structure of a human face and in the lengths of a finger's phalanges. A botanist sees it the pattern of seeds in a sunflower and in the positioning of leaves in a plant. A zoologist sees it in the population growth of rabbits. There are still a lot more examples that naturally follow the divine proportion. It is as if there is a divine power that guides all things to be created in such a beautiful manner. Truly, \(\varphi\) is such a divine number.

ABOUT THE AUTHOR:
Janree Ruark Gatpatan obtained his B.S. in Applied Mathematics at the University of the Philippines - Visayas in 2008 and is currently taking his Ph. D. in Mathematics at the Ateneo de Manila University.

REFERENCES:

[1] Image taken from http://en.wikipedia.org/wiki/Fibonacci_number.
[2] Image taken from http://emptyeasel.com/wp-content/uploads/2009/01/golden-parthenon.jpg.

OLYMPIAD CORNER

from the Putnam Competition, 2002

Problem: Let \(n\geq 2\) be an integer and \(T_n\) be the number of non-empty subsets \(S\) of \(\{1,2,3,...,n\}\) with the property that the average of the elements of \(S\) is an integer. Prove that \(T_n-n\) is always even.

Solution: 

We first note that each of the singletons \(\{1\},\{2\},...,\{n\}\) has the necessary property, for a total of \(n\) sets. 

Now suppose \(S\) is a set of more than one element with an integer average \(m\). Then \(S\) either contains \(m\) or does not contain \(m\). If \(S\) does not contain \(m\), then the set \(S\cup \{m\}\)  has an average \(m\). If \(S\) contains \(m\), then the set \(S\setminus \{m\}\) also has an average \(m\). This means that there is a one to one correspondence between sets that contain \(m\), and sets that don't contain \(m\). 

Therefore sets with the desired property other than \(\{1\},\{2\},...,\{n\}\) can be grouped into pairs, and hence \(T_n-n\) is always even.

PROBLEMS

13. (The previous question was scrapped due to its ambiguity; submissions to this new question will be accepted until January 26, 2013.)
    Let \(n \geq 2\) be an integer and \(a_1, a_2, \ldots, a_n\) be real numbers. Prove that for any nonempty subset \(S \subset \{1, 2, \ldots, n\}\), the following inequality holds: \[ \left(\sum_{i \in S}a_i\right)^2 \leq \sum_{1 \leq i \leq j \leq n}(a_i + \ldots + a_j)^2. \]
14. There are \(n\) straight lines in a plane, such that every two intersect with each other. Prove that among the angles formed there is at least one angle which is not greater than \(\frac{180^\circ}{n}\).
15. In the figure above, \(ABCDEF\) is a regular hexagon. The midpoint of \(\overline{AB}\) is \(W\) and \(X\) and \(Y\) are intercepts as shown. What is the value of \(\frac{[XYDE]}{[WXF]}\)?

We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions may be submitted to the PEM facilitators on the deadline date or online via mactolentino@math.admu.edu.ph. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is 12:00 PM January 19, 2013. 

SOLUTIONS

(for December 8, 2012)

10. Sherlock and Mycroft play a game which involves flipping a single fair coin. The coin is flipped repeatedly until one person wins. Sherlock wins if the sequence TTT (tails-tails-tails) shows up first while Mycroft wins if the sequence HTT (heads-tails-tails) shows up first. Who among the two has a higher probability of winning? (13th Philippine Mathematical Olympiad, Area Stage)
    
    (Solved by Clyde Wesley Ang [Chiang Kai Shek College], Kimberly Co [St. Stephen's], Jecel Manabat [Valenzuela City Sci HS], Terence Tsai [Chiang Kai Shek College], Hans Jarett Ong [Chiang Kai Shek College], Jan Kendrick Ong [Chiang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Christian Academy])
    
    
    SOLUTION: 
    Sherlock wins if the sequence TTT comes up first. This sequence has probability \(P(\left\{TTT\right\})=\frac{1}{8}\). Mycroft wins as long as the sequence HTT comes up, which has probability \(P(\left\{HTT, THTT, TTHTT, \ldots \right\})>\frac{1}{8}\). Therefore Mycroft has a higher probability of winning.
11. In \(\Delta ABC\), \(D\) is the midpoint of \(AB\), \(E\) and \(F\) are on \(AC\) and \(BC\) respectively. Prove that the area of \(\Delta DEF\) is not greater than the sum of the areas of \(\Delta ADE\) and \(\Delta BDF\). (All-Russia Olympics Mathematical Competitions, 1983)
    (Solved by Clyde Wesley Ang [Chiang Kai Shek College], Hans Jarett Ong [Chiang Kai Shek College], Jan Kendrick Ong [Chiang Kai Shek College]; partial credit for Farrell Eldrian Wu [MGC New Life Christian Academy])
    
    
    SOLUTION: 
    

    We denote the area enclosed by the convex polygon by \([\cdot]\). Extend \(\overline{ED}\) to \(E_1\) such that \(\overline{E_1D} = \overline{ED}\). Connecting \(E_1\) to \(B\) and \(F\), we note that \(\Delta ADE \cong \Delta DE_1B\) by SAS and \([DE_1F] = [EDF]\) since they have equal height. Thus \[[ADE]+[DBF]=[DE_1B]+[DBF]=[DE_1BF]>[DE_1F]=[DEF]. \]
12. Let \(n \geq 3\) be an odd number. Show that there is a number in the set \[ \{ 2^1 - 1, 2^2 - 1, \ldots, 2^{n-1} - 1 \} \]which is divisible by \(n\). (USSR MO, 1980)
    (Solved by Farrell Eldrian Wu [MGC New Life Christian Academy])
    
    
    SOLUTION: 
    Let \(S = \left\{2^1-1, 2^2-1, \ldots, 2^{n-1}-1\right\}\) and \(R = \left\{1, 2^1, 2^2, \ldots, 2^{n-1}\right\}\). Then \(|R| = n\) where none of its elements are divisible by \(n\) since \(n \geq 3\) is an odd positive integer. This implies that none of the elements of \(S\) are divisible by \(n-1\), since \[ \begin{align*} 2^i-1 &\equiv (n-1) \pmod{n} \\ 2^i &\equiv 0 \pmod{n}, \end{align*} \] which we know is false. Hence by the Pigeonhole Principle there are \(\left\lceil\frac{|S|}{|R|}\right\rceil = 2\) numbers in \(S\) that have the same remainder when divided by \(n\), say \(2^i - 1\) and \(2^j - 1\) for some \(1 \leq i < j \leq n-2\). We now have \[ \begin{align*} 2^j - 1 &\equiv 2^i - 1 \pmod{n} \\ 2^j - 2^i &\equiv 0 \pmod{n} \\ 2^i(2^{j-i} - 1) &\equiv 0 \pmod{n}, \end{align*} \] and since \(n \nmid 2^i\), \(n \mid 2^{j-i} - 1\), and we are done.

ERRATA

There were several typographical errors in the article "Winning the Ultimatum Game"

  - par. 2, "\pi" was written as "pi"

  - par. 3, "...leave the money where you found..." was written as "...leave the money here you found...".)

There was a typographical error in Problem #8 that read \(2^n-1, 2^2-1, \ldots, 2^{n-1}-1\), which has been corrected to read  \(2^1-1, 2^2-1, \ldots, 2^{n-1}-1\) .

Jazzrine Tagle [Valenzuela City High School] was not credited for answering problem 8.