Equations that Shook the World, Part II (Tuklas Vol. 15, No. 4 - December 7, 2013)

EQUATIONS THAT SHOOK THE WORLD, PART II: $t' = t/\sqrt{1-v^2/c^2}$

We have always believed that time is an absolute quantity, an independent variable against which any other changing quantity is measured. We respect time in that manner. We accept that we cannot control it and that we are slaves to it. We age as though we are prisoners of time -- our existence bounded by its chains, never escaping that inevitable march towards non-existence. Time is our master indeed. We talk about standard time, and standards of time. But what do we really mean by all this? Is time really unchangeable?

Another equation that shook the world, particularly our beliefs about time, is part of a system defined in Einstein's famous "Principle of Special Relativity." It reconciled a particular quirk about time that 19th century mathematicians and physicists wondered about. Namely, if time were really an absolute quantity, how come light does not seem to obey it?

Consider two cars A and B whose speeds are actually the same (as indicated by their respective speedometers). Suppose you are riding in car A, and you see car B approaching from the other end of the road. How fast does car B seem? Consider another situation. Suppose car B is running exactly by your side and toward the same direction. How fast does car B seem now? 

In the first situation, it would appear that car B ran faster although the speedometer inside car B registered the same speed as car A. In the second situation, and with the same agreed speeds, car B does not seem to be moving from your point of view. Yet nothing has changed with the number indicated in the speedometers. The two situations seem different because of what you perceive is the relative difference in speeds. In the first, $v_A = v$ and $v_B = -v$ so that $\left|v_B - v_A\right| = |-v - v| = 2v$ (car B apparently moved two times faster than it should be to the opposite direction). In the second situation, $v_A = v_B = v$; hence, $\left|v_B - v_A\right| = |v - v| = 0$ (car B does not seem to be moving from your point of view).

Now replace car B with a light beam. Suppose light speed is $\left|v_B\right| = c$. You would be surprised that  your measurements in both situations would actually be the same: $\left|v_B - v_A\right| = c$. Wait! That could not be right. Let us check the math. In the first situation, light runs against you so we should perceive its speed as: $\left|v_B - v_A\right| = |-c - v| = c+v$. But in the second situation, it should have been: $\left|v_B - v_A\right| = |c - v| = c-v$. Does it mean, mathematically, that:  $c = c+v = c-v$? We would come to the absurd conclusion: $c = 0$ and $v=0$, which are obviously inconsistent with the premise that light and car are moving. 

The solution to the puzzle came from a group of physicists, most notable of which are George Fitzgerald and Hendrik Lorentz. They were trying to come up with an explanation of a baffling result from an experiment on light quite similar to the hypothetical situation above described. 

The solution shook the beliefs of the entire world until that period, and involved a certain transformation of coordinates. Let $t'$ be the time it took light to travel a distance $x'$ for a non-moving observer. Its speed should be the ratio between distance and time traveled; hence, $x'/t' = c$ The transformation addresses what you (in car A) are supposed to observe; namely, a generally different light speed, $x/t = v$. Consider $\tilde{x} = x-ct$ and $\tilde{t} = t-\dfrac{vx}{c^2}$. Then, $$\begin{eqnarray*} \frac{x'}{t'} = c &=& \frac{x-vt}{t-vx/c^2} \\ &=& \frac{c^2x-c^2vt}{c^2t-vx} \\ &=& \left(\frac{cx - cvt}{c^2t - vx}\right)c \end{eqnarray*}$$ In order for the equation to be true, then we must impose $cx - cvt = c^2t - vx$. Consequently, $$\begin{eqnarray*} cx + vx &=& c^2t + cvt\\ (c+v)x &=& (c+v)ct\\ x &=& ct\\ \frac{x}{t} &=& c = v \end{eqnarray*}$$ which leads to the conclusion that the observer must be seeing the same speed regardless of its motion relative to the light beam. But the coordinate transformation (known today as the Lorentz transformation) implies that we must give up the concept of absolute time. Now we have to accept that time depends on the relative motion of the observer: $t' = t/\sqrt{1 - v^2/c^2}$. If they are moving fast enough, then they would see everything around them slow down, i.e., the inequality $t' > t$ becomes perceptible. Fortunately, in most typical speeds we experience in our travels, $v\ll c$ so that $1-v^2/c^2 \approx 1$ and $t' \approx t$. Thus, the illusion that time is absolute under normal circumstances.  

One particular hypothetical consequence of the Lorentz transformation is the following. When you only have $5$ minutes left to go to class, try running very fast so that everyone around seems slower. Your $5$ minutes will seem less than $5$ minutes to your teacher. But isn't running fast what we usually do when we only have $5$ minutes left to be somewhere on time?

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 Asian Pacific Mathematics Olympiad, 2013

Problem: For $2k$ real numbers $a_{1},a_{2},....,a_{k},b_{1},b_{2},...,b_{k},$ define the sequence of numbers $X_{n}$ by $$X_{n}=\sum_{i=1}^{k}\,\left\lfloor a_{i}n+b_{i}\right\rfloor, n=1,2,\ldots .$$ If the sequence $X_{n}$ forms an arithemetic progression, show that $\sum_{i=1}^{k}a_{i}$ must be an integer.

Solution: 

Let $A=\sum_{i=1}^{k}a_{i}$ and $B=\sum_{i=1}^{k}b_{i}$. Moreover, suppose that $\{X_{n}\}$ is an arithmetic progression with the common difference $d$; that is, $X_{n+1}=X_{1}+nd$.

Note that for all $i$, $$a_{i}\left( n+1\right) +b_{i}-1<\left\lfloor a_{i}\left( n+1\right) +b_{i}\right\rfloor \,\ \leq a_{i}\left( n+1\right) +b_{i}.$$ Hence $$\begin{eqnarray*} \sum_{i=1}^{k}\left(a_{i}\left(n+1\right)+b_{i}-1\right) & < & \sum_{i=1}^{k}\left\lfloor a_{i}\left(n+1\right) + b_{i}\right\rfloor \leq \sum_{i=1}^{k}\left(a_{i}\left (n+1\right) + b_{i}\right) \\ A\left(n+1\right)+B-k & < & X_{n+1} \leq A\left(n+1\right) + B \\ An + \left(A+B-X_{1}\right)-k & < & nd \leq An+\left(A+B-X_{1}\right) \\ \end{eqnarray*}$$ However, note that $A+B-k<X_{1}\leq A+B$. Therefore $A+B-X_{1}\geq 0$, which implies that $$An-k\leq An+\left( A+B-X_{1}\right) -k$$ Furthermore, $A+B-X_{1}<k$, and therefore $$An+\left( A+B-X_{1}\right) <An+k.$$ Combining these inequalities, we have $$\begin{eqnarray*} An-k &<&nd<An+k \\ -k &<&-An+nd<k \\ \left| \left( -A+d\right) n\right|  &<&k \\ \left| A-d\right|  &<&\frac{k}{n}. \end{eqnarray*}$$ But since $\left| A-d\right| <\frac{k}{n}$ holds true for all positive integers $n$, it follows that $A=d$. Since $\{X_{n}\}$ is a sequence of integers, then $d$ is also an integer. Hence $A=\sum_{i=1}^{k}a_{i}$ must be an integer.

PROBLEMS

13. Suppose $n \geq 3$. Show that an even number of the fractions $$\frac{1}{n},\frac{2}{n},\frac{3}{n},\ldots,\frac{n-1}{n},$$ are in lowest terms.
    
14. For every positive integer $n > 1$, prove that there exists $n$ factorials, each $> 1$, whose product is also a factorial, i.e. $x_{1}!x_{2}! \ldots x_{n}! = y!$.
15. Find all prime numbers $a$ and $b$ such that $a^{b}+b^{a}$ is also a prime and show that there are no other prime pairs $\left(a,b\right)$ satisfying this property other than the pairs that you have.
16. Suppose that $A$ is a finite set of points in the plane with the property that evey line determined by two points of $A$ contains a third point of $A$. Prove that $A$ is a set of collinear points.
17. Find the real roots of the equation $$x^{3}+2ax+\frac{1}{16}=-a+\sqrt{a^{2}+x-\frac{1}{16}}\quad\left(0 < a < \frac{1}{4}\right).$$
18. Let $ABCD$ be a square, and let $k$ be the circle with center $B$ passing through $A$, and let $l$ be the semicircle inside the square with diameter $AB$. Let $E$ be a point on $l$ and let the extension of $BE$ meet circle $k$ at $F$. Prove that $\angle DAF\cong\angle EAF$.

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 vantonio1992@gmail.com. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is 12:30 PM January 11, 2014. 

SOLUTIONS

(for November 23, 2013)

10. Two circles have exactly two points $A$ and $B$ in common. Find a straight line $L$ through $A$ such that the circles cut chords of equal lengths out of $L$. How many solutions can the problem have? (Taken from Mathematics as Problem Solving by Alexander Soifer)
    (solved by Farrell Eldrian Wu [MGC New Life Academy]; partial credit for Jan Kendrick Ong [Chang Kai Shek College])
    
    
    SOLUTION: 
    Trivially, $AB$ is a solution.
    
    We denote the first circle by $C_{1}$ and the second by $C_{2}$.
    
    Suppose the line $L$ does not pass through $B$. Based from the conditions, $\left|AC\right|=\left|AD\right|$, where $C$ is the points where $L$ intersects with $C_{2}$.
    
    

    

    
    Construct a symmetric image of $C_{1}'$ of $C_{1}$, as in the figure. Now, since $\left|AC\right| = \left|AD\right|$, then $C$ must be on $C_{1}'$.
    
    Now, note that there must exist $X$, a point in $C_{1}$, which is inside $C_{2}$. This is true because they intersect at exactly two points. Combining this with $A$ in $C_{2}$, then we can say that the circle $C_{1}'$ will have a point $X'$ outside of $C_{2}$. This point is actually the symmetric image of $X$ with respect to $A$. Therefore, $C_{1}'$ and $C_{2}$ have exactly one more point of intersection in addition to $A$. Thus, $L$ is another solution to the problem. So there are two solutions.
11. For nonnegative real numbers $a$, $b$, $c$ with $a+b+c=1$, prove that $$\sqrt{a+\frac{\left(b-c\right)^{2}}{4}}+\sqrt{b}+\sqrt{c}\leq3.$$
    (solved by Farrell Eldrian Wu [MGC New Life Academy]; partial credit for Jan Kendrick Ong [Chang Kai Shek College])
    
    
    SOLUTION: 
    Let $$x =\frac{\left(\sqrt{b}+\sqrt{c}\right)^{2}}{2},\quad y=\frac{\left(\sqrt{b}-\sqrt{c}\right)^{2}}{2}.$$ From the given expression for $x$, we have $$\begin{align} 2x & = \left(\sqrt{b}+\sqrt{c}\right)^{2}\\ \sqrt{2x} & = \sqrt{b}+\sqrt{c}, \end{align}$$ since $\sqrt{b}$ and $\sqrt{c}$ are nonnegative. In addition, $$\begin{align} x+y & = \frac{\left(\sqrt{b}+\sqrt{c}\right)^{2}}{2}+\frac{\left(\sqrt{b}-\sqrt{c}\right)^{2}}{2}\\ & = \frac{\left(b+2\sqrt{bc}+c\right)+\left(b-2\sqrt{bc}+c\right)}{2}\\ & = b+c. \end{align}$$ This means that $a+b+c=a+x+y=1$.
    
    Lastly, $$\begin{align} xy & = \frac{\left(\sqrt{b}+\sqrt{c}\right)^{2}}{2}\cdot\frac{\left(\sqrt{b}-\sqrt{c}\right)^{2}}{2}\\ & = \frac{\left[\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}-\sqrt{c}\right)\right]^{2}}{4}\\ & = \frac{\left(b-c\right)^{2}}{4}. \end{align}$$ So the left-hand side of the inequality reads $$\begin{align} \sqrt{a+\frac{\left(b-c\right)^{2}}{4}}+\sqrt{b}+\sqrt{c} & = \sqrt{a+xy}+\sqrt{2x}\\ & = \sqrt{a\left(1\right)+xy}+\sqrt{2x}\\ & = \sqrt{a\left(a+x+y\right)+xy}+\sqrt{2x}\\ & = \sqrt{\left(a+x\right)\left(a+y\right)}+\sqrt{2x}\\ & = \sqrt{3}\left[\sqrt{\frac{\left(a+x\right)}{3}\left(a+y\right)}+\sqrt{\frac{2}{3}x}\right]. \end{align}$$ By AM-GM Inequality, $$\sqrt{\frac{\left(a+x\right)}{3}\left(a+y\right)}\leq\frac{1}{2}\left(\frac{a+x}{3}+a+y\right)$$ and $$\sqrt{\frac{2}{3}x}\leq\frac{1}{2}\left(\frac{2}{3}+x\right).$$ So $$\begin{align} \sqrt{a+\frac{\left(b-c\right)^{2}}{4}}+\sqrt{b}+\sqrt{c} & \leq \sqrt{3}\left(\frac{1}{2}\right)\left(\frac{a+x}{3}+a+y+\frac{2}{3}+x\right)\\ & = \sqrt{3}\left(\frac{1}{2}\right)\left(\frac{a+x}{3}+1+\frac{2}{3}\right)\\ & = \sqrt{3}\left(\frac{1}{2}\right)\left(1+\frac{a+x+2}{3}\right)\\ & = \sqrt{3}\left(\frac{1}{2}\right)\left(1+\frac{a+x-1+3}{3}\right)\\ & = \sqrt{3}\left(\frac{1}{2}\right)\left(2-\frac{y}{3}\right)\\ & = \sqrt{3}\left(1-\frac{y}{6}\right)\\ & \leq \sqrt{3}. \end{align}$$
12. If $2n-1$ is a prime number, then for any group of distinct positive integers $a_{1}$, $a_{2}$, $\dots, a_{n}$ there exist $i,j\in\left\{ 1,2,\dots,n\right\}$ such that $${\displaystyle \frac{a_{i}+a_{j}}{\left(a_{i},a_{j}\right)}}\geq2n-1.$$
    (solved by Farrell Eldrian Wu [MGC New Life Academy]; partial credit for Jan Kendrick Ong [Chang Kai Shek College])
    
    
    SOLUTION: 
    Let $2n-1$ be a prime $p$. Without loss of generality, we can assume $\left(a_{1},a_{2,}\dots,a_{n}\right) = 1$. Otherwise, $d=\left(a_{1},a_{2,}\dots,a_{n}\right)$ can be divided from all terms, leaving ${\displaystyle \frac{a_{i}+a_{j}}{\left(a_{i},a_{j}\right)}}$ unchanged.
    
    Let us first consider the case where $p|a_{i}$ for some $i$ from $1$ to $n$. If this is the case, then there must exist $j$ such that $p\nmid a_{j}$. This means that $p\nmid\left(a_{i},a_{j}\right)$. So $$\frac{a_{i}+a_{j}}{\left(a_{i},a_{j}\right)}\geq\frac{a_{i}}{\left(a_{i},a_{j}\right)}\geq p=2n-1.$$
    Now suppose $\left(a_{i},p\right)=1$ for all $i$ from $1$ to $n$. Then $p\nmid\left(a_{i},a_{j}\right)$. Note that there are $n$ numbers from $a_{i},\dots,a_{n}$.
    
    Suppose there are no two numbers in this set such that $a_{i}+a_{j}\equiv0(\mod p)$.
    
    By Pigeonhole Principle, there are two numbers from $\left\{ a_{1},\dots,a_{n}\right\}$ such that their difference is divisible by $p$.
    
    This means that there are two numbers, suppose $a_{i}$ and $a_{j}$, such that their sum or difference is divisible by $p$.
    
    If $a_{i}-a_{j} \equiv 0(\mod p)$, then $a_{i}\equiv a_{j}(\mod p)$ and $$\frac{a_{i}+a_{j}}{\left(a_{i},a_{j}\right)}\geq\frac{a_{i}-a_{j}}{\left(a_{i},a_{j}\right)}\geq p = 2n-1.$$
    If $a_{i}+a_{j}\equiv0(\mod p)$, then $$\frac{a_{i}+a_{j}}{\left(a_{i},a_{j}\right)}\geq p=2n-1.$$

ERRATA

Problem-solvers for Week 3 were incorrectly attributed.

Another Take on Averages (Tuklas Vol. 15, No. 3 - November 23, 2013)

ANOTHER TAKE ON AVERAGES

One day, Carlo and Hanz decided to perform an experiment. They each had an opaque box in which they placed a red ball and a white ball. They would then draw from their respective boxes, record the color of the drawn ball, then place the ball back. They can draw as many times as they prefer. The goal is to have the highest percentage of red balls drawn.

Carlo's strategy was to draw significantly more than Hanz, and it paid off. In their first session, Carlo was able to draw the red ball 7 out of 19 times to get a percentage of 0.368. Hanz only made 9 attempts and drew the red ball 4 times to get a percentage of 0.364.

For the second session, Hanz made sure he will draw more times than Carlo, but things would still not go Hanz's way. Carlo drew the red ball 17 times out of 39 (43.6%), while Hanz drew 32 out of 74 (43.2%).

Hanz almost admitted defeat, but the statistician in him found hope by taking the aggregate draws as follows:

PLAYER	1ST SESSION		2ND SESSION		TOTAL
Carlo	7/19	0.3684	17/39	0.4360	24/58	0.4138
Hanz	4/11	0.3636	32/74	0.4324	36/85	0.4235

Using the total data, Hanz argued that he won.

We now ask two questions. First, who ACTUALLY won? And more importantly, how did this happen?

This story is an example of a phenomenon called the Simpson's Paradox. This states that if a group of data were separated into several categories, the population which exhibits the most successes may not exhibit less in each of the categories. What this means is that when comparing two groups of data, the successes of one group can still be ahead when we look at the aggregate, even if that group is behind percentage-wise in all the specific categories. This phenomenon had been observed as early as the 1920's through medical and sociostatistical data. However, Edward H. Simpson was one of the first who explicitly pointed it out in 1951, and it was not until 1972 that a paper formalized its definition.

A CLASSIC EXAMPLE

Before we go deeper into the situations that make up the paradox, let us introduce one of the most classic examples. A popular scenario where Simpson's Paradox occurs is through a sport -- baseball. Although this was not the first occurrence of the paradox on record, it gained the interest of statisticians and baseball junkies alike, especially those who utilized Sabermetrics (the Math of Baseball). These people, called sabermetricians, use baseball data in order to analyze the potential of a new player, team chemistry, even the use of performance-enhancing drugs. Sabermetricians may find this particular set of data interesting, since the presence of this phenomenon will make them rethink their traditional ways of looking into data.

For 1995-1996, the batting averages of Derek Jeter and David Justice (both of whom were very popular at the time) were compared, and the summary is in the next table.

PLAYER	1995		1996		TOTAL
Jeter	12/48	0.250	183/582	0.314	195/630	0.310
Justice	4/11	.253	32/74	0.321	149/551	0.270

Similar to the Carlo-Hanz experiment, Justice had better averages if we look at the seasons independently. Yet when we take the aggregate, Jeter is the actual winner. Because of this, baseball junkies should consider the yearly averages or even the aggregate (more commonly called the "career" statistic) separately, and not take one or the other alone.

REASONING

So after the two scenarios, you might be thinking, "Is there something wrong with the computation of the data?", or "Is there something that can be done to make the data make sense?"

To answer these questions, we go back to the data. One of the main reasons for the paradox is the discrepancy in the trials. To see this, let us try to visualize what is happening. We assign to the y-axis the number of successful trials per category, and to the x-axis the total number of trials per category. The success rate per category per group can be thought of as the ratio $m = \frac{p}{q}$, with $p$ successes out of $q$ trials done by a particular group. On the Cartesian plane, this can be represented as a vector with slope equal to $m$. We assign trials $a$ and $b$ (in black) to Group A and trials $c$ and $d$ to Group B (in blue). Trials $a$ and $c$ belong to the same category, as do trials $b$ and $d$. The total success rate for Groups A and B can then be represented by the slope of the vectors $a+b$ and $c+d$, respectively (see figure below).

Let us now compare the slopes. Note that $m_a > m_c$ and $m_b > m_d$. That is, Group A performed better than Group B under each individual category. But upon forming $a+b$ and $c+d$ (using the Parallelogram Law), we can see that $m_{a+b} < m_{c+d}$, despite the fact that $a+b$ is longer than $c+d$. This indicates that the trials themselves are not considered, only the ratios.

Given all of this, is there any way for us to make sense out of the separate data and try to make the conclusion consistent? To do so, we need to do what we call a "normalization process". For the Hanz-Carlo experiment, we will "normalize" the data by making the percentages consistent with their respective number of trials. That is, we give their trials weights so that the aggregate percentage would be a reflection of the same number of data points. This removes the paradox by removing the inconsistencies that it hinges on. In doing so, we have the following table:

PLAYER	1ST SESSION		2ND SESSION		TOTAL
Carlo	7/19	0.3684	(17/39)*(74/74)	0.4360	39.2564/93	0.4221
Hanz	(4/11)*(19/19)	0.3636	32/74	0.4324	38.9091/93	0.4184

CONCLUSION

On the practical side, the statistical anomaly shown by Simpson's Paradox is a good reminder to be mindful of the computed numbers. For instance, sabermetrics is a young field, and Simpson's Paradox is a good example on how to improve upon current methods.

There is a certain viewpoint from psychology that goes by the mantra "the whole is more than the sum of its parts." Looking at things from this perspective, Simpson's Paradox would be more in line with the viewpoint "the whole is not the same as its parts." That is, we can also say that there are situations wherein the sum or aggregate says something different from the individual parts.

Now that we have a clearer picture of what has happened, we go back to the story of Carlo and Hanz. The question of who won among the two is still open to debate. What about you? Do you believe that the individual sessions have more weight, or that the overall performance is the best indicator?

ABOUT THE AUTHOR:
Victor Andrew Antonio is a Lecturer at the Ateneo de Manila University. He obtained his B.S. in Mathematics at the Ateneo de Manila University in 2012 and is currently taking his M.S. in Mathematics at the same university.

REFERENCES:

[1] Bogomolny, Alexander. Simpson Paradox. Retrieved from http://www.cut-the-knot.org/Curriculum/Algebra/SimpsonParadox.shtml.

[2] Pearl, Judea. Simpson's Paradox: An Anatomy. 1999.
[3] Wagner, Clifford H. Simpson's Paradox in Real Life. The American Statistician, Vol. 36 No. 1 (Feb 1982), 46-48.

OLYMPIAD CORNER

from the Asian Pacific Mathematics Olympiad, 2011

Problem: Let $a,b,c$ be positive integers. Prove that it is impossible to have all of the three numbers $a^{2}+b+c,b^{2}+c+a, c^{2}+a+b$ to be perfect squares.

Solution: 

Suppose on the contrary that all three numbers are perfect squares. Since $a^{2}+b+c$ is a perfect square larger than $a^{2}$, it follows that $a^{2}+b+c\geq \left( a+1\right) ^{2}$, which is equivalent to $$b+c\geq 2a+1.$$ Using the same argument, we also obtain $$c+a\geq 2b+1$$ and $$a+b\geq 2c+1.$$ Combining the above inequalities, we have $$2\left( a+b+c\right) \geq 2\left( a+b+c\right) +3$$ which results in a contradiction. 

This proves that it is impossible to have all of the three numbers to be perfect squares.

PROBLEMS

10. Two circles have exactly two points $A$ and $B$ in common. Find a straight line $L$ through $A$ such that the circles cut chords of equal lengths out of $L$. How many solutions can the problem have?
    
11. For nonnegative real numbers $a$, $b$, $c$ with $a+b+c=1$, prove that $$\sqrt{a+\frac{\left(b-c\right)^{2}}{4}}+\sqrt{b}+\sqrt{c}\leq\sqrt{3}.$$
12. If $2n-1$ is a prime number, then for any group of distinct positive integers $a_{1}$, $a_{2}$, $\dots, a_{n}$ there exist $i,j\in\left\{ 1,2,\dots,n\right\}$ such that $${\displaystyle \frac{a_{i}+a_{j}}{\left(a_{i},a_{j}\right)}}\geq2n-1.$$

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 vantonio1992@gmail.com. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is 12:30 PM December 7, 2013. 

SOLUTIONS

(for November 16, 2013)

7. Prove that if $a$ and $b$ are two sides of a triangle and $m_{c}$ is the median drawn to the third side, then $$\left|m_{c}\right|\leq\frac{\left|a\right|+\left|b\right|}{2}.$$  (Taken from Mathematics as Problem Solving by Alexander Soifer)
   (solved by Jan Kendrick Ong [Chang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Academy]; partial credit for Jayson Dwight S. Catindig [Ateneo de Manila HS])
   
   
   SOLUTION: 
   Let $O$ be the midpoint of $AB$. Construct $D$ on line $\left|OC\right|$ such that $\left|OC\right| = \left|OD\right|$.
   
   

   

   
   Then the quadrilateral $ABCD$ is a parallelogram, and $\left|BD\right|=\left|CA\right|=\left|b\right|$. In addition, by the Triangle Inequality, $\left|CD\right|<\left|CB\right|+\left|BD\right|$, which means $$\begin{eqnarray*} 2\left|m_{c}\right| & < & \left|a\right|+\left|b\right|\\ \left|m_{c}\right| & \leq & \frac{\left|a\right|+\left|b\right|}{2}. \end{eqnarray*}$$
   Now, when $n$ is even, we can write $n$ as $4^km^2$, where $m$ is odd. Then look at all pairings of divisors. There is an odd number of pairs whose sum is odd. They are $$\left(4^k\left(1\right)\cdot m^2, \dots, 4^k\left(i\right)\cdot \frac{m^2}{i}, \dots, 4^km \cdot m, \dots, 4^k\left(i\right)\cdot \frac{m^2}{i}, \dots, 4^km^2\cdot 1\right).$$ So, $\sigma\left(n\right)$ is odd since we are adding an odd number of odd numbers.
8. Prove that, for all integers $n\geq2$, $$\frac{1}{1^{2}}+\frac{1}{2^{2}}+\cdots+\frac{1}{n^{2}}\leq2.$$
   (solved by Farrell Eldrian Wu [MGC New Life Academy]; partial credit for Jayson Dwight S. Catindig [Ateneo de Manila HS] and Jan Kendrick Ong [Chang Kai Shek College])
   
   
   SOLUTION: 
   This is equivalent to proving $$\frac{1}{2^{2}}+\cdots+\frac{1}{n^{2}}\leq1.$$ Note that $$\frac{1}{\left(i-1\right)i}\leq\frac{1}{i^{2}}$$ for all $i\geq2$. This means that $$\begin{eqnarray*} \frac{1}{2^{2}}+\cdots+\frac{1}{n^{2}} & \leq & \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{\left(n-1\right)n}\\ & = & \left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \cdots + \left(\frac{1}{n-1}-\frac{1}{n}\right) \\ & = & 1-\frac{1}{n}\\ & = & \frac{n-1}{n}\\ \frac{1}{2^{2}}+\cdots+\frac{1}{n^{2}} & < & 1. \end{eqnarray*}$$
9. Find all solutions of $2^{n}+7=x^{2}$ where $n$ and $x$ are integers. (Taken from International Mathematics: Tournament of the Towns, Questions, and Solutions,Tournaments 6 to 10 (1984 to 1988) by P. J. Taylor)
   (solved by Jayson Dwight S. Catindig [Ateneo de Manila HS] and Farrell Eldrian Wu [MGC New Life Academy]; partial credit for Marielle Macasaet [St. Theresa's College] and Jan Kendrick Ong [Chang Kai Shek College])
   
   
   SOLUTION: 
   Since we are required to make $x$ an integer, then $2^{n}$ must be an integer, and so $n\geq0$. We consider all possible cases when we rewrite the equation modulo $4$.
   
   If $n>1$, then $2^{n}\equiv0\mod4$. On the other hand, if $n=1$, then $2^{n}\equiv2\mod4$, and finally, if $n=0$, then $2^{n}\equiv1\mod4$. We look at those three possibilities and deduce solutions for $2^{n}+7\equiv x^{2}\mod4$. Note that the only possible quadratic residues modulo $4$ are $0$ and $1$.
   

   • CASE 1: $n>1$.
     We start with $$\begin{eqnarray*} x^{2} & \equiv & \left(2^{n}+7\right)\mod4\\ & \equiv & 3\mod4, \end{eqnarray*}$$ which is impossible.
   • CASE 2: $n=1$. $$\begin{eqnarray*} x^{2} & \equiv & \left(2^{n}+7\right)\mod4\\ & \equiv & 1\mod4, \end{eqnarray*}$$ which is acceptable. Going back to the original equation, it can be seen that $x=\pm3$.
   • CASE 3: $n=0$. $$\begin{eqnarray*} x^{2} & \equiv & \left(2^{n}+7\right)\mod4\\ & \equiv & 0\mod4, \end{eqnarray*}$$ which is also acceptable. But this means $x=\pm2\sqrt{2}$, which are not integers.

   Therefore, the only solution is $n=1$, $x=\pm3$.

ERRATA

The right-hand side of Problem #11 was changed from "$3$" to "$\sqrt{3}$."

Equations that Shook the World, Part I (Tuklas Vol. 15, No. 2 - November 16, 2013)

EQUATIONS THAT SHOOK THE WORLD, PART I: $1 = 0.999\ldots$

Counting may perhaps be our first ever encounter with mathematics. The anatomic ``digitus'' (i.e., fingers and toes) serves as our most basic counting tool. Perhaps that is the reason we refer to a number $x$ in the set $\left\{x|x\in\mathbb{N}, 0\leq x \leq 9\right\}$ as a ``digit.'' In turn, that set serves as the basis of our decimal number system.

Part of being human is the desire to transcend limits imposed by our beliefs or by our ignorance. Counting is one of those activities wherein we love to exercise that desire. ``Count as far as you possibly can'' is perhaps a challenge we subsconsciously tackle when we attempt to break world Olympic records, generate as much wealth as we can, or enumerate as many stars as possible on a clear night. The number just seems endless, without bound!

Interestingly enough, the equation $1 = 0.99999\ldots$ is an elegant way to express the notion of ``endlessness'' or ``unboundedness'' in an activity as finite as counting. Yet therein lies the mystery. For the longest time in antiquity, dating back to ancient India and the ancient Greeks (from where our present counting system originates), there had been no formal way to express the concept we now call ``infinity.'' Hence, the infinite was discussed more from a philosophical viewpoint, in the sense of being an ideal one can only strive for. The ancient Greeks defined the circle as a perfect shape which can be made only by creating a polygon with an endless number of sides. In ancient India, infinity had been a monster of innumerable multitude.

Only between the late 19th and early 20th centuries did human society come to grips with the concept of infinity through the efforts of Georg Cantor in his set theory about cardinalities. But at an earlier time, around the 17th century, the proponents of analytic geometry had pieces of the puzzle laid out. That era also saw ``infinity'' taking on the symbol of a lemniscate ``$\infty$'' (which looks a lot like a horizontal digit $8$). That symbol was formally introduced in 1655 by John Wallis, an English mathematician credited for his marked contributions to calculus. In fact, the German mathematician and co-inventor of calculus Gottfried Leibniz treated $\infty$ as an ``ideal quantity'' which is by nature different from but seems to resemble the properties of very large numbers.   

Undoubtedly, the equation $1=0.99999\ldots$ can be proven in many different ways, none of which would make any sense in the absence of an understanding of the notion of $\infty$. For example, a very common proof starts with denoting $x=0.99999\ldots$.  So $10x = 9.9999\ldots = 9 + 0.9999\ldots = 9 + x$. The interpretation made in the last part of the equation would of course only make sense if one considers that the digit $9$ repeats itself endlessly towards the right. Finally, the proof is simply a matter of algebra with $10x = 9 + x$, which of course yields $x=1$. Therefore, one shows that $1 = 0.99999\ldots$. Truly it is one among a few equations that shook the human world by putting mathematical substance to what only used to be a philosophical ghost - the concept of infinity.

Now try typing in $0.99999\ldots$ in your home calculator until it spans from one end of the screen to the other; then, press the [ = ] or [ Enter ] button. What does it say?

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 Iboamerican Math Olympiad, 1997

Problem: Let $H$ be the orthocenter of the non-equilateral triangle $ABC$ and $O$ its circumcenter. Let the lines $AH$ and $AO$ intersect the circumcircle at the points $M$ and $N$, respectively. Denote $P$ ,$Q$, and $R$ to be the intersection points of the lines $BC$ and $HN$, $BC$ and $OM$, $HQ$ and $OP$, respectively. Prove that $AORH$ is a parallelogram.

Solution: 

We need to show that $AO\Vert HR$ and $OR\Vert AH$.

Let $E$ be the intersection of the segments $AM$ and $BC$. At this point, we will prove a property relating to the orthocenter.

Claim: $E$ is the midpoint of $HM$, that is $HE = EM$.

Since $\angle CMA$ and $\angle ABC$ are inscribed angles intercepting the same arc, then $m\angle CMA=m\angle ABC=\alpha$. Given that $CD$ and $AE$ are altitudes, then focusing on the quadrilateral $BEHD$, we have  $$m\angle EHD=360^{\circ }-\left(90^{\circ }+90^{\circ }+\alpha \right) =180-\alpha.$$ Hence it follows that $m\angle CHM=\alpha$. This implies that the two right triangles $\Delta CHE$ and $\Delta CME$ are congruent, and therefore $HE = EM$. $\blacksquare$

Since $E$ is the midpoint of $HM$, it follows that the right triangles $EQM$ and $EQH$ are congruent, so $m\angle EMQ=m\angle QHE$. Furthermore, $AO=OM$, which implies that $m\angle OAM=m\angle AMO$. Thus $\angle QHE=m\angle OAM$, and therefore $AO\Vert HR$.

Next, we note that $m\angle AMN=m\angle HEQ=90^{\circ }$, hence $MN\Vert EQ$. Moreover, since $E$ is the midpoint of $HM$, then $P$ is the midpoint of $HN$. Since $\Delta HNM$ is a right triangle, then $P$ the center of the circumcircle of $\Delta HNM$. This means that $PM=PN$ and $\Delta PNM$ is isosceles. As a result, $P$ lies on the perpendicular bisector of base $MN$. However, the vertex $O$ of the isosceles triangle $ONM$ also lies on this perpendicular bisector. Thus $OP\perp NM$, which implies $OR\Vert AH$. Hence we have established that $AORH$ is a parallelogram.

PROBLEMS

7. Prove that if $a$ and $b$ are two sides of a triangle and $m_{c}$ is the median drawn to the third side, then $$\left|m_{c}\right|\leq\frac{\left|a\right|+\left|b\right|}{2}.$$
   
8. Prove that, for all integers $n\geq2$, $$\frac{1}{1^{2}}+\frac{1}{2^{2}}+\cdots+\frac{1}{n^{2}}\leq2.$$
9. Find all solutions of $2^{n}+7=x^{2}$ where $n$ and $x$ are integers.

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 vantonio1992@gmail.com. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is 12:30 PM November 23, 2013. 

SOLUTIONS

(for October 19, 2013)

1. Show that if $n$ is a perfect square, then the sum of the divisors of $n$ is odd.
   (solved by Jayson Dwight S. Catindig [Ateneo de Manila HS] and Farrell Eldrian Wu [MGC New Life Academy]; partial credit for Hans Jarrett Ong [Chang Kai Shek College])
   
   
   SOLUTION: 
   We prove this first for the case when $n$ is odd. If $n$ is odd, then all divisors of $n$ are odd. The pairs of divisors will add up to form even numbers, except for $\sqrt{n}$, an odd number which will come up in the list of divisors only once. Hence, $\sigma\left(n\right)=2k+\sqrt{n}$, where $2k$ is the sum of the divisors of $n$ excluding $\sqrt{n}$.
   
   Now, when $n$ is even, we can write $n$ as $4^km^2$, where $m$ is odd. Then look at all pairings of divisors. There is an odd number of pairs whose sum is odd. They are $$\left(4^k\left(1\right)\cdot m^2, \dots, 4^k\left(i\right)\cdot \frac{m^2}{i}, \dots, 4^km \cdot m, \dots, 4^k\left(i\right)\cdot \frac{m^2}{i}, \dots, 4^km^2\cdot 1\right).$$ So, $\sigma\left(n\right)$ is odd since we are adding an odd number of odd numbers.
2. Let $A$ be any set of $12$ distinct integers chosen from the arithmetic progression $9, 13, 17 \ldots, 77$. Prove that there must be two distinct integers in $A$ whose sum is $90$.
   (solved by Jayson Dwight S. Catindig [Ateneo de Manila HS], Luis Salvador R. Diy [Xavier], Hans Jarett Ong [Chang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Academy])
   
   
   SOLUTION: 
   Consider the set $$S=\left\{\left(9\right),\left(13,77\right),\left(17,73\right),\dots,\left(41,49\right),\left(45\right)\right\}.$$ Note that $\left|S\right|=10$. Suppose we choose $9$ and $45$, so that we can still avoid having a sum of $90$. We still have to choose $10$ numbers. By the Pigeonhole Principle, there is at least one pair in $S$ wherein both numbers will be chosen, which means there are $2$ numbers whose sum is $90$.
3. Let $p\left(x\right)$ be a polynomial with integer coefficients satisfying $$p\left(0\right) = p\left(1\right) = 2011.$$ Show that $p$ has no integer zeros.
   (solved by Hans Jarett Ong [Chang Kai Shek College]; partial credit for Farrell Eldrian Wu [MGC New Life Academy])
   
   
   SOLUTION: 
   We prove by contradiction.
   
   Let $$p\left(x\right)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0},$$ where $a_{i}\in\mathbb{Z}$. If $p\left(0\right)=2011$, then this means $a_{0}=2011$. At the same time, since $p\left(1\right)=2011$, then $\begin{eqnarray*} a_{n}+a_{n-1}+a_{1}+a_{0} & = & 2011\\ a_{n}+a_{n-1}+a_{1}+2011 & = & 2011\\ a_{n}+a_{n-1}+\cdots+a_{1} & = & 0. \end{eqnarray*}$Suppose there is another integer $z$ such that $p\left(z\right)=0$.
   
   Then $\begin{eqnarray*} a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z+2011 & = & 0\\ a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z & = & -2011 \end{eqnarray*}$From this, we can see that $z$ cannot be even, because then, each terms in the sum must be even. So we consider the case where $z$ is odd.
   
   Since we already know that $a_{n}+a_{n-1}+\cdots+a_{1}=0$ is even, and multiplying an odd number (powers of $z$) to each of the terms will not change the parity, then $a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z$ is even. This is a contradiction. Hence, $p$ cannot have integer zeroes.
4. Evaluate $$\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}\right)\left(\frac{1}{n+2}\right).$$
   (solved by Hans Jarett Ong [Chang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Academy])
   
   
   SOLUTION: 
   It can be seen that $$\begin{eqnarray*} \frac{1}{n}\left(\frac{1}{n+1}\right)\left(\frac{1}{n+2}\right) & = & \frac{\frac{1}{2}}{n}+\frac{-1}{n+1}+\frac{\frac{1}{2}}{n+2}\\ & = & \frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2\left(n+2\right)} \end{eqnarray*}$$ So $$\begin{eqnarray*} S=\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}\right)\left(\frac{1}{n+2}\right) & = & \sum_{n=1}^{\infty}\left(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2\left(n+2\right)}\right)\\ & = & \sum_{n=1}^{\infty}\frac{1}{2n}-\sum_{n=1}^{\infty}\frac{1}{n+1}+\sum_{n=1}^{\infty}\frac{1}{2\left(n+2\right)}\\ & = & \left(\frac{1}{2}+\frac{1}{4}+\cdots\right)-\left(\frac{1}{2}+\frac{1}{3}+\cdots\right)\\ & & +\left(\frac{1}{6}+\frac{1}{8}+\cdots\right). \end{eqnarray*}$$ Rearranging some of the addends we have $$\begin{eqnarray*} S & = & \left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots\right)+\left(\frac{1}{6}+\frac{1}{8}+\cdots\right)-\left(\frac{1}{2}+\frac{1}{3}+\cdots\right)\\ & = & \left[\frac{1}{2}+\frac{1}{4}+\left(\frac{1}{3}+\frac{1}{4}+\cdots\right)\right]-\left(\frac{1}{2}+\frac{1}{3}+\cdots\right)\\ & = & \frac{1}{4}. \end{eqnarray*}$$
5. Let $ABC$ be an equilateral triangle and $P$ a point in its interior. Consider $XYZ$, the triangle with $XY=PC$, $YZ=PA$, and $ZX=PB$, and $M$ a point in its interior such that $\angle XMY=\angle YMZ=\angle ZMX=120^{\circ}$. Prove that $XM+YM+ZM=AB$. (Taken from Mathematical Olympiad Challenges by Andreescu and Gelca)
   (solved by Jayson Dwight S. Catindig [Ateneo de Manila HS], Hans Jarett Ong [Chang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Academy])
   
   
   SOLUTION: 
   Rotate triangle $ZMY$ through $60^{\circ}$ counterclockwise about $Z$ to $ZNW$.
   

   

   First note that triangles $ZMN$ and $ZYW$ are equilateral. Hence $MN=ZM$ and $YW=YZ$. Now $XMN$ and $MNW$ are straight angles, both being $120^{\circ}+60^{\circ}$ , so $XW=XM+YM+ZM$. On the other hand, when constructing backwards the triangle $ABC$ from triangle $XYZ$, we can choose $A=W$ and $C=X$. Then the side length of the equilateral triangle is $XW$, which is equal to $XM+YM+ZM$.
6. An interior point $P$ is chosen in the rectangle (ABCD) such that $\angle APD+\angle BPC=180^{\circ}$. Find the sum of the angles $\angle DAP$ and $\angle BCP$. (Taken from Mathematical Olympiad Challenges by Andreescu and Gelca)
   (solved by Marielle Macasaet [St. Theresa's College, QC], Hans Jarett Ong [Chang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Academy])
   
   
   SOLUTION: 
   Translate triangle $DCP$ to triangle $ABP'$, as below.
   

   

   This way we obtain the quadrilateral $APBP'$, which is cyclic, since $\angle APB+\angle AP'B=360^{\circ} - \angle APD - \angle BPC=180^{\circ}$. Let $Q$ be the intersection of $AB$ and $PP'$. Then since the lines $AD$, $PP'$, and $BC$ are parallel, we have $\angle DAP+\angle BCP=\angle APQ+\angle QP'B$. The latter two angles have measures equal to half of the arcs $AP'$ and $BP$ of the circle circumscribed to the quadrilateral $APBP'$. On the other hand, the angle $\angle BQP$, which is right, is measured by half the sum of these two arcs. Hence $\angle DAP+\angle BCP=\angle BQP=90^{\circ}$.

PEM session for November 9, 2013 cancelled

In light of supertyphoon Yolanda, tomorrow's PEM session (November 9, 2013) will be cancelled. The Tuklas problems deadline will also be extended accordingly.

We will be discussing when to hold possible make-up sessions next week.

Stay safe everyone! :]

A Paradox in an Ellipsis (Tuklas Vol. 15, No. 1 - October 19, 2013)

A PARADOX IN AN ELLIPSIS

Are you familiar with infinite decimals? Perhaps you have already encountered a number like this on your calculator screen:  $$0.333333\ldots .$$ Recall that to express this infinite decimal as a fraction, we simply represent this value as $x$ and do the following computations:  $$\begin{align*} x &= 0.333333\ldots \\ 10x &= 3.333333\ldots \\ 10x - x &= 3.333\ldots - 0.333\ldots \\ 9x &= 3 \\ x &= \frac{1}{3}. \end{align*}$$ Another method for this would be to consider the geometric series  $$a_1 = \frac{3}{10}, r = \frac{1}{10},$$  which yields  $$0.333\ldots = 0.3 + 0.03 + 0.003 + \ldots = \sum_{i=1}^{\infty}{\frac{3}{10^i}} = \frac{3/10}{1-1/10} = \frac{1}{3}.$$ The problematic part of the number $0.333\ldots$ lies in the fact that it is composed of a sequence of $3$'s repeated infinitely, and yet we somehow paradoxically resolve this as a finite value. Recall that a paradox is a statement that is seemingly contradictory by nature but might actually be true. In our context, our problem is that we are dealing with infinity (which is, well, infinite) as a complete object (that is, finite). Let us see how this tricky concept was addressed, and gain some perspective on the importance of paradoxes in mathematics.

We begin with the ancient Greek philosopher Parmenides. He believed that there was only one indivisible unchanging reality. To that end, he said that plurality and change were illusions and our senses were lying to us. In particular, he said that motion was an illusion. One of his students, Zeno of Elea, proposed a series of famous paradoxes that demonstrated Parmenides' claims. We discuss one of them below.

Suppose that Achilles and the tortoise are in a 100-meter footrace, where both racers run at a constant speed. Suppose further that Achilles is faster and gives the tortoise a head start, say fifty meters. Zeno argues that by the time Achilles covers the distance from his starting position to the starting position of the tortoise (which we call $P_1$), the tortoise will have moved ahead to a new position $P_2$. When Achilles reaches the position $P_2$, the tortoise will have moved to a new position further ahead, say $P_3$. This will go on forever; in effect, Achilles will never catch up to the tortoise. However, if we compute the time it takes for each racer to finish the race, it is clear that Achilles will win.

Let us plug some numbers into the example to make the comparison more obvious. Say Achilles runs at a speed of 10 meters/minute and the tortoise runs at a speed of 1 meter/minute. From there, we can see that Achilles completes the race in ten minutes whereas the tortoise completes it in 100 minutes. More explicitly, we can use algebra to compute that Achilles and the tortoise meet $5\frac{5}{9}$ minutes after the race begins at the $55\frac{5}{9}$-meter position. Since Achilles always runs faster, this means that after $5\frac{5}{9}$ minutes he will pass the tortoise and eventually finish the race ahead of the tortoise.

What is the problem here? Remember, Parmenides (and many of his contemporaries) said that reality was unchanging and indivisible. The computations above yielded the values $55\frac{5}{9} = 5.555\ldots$ meters and $5\frac{5}{9} = 5.555\ldots$ minutes. In a sense, what we have done is slice our observations of distance and time into smaller and smaller segments under the assumption that these things are infinitely divisible. This process is made explicit in our geometric series representation, where we keep adding a new addend that is a fraction of the previous value. The problem is that there seems to be a disconnect from one side of the equation (the fraction) to the other side (the infinite decimal). 

How do we solve this problem? Practically speaking, it is absurd to think of taking the sum of an infinite number of terms. Look back to the number $0.333\ldots$ and its geometric series representation. We are applying a finite process (taking the sum) an infinite number of times. The best that we can do is attempt to approximate the value of $0.333\ldots$ by adding the value $\frac{3}{10^k}$ to our approximation $\widehat{x}$, which will always be slightly smaller than $0.333\ldots$. That is, we have  $$\begin{eqnarray*} k = 1, & \widehat{x} = 0.3 &< 0.33\ldots \\ k = 2, & \widehat{x} = 0.33 &< 0.333\ldots \\ k = 3, & \widehat{x} = 0.333 &< 0.3333\ldots \\ \vdots \\ \end{eqnarray*}$$ We now make the observation that as we continue this process, the values seemingly get closer and closer to the value $\frac{1}{3}$. However, the formal argument hinges largely on a particular assumption of a property of real numbers (called an axiom) which says that every increasing sequence of (real) numbers that is bounded above must approximate a (unique) real number. Yet we see that the equation does make sense, if viewed in this light.

Thus, we are able to combine two paradoxical concepts under one simple equation. It becomes clear that the summation does not only signify a single process but an infinite continued process that is bounded by a natural property of the numbers used. Hence, it is natural that the other side could be written as a finite value. While some philosophical arguments still debate the merits of Zeno's paradox, this particular resolution by mathematics leads to many beautiful results. In particular, it is this construction of infinite decimals (and by extension, the real numbers) from which one can intuit the notion of a limit. The limit is the central concept of calculus, and leads into further mathematics that surprisingly enough finds applications in fields as varied as engineering and economics.

We now see how the paradox, far from highlighting the pitfalls of mathematics, in fact forces it to grow and enhance its potential. Furthermore, it is able to do so in a beautiful manner, one that makes sense to the point that when we look at it afterwards, we cannot imagine how it could have been otherwise.

ABOUT THE AUTHOR:
Jesus Lemuel L. Martin, Jr. is an Instructor at the Ateneo de Manila University. He obtained his B.S. in Applied Mathematics major in Computational Science at the Ateneo de Manila University in 2010 and his M.S. in Mathematics at the same university in 2013.

REFERENCES:

[1] Byers, William. How Mathematicians Think.
[2] Rademacher, H., Toeplitz, O. The Enjoyment of Mathematics.

OLYMPIAD CORNER

from the 22nd Irish Mathematical Olympiad, 2009

Problem: Let $ABCD$ be a square. The line segment $AB$ is divided internally at $H$ so that $$\left|AB\right| \left| BH\right| = \left| AH\right| ^{2}.$$ Let $E$ be the midpoint of $AD$ and $X$ be the midpoint of $AH$. Let $Y$ be a point on $EB$ such that $XY$ is perpendicular to $BE$. Prove that $\left|XY\right| = \left| XH\right|$.

Solution: 

Let the square be of length $2k$ for some $k>0$, and $\left| AH\right| =x$. Then
$$\begin{eqnarray*} \left| AB\right| \left| BH\right|  &=&\left| AH\right| ^{2} \\ 2k\left( 2k-x\right)  &=&x^{2} \\ x^{2}+2kx-4k^{2} &=&0. \end{eqnarray*}$$ Solving for $x$, we have $$\begin{eqnarray*} x &=&\frac{-2k\pm \sqrt{4k^{2}+16k^{2}}}{2} \\ &=&\frac{-2k\pm 2\sqrt{5}k}{2} \\ &=&\left( \pm \sqrt{5}-1\right) k \end{eqnarray*}$$ But since $x>0$, then $x=\left( \sqrt{5}-1\right) k$.

Note that $$\begin{eqnarray*} \left| BE\right|  &=&\sqrt{\left( 2k\right) ^{2}+a^{2}}=k, \\ \left| BH\right| &=&2a-\left( \sqrt{5}-1\right) k=\left( 3-\sqrt{5}\right) k, \\ \left| XH\right|  &=&\frac{\left( \sqrt{5}-1\right) k}{2}. \end{eqnarray*}$$ Moreover, $\Delta BXY$ and $\Delta BEA$ are similar. It then follows that $$\begin{equation*} \frac{\left| XY\right| }{\left| BX\right| }=\frac{\left| EA\right| }{\left|BE\right| }\Rightarrow \frac{\left| XY\right| }{\left| BX\right| }=\frac{k}{\sqrt{5}k}\Rightarrow \left| XY\right| =\frac{\left| BX\right| }{\sqrt{5}}\end{equation*}$$ and $$\begin{eqnarray*} \left| XY\right| &=&\frac{\left| BH\right| +\left| XH\right| }{\sqrt{5}} \\ &=&\frac{\left( 3-\sqrt{5}\right) k+\frac{1}{2}\left( \sqrt{5}-1\right) k}{\sqrt{5}} \\ &=&\frac{\left( 5-\sqrt{5}\right) k}{2\sqrt{5}}=\frac{\left( \sqrt{5}-1\right) k}{2}. \end{eqnarray*}$$ Therefore, $\left| XH\right| =\left| XY\right|$.

PROBLEMS

1. Show that if $n$ is a perfect square, then the sum of the divisors of $n$ is odd.
2. Let $A$ be any set of $12$ distinct integers chosen from the arithmetic progression $9, 13, 17 \ldots, 77$. Prove that there must be two distinct integers in $A$ whose sum is $90$.
3. Let $p\left(x\right)$ be a polynomial with integer coefficients satisfying $$p\left(0\right) = p\left(1\right) = 2011.$$ Show that $p$ has no integer zeros.
4. Evaluate $$\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}\right)\left(\frac{1}{n+2}\right).$$
5. Let $ABC$ be an equilateral triangle and $P$ a point in its interior. Consider $XYZ$, the triangle with $XY=PC$, $YZ=PA$, and $ZX=PB$, and $M$ a point in its interior such that $\angle XMY=\angle YMZ=\angle ZMX=120^{\circ}$. Prove that $XM+YM+ZM=AB$.
6. An interior point $P$ is chosen in the rectangle (ABCD) such that $\angle APD+\angle BPC=180^{\circ}$. Find the sum of the angles $\angle DAP$ and $\angle BCP$.

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 vantonio1992@gmail.com. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is 12:30 PM November 9, 2013.

PEM session for October 12, 2013 cancelled

In light of the current weather, tomorrow's PEM session (October 12, 2013) will be cancelled. Stay safe everyone! :]

Patterns and the Danger of Assuming (Tuklas Vol. 14, No. 8 - Feb. 10, 2013)

PATTERNS AND THE DANGER OF ASSUMING

Human beings have a tendency to see patterns in everything. It is a skill we have unconsciously mastered to survive. Every observation we make, whether consciously or unconsciously, relies on perceiving whether a pattern is developing.

This is not necessarily a bad thing. Skills would not have been developed if they were not essential to survival. If you know that a particular route is usually clogged by traffic after class and you need to go home quickly, then that particular route is out of the question. If one knows that a particular teacher does not regularly check homework and there is another project due for tomorrow, then one might choose to focus on finishing that project first. Of course, if the teacher does check for homework the next day, then we might not have anything to show for it. The pattern fails.

Let’s see how our fondness for patterns might be a little misleading. Consider the sequence of numbers $$0 \,\, 3 \,\, 6 \,\, 12 \,\, 24 \,\, 48 \,\, 96 \,\, 192 \,\, 384$$ which we construct by doubling the number (starting with 3). Let’s now add 4 to each number and divide by 10. This gives us $$0.4 \,\, 0.7 \,\, 1.0 \,\, 1.6 \,\, 2.8 \,\, 5.2 \,\, 10.0 \,\, 19.6 \,\, 38.8$$ What is so special about this sequence? In the 18th century, a couple of astronomers, notably Johann Daniel Titius and Johann Elert Bode, noticed a quirk in the distances between planets in the Solar System. Specifically, if one measures the mean distances (measured in astronomical units, or A.U.) of the planets from the Sun, it seems that one planet is twice as far from the Sun as the previous planet. This sequence of measurements happens to mirror the sequence we had just constructed.

When the Titius-Bode law (as it would later be known) was proposed, the known planets in the Solar System were Mercury, Venus, Earth (of course), Mars, Jupiter and Saturn. Astronomers found out that the sequence predicted the distances of each planet from the Sun accurately, except that there was a gap between Mars and Jupiter.

Planet	Actual distance (A.U.)	T-B Law
Mercury	0.39	0.4
Venus	0.72	0.7
Earth	1.00	1.0
Mars	1.52	1.6
		2.8
Jupiter	5.20	5.2
Saturn	9.54	10.0

Interesting, but not really something that astronomers considered important until the discovery of Uranus in the tail-end of the century, whose mean distance from the Sun eerily fitted the next number in the sequence. Spurred by this, astronomers discovered the heavenly body Ceres (the largest object on the asteroid belt) which fitted the 5th number in the sequence. Note that this was the blank entry in the previous table.

It seemed like a triumphant example of universal order, discovered by mathematics. The discovery of the planet Neptune in the mid-19th century, however, showed that the law didn’t work. Furthermore, Ceres was not really considered a planet after the discovery of more asteroids in the asteroid belt. Some scientists took it as an example of fallacious reasoning and hasty generalization. Others sought to show that the law just needed some adjustment and that the planets really behaved in some predictable manner. The most important effect of the Titius-Bode rule (as some now call it) though is that it drove astronomers to discover why a pattern could (or couldn’t) develop, and whether there were objects that would fit this pattern. The discovery of Ceres was prompted by a guess that the rule could work.

Sometimes, we tend to look for things that aren’t there. At other times, we see things but misinterpret them to mean something else. The Titius-Bode rule is a reminder that the appearance of a pattern is not enough. To discover why something works, we must delve deeper into the underlying factors that cause a particular phenomenon to exist.

ABOUT THE AUTHOR:
Jesus Lemuel Martin, Jr. obtained his B.S. in Applied Mathematics major in Computational Science at the Ateneo de Manila University in 2010 and is currently taking his MS in Mathematics at the same university.

REFERENCES:

[1] Haynes, M. "Bode's Law." http://www.astro.cornell.edu/academics/courses/astro201/bodes_law.htm. Online, accessed February 8, 2013.

OLYMPIAD CORNER

from the Canadian Mathematical Olympiad, 2009

Problem: Two circles of different radii are cut out of cardboard. Each circle is subdivided into 200 equal sectors. On each circle 100 sectors are painted white and the other 100 are painted black. The smaller circle is then placed on top of the larger circle, so that their centers coincide. Show that one can rotate the small circle so that the sectors on the two circles line up and at least 100 sectors on the small circle lie over sectors of the same color on the big circle. 

Solution: 

Let $x_i, i=0,1,...,199$ be defined as follows: $x_i=1$ when the $(i+1)$-th sector of the larger circle is black, and $x_i=-1$ when the $(i+1)$-th sector of the larger circle is white (counterclockwise). The variables $y_i, i=0,1,...,199$ are similarly defined:, i.e. $y_i=1$ when the $(i+1)$-th sector of the smaller circle is black, and $y_i=-1$ when the $(i+1)$-th sector of the smaller circle is white.

To show the required condition, we need to prove that there exists some $j=0,1,...,199$ such that $S_j=\sum_{i=0}^{199}x_iy_{i+j}$ is nonnegative. We note that the subscript $i+j$ is taken modulo 200, and $j$ indicates the number of times the smaller circle rotated by 1.8 degrees (counterclockwise) from its original position. 

Note that $y_0+\cdots +y_{199}=0$ as there's an equal number of black and white sectors. We then take the sum $$\begin{equation*} S_0+S_1+\cdots +S_{100}=\sum_{i=0}^{199}x_i\left(y_0+y_1+\cdots+y_{199}\right)=0 \end{equation*}$$ Hence, there exists a $j=0,1,2,...,199$ such that $S_j\geq 0$.

SOLUTIONS

(for January 19, 2013)

19. Find all pairs $(x,y)$ of two positive integers such that $N = 11x + 99y$ is a perfect square number less than or equal to $1199$ and $x+y$ is also a perfect square. (Modified problem from the China Mathematical Competition for Secondary Schools, 2004)
    (Solved by Clyde Wesley Ang [Chiang Kai Shek College], Hans Jarett Ong [Chiang Kai Shek College], Jan Kedrick Ong [Chiang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Christian Academy]; partial credit for Jecel Manabat [Valenzuela City Sci HS])
    
    
    SOLUTION: 
    Since $N = 11x + 99y = 11(x + 9y)$ we can conclude that $x + 9y = 11m^2$ for some positive integer $m$, so $$N = (11m)^2 \leq 1199 \Leftrightarrow m^2 \leq \frac{1199}{121} < 10.$$ Hence $m^2$ is equal to $1$, $4$ or $9$. We now have the three cases $x+9y = 11$, $x+9y = 44$ and  $x+9y = 99$. Since we want $x+y$ to be a perfect square, we get the ordered pair $(35,1)$.
20. Show that for all positive integers $x, y, z$ $$\left(\frac{x+y}{x+y+z}\right)^{\frac{1}{2}} + \left(\frac{x+z}{x+y+z}\right)^{\frac{1}{2}} +\left(\frac{y+z}{x+y+z}\right)^{\frac{1}{2}} \leq 6^{\frac{1}{2}}.$$  (Taken from The Cauchy-Schwartz Master Class by J. Michael Steele)
    (Solved by Clyde Wesley Ang [Chiang Kai Shek College], Jecel Manabat [Valenzuela City Sci HS], Hans Jarett Ong [Chiang Kai Shek College], Jan Kedrick Ong [Chiang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Christian Academy])
    
    
    SOLUTION: 
    Without loss of generality, we can assume that $x \leq y \leq z$. Note that $x+y \leq 2z$ and $x+y+z \leq 3z$ give us $$\frac{x+y}{x+y+z} \leq \frac{2}{3},$$ and by symmetry we have $$\frac{x+z}{x+y+z} \leq \frac{2}{3} \text{ and } \frac{y+z}{x+y+z} \leq \frac{2}{3}.$$ Taking the sum of the roots, we now have $$\left(\frac{x+y}{x+y+z}\right)^{\frac{1}{2}} + \left(\frac{x+z}{x+y+z}\right)^{\frac{1}{2}}  +\left(\frac{y+z}{x+y+z}\right)^{\frac{1}{2}} \leq 3\sqrt{\frac{2}{3}} = 6^{\frac{1}{2}}.$$
21. For a fixed positive integer $n$, find the minimum value of the sum $$x_1 + \frac{{x_2}^2}{2} + \frac{{x_3}^3}{3} + \ldots + \frac{{x_n}^n}{n},$$ given that $x_1, x_2, \ldots, x_n$ are positive numbers satisfying the property that the sum of their reciprocals is $n$. (Polish Mathematical Olympiad, 1995)
    (Solved by Clyde Wesley Ang [Chiang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Christian Academy])
    
    
    SOLUTION: 
    By the AM-GM inequality, we get $kx_k \leq {x_k}^k + (k-1)$ for all $k$. Thus, we have $$\begin{align*} x_1 + \frac{{x_2}^2}{2} + \frac{{x_3}^3}{3} + \ldots + \frac{{x_n}^n}{n} &\geq x_1 + x+2 - \frac{1}{2} + \ldots + x_n - \frac{n-1}{n} \\ &= x_1 + x_2 + \ldots + x_n -n + M, \end{align*}$$ where $M = 1 + \frac{1}{2} + \ldots + \frac{1}{n}$. On the other hand, by the AM-HM inequality we have $$\frac{x_1 + x_2 + \ldots + x_n}{n} \geq \frac{n}{\frac{1}{x_1} + \ldots + \frac{1}{x_n}} = 1.$$ Thus the previous expression is at least $n - n + M = M$. This implies that $M$ is the desired minimum value, and it is obtained by setting $x_1 = x_2 = \ldots = x_n = 1$.