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 \(\angle DAP+\angle BCP=\angle BQP=90^{\circ}\).