The handouts for the Week 2 PEM sessions can be downloaded via the links below.
Wednesday, November 28, 2012
Saturday, November 24, 2012
Reflections on Light, of Light (Tuklas Vol. 14, No. 2 - Nov. 24, 2012)
REFLECTIONS ON LIGHT, OF LIGHT
The theory of light is an interesting phenomenon to study because of its curious mathematics (such as in this article [1]) and the many discoveries it leads to. Light has many uses, ranging from the prehistoric to the modern: keeping us warm, cooking the food of our neighborhood caveman, and decorating our Christmas trees with dancing colors. Let us now take a look at some of light's quirks and behaviors.
Images are formed when light rays strike an object from all directions and then come together to converge at a single point. The image is formed at the point where the light rays converge. A very common example is found in a very unlikely location: the bathroom, where mirrors are placed so that ladies and gents can spend countless hours preserving their vanity! A movie projector, on the other hand, uses light to broadcast the image on a screen. Note that the most common light that we see is one that is incoherent in nature. It comes from all directions, like light from the sun, or the common fluorescent light. Since the light comes from everywhere without a single direction, we naturally need to have a screen where the image is to be seen, otherwise there is no image at all!
So is it possible to form an image without a screen? Is it possible to bring light rays that travel in all direction to converge at one point?
Welcome the parabolic mirror. This novelty device sold in many toy shops is shaped like a three-dimensional parabola (for which the proper term is a paraboloid). This device is made of two mirrors: one opening up and one facing down. They are shaped in such a way that the focus of the lower mirror is the vertex of the upper mirror, and vice versa. The construction of these mirrors are very meticulous so that the entire setup works fabulously!
Let the lower mirror have an equation \(y=ax^2\). The vertex of the parabola is conveniently located at the origin. The focus of the parabola is located at the point \(\left (0, \frac{1}{4a} \right )\). This focus should be the vertex of the inverted parabola. So the upper mirror must have an equation \(y-\frac{1}{4a} = -x^2\). The line segment containing the intersections of the parabola is the width of the mirror. This is how wide any parabolic mirror should be for a given value of \(a\); otherwise, the desired 'wow-factor' cannot be achieved.
 |
| Figure 1: A schematic of the parabolic mirror. |
 |
| Figure 3: In this image [2], the espresso cup and the coffee beans are just an image. It floats at the point where the focus of the lower parabola coincide. |
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.
REFERENCES:
[1] Clark Kendrick Go, "Some Mathematics of Light", Tuklás Matemátika (Vol. 13, No. 8).[2] Image taken from http://en.wikipedia.org/wiki/File:Mr._Espress_Cup_Web_copy.jpg.
OLYMPIAD CORNER
from the Canadian Mathematical Olympiad, 2002
Problem: Call a positive integer \(n\) practical if every positive integer less than or equal to \(n\) can be written as the sum of distinct divisors of \(n\). For example, the divisors of 6 are 1, 2, 3, and 6 . Since \[ 1=1,\quad 2=2,\quad 3=3,\quad 4=1+3,\quad 5=2+3,\quad 6=6, \] we see that 6 is practical.
Prove that the product of two practical numbers is also practical.
Since \(M\) and \(N\) are practical, we can write \[p=a_1+\cdots+a_m,\quad q=b_1+\cdots+b_n, \] where the \(a_i\)'s are distinct divisors of \(M\) and the \(b_j\)'s are distinct divisors of \(N\). Hence \[ \begin{align*} k &= \left( a_1+\cdots+a_m\right)N+b_1+\cdots+b_n \\ &= a_1 N+\cdots +a_m N+b_1+\cdots +b_n \end{align*} \] Note that each \(a_i N\) and \(b_j\) are divisors of \(MN\). Since \(b_j\leq N\leq a_i N\) for any \(i, j\) and the \(a_i N\)'s and \(b_j\)'s are distinct, then the product \(MN\) is also practical.
Prove that the product of two practical numbers is also practical.
Solution:
Let \(M\) and \(N\) be practical. Any \(k\leq MN\) can be expressed as \[ k=pN+q \] where \(0\leq p\leq M\) and \(0\leq q\leq N\).Since \(M\) and \(N\) are practical, we can write \[p=a_1+\cdots+a_m,\quad q=b_1+\cdots+b_n, \] where the \(a_i\)'s are distinct divisors of \(M\) and the \(b_j\)'s are distinct divisors of \(N\). Hence \[ \begin{align*} k &= \left( a_1+\cdots+a_m\right)N+b_1+\cdots+b_n \\ &= a_1 N+\cdots +a_m N+b_1+\cdots +b_n \end{align*} \] Note that each \(a_i N\) and \(b_j\) are divisors of \(MN\). Since \(b_j\leq N\leq a_i N\) for any \(i, j\) and the \(a_i N\)'s and \(b_j\)'s are distinct, then the product \(MN\) is also practical.
PROBLEMS
- Prove that the sum of squares of 3, 4, 5, or 6 consecutive integers is not a perfect square.
- If it is possible to construct a triangle with side length \(a<b<c\), prove that it is possible to construct a triangle with side lengths \(\sqrt{a}<\sqrt{b}<\sqrt{c}\); also, show that the converse is false.
- Prove that among any seven distinct integers, there must be two such that their sum or difference is divisible by 10.
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 December 1, 2012.
SOLUTIONS
(for November 17, 2012)
- Find the solution set in \((-\frac{\pi}{2},\frac{\pi}{2})\) of the equation \[\log_2(1+\sin^2\theta+\sin^4\theta+\sin^6\theta+\ldots) = 1.\](solved by Clyde Wesley Ang [Chiang Kai Shek College], Jecel Manabat [Valenzuela City Sci HS], Andrew Brandong Ong [Chiang Kai Shek College], Hans Jarett Ong [Chiang Kai Shek College] and Jan Kedrick Ong [Chiang Kai Shek College]; partial credit for Hanz Manguiat [Ateneo HS], Christian Jon Patagan [Regional Science HS III], Iliana Tan [Jubilee Christian Academy] and Farrell Eldrian Wu [MGC New Life Christian Academy])SOLUTION:We have \[ \log_2(1+\sin^2\theta+\sin^4\theta+\sin^6\theta+\ldots) = \log_2\left(\sum_{i=0}^{\infty}\left[\sin^2\theta\right]^i\right) \] which (restricting \(\left|\sin^2\theta\right|<1\)) is an infinite geometric series thus \[ \log_2\left(\sum_{i=0}^{\infty}\left[\sin^2\theta\right]^i\right) = \log_2\left(\frac{1}{1-\sin^2\theta}\right) = 1, \] giving us \[ \frac{1}{1-\sin^2\theta} = 2 \] which yields \(\theta = \pm\frac{\pi}{4}\).
Alternatively, we have \[ \begin{align*} 1+\sin^2\theta+\sin^4\theta+\sin^6\theta+\ldots &= 2 \\ \sin^2\theta+\sin^4\theta+\sin^6\theta+\ldots &= 1 \end{align*} \] hence \[ \begin{align*} 1 &= \sin^2\theta+\sin^4\theta+\sin^6\theta+\ldots \\ &= \sin^2\theta(1+\sin^2\theta+\sin^4\theta+\sin^6\theta+\ldots) \\ &= \sin^2\theta(2), \end{align*} \] which yields the same answer. - Given that \(R\) is an inner point of \(\Delta ABC\). Prove that \[\frac{1}{2}(AB+BC+CA)<RA+RB+RC<AB+BC+CA.\] (Taken from Lecture Notes on Mathematical Olympiad Courses by Xu Jiagu)(solved by Clyde Wesley Ang [Chiang Kai Shek College], Hans Jarett Ong [Chiang Kai Shek College], Jan Kedrick Ong [Chiang Kai Shek College], Christian Jon Patagan [Regional Science HS III] and Farrell Eldrian Wu [MGC New Life Christian Academy]; partial credit for Kimberly Co [St. Stephen's] and Hanz Manguiat [Ateneo HS])SOLUTION:We make extensive use of the triangle inequality. Extend \(BR\) to intersect \(AC\) at \(S\). Then \[ AB+AC=AB+(AS+SC)>BS+SC=BR+(RS+SC)>BR+RC. \] This result can also be obtained by noting that since Heron's formula is a product of the side lengths, it is directly proportional to the side lengths (and thus the semiperimeter) giving us \[ \frac{AB+AC+BC}{2}>\frac{RB+RC+BC}{2}. \] Similarly, we have \(BA+BC>RC+RA\) and \(BC+CA>RA+RB\). Adding them together, we have \[ 2(AB+BC+CA)>2(RA+RB+RC), \] giving us \[ RA+RB+RC<AB+BC+CA. \] By the triangle inequality we have \(RA+RB>AB\), \(RB+RC>BC\), and \(RC+RA>CA\). Adding them together, we have \[ \frac{1}{2}(AB+BC+CA)<RA+RB+RC. \]
- Find the probability of obtaining two numbers \(x\) and \(y\) in the interval \([0,1]\) such that \[x^2-3xy+2y^2>0.\] (13th Philippine Mathematical Olympiad, Area Stage)(solved by Clyde Wesley Ang [Chiang Kai Shek College], Andrew Brandong Ong [Chiang Kai Shek College], Hans Jarett Ong [Chiang Kai Shek College], Jan Kedrick Ong [Chiang Kai Shek College]; partial credit for Jecel Manabat [Valenzuela City Sci HS] and Farrell Eldrian Wu [MGC New Life Christian Academy])SOLUTION:Obtaining two numbers \(x\) and \(y\) in the interval \([0,1]\) is equal to obtaining ordered pairs \((x,y)\) in the unit square \([0,1]^2\).
Furthermore, we note that \(x^2-3xy+2y^2 = (x-2y)(x-y) > 0\) implies that \[ P(x^2-3xy+2y^2 > 0) = P\left(\{x>2y,x>y\}\cup\{x<2y,x<y\}\right). \] Graphing the lines \(x=2y\), \(x=y\) and the unit square on the Cartesian plane, we can see that the shaded regions (see above) contain the points that satisfy our condition. Hence, the probability is \[ \frac{\frac{1}{2}+\frac{1}{4}}{1} = \frac{3}{4}. \]
ERRATA
The previous issue (released November 17, 2012) incorrectly numbered the Problems section.
Monday, November 19, 2012
Announcements (for Weeks 2 and 3)
The venue for the PEM session for teachers will be at the JG School of Management building, room SOM 111 (8:30AM-12:30PM).
The first PEM session for provincial students is moved to December 1, 2012 (instead of November 24, 2012). The venue is at the SEC C building, room SEC-C 201.
Please be guided accordingly.
Saturday, November 17, 2012
The Attraction of Mathematics (Tuklas Vol. 14, No. 1 - Nov. 17, 2012)
THE ATTRACTION OF MATHEMATICS
For those of us without much mathematical preparation, mathematics can also be a source of awe, similar to what Wigner had experienced. To see what I mean, we look at a few aspects of mathematics (albeit superficial in such a short note) which hold some of its characteristic attractions for many people.
Math is a source of understanding and illumination. For instance, what does it mean for an Olympic sprinter to run a world record of 9.87 seconds in the 100-meter dash? If we divide both numbers by ten, we see that, on the average, the athlete ran 10 meters in less than one second. Using a meter tape, one can see that 10 meters is a significant distance. To take another example, around 6 million Jews were killed in the Second World War, from 1939 to 1945. As the mind is unused to such a large number, try dividing 6 million by 6, then by 364. This comes out to 2747 Jews killed everyday for six years! As a last example, try to gain an understanding of the following phenomena: at the height of Michael Jordan's power and popularity in the NBA, his salary was 36 million dollars in one year.
These examples, simple as they are, show the great aid mathematics adds to our understanding. In the second example, our emotional response, one would hope, was heightened because of the real horror of more than a couple of thousand of murders daily. In the first and third, in addition to illumination, we experience that element which is common to good stories, that of surprise. Lest we belittle these small demonstrations, employing as they are only the basic operations, we can imagine that schoolchildren and accountants alike in the times of the Roman Empire had a grand time multiplying, say, XXIV (24) by XIX (19). To keep track of the Empire's population must have been quite difficult.
More than simple understanding, mathematics shed deeper light on a few things we take for granted such as counting, the computation of areas, and the twin concepts of being outside and being inside a closed curve.
To count the number of persons in a room, we mentally tick off one, two, three, etc. while pointing at each person in turn. We know that there are sufficient chairs in the room if there are empty chairs after everyone is seated, that is, there are chairs not corresponding to any person in the room. This one-to-one correspondence between objects counted and the abstract set of natural numbers is the essence of counting. That is why it takes a very little leap of the imagination to count to "infinity": after ticking off any number, however large, we just add 1. If we are now asked if there are many more natural numbers than positive even integers, it should not be too difficult to observe that the answer is no. One set has as much elements as the other. Indeed, the correspondence \(1\leftrightarrow 2,2\leftrightarrow 4,3\leftrightarrow 6\), etc. demonstrates the equality of the number of elements between the two sets. Similarly, there are as much natural numbers as there are odd numbers, and as much natural numbers as the squares 1, 4, 9, 16, and so on. Much more surprising is the fact that there are the same number of elements in the set of natural numbers as in the set of positive fractions. This is made much more counter-intuitive by the fact that the integers are "far" away from each other while between any two fractions, however close, there is a fraction in the middle of them.
In calculus, a very powerful result (the Fundamental Theorem) and a very simple algorithm (subtract the values of the antiderivative at the endpoints) will tell us that a figure such as the first curve shown below has a finite area, although it is not "closed" or "bounded." This considerably generalizes our notion of area and is one source of the power of the calculus.
Mathematics, just like art, is a source of great creativity from its practitioners. Just witness the inexorable logic and beauty of two of the earliest mathematical proofs known to the ancient Greeks. The first is the incommensurability of the number \(\sqrt{2}\). This just means that \(\sqrt{2}\) is not a fraction. To begin the proof, suppose it is equal to the fraction \(\frac{p}{q}\) written in lowest terms. Simple manipulation will give \(2p^2=q^2\). This implies that \(q^{2}\), hence also \(q\), is an even integer. Write \(q=2k\) and substitute it into our first equation to obtain \(2p^2=4k^2\), or \(p^{2}=2k^{2}\). This, in turn, means that \(p^{2}\), hence \(p\), is an even integer. The fact that both \(p\) and \(q\) are even contradicts our initial assumption that the fraction \(\frac{p}{q}\) is in lowest terms.
Now take a look at the second proof, this time on the fundamental result claiming the infinitude of primes. The prime numbers are the basic building block of all integers and hence of all numbers. They are the integers greater than 2 which are not divisible by any integer except 1 and itself. To show that the set of prime numbers has infinitely many elements, assume the contrary. To this end, suppose we have the complete list of all primes: \(p_1, p_2, p_3, \ldots, p_n\). Consider the number \(N = p_1p_2\cdots p_n + 1\). Then this number is not divisible by any of the primes \(p_{k}\) in our list, as the division always leaves the remainder 1. This indicates that \(N\) is itself prime. We have reached a contradiction to our assumption that we already have the complete list of all primes.
The infinitude of primes is one of those simple statements in the Theory of Numbers whose proofs are either of great complexity or as yet undiscovered. Some of these are the Prime Number Theorem on the number of primes less than a given number, the twin prime conjecture on the infinitude of twin primes such as 11 and 13, 17 and 19, 29 and 31, etc., and Goldbach's conjecture which states that all large odd numbers can be written as the sum of two primes. Fairly recent mathematical achievements are the Fermat-Wiles Theorem (formerly Fermat's Last Theorem) and the solution by G. Perelman of the Poincare Conjecture on the possible shapes of 4-dimensional objects with given special properties. Mathematics is thus full of beautiful and immensely difficult problems the solution of which will bestow more than passing fame to those lucky and persistent enough to solve them.
ABOUT THE AUTHOR:
Job A. Nable is an Assistant Professor at the Ateneo de Manila University. He obtained his Ph.D. in Mathematics at the University of the Philippines.
OLYMPIAD CORNER
from the 51st National Mathematical Olympiad of Romania, 2000
Problem: Let \(p\) and \(q\) be positive integers such that \(1<q<p\), and \(a = \left( p + \sqrt{p^2+q} \right)^2\). Prove that \(a\) is an irrational number and that its fractional part is greater than \(0.75\).
Solution:
It is easy to see that \(p^{2}<p^{2}+q<(p+1)^{2}\) therefore \(p^{2}+q\) is not a square, so \(\sqrt{p^{2}+q}\) is an irrational number. It follows that \[a=\left(\sqrt{p^2+q}+p\right)^2=2p^2+q+2p\sqrt{p^2+q}\] is also irrational. Let us consider the number \(b=\left(\sqrt{p^2+q}-p\right)^2\). Clearly, \(b\) is also irrational, hence, we have \(b>0\). By AM-GM inequality, \[\sqrt{p^2\left(p^2+q\right)}<\frac{p^2+\left(p^2+q^2\right)}{2}\] which leads to \[\sqrt{p^2+q}<\frac{2p^2+q}{2p}=p+\frac{q}{2p}\leq p+\frac{1}{2}.\] Then \(\sqrt{p^2+q}-p<\frac{1}{2}\) so that \(b=\left(\sqrt{p^2+q}-p\right)^2<\frac{1}{4}\). This means that \(b\in (0,1/4)\) and since \(a+b\) is obviously a positive integer, it follows that the fractional part of \(a\) is greater than \(3/4=0.75\).
PROBLEMS
- Find the solution set in \((-\frac{\pi}{2},\frac{\pi}{2})\) of the equation \[\log_2(1+\sin^2\theta+\sin^4\theta+\sin^6\theta+\ldots) = 1.\]
- Given that \(R\) is an inner point of \(\Delta ABC\). Prove that \[\frac{1}{2}(AB+BC+CA)<RA+RB+RC<AB+BC+CA.\]
- Find the probability of obtaining two numbers \(x\) and \(y\) in the interval \([0,1]\) such that \[x^2-3xy+2y^2>0.\]
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 November 24, 2012.
Subscribe to:
Posts (Atom)