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.
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| Figure 1: A schematic of the parabolic mirror. |
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| 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.
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