Extended Essay Mathematics. Alhazen s Billiard Problem

Transcription

1 Extended Essay Mathematics Alhazen s Billiard Problem Antwerp International School May 007 Word Count:

2 Abstract The research question o this Mathematics Extended Essay is, on a circular table there are two balls; at what point along the circumerence must one be aimed at in order or it to strike the other ater rebounding o the edge. In investigating this question, I irst used my own initial approach (which involved measuring various chord lengths), ollowed by looking at a number o special cases scenarios (or example when both balls are on the diameter, or equidistant rom the center) and inally orming a general solution based on coordinate geometry and trigonometric principles. The investigation included using an idea provided by Heinrich Dorrie and with the use o diagrams and a lengthy mathematical analysis with a large emphasis on trigonometric identities, a solution was ound. The conclusion reached is, i we are given the coordinate plane positions o billiard ball A with coordinates (x A, y A ) and billiard ball B with coordinates (x B, y B ), and also the radius o the circle, the solution points are at any o the points o intersection o b c the circular table with the hyperbola, y P + r b = y A + y B A x A c `, M = y A A y x A a b, p = x A + x B b c` xm a b c + xym, where P c b c, m = y A + y B and r is the radius. The solution was veriied by considering speciic examples through technology such as Autograph sotware and a TI-84 graphing calculator. Finally I briely looked at various other solutions to the problem and also considered urther research questions. Word Count :

3 Table o Contents Heading Page Introduction 3 Pre-examination o the problem 4 Initial approach 7 Analysis o speciic scenarios 8 Forming a general solution 10 Solution 15 Veriication o solution 16 Other possible solutions 18 Further investigation 0 Bibliography 1 - -

4 Introduction: Extended Essay Mathematics Alhazen s Billiard Problem Regarded as one o the classic problems rom two dimensional geometry, Alhazen s Billiard Problem has a truly rich history. The problem is believed to have been irst introduced by Greek astronomer Ptolemy back in 150 AD 1 and then eventually noticed by 17 th century Arabic mathematician Abu Ali al Hassan ibn Alhaitham (whose name was later Latinized into Alhazen). Alhazen made reerence to this problem in one o his published works entitled Optics and presented it in the orm, Find the given point on a spherical mirror at which a ray o light coming rom a given point must strike in order to be relected 3. Nowadays, this problem is oten reerred to as the Billiard Problem because it involves locating the point on the edge o a circular billiard table at which a cue ball at a given point must be aimed in order to carom (bounce) once o the edge and strike another ball at a second given point. 4 The ocus question o this extended essay will be: On a circular billiards table there are two balls; at what point along the circumerence must one be aimed at in order or it to strike the other ater rebounding o the edge? Heinrich Dörrie also described the problem as ind in a given circle an isosceles 1 Jack Kla, The World May be Divided into Two Types o People Alhazen s Billiard Problem. Viewed 19 February 005. < Heinrich Dörrie, 100 Great Problems o Elementary Mathematics: Their History and Solutions. Dover Publications New York, Dörrie Eric W Weisstein, Alhazen s Billiard Problem. Mathworld. Dated Viewed February <

5 triangle whose legs pass through two given points inside the circle. 5 My primary reason or choosing to investigate this ocus question is that the I.B Higher Level Mathematics Programme at our school is at times limited with regards to the study o geometry and trigonometry. Investigating this problem gave me an opportunity to ill this void. That being said, the problem was in itsel also very appealing to me as I personally enjoy playing billiards or pool and was eager to ind out about the mathematics o the game. The problem appeared in the Daily Telegraph news in 1997 when Dr Peter Nueman, an Oxord don o Queen s Collage, managed to provide a new solution to the problem. Inspired by early mathematician Descartes, Nueman cleverly translated the billiards table geometry simply into x and y coordinates on two axes. 6 This is a method I intend to use urther into my extended essay. Please note that this essay (and the solution to the ocus question) is narrowed down to emphasize the algebraic solution to Alhazen s Problem - however in the conclusion, other methods are briely discussed. Pre-examination o the problem: The great diiculty with this investigation lies within two concepts. First o all, the balls in question are randomly scattered on the table with no speciic locations in other words our solution would need to be generalized or any set o billiard balls. Second o all, the balls need to be treated as ixed points. To begin this investigation one should irst consider where and how many possibilities there can be on a circular pool table that would allow or a ball to strike once o the edge and then hit another ball. Moreover, what exactly characterizes the direction o a ball bouncing o a circular table 5 Dörrie 17 6 Highield, Roger. Don Solves the Last Puzzle Let by Ancient Greeks. Daily Telegraph. April 1, 1997, Issue

6 border? The law o relection states that that the angle o relection and angle o incidence are equal, with each angle being measured rom the normal to the boundary (line indicating the border) 7. In igure 1, the incident path θ i must have an angle equal to the relected path θ r. θ r θ i Figure 1) Law o Relection, θ i = θ r The boundary in our case would be a tangent line drawn to the point on the border o the circle where the ball A bounces o the circular side to ball B (Figure ). Figure ) B C A θ r θ i Another way to express this problem is, to describe in a given circle an isosceles triangle whose legs pass through two given points made inside the circle 8. This is useul because it allows us to relate the billiard balls to chords within the circle. Observe Figure 7 Henderson, Tom. Relection and Its Importance. The Physics Classroom. Dated 004. Viewed 1 March 005. < 8 Heinrich Dörrie, 100 Great Problems o Elementary Mathematics: Their History and Solutions. Dover Publications New York,

7 3: ball A and ball B are located within the circular billiard table with the table s center at point C. Ball A needs to make contact with the border at point T in order to strike ball B. I we extend the path that ball A must take to the opposite side o the circle, we have a chord the same can be done or ball B. The points A' and B' are the second points o intersection o the circle with the respective chords. Figure 3) A' A chord a C T B chord b B' I two radii are drawn to the centre o the circle rom the points A' and B', we have essentially two triangles CTA' and CTB'. The length o chord a is equal to that o chord b or the ollowing reason: Claim: A T B T Proo: 1) CT CA CB (radii to circle) ) CT A and CT B are isosceles 3) C T A C A T and C T B C B T 4) C TA CT B (angle o incidence = angle o relection) 5) A CT B C T 6) A CT B C T (Side-angle-side property) 7) A T B T (corresponding parts o congruent triangles are congruent) - 6 -

8 This explains why instead o looking at how one ball must be struck in order or it to strike the other ater rebounding o the edge, we can look or an inscribed isosceles triangle whose legs pass through ball A and ball B. Initial approach: Now in order or us to have a rough idea o the range o possible solutions, we must consider several general cases and see what results we get. Consider Figure 4, here I have randomly chosen two points to be my locations or ball A and ball B. Then I divided the circle into 1 equal parts around the circumerence although the more you divide the circle the more accurate your indings will likely be. Figure 4) A 3 10 C 4 B

9 Chords are drawn going through ball A to each o the 1 points, and the same or ball B. The lengths o the chords are measured and recorded in a table (below). The solutions to where ball A must hit to bounce o and hit ball B can possibly be ound by looking at where the corresponding chords are equal to one another, in other words where chord a chord b = 0. By making a table showing chord a chord b we could perhaps ind possible solutions chord a (cm) chord b (cm) chord a - chord b (cm) Solution * * * * From looking at the changing chord a chord b we can see that solutions should be at point 4, next to point 7, between points 10 and 11, and between points 11 and 1. However there appears to be an apparent paradox as although our results suggest that there is a solution between the points 11 and 1 and also between the points 6 and 7 on the circumerence, by looking at the graph one can see 1 Figure 5) that these chords leading to the points are in act the same chords and the points would thereore Although both chords are o same length, the points 1 and would clearly not be solutions points because the chords are in act the same. deinitely not work as solutions (unless these chords are in act the diameter, as we will see in the ollowing example)

10 Analysis o speciic scenarios: Let us analyze another more speciic scenario. In igure 6, ball A and ball B both lie on the diameter, and are equidistant rom the centre o the circle at point C. Figure 6) 1 A 4 c B 3 Possible solutions can be ound at points 1,, 3 and 4 as shown on the diagram. In this case we have our places where we can strike one o the balls so that it rebounds and hits the other. However that being said, i ball A was aimed at point 3, then ball A would go through ball B and then bounce o the border to hit the ball back. In other words or ball A, mathematically points 1,, 3 and 4 are all solutions, but realistically only points 1,, and 4 are solutions because ball B would block the path o ball A beore it can reach point 3. There is another scenario where we arguably only have solutions. In Figure 7 ball A is located exactly at the centre o the circle and ball B is located along the diameter

11 Figure 7) 1 B A c I we were striking ball A, the place where it would rebound to then hit ball B is located at point. However that being said, i ball A was aimed at point 1, then point 1 would also be a solution. In other words, mathematically speaking both point 1 and point are solutions, but realistically only point is a solution because ball B would block the path o ball A beore it can reach point 1. Forming a general solution: To orm a general solution to the problem, I have used an idea provided by Heinrich Dörrie in 100 Great Problems o Elementary Mathematics: Their History and Solutions 9. This method is based on using coordinate geometry to orm a general solution. I know that the solution points must be on the circumerence o the circle (they satisy the equation o the circle). With this in mind I will attempt to ind another equation which also includes the points, using ideas rom geometry and trigonometry. 9 Dörrie

12 y (0, r) A (x a, y a ) B (x b, y b ) x (r, 0) Figure 8) Consider a given circular billiard table with the tables centre at point C and its radius r. Ball A and ball B are randomly scattered on the table. I we then translate this onto a perpendicular coordinate system and make C the origin, then ball A will have the coordinates (X a, Y a ) and ball B will have the coordinates (X b, Y b ) as can be seen on Figure 8. From here we can orm an isosceles triangle that includes ball A and ball B (Figure 9 on the next page). Point O is where one o the balls must be aimed at in order to bounce o and hit the other

13 O α β φ A C µ B λ B' A' Figure 9) Extending the line OA and OB, we obtain the chords OA' and OB'. Consider the angles that these legs orm with the radius OC, angles α and β i ball A is to rebound and strike ball B then α must equal β (angle o incidence = angle o relection). Furthermore, i we extend lines OA, OC, and OB to the circle at A' and B' then we orm 3 new angles. The angle between OA' and the x-axis is noted φ, the angle between OC and the x-axis is µ, and the angle between OB' and the x-axis is λ

14 Figure 10) O (x,y) µ = α + φ α = µ - φ α β λ = β + µ β = λ - µ A (x a, y a ) B (x b, y b ) φ µ λ C In Figure 10, it is apparent that α = µ φ and that β = λ µ. Also, rom the diagram we can derive that: tan ` ϕ a = y a x a tan ` µ a = y x Since angle α = β Then tan ` α a = tan b β c ` a y tan γ = b x b We know that α = µ φ and β = λ µ Using the compound angle identities, ` a # tan µ@ ϕ = tan µ ` tan B a tan ` A = b c tan ` B a 1 + tan ` A a tan ` B a We arrive at, b ` a tan ϕ = tan µ c tan ` µ tan ` ϕ 1 + tan ` µ a tan ` ϕ a tan ` λ tan ` µ a = 1 + tan ` λ a tan µ a ` a

15 now since: tan ` ϕ we have: a = y a x a tan ` µ a = y x tan ` λ a = y b x b d 1 + e y x y x y A g x A e g y A x A = 1 + y B g x B d e y d y x y B x B e g or, ` a ` y x@ x x y@ y ` a A x x@ x A ` a ` x x@ x A + y y@ y ` a A x x@ x A a a = ` a ` x y@ y y x@ x ` a B x x@ x B ` a ` x x@ x B + y y@ y ` a B x x@ x B a a or, yx xy + xy A x A x A + y A y A xy = xy + yx B x + y A y B or, xy yx A x + x A x y A y A yx = xy B x + y A y B Simpliying, LHS: RHS: x xy A + y xy x y yxy A A y x yx y yx A + yx A x A + y x A A y B = x yx B + y yx xyx y y x xy xy A y B + x A x A A y B + xy A y A A y B Bringing everything to one x A xy y xy A + x y A + yx A y A A y B + x yx A + y yx yx A x y x A A y B + x yx B + y yx xy A x y y x xy y xy B + x x A A y B + xy A y A A y B =

16 or, Alexander Zouev b c b c b x y A + y B x y y A + y B A x A + y yx A + yx xy xy B b c b c + x yx A + yx xy xy B + yx ya A y x x A + y A A y B = 0 Factoring out, b cb c b y y A + y B A x A + y + x cb yxa + yx xy xy B I we assign the ollowing variables: b c ` a P = y A + y B A x A, M = y A A y x A c ` a + xy ya A y x A = 0 c Then we have b c b cb c b c y P + y + x yxa + yx xy xy B + xym = 0 Further expanding the second bracket o the second sum, b c b cd b c b y P + y + x y xa + x x ya + y B ce b c + xym = 0 Assigning the ollowing variables: We have the ollowing: b c b c p = x A + x B, m = y A + y B b c b c y P + y + x ` a b c xm + xym = 0 Solution: Because the point O(x,y) - where one o the balls must be aimed at in order to bounce o and hit the other - lies on the circle, we can apply the circle equation: x + y = r to our equation. Thereore our solution has the orm o: b c b c y ` a b c P + r xm + xym = 0 This equation represents a hyperbola (the standard Cartesian hyperbola equation is Ax + Bxy + Cy + Dx + Ey + F = 0), thereore our solution is:

17 Given the position o both o the billiard balls in a circle and the radius o the table, one can ind the solution points at any o the points o intersection o the circle b c b c with the hyperbola, y ` a b c P + r xm + xym There are at most our places where a hyperbola intersects with a circle, thereore in general there should be our places where one could strike a ball so that it rebounds and hits the other. However given the nature o a hyperbola, intersections with a circle can occur at only one, two, or even three points (i one o the disconnected curves (arms) o the hyperbola is tangent to the circle) meaning there will not always be our valid solutions on the circular billiard table. Relating back to our ocus question - i we had two balls in a circular billiard table, and we were able to relate their positions into (x,y) coordinates on a plane and also measure the radius o the table, then we would be able to ind the exact point(s) along the circumerence where one ball must be aimed in order or it to strike the other ater rebounding o the edge. Veriication o Solution Now to veriy our solution, I will chose a circular table with a radius o 4 units, and two balls with somewhat simple coordinates; ball A with the coordinates (0,1) and ball B with the coordinates (0, -1). The variables we have identiied are as ollows: X a = 0, Y a = 1, X b = 0, Y b = -1, r = Plugging these into the hyperbola equation we have ound, and also into the equation o the circle: ( x y )(0 + 0) + ( )( y(0) x(0)) + ()( 1) xy) = 0 xy = 0 the equation is zero when y = 0, and when x =

18 Now we will plug these into the equation o the circle: x + y = 4 Substituting x = 0 y = 4 Figure 11) (0, ) y A (0, 1) y = ± Substituting y = 0 (-, 0) B (0, -1) (, 0) x x = 4 x = ± (0, -) There are our points along the circumerence where one could strike either ball A or ball B and have it rebound to then strike the other ball. These solution points are: (0, -), (0, ), (, 0), (-, 0) as shown in igure 11 I we are to choose points that are slightly more complicated, such as ball A at (-1, 1.5) and ball B at (0.5, 1), then the equation becomes more interesting: X a = -1, Y a = 1.5, X b = 0.5, Y b = 1, r = Plugging all the variables into our ound hyperbolic equation, we obtain: { (x² - y²)((1.5)(0.5) + (1)(-1)) + 4(y( ) - x( )) + (((1.5)(1) - (- 1)(0.5))xy) = 0 } 0.5y 0.5x y 10x + 4xy = 0 The equation o the circle is: x + y = 4 Using the equation o the circle: y = ± 4 x Now plugging this into the hyperbolic equation (since we are looking or the intersections between the circle and the hyperbola), we obtain two equations:

19 Using positive square root: 0.5(4 x ) 0.5x 4 x 10x + 4x 4 x = 0 Using negative square root: 0.5(4 x ) 0.5x + 4 x 10x 4x 4 x = 0 Looking at these equations, it becomes evident that we could ind possible solutions using a simple graphing calculator such as the TI-84. The zeros o the unctions should give us the x-values o the solution. Solution 1 at x = to 4 signiicant igures and y = 4 (0.747) = (4s.), thereore solution 1 at (0.747, ) Solution at x = to 4 decimal places and y = 4 ( ) = (4s.), thereore solution 1 at ( , 1.743) Using computer graphing sotware Autograph 3, we can graph both equations and see how the intersections occur and also ind the points o intersection: Figure 1) Indeed both methods give the same result, and using Autograph 3 we can clearly see how the hyperbola intersects the circle in only two places

20 Other possible solutions: This analytical solution just described is only one o many methods known. It answers our ocus question and works or any randomly located and ininitely small billiard balls. When plugging in all the known values (such as the radius r and the ball locations P, p, M and m) one will be able to solve the equation and arrive at one, two, three or our solutions. This will give the point along the circumerence must one ball must be aimed at in order or it to strike the other ater rebounding o the edge. Another method introduced by Michael Drexler and Martin J. Gander in their essay Circular Billiards involves using a dierent geometric derivation 10. They emphasize the act that or any given ellipse, i a ball is placed in each ocus then any point on the rim would be a solution (such is the nature o the oci points). Thereore to solve Alhazen s problem one would need to derive an ellipse that touches the circle (tangent to the circle) and has the two given billiard balls as ocal points. Alhazen s Problem has also been solved using a trisection, as done by Roger C. Alperin in his paper Trisections and Totally Real Origami. Alperin argues that Alhazen s problem can be solved by Euclidean tools and trisections or equivalently using Origami constructions 11. Moreover, modern mathematicians amiliar with mathematical computer programs and dynamic geometry sotware have ormulated simple computations that allow one to ind solutions to Alhazen s Problem. Luiz Carlos Guimarães and Franck Bellemain argue in their paper entitled Relections on the Problema Alhazen that dynamic geometry sotware gives an extra touch to geometric 10 M Drexler and M Gander. Circular Billiard. SIREV, 1998 volume 40 issue SIREV Journal Alperin, Roger C. Trisections and Totally Real Origami. MAA Monthly 005. Viewed August <

21 manipulations, especially with regards to Alhazen s Problem 1. Although I mysel am not greatly amiliar with computer geometry sotware, I can imagine how one could use it with regards to Alhazen s problem. For instance, i a program was instructed to search or same length chords that pass through two given points within a given sized circle, then it would run through all the possibilities, measure them, and ind the correct ones (similar to the method I used mysel by hand in the early stages o this essay). Interestingly enough, an easy to use application addressing Alhazen s Problem has been created and can be ound reely on the internet. Created by A.I Sabra, the Java application allows one to place two colored dots in any given situation within a spherical concave mirror, and ind the maximum amount o points o relection 13. Further Investigation: Although Alhazen s Problem might have already been solved and analyzed in a numerous amount o creative ways, one cannot help but wonder how many extended study questions can be proposed relating to Alhazen s Problem. For instance, what would happen i the table were triangular, or hexagonal? Perhaps one could try to interpret Alhazen s Problem not as two points on a two-dimensional circle, but as two points in a three-dimensional sphere. 1 L. Guimarães ; F. Bellemain.. Relections on the Problema Alhazeni. The 10 th International Congress on Mathematical Education, 004. Viewed Aug << 13 Sabra, Abdelhamid I. 4 Points o Relection Applet, Alhazen s Applet. Viewed August Harvard University. <

22 Bibliography Alperin, Roger C. Trisections and Totally Real Origami. MAA Monthly 005. Viewed August < Drexler, M Gander, M. Circular Billiard. SIREV, 1998 volume 40 issue SIREV Journal Dörrie, Heinrich. 100 Great Problems o Elementary Mathematics: Their History and Solutions. Dover Publications New York, Guimarães, L. C. ; Bellemain, F.. Relections on the Problema Alhazeni. The 10 th International Congress on Mathematical Education, 004. Aug < er.doc> Henderson, Tom. Relection and Its Importance. The Physics Classroom. Dated 004. Viewed 1 March 005. < Highield, Roger. Don Solves the Last Puzzle Let by Ancient Greeks. Daily Telegraph. April 1, 1997, Issue 676. Kla, Jack. The World May be Divided into Two Types o People Alhazen s Billiard Problem. Viewed 19 February 005. < Klingens, Dick. Relectie Binnen een Cirkelvorminge Spiegel. Wisaq!. Dated 1 October 003. Viewed February Sabra, Abdelhamid I. 4 Points o Relection Applet, Alhazen s Applet. Viewed August Harvard University. <

23 Weisstein, Eric W. Alhazen s Billiard Problem. Mathworld. Dated Viewed February < - -