AA LIGHTHOUSE ILLUMINATION PROBLEM
ERHAN TEZCAN
Abstract.
This paper discusses a problem that consists of n “light-houses” which are circles with radius 1, placed around a common cen-ter, equidistant at n units away from the placement center. Consecutivelighthouses are separated by the same angle: 360 ◦ /n which we denote as α . Each lighthouse “illuminates” facing towards the placement centerwith the same angle α , also called “Illumination Angle” in this case.As for the light source itself, there are two variations: a single pointlight source at the center of each lighthouse and point light sources onthe arc seen by the illumination angle for each lighthouse. The problem:what is the total dark (not illuminated) area for a given number of light-houses, and as the number of lighthouses approach infinity? We showthat by definition of the problem, neighbor lighthouses do not overlapor be tangent to each other. We propose a solution for the center pointlight source case, and discuss several small cases of n for the arc lightsource case. Keywords:
Euclidean, Geometry, Circle, Illumination, Lighthouse,Problem
MSC:
Primary 52C05, Secondary 51M05; 51M04 Introduction
This problem originated to the author during a 17-hour long bus trip fromWarsaw to Tallinn, in line with author’s inability to sleep in a bus. At night,with only visual input being the bus’ indoor ceiling lights that are circular ,the author tried to pass time by thinking of a random problem originatingfrom aforementioned light sources. This paper discusses that problem, whichthe author initially called “Lighthouse Problem” but to differentiate it fromanother problem of the same name that takes place in [1], we would like tocall our problem “A Lighthouse Illumination Problem”.In the next section we will be defining the problem, afterwards, we willdiscuss the boundaries and then delve into the problem itself. Finally, wewill be recapitulating our questions and question marks. We created all ourfigures using a software called “GeoGebra” [2].2.
The Problem
Definition 2.1 (The Lighthouse Illumination Problem) . Suppose we have n circles with radius in an infinite plane, placed around a common centerpoint (placement center) where the distance between any circle’s center andthe placement center is n . Placing n circles like this divides the ◦ ofthe center point into n angles of ◦ /n . We will denote this angle as α .Each circle acts as a “lighthouse”, illuminating towards the center, looking Date : March 3, 2019. a r X i v : . [ m a t h . HO ] M a r ERHAN TEZCAN
Figure 1.
Point Light Sourceat Center
Figure 2.
Point Light Sourceat Arc / Edge directly at the placement center point with the illumination angle α . Thelight source has two variations in this problem: (1) The center point of each lighthouse is a point light source. (Figure1) (2)
Every point on the intercepted minor arc of the illumination angle α act as a point light source. (Figure 2)For both variations of the light source, what is the total dark area for a givennumber of lighthouses n ? To further explain figures 1 and 2, we shall explain the three shades:white, dark gray and light gray. • White denotes an illuminated area. The circular sector defined bythe illumination angle α is also white. • Dark gray denotes a dark area, these are not illuminated by anylighthouse. • Light gray denotes the non-illuminating part of the lighthouse.One might wonder whether the light gray counts as a dark area or not. Wemade our calculations with the light gray area excluded, because it is easyto include it if we want to. The light gray area for every lighthouse is givenby π − π α ◦ = π (cid:32) − ◦ n ◦ (cid:33) = π (cid:18) − n (cid:19) (1)Multiplying (1) by n gives the total light gray area nπ (cid:18) − n (cid:19) = π ( n −
1) (2)If we want to include the light gray area as darkness we just add π ( n − n . As for LIGHTHOUSE ILLUMINATION PROBLEM 3
On figure 3 we have three points: L i is a lighthouse, L i +1 is the light-house placed next to it and P istheir common center which they areplaced around of. Recall that α =360 ◦ /n = 2 π/n . This figure is truefor any pair of neighbor lighthousesfor any n . Figure 3.
Distance be-tween 2 lighthouses.notation, we will use D ( n ) for total dark area, d ( n ) for the dark area behinda single lighthouse. D ( n ) = n × d ( n ) (3)The dark area behind a lighthouse is defined by two light rays coming fromtwo other lighthouses. We will call the lighthouse we are calculating the darkarea behind of as the “target lighthouse”. The lighthouse bearing the lightsource of the aforementioned light ray will be called “source lighthouse”.3. Boundary Cases
Before we actually delve into the dark area calculations, we want to discusstwo cases. For some number of lighthouses m is it possible that(1) Lighthouses are tangent to each other?(2) Lighthouses are overlapping? Theorem 3.1.
Neighbor lighthouses never overlap or touch for n ≥ .Proof. We can answer both by looking at the triangle defined by a pair ofneighbor lighthouse centers and the placement center. Each lighthouse hasa radius of 1. Looking at figure 3, if the neighbor lighthouses are tangent,then d = 2. Similarly, if the neighbor lighthouses are overlapping then d < d in terms of n using Cosine Theorem. d = n + n − n cos( α )This reduces to d = 2 n (1 − cos( α )). Recall that 1 − cos( φ ) = 2 sin ( φ ) sowe get d = 2 n (2 sin ( α )) = 4 n sin ( α ). Taking the root yields d = 2 n sin( π/n ) (4)Now we can prove the cases by showing that d >
2. Note that d only existswhen there are 2 or more lighthouses, so n ≥ d > n sin( π/n ) > n = 2, we have 4 sin(90 ◦ ) = 4 > n sin( π/n ) is monotonically increasing,and since the first value was 4 the remaining values will never be less than4. (cid:3) ERHAN TEZCAN
Figure 4.
Figure 5.
Remark 3.2.
The case of tangent lighthouses d = 2 can be also studied byreferring to Steiner Chain [3] . Regarding a Steiner Chain: sin (cid:16) πn (cid:17) = a − ba + b (5) where there are n circles packed between a central circle of radius b and anouter concentric circle of radius a . Relating this to our problem, we have b = n − and a = n + 1 . Plugging them in equation (5) yields: sin (cid:16) πn (cid:17) = n + 1 − n + 1 n + 1 + n − n = 1 n This is the equation we would get by plugging d = 2 in equation (4) . Distance between neighbor lighthouses at infinity.
We can findthe distance between neighbor lighthouses at infinity just by looking atlim n →∞ n sin( π/n ) = lim n →∞ π sin( π/n ) π/n Substituting θ = π/n gives uslim θ → π sin( θ ) θ = 2 π This means that as n approaches infinity the distance between neighborlighthouses d becomes 2 π . Now that we have shown lighthouses will remainunscathed without colliding with one another, we can continue with the darkarea calculations.4. Point Light Source at the Center
Our first variation of the problem has each lighthouses having a singlepoint light source at their center. We will be giving examples for 1, 2, 3, 4,5 and 6 lighthouses. We will then discuss the dark area for any n .4.1. Our first case is a single lighthouse. Illumination angle α was defined to be 360 ◦ /n . Now that n = 1 we have 360 ◦ illumination angle.This basically describes a single lighthouse illuminating in all directions. Onfigure 4 we can see two points. The point on the left is the placement center,point on the right is the center of the lighthouse. Total dark area is 0, inother words, D (1) = 0. LIGHTHOUSE ILLUMINATION PROBLEM 5
Figure 6.
On figure 5 we see the case of n = 2. Two lighthousesare facing each other and evidently they can not illuminate behind eachother. The result is an infinite dark area, D (2) = ∞ .4.3. The first visually appealing case is n = 3. It is alsothe first non-zero finite value for the dark area. Our eyes could measure 0and ∞ but now we will have to do some calculations. We show the n = 3case on figure 6. Recall that in section 2 we talked about the fact that itis possible to focus on just a single lighthouse and the dark area behind it.That is exactly what we will be doing. On figure 7 we can see the targetlighthouse and a way to calculate the dark area behind it. Note that onlyhalf of the area behind it is shaded dark gray. Notice the symmetry, | EB | , | F B | and | P B | all intersect at the same point behind the target lighthouse. | EB | is the ray coming from top lighthouse, | F B | is the ray coming frombottom lighthouse. This enables us to split the dark area behind the targetlighthouse in half. The dark gray shaded area in this case is d (3) / (cid:52) ACB minus the area of the circular sectorformed by
A, C, G . Area ( ACB ) = x φ but we can overcome this byseeing that tan( φ ) = x/ x . This tells us φ = arctan( x ). Then, Area ( ACG ) = π φ ◦ = π arctan( x )2 π = arctan( x )2 d (3)2 = Area ( ACB ) − Area ( ACG ) = x − arctan( x )2 (6) ERHAN TEZCAN
Figure 7. D ( n ) = d ( n ) × n we get D (3) = d (3) × d (3)2 × × D (3) = x − arctan( x )2 × × x − arctan( x )) (7)Now all that is left to do is find x . In this case 120 ◦ and its complementary60 ◦ are beautiful angles, therefore could be useful for us. We already know | EP | = | P C | = 3 and | AC | = 1 because that is how the problem is defined.First we can find | EC | = 3 √ (cid:52) EAC trianglefind | EA | = √
26. Then, we draw the triangle (cid:52)
EDP and find | ED | to be3 √ /
2. We now have two similar triangles, notice that ∠ CAB = ∠ EDB =90 ◦ , ∠ ABC = ∠ EBD and ∠ ACB = ∠ DEB . Thanks to this similaritybetween (cid:52)
ABC and (cid:52)
EBD we can say | AB || DB | = | BC || EB | = | AC || ED | Looking at | AC | / | ED | = | BC | / | EB | we have1 √ = √ x √
26 + x Squaring both sides 427 = 1 + x x + 2 x √
26 + 264 x + 8 x √
26 + 104 = 27 + 27 x This gives the equation 23 x − x √ −
77 = 0. Using the quadratic formula: x , = 8 √ ± √ ×
26 + 4 × × x is the length of | AB | , since 8 √ < √ ×
26 + 4 × ×
77 we will choose + in place of ± . Then we have x = √ √ LIGHTHOUSE ILLUMINATION PROBLEM 7
Figure 8. x using GeoGebra, where we draw these figures, yields x ≈ . . D (3) = 3 (cid:32) √ √ − arctan (cid:32) √ √ (cid:33)(cid:33) (9)The result is D (3) ≈ . On figure 8 we can see the case for n = 4.Similar to n = 2 case we have D (4) = ∞ because the dark area behind eachlighthouse goes to infinity. It is important to note that this is because notwo rays can meet behind a lighthouse. Looking at a target lighthouse for n = 6 on figure 9 we see the same thing happening, D (6) = ∞ Even Number of Lighthouses.
For even lighthouses, we were notable to draw a tangent to define a finite dark area behind the target light-house. This causes the dark area to be infinite. We can show that this isalways true, using proof by contradiction.
Theorem 4.1. D ( n ) = ∞ , n ≡ Proof.
Imagine an even number of lighthouses placed around a center. Placethe first lighthouse directly n units to the right of the placement center (aswe always did in our figures). Let L be the first and target lighthouse,number the rest of the lighthouses as L , L , ..., L n − counter-clockwise. Ifwe number like this, the lighthouse on the opposite side of L is L n/ . Wehave to show that none of the lighthouses numbered L n/ − , L n/ − , ..., L Figure 9. L . Note how they are all num-bered in reference to L n/ . Now to use proof by contradiction, we assumethat it is possible to draw a finite dark area defining tangent from one ofthe L n/ − , L n/ − , ..., L giving us the figure 10. In this figure, E is thecenter of lighthouse L n/ − k , k ∈ N , P is the placement center and C is thecenter of target lighthouse L . By definition | EP | = | P C | , but if k = 1then ∠ EBP = ∠ P EB which requires | EP | = | P B | . | P B | (cid:54) = | P C | therefore k = 1 is not possible. For k > ∠ EBP > ∠ P EB . Now lookingat | EP | = n and | P B | = n + | CB | , if ∠ EBP > ∠ P EB this would require | EP | > | P B | , | EP | > | P B | n > n + | CB | > | CB | (10)Thus showing that this is not possible. Another case to consider is whenthere is such a tangent that ∠ P EB < α/
2, but again this causes ∠ EBP to be bigger and inevitably ∠ EBP > ∠ P EB , which we have shown to becontradictory in 10. (cid:3)
Now we can safely say that dark area is infinite for even number of light-houses.4.6.
Odd Number of Lighthouses.
Looking at 5 lighthouses case in fig-ure 11 it seems as if the furthest lighthouses define the dark area behind thetarget lighthouse. We will show that it is indeed the furthest lighthousesthat define the dark area behind a target lighthouse and then give a formulato find the dark area itself.
LIGHTHOUSE ILLUMINATION PROBLEM 9
Figure 10. D ( n ) = ∞ , n ≡ Figure 11.
Theorem 4.2.
The finite dark area d ( n ) behind a target lighthouse is definedby the tangents coming from two furthest lighthouses when n ≡ .Proof. Our proof is similar to the proof we had for theorem 4.1. Imagine anodd number of n lighthouses. Place the first lighthouse directly n units tothe right of the placement center. Let L be the first and target lighthouse.Draw a line that passes through the placement center and the center of thetarget lighthouse. This divides the plane in half, ( n − / n − / n is an odd number we cansay n = 2 m + 1. So in other words, we have m lighthouses on one side and m on the other. Starting from the right, number the lighthouses onthe upper half as L , L , ..., L m , counter-clockwise. Our claim is that L isilluminated by L m and L m only. We said “furthest lighthouses” which isplural, the other lighthouses is the lighthouse that is L m ’s reflection on theline we just drew. Like we demonstrated on n = 3 case the dark area behindthe lighthouse is divided in two by this line, calculating on one half suffice.We can use proof by contradiction, assuming that some other lighthouse L i , < i < m illuminates L . Drawing the figure 12 gives us an idea. In this Figure 12. L and the two furthest lighthouses.figure, E is the center of lighthouse L m − k , k ∈ N ∪ { } , P is the placementcenter, C is the center of the target lighthouse L . For positive values of k we see that αk > α/
2. This would require | EP | > | P B | but that isnot possible, exactly the same contradiction on (10). We now know that k = 0, so ∠ EP B = 180 ◦ − α/ L m , but whatis ∠ EP B ? Let us say that ∠ EBP = β and P EB = α/ − β such that ∠ EP B < ∠ P EB . This gives us β < α/ − β which results in β < α/ ∠ EP B approaches 180 ◦ because α gets smaller, and now we also see that β gets smaller too, suchthat β < α/
4. In conclusion, L m does illuminate L and it is the only onedoing so. (cid:3) Theorem 4.3. D ( n ) = n ( x n − arctan( x n )) , n ≡ , n > where x n = (cid:113) n cos (cid:0) π n (cid:1) − n sin( πn ) cos ( π n ) n sin (cid:0) πn (cid:1) − Proof.
This is going to be a pretty straightforward proof. To start, let usagain imagine the same figure we did for the previous proof 4.6. We have L to the right and L , L , ..., L m going counter-clockwise. We just showedthat we only need to care about L m so that is exactly what we are going todo. We get figure 13 as a result. We have some new variables in our figure,namely y, z and t . We will describe all three of them in terms of n and thenwork our way towards x . x in this figure is x n in equation (11). Like we LIGHTHOUSE ILLUMINATION PROBLEM 11
Figure 13. n Lighthouses, n ≡ L ,the lighthouse to the upper-left is L m have done for n = 3, if we find x we can find d ( n ), and consequently D ( n ).So, let us describe the uninvited guests one by one: • y can be found by looking at sin( α/
2) = y/n , which gives y = n × sin (cid:18) ◦ n (cid:19) = n × sin (cid:18) ◦ n (cid:19) (12) • t can be found using Cosine Theorem on | EC | seen by the angle ∠ EP C . The theorem gives us t = n + n − n cos(180 ◦ − α/ n + n − n cos(180 ◦ − α/
2) can be reduced. n + n − n cos(180 ◦ − α/
2) = 2 n (1 − cos(180 ◦ − α/ − cos( φ ) = 2 sin ( φ/ n (1 − cos(180 ◦ − α/ n (2 sin ((180 ◦ − α/ / n (cid:18) sin (cid:18) ◦ − α (cid:19)(cid:19) = 4 n (cid:32) sin (cid:32) ◦ − ◦ n (cid:33)(cid:33) = 4 n (cid:18) sin (cid:18) ◦ (cid:18) − n (cid:19)(cid:19)(cid:19) Finally we get t = (cid:115) n (cid:18) sin (cid:18) ◦ (cid:18) − n (cid:19)(cid:19)(cid:19) = 2 n × sin (cid:18) ◦ (cid:18) − n (cid:19)(cid:19) (13) • z can be written in terms of t using Pythagoras rule at (cid:52) EAC . Therule gives us t = 1 + z therefore z = t −
1. Since we alreadywrote t in terms of n we get z = 4 n (cid:18) sin (cid:18) ◦ (cid:18) − n (cid:19)(cid:19)(cid:19) − Now that we have y, z and t in terms of n we can work on x . We will be doingthe same thing we did for n = 3, thanks to the similarity (cid:52) EDB ∼ (cid:52)
CAB . | AB || DB | = | BC || EB | = | AC || ED | Looking at | AC | / | ED | = | BC | / | EB | we have1 y = √ x z + x Squaring both sides 1 y = 1 + x x + 2 xz + z This gives us x + 2 xz + z = y + x y which yields the equation x ( y −
1) + x ( − z ) + ( y − z ) = 0Plugging this into quadratic formula gives us x , = 2 z ± (cid:112) z − y − y − z )2 y − y < z and because of this the ± will have tobe + sign, otherwise x will be negative and | AB | can’t be negative. x = 2 z + (cid:112) z − y − y − z )2 y −
2= 2 z + (cid:112) z − y + 4 y z + 4 y − z y −
2= 2 z + (cid:112) y ( z − y + 1)2 y −
2= 2 z + 2 y (cid:112) z − y + 12 y − z + y (cid:112) z − y + 1 y − y and z are using radians insteadof degrees. y = n sin (cid:16) πn (cid:17) z = 4 n sin (cid:16) π − π n (cid:17) − n cos (cid:16) π n (cid:17) − (cid:112) z − y + 1 is (cid:112) z − y + 1 = (cid:114) n sin (cid:16) π − π n (cid:17) − − n sin (cid:16) πn (cid:17) + 1= (cid:113) n (cid:0) ( π/ − π/ n ) − sin ( π/n ) (cid:1) = (cid:113) n (cid:0) ( π/ n ) − sin ( π/n ) (cid:1) LIGHTHOUSE ILLUMINATION PROBLEM 13
Remember that sin 2 φ = 2 sin φ cos φ . If φ = π/ n we get sin( π/n ) =2 sin( π/ n ) cos( π/ n ) therefore sin ( π/n ) = 4 sin ( π/ n ) cos ( π/ n ). Wefurther reduce: (cid:112) z − y + 1 = (cid:113) n (4 cos ( π/ n ) − ( π/ n ) cos ( π/ n ))= (cid:113) n cos ( π/ n )(1 − sin ( π/ n ))Also remembering sin φ + cos φ = 1 we can further reduce (cid:112) z − y + 1 = (cid:112) n cos ( π/ n )(cos ( π/ n ))= 2 n cos ( π/ n ) y (cid:112) z − y + 1 is then 2 n sin( π/n ) cos ( π/ n ). Finally we have the formulafor x : x = (cid:113) n cos (cid:0) π n (cid:1) − n sin( πn ) cos ( π n ) n sin (cid:0) πn (cid:1) − (cid:3) Giving n = 3 in equation (15) and using calculator yields x = √
416 + √ √ √ n = 5 case, measuring x in GeoGebra yields x ≈ . n = 5 to equation (15) returns x ≈ . d ( n )we can do what we did in (6) and by looking at figure 13 we can say that d ( n ) = x − arctan x . Taking x = 4 . d (5) yields d (5) = 4 . − arctan 4 . ≈ . D (5) = d (5) × ≈ . Point Light Sources at the Arc
The second variation of the problem is when the points on the arc seenby the illumination angle α act as point light sources. This case is a bitmore complex than the “Point Light Source at the Center” variation, whichwe have solved. We will be giving examples for 1, 2, 3, 4 and 5 lighthouses,then try to generalize it, which is where the problem passionately slaps usfor trying to do so.5.1. This is the same case as it was for point light sourceat the center, because the illumination angle is 360 ◦ . To see how this lookslike, refer to figure 4. For the record, D (1) = 0. Figure 14.
Figure 15.
This case is shown on figure 14. Again, the dark areais infinite, but in a slightly different way. It is literally “barely” infinite.The point light sources 180 ◦ apart from each other in both arcs make aparallel light ray tangent to the lighthouses facing each other, resulting ina rectangular-like dark area that extends to infinity. We are safe to say D (2) = ∞ .5.3. Figure 15 shows us the case for n = 3. Like we alwaysdid so far we will be focusing on a single lighthouse. On figure 16 we cansee a way to approach x . We should immediately warn that though it looksas if ∠ ACB = 60 ◦ that is not correct, it is just that the angle is quite closeto 60 ◦ . To find x we draw the points D and G , thereby finding | DG | = 1and | GP | = 1 /
2. We already know | P C | = 3. Since the 30 ◦ angle ∠ DEP isseeing | DP | = 3 / ◦ angle ∠ EP D sees | ED | = 3 √ /
2. Also because | ED | = | F G | we have | F G | = 3 √ /
2. We the notice the similarity (cid:52)
F GB ∼ LIGHTHOUSE ILLUMINATION PROBLEM 15
Figure 16. (cid:52)
CAB which tells us that there is a ratio | AC | / | F G | = | AB | / | GB | . Thisgives us the equation below. 1 √ = x + √ x The author could not reduce this to a quadratic equation and sought helpfrom Wolfram | Alpha [4], which gave x = 3(4 √ √ / ≈ . | AB | which gives approximately2 . x = 2 . D (3) = 3(2 . − arctan(2 . ≈ . Unlike the first variation, we actually have a finitedark area in this case. Figure 17 shows the case for n = 4. Now, the closestneighbors of a target lighthouse can illuminate behind it. In figure 18 wewill try to find x . We find | F G | = 1 / √ (cid:52) EF G .Thanks to the right-angled triangle (cid:52)
GDC where | GD | = | DC | we can usePythagoras rule to find | GC | = 4 √ −
1. Once again using Pythagoras rulewe find | GA | = (cid:112) − √
2. A final Pythagoras rule | GD | + | DB | = | GB | gives us (cid:18) − √ (cid:19) + (cid:18) − √ (cid:112) x (cid:19) = (cid:18)(cid:113) − √ x (cid:19) After unwrapping and reducing both sides we get1 + (cid:18) − √ (cid:19) (cid:112) x = x (cid:113) − √ | Alpha[5] to find x which returns x ≈ . Figure 17. | AB | also yields approximately 1 . x = 1 . D (4). D (4) = 4(1 . − arctan(1 . ≈ . Looking at n = 5 on figure 19, again it appears thatthe closest neighbors define the dark area behind a target lighthouse. Look-ing closely to the target lighthouse as shown in figure 20 Immediately weare blessed with | EP | = | EG | = 5, therefore | DG | = 4. We want to use thesimilarity (cid:52) DBF ∼ (cid:52)
ABC like we did on previous cases, so we have to find √ b − √ − a before we actually come to x . To find a we can usethe Cosine Theorem as | F G | = | DF | + | DG | − | DF || DG | cos(18 ◦ ) andleave a out to find its value. Remembering that cos(18 ◦ ) = (cid:112)
10 + 2 √ / a = 16 + 16 − a − (cid:112) − a ) (cid:112)
10 + 2 √ − a ) = 2( (cid:112) − a ) (cid:113)
10 + 2 √ − a √ − a = (cid:113)
10 + 2 √ LIGHTHOUSE ILLUMINATION PROBLEM 17
Figure 18. (cid:112) − a = (cid:113)
10 + 2 √ − a = 10 + 2 √ a = 6 − √ a = (cid:113) − √ | GC | . | GC | = 5 − | P G | and | P G | can be found by CosineTheorem as | P G | = 5 + 5 − )(5 ) cos(36 ◦ ). This is reduced to | P G | =50(1 − cos(36 ◦ )) = 50(1 − (1 + √ / | P G | = 504 (3 − √ | P G | = 52 (cid:113) − √ | GC | = 5 − (cid:113) − √ | GC | and a we can find | F C | = a + | GC | . | F C | = 5 − (cid:113) − √ (cid:113) − √ − (cid:113) √ Figure 19. b by doing Pythagoras rule as b = | F C | + | DF | . b = (5 − (cid:113) − √ + 16 − a b = 25 + 94 (6 − √ − (cid:113) − √ − √ b = 41 + 54 (6 − √ − (cid:113) − √ | DA | = √ b − (cid:112) b − (cid:114)
40 + 54 (6 − √ − (cid:113) − √ x . We will do it by using the (cid:52) DBF ∼(cid:52)
ABC . The similarity gives us the equation | AC | / | DF | = | CB | / | DB | .1 √ − a = √ x x + √ b − − a = 1 + x x + b − x √ b − LIGHTHOUSE ILLUMINATION PROBLEM 19
Figure 20. − √ x x + b − x √ b − − √ x x + b − x (cid:113)
40 + (6 − √ − (cid:112) − √ | Alpha [6] to calcu-late x ≈ . x using GeoGebra also gives x ≈ . D (5)is then given as D (5) = 5(1 . − arctan(1 . ≈ . A general rule for any number of lighthouses?
So far, we havefailed to reduce the equation obtained from the similarity of the triangles,we sought help from Wolfram | Alpha and compared the result to the mea-surement using GeoGebra. Nevertheless, we could try to find a general formfor x , basing a claim that the dark area is defined by the closest lighthousesof the target lighthouse, like we did for the previous variation where thedark area was defined by the furthest lighthouses when n was odd. Butwe realize this is not the case. Upon generalization, we might think that maybe the illumination angle α becomes so small that the closest neighboris no more able to illuminate behind the target lighthouse, instead, someother lighthouse pair does the job. To approach this case, consider a targetlighthouse L and a source lighthouse L s . At most, the light ray comingout of the point on the arc of L s would be tangent to L s . More than thatwould mean that the light ray actually passes through L s to be tangent to L . This is exactly the case for n = 20. At n = 19 the angle is slightly morethan 90 ◦ but at n = 20 the angle is less than 90 ◦ , so perhaps some otherlighthouse is illuminating the target lighthouse. For n = 20, the 3 rd closestlighthouse is the one illuminating the target lighthouse, as shown in figures21 and 22. So even if we had a formula to calculate x based on the cases Figure 21.
20 Lighthouses, 9 ofthem are drawn.
Figure 22.
20 Lighthouses,zoomed on target lighthouse.Red striped line is coming fromthe closest lighthouse, bluestriped line is coming from thesecond closest lighthouse, thegray striped line, which actuallydefines the dark area, is comingfrom the third closest lighthouse. n = 3 , , n = 19. The general figure is givenby 23. Knowing that α = 360 ◦ /n , describe x and k in terms of n , that is thetask! k is the number of the lighthouse, k = 1 means the closest lighthouse, k = 2 means the second closest lighthouse, k = 3 the third closest and soon... For example, k = 1 for values n = 1 , , ...,
19 and then k = 3 for n = 20.After we actually find a way to deal with k we will find x (which the authorhad trouble even without k ). The case of n = 20 was found empirically, andwe do not have a formula that tells us which lighthouse will illuminate thetarget lighthouse for a given number of lighthouses. LIGHTHOUSE ILLUMINATION PROBLEM 21
Figure 23.
The geometry between a target lighthouse andthe source lighthouse that illuminates behind it6.
Conclusion & Closing Remarks
In this section we will recapitulate our findings on both variations of theproblem.If the lighting is done using a single point light source at the center ofeach lighthouse, we can define a piece-wise formula for the dark area.
Definition 6.1 (Total Dark area when there is a point light source at thecenter) . D ( n ) = n = 1 ∞ n ≡ n ( x n − arctan( x n )) n ≡ , n > x n = (cid:113) n cos (cid:0) π n (cid:1) − n sin( πn ) cos ( π n ) n sin (cid:0) πn (cid:1) − n →∞ n ( x n − arctan( x n ))? Furthermore, as n goes to infinity we have shown that the distance between two lighthousesis 2 π , on the other hand α = 2 π/n so the illumination angle would beapproaching 0. Reconsidering figure 13 we would say that y = π but thenwe have the angle ∠ CP E approaching 180 ◦ , so does the lighthouse on theleft illuminate the target lighthouse on the right at all? We believe that itdoes illuminate the target lighthouse but the tangent light rays are almostparallel to each other so the dark area approaches infinity, but we do nothave a proof for this yet. Furthermore, there could be a better way tocalculate x n instead of brute-forcing our way to x with Cosine theorems andPythagoras rules.As for the point light source at arc case, there is another problem regard-ing which lighthouse is illuminating the target lighthouse. We have to finda rule regarding which lighthouse illuminates the target lighthouse for given n , which enables us to write k in terms of n , then we could find x in termsof n .The results obtained in this paper can be seen on table 1. The questionremains: What is the total dark area for any n for both variations of theproblem, and what is it as n approaches infinity? Table 1.
Results obtained in this paper for the total darkarea for a given number of lighthouses.Number oflighthouses Point Light Sourceat the Center Point Light Sourcesat the Arc1 0 02 ∞ ∞ . . ∞ . . . n Equation (23) for odd n ∞ for even n ? ∞ Possibly ∞ ? References [1] Gull S.F. (1988) Bayesian Inductive Inference and Maximum Entropy. In: EricksonG.J., Smith C.R., eds.
Maximum-Entropy and Bayesian Methods in Science and En-gineering. Fundamental Theories of Physics (An International Book Series on TheFundamental Theories of Physics: Their Clarification, Development and Application) ,vol 31-32. Springer, Dordrecht, pp. 53–74[2] Hohenwarter, M., Borcherds, M., Ancsin, G., Bencze, B., Blossier, M., ´Eli´as, J., Frank,K., G´al, L., Hofst¨atter, A., Jordan, F., et al. (2019),
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Steiner Chain
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