Holy Halved Heaquarters Riddler

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Matilda Blankenship
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1 Holy Halve Heaquarters Riler Anonymous Philosopher June 206 Laser Larry threatens to imminently zap Riler Heaquarters (which is of regular pentagonal shape with no courtyar or other funny business) with a high-powere, vertical planar ray that will slice the builing exactly in half by area, as seen from above. The builing is quickly evacuate, but not before in-house mathematicians move the most sensitive riling equipment out of the places in the builing with an extra high risk of getting zappe. Where are those places, an how much riskier are they than the safest spots? (It is sufficient to escribe those places qualitatively. For extra creit: get quantitative. Seen from above, how many high-risk points are there (where even a point-size object woul be at higher-than-necessary risk)? If infinitely many, what is their total area? What if the shape is an n-sie polygon?) Solution (Thanks to Daniel Tello for valuable (an prouctive) bluner-hunting!) Keep away from the center, because area bisectors intersect near that point, raising the os that central regions will be zappe. The perimeter is safest (an if Larry ranomizes uniformly over perimeter points rather than angle, the miles of the sies are the safest of the safe). In an even-sie regular polygon, there is not much more to say than that the center (point) itself, where all area bisectors intersect, is the most angerous place an anger ecreases with istance from it for all but point-size objects (which are at ae risk only at the center itself). But for an o-sie polygon, while it remains true that the most angerous places (even for point-size objects) are where bisectors intersect, it is not true that all bisectors intersect at a single point. Among the area bisectors of a regular pentagon are the five that join vertices to mipoints; call these meians. Choose some vertex an neighboring mipoint, an consier the area bisectors that pass between them. Each passes also through a non-neighboring sie, between another vertex an mipoint, an together they comprise a fifth of all bisectors. The area swept out by these line segments clearly inclues the two triangles forme by the intersections of the two meians an two sies, an every point not on a meian is on a bisector running through its home triangle in this way its home bisector, we ll say (an

Figure : An area bisector. we ll let a meian be the home bisector of the points on it). But the area swept out inclues a bit of terrain on the outsie of those triangles.

$Five of these together prouce a slener, concave-curvy five-pointe star that occupies a fraction of just 0.00536 of the pentagon s area. The curves also generate an inner curvy pentagon insie the star.$

2 Figure : An area bisector. we ll let a meian be the home bisector of the points on it). But the area swept out inclues a bit of terrain on the outsie of those triangles. Figure 2: The curve trace by the mipoints of area bisectors. The extra terrain forms a curve crotch between the two meians forming those triangles. It turns out that these curves are segments of hyperbolae, an the ens of the segments are the mipoints of the meians; inee the curve is compose of bisector mipoints. Five of these together prouce a slener, concave-curvy five-pointe star that occupies a fraction of just of the pentagon s area. The curves also generate an inner curvy pentagon insie the star. How o the crotches affect the number of bisectors a point is on? A point insie a crotch is on (in aition to its home bisector) two of the bisectors that form that crotch, namely the two tangents to the hyperbola from that point. A point on the crotch curve itself is on only one, the tangent at that point. An a point on a meian borering a crotch is on (aitionally to that meian) one other crotch-forming bisector, nameley the other tangent, besies that meian, containing that point. This allows us to tally the number of bisectors through 2

On two: the perimeter of the star (home bisector plus another ue to being on a crotch curve).

3 Figure 3: The five hyperbolae containing the curves. Figure 4: The inner star. every point in the star. On one bisector: points outsie the star, plus the cusps of the star. On two: the perimeter of the star (home bisector plus another ue to being on a crotch curve). On three: the interior of the part of the star outsie of the inner pentagon (home plus two for being insie a crotch; points along a meian are not insie a crotch but borer two), an the vertices of the inner pentagon (home 3

4 plus two for being on curves). On four: the sies of the inner pentagon (home plus one for being on a curve plus two for being insie a crotch; the mipoints are not insie a crotch but borer two). On five: the interior of the inner pentagon (the center is on all five meians; points on meians are on meians, on bounaries of two crotches, an insie another; an each other point is on its home bisector an four others thanks to being insie two crotches). 2 Extra Creit: Fin the Star s Area The extra creit problem is to quantify the most angerous points, which means fining the area of the star, the region of points on multiple area bisectors. Take a polygon of any o number (an even-sie polygon has only one point on multiple area bisectors) n 3 of sies. Our figures will feature a heptagon an a pentagon, but the reasoning will be inepenent of the number of sies. Take any two sies of this polygon between which there are area bisectors. These bisectors teeter along the crotch curve between the meians to the two sies. Exten the sies to where the lines they are on intersect, forming a triangle with each of these bisectors. The bisectors, preserving as they o the area on each sie of the polygon, preserve also the area of this triangle. This is a efining feature of hyperbolae ( line segments that yiel the same triangular area in that way are tangent, at their mipoints, to a single hyperbola with the extene sies as asymptotes. So we know the shape of our curvy star: its sies are segments of hyperbolae whose asymptotes exten the polygon s opposing sies. There are n such hyperbolae, each tangent to two meians, the points of tangency forming the cusps of the star. The curves are the locus of the bisector mipoints, an the cusps occur exactly halfway along the meians, which is very close to the center of the polygon even for n = 5 (.0955 of the istance to a vertex). I have erive the star s area (for any o n 3) in two ways, à la Eucli (without specifying axes or coorinates) an à la Descartes. My preferre version is the Eucliean, because at least in my hans it has prouce a tiier expression for the area (an it also makes it easy to show that the result applies also to affine transformations of regular polygons; open question: oes it apply to an even larger class of polygons?). I i a secon erivation to check the results, an I inclue the Cartesian version here because the math will be more familiar to many reaers. 2. Eucliean Approach Consulting the figure, an assuming an out-raius (istance from the center P of our polygon ABCDE to its vertices) of : EOC = π n The apothem = P F = cos π n 4

Figure 5: A polygon with opposing sies extene. a = OE = + sin π n b = OF = + tan π n = cot π The half-sie c = F E = sin π n Let the vectors v E an v c be (E O) an (C O), respectively. Note that = b a.

5 Figure 5: A polygon with opposing sies extene. a = OE = + sin π n b = OF = + tan π n = cot π The half-sie c = F E = sin π n Let the vectors v E an v c be (E O) an (C O), respectively. Note that = b a. EH bisects the polygon an the area of EOH is 2 ab sin π n. Let GJ be an area bisector of the polygon with enpoints on EF an CH. Then, by constancy of area: OG OJ = ab We will parameterize the area bisectors that have enpoints on EF an CH as a function of t for t [, ]: G t = O + tv E J t = O + t v C 5

6 The mipoint of GJ is: M t = G t + J t 2 = O + t 2 v E + 2t v C Because H = O + b a v C, the center P of the polygon is: P = + E + + H = O + + v E + Their ifference is the vector v M from P to M t : ( t v M = 2 ) ( v E + + 2t + + v C ) v C We will integrate the area of the section of of the central star consisting of triangles with area 2 v M t v M t (the magnitue of the cross-prouct of two vectors is the area of the parallelogram of which they form two sies). Figure 6: Visualizing the integral. t v M = 2 v E 2t 2 v C 2 v M t v M = ( ( t 2 EOC 2t 2 2 ) + ( )) + 2 2t ( + ) = 4 a2 sin π ( ) n t ( + )t 2 ( + ) 6

7 The area of the star is times the integral of this for [t, t ], where t is the t such that v M = k(p A) for some k (i.e., M t is on the meian containing AP ). To fin t we first fin an expression for A: A = O + a 2c v E = O + (2 )v E a (That last step follows from trigonometric ientities.) P A = 22 + v E + + v C M t is on the meian containing AP when P A is a scalar multiple of v M : ( t + 2 ) ( ) ( 2 2 = + + 2t ) + This simplifies algebraically to: The quaratic formula gives: t 2 + 2( 2)t + (2 2 ) = 0 t = 2 ± 2( ) Only the plus solution falls between an, an so: The area of the star is: 2 na2 sin π t n t= = 2 na2 sin π ( t (ln n = ( n sin π n cos π ) n π 2 t = (2 2) + 2 ( t ) + cot2 π 2 ) ( + )t 2 t ( + ) ) ( + )t ( + ) t + ( ( ) t ln (t + )(t ) ) ( + )t Since n sin π n cos π n is the area of the polygon, the ratio of the star s area to that of the polygon is: cot 2 ( ( ) t ln (t + )(t ) ) ( + )t [ ( = cot2 π ln 2 ) cos π n (4 3 2) cos 2 π n + (4 2 6) cos π n + 2 ] 2 (2 2) cos 2 π n + cos π n + 2 Here are some values of this ratio: 7

n Ratio 3 0.0986038542 5 0.0053624226 7 0.0003544589389 9 0.00023687743 0.000054373298 3 0.000027377553 5 0.00005365375 7 0.0000092379205 9 0.

8 n Ratio In the case of n = 3 (where the inner shape is not so much a star as a curvy triangle) this reuces to a tiy: 2.2 Cartesian Approach 3 4 ln(2) 2 Figure 7: Heptagon. Let the origin be the center of the hyperbola whose asymptotes exten two opposing sies the polygon, for o n 3 (for n = 3, the origin is a vertex of 8

9 the triangle), with out-raius (center to vertex) r =. The asymptotes meet at a π/n angle to each other. We are going to fin equations for two meians an a hyperbola that boun a region (too small to be seen in the figure above, so there is a close-up below) that forms of our star. Because CAO is a right triangle an COA is π (where is the apothem length cos π n ): ( ) C = 0, sin ( ) π AB is at a slope of m = tan ( π ), an so: ( ( π ( π A = cos, y C sin ) )) ( ( π ( π B = cos, y C + sin ) )) CD is at a slope of m = tan ( 3 ) π, an so: ( ( ) ( )) 3π 3π D = cos, y C sin Figure 8: Close-up of /() of the inner star. The cusp of our region is at the mipoint M of AB: ( xa + x B M =, y ) A + y B 2 2 H has asymptotes with slopes ± tan( ) π an passes through M, which etermines an equation for H by settling the values of a an b in the stanar format for an upright hyperbola (where the slope of the asymptotes is always a b ): H : y2 a 2 x2 b 2 = a b = tan ( ) π 9

10 x 2 M a = ym 2 tan ( ) 2 π ( π b = a tan ) We will ivie R into the part where it is boune above an below by DE an AB, an the rest, where it is boune by H an AB. Where I is the intersection of DE an H, we want to ivie at the line x = x I, which we will fin by substituting m x + c (where c = y C ) for y in the equation for H. We obtain a quaratic equation an so use the quaratic formula: x 2 I(b 2 m 2 a 2 ) + x I (2cm b 2 ) + (c 2 b 2 a 2 b 2 ) = 0 x I = 2cm b 2 + (2cm b 2 ) 2 4(b 2 m 2 a 2 )(c 2 b 2 a 2 b 2 ) 2(b 2 m 2 a 2 ) First we fin the area of the small triangle boune by AB, DE an x = X I : Then the remaining area: = [ xm A 2 = ( 2 a x x=x I A = 2 x I(m x I mx I ) xm The area of the polygon is: x=x I a + x2 b 2 (mx + y c)x ( + x2 b 2 + b x ) ) ( m ) ] sinh b 2 x2 + y c x n A = 4 tan π n an so the ratio of the area of the star to that of the polygon is: 3 Hypocycloi? R n = (A + A 2 ) A A natural hunch for Spirograph aficionai is that the star s perimeter is a selfintersecting hypocycloi, a curve trace by a point on a smaller circle rolling within a larger circle. There is a single five-point self-intersecting hypocycloi, which can be forme ientically (an interesting fact) with circles whose raii are in a ratio of 5:2 or 5:3. It looks promising, oesn t it? The hypocycloi s raius (center to cusp) is that of the larger circle. If our star is a 5-2 hypocycloi, then it is forme from circles of raii.0955 an 2/5 times that, or Then the istance from the pentagon s center to the 0

Figure 9: 5-2 hypocycloi. nearest points on the curves, which are where the bisectors parallel to sies are tangent to them, is the larger raius minus twice the smaller, or.09.

0404, which of course is not an integral ivisor of.0955 an so woul not make a five-pointe star.

11 Figure 9: 5-2 hypocycloi. nearest points on the curves, which are where the bisectors parallel to sies are tangent to them, is the larger raius minus twice the smaller, or.09. But in fact that istance is.047, so the hunch is incorrect; our star is significantly more svelte. To reach.047 from the center, the inner circle woul have to have a raius of ( )/2 =.0404, which of course is not an integral ivisor of.0955 an so woul not make a five-pointe star. A natural revise hunch is that our star s shape can be forme with an elliptical inner wheel of semi-major axis.0382 an circumference 2/5 of the outer circle. However, espite being close there is no cigar to be ha; the resulting shape is clearly not hyperbolic. Figure 0: A 5-2 elliptical hypocycloi. There is, however, an inner wheel that woul prouce our hyperbolic star, an so the star is hypocyclic in a broa sense. Here s a thought-experiment to show that. Start with an inner circle just large enough to reach from the outer circle to the points on the hyperbolae nearest the center. It s too big, so we nee to shave a bit off the sies. To see where we nee to cut we measure from each point on the outer circle to the nearest point on the hyperbola (on a line perpenicular to its tangent). That s how far the rawing point of the inner wheel nees to be from this point on the outer circle when the inner wheel is

tangent to the outer at that point, because the tangent to a hypocyclic curve is perpenicular to the line from the rawing point to the point of tangency between the wheels.

12 tangent to the outer at that point, because the tangent to a hypocyclic curve is perpenicular to the line from the rawing point to the point of tangency between the wheels. Relying on these measurements from several points, we can construct an inner wheel as in the figure below. It will raw a pretty close approximation, compose of circular arcs, of the hyperbola segment. As we increase the number of points, we have our esire shape at the limit. Figure : A custom hyperbolic hypocycloi. 2

CONSTRUCTION AND ANALYSIS OF INVERSIONS IN S 2 AND H 2. Arunima Ray. Final Paper, MATH 399. Spring 2008 ABSTRACT

CONSTRUCTION AND ANALYSIS OF INVERSIONS IN S 2 AND H 2. Arunima Ray. Final Paper, MATH 399. Spring 2008 ABSTRACT CONSTUCTION AN ANALYSIS OF INVESIONS IN S AN H Arunima ay Final Paper, MATH 399 Spring 008 ASTACT The construction use to otain inversions in two-imensional Eucliean space was moifie an applie to otain