Eight points are placed on the surface of a sphere with a radius of 1. The shortest distance between any two points is greater than 1.2. How can the points be arranged?
Hint: They are not arranged as a cube. The cube would have an edge length of only 2/sqrt(3) = 1.1547.
(In reply to
re: Solution by brianjn)
The below program has an error: It uses pi/190 rather than pi/180 to convert degrees to radians. See correction in my later comment.
For ease of discussion treat the earth as having radius 1 unit, so we can talk about latitude and longitude.
Taking brianjn's idea of one square twisted with regard to the other, place the vertices of one square at the same north latitude, and those of the other square at the latitude opposite in sign but equal in absolute value (i.e., in the southern hemisphere, as far south as the first one was north). That way they'll have the same side lengths. Make the twist 45 degrees so that no sides of the triangle are larger than any other (at least extending from one square to the other). But we'll also want to make the sides of the squares equal to the sides of the triangles, and so the triangles will also be equilateral, not just isosceles.
We use spherical trigonometry, and as the latitude is the complement of the side of the great circle arc from the north pole, when using latitude, sines replace cosines and vice versa.
cos(arcdist) = sin(lat1)*sin(lat2) + cos(lat1)*cos(lat2)*cos(lon2 - lon1)
We maximize the minimum distance by finding equality, and just tabulate, with this program:
DECLARE FUNCTION acos# (x#)
DEFDBL A-Z
DIM SHARED dr
pi = ATN(1) * 4
dr = pi / 190 'deg to radians
CLS
FOR lat = 32.9729# TO 32.97295# STEP .000001#
lonDiff = 90
sinLat = SIN(lat * dr): cosLat = COS(lat * dr)
distOnSq = acos(sinLat * sinLat) ' cos(lonDiff)=0
distOnSq = 2 * SIN(distOnSq * dr / 2)' convert arc degrees to chord length
lonDiff = 45
sinLat2 = -sinLat
distOnTri = acos(sinLat * sinLat2 + cosLat * cosLat * COS(lonDiff * dr))
distOnTri = 2 * SIN(distOnTri * dr / 2)' convert arc degrees to chord length
PRINT USING "##.####### ##.####### ##.#######"; lat; distOnSq; distOnTri
NEXT lat
FUNCTION acos (x) ' works in degrees
a = ATN(SQR(1 - x * x) / x) / dr
IF a < 0 THEN a = a + 180
acos = a
END FUNCTION
where the 32.9729# TO 32.97295# limits were the result of further and further refining the results. It started out as 2 to 88 step 2:
lat side side
of square of triangle
2 1.4134404 0.7296212
4 1.4111216 0.7373706
6 1.4072597 0.7500714
8 1.4018591 0.7674242
10 1.3949255 0.7890491
12 1.3864666 0.8145156
14 1.3764917 0.8433700
16 1.3650115 0.8751578
18 1.3520388 0.9094401
20 1.3375876 0.9458044
22 1.3216737 0.9838700
24 1.3043147 1.0232901
26 1.2855294 1.0637516
28 1.2653384 1.1049729
30 1.2437637 1.1467018
32 1.2208290 1.1887120___
34 1.1965594 1.2308004
36 1.1709813 1.2727844
38 1.1441228 1.3144991
40 1.1160132 1.3557954
42 1.0866833 1.3965377
44 1.0561651 1.3167895
46 1.0244919 1.3593756
48 0.9916986 1.4011926
50 0.9578208 1.4421414
52 0.9228956 1.4821304
54 0.8869613 1.5210746
56 0.8500571 1.5588946
58 0.8122234 1.5955167
60 0.7735015 1.6308720
62 0.7339338 1.6648962
64 0.6935636 1.6975295
66 0.6524349 1.7287160
68 0.6105929 1.7584034
70 0.5680831 1.7865435
72 0.5249522 1.8130914
74 0.4812472 1.8380053
76 0.4370160 1.8612472
78 0.3923070 1.8827818
80 0.3471689 1.9025771
82 0.3016512 1.9206043
84 0.2558037 1.9368373
86 0.2096765 1.9512533
88 0.1633199 1.9638320
(the latitudes are to be considered north and south in each case.)
the underscore line to the right above shows where the result should lie.
Further refining
32.0 1.2208290 1.1887120
32.1 1.2196470 1.1908162
32.2 1.2184617 1.1929205
32.3 1.2172730 1.1950250
32.4 1.2160811 1.1971296
32.5 1.2148858 1.1992343
32.6 1.2136871 1.2013390
32.7 1.2124852 1.2034438
32.8 1.2112799 1.2055487
32.9 1.2100713 1.2076535_____
33.0 1.2088594 1.2097583
33.1 1.2076443 1.2118631
33.2 1.2064258 1.2139678
33.3 1.2052040 1.2160724
33.4 1.2039789 1.2181769
33.5 1.2027505 1.2202813
33.6 1.2015189 1.2223855
33.7 1.2002839 1.2244896
33.8 1.1990457 1.2265934
33.9 1.1978042 1.2286971
32.9000 1.2100713 1.2076535
32.9025 1.2100411 1.2077061
32.9050 1.2100108 1.2077587
32.9075 1.2099806 1.2078114
32.9100 1.2099503 1.2078640
32.9125 1.2099200 1.2079166
32.9150 1.2098898 1.2079692
32.9175 1.2098595 1.2080219
32.9200 1.2098292 1.2080745
32.9225 1.2097989 1.2081271
32.9250 1.2097687 1.2081797
32.9275 1.2097384 1.2082323
32.9300 1.2097081 1.2082850
32.9325 1.2096778 1.2083376
32.9350 1.2096475 1.2083902
32.9375 1.2096173 1.2084428
32.9400 1.2095870 1.2084954
32.9425 1.2095567 1.2085481
32.9450 1.2095264 1.2086007
32.9475 1.2094961 1.2086533
32.9500 1.2094658 1.2087059
32.9525 1.2094355 1.2087585
32.9550 1.2094052 1.2088112
32.9575 1.2093749 1.2088638
32.9600 1.2093446 1.2089164
32.9625 1.2093143 1.2089690
32.9650 1.2092840 1.2090216
32.9675 1.2092537 1.2090743
32.9700 1.2092234 1.2091269
32.9725 1.2091930 1.2091795 _______
32.9750 1.2091627 1.2092321
32.9775 1.2091324 1.2092847
32.9800 1.2091021 1.2093374
32.9825 1.2090718 1.2093900
32.9850 1.2090414 1.2094426
32.9875 1.2090111 1.2094952
32.9900 1.2089808 1.2095478
32.9925 1.2089505 1.2096005
32.9950 1.2089201 1.2096531
32.9975 1.2088898 1.2097057
33.0000 1.2088594 1.2097583
32.97270 1.2091906 1.2091837
32.97275 1.2091900 1.2091848
32.97280 1.2091894 1.2091858
32.97285 1.2091888 1.2091869
32.97290 1.2091882 1.2091879_____
32.97295 1.2091876 1.2091890
32.97300 1.2091870 1.2091900
The final refinement:
32.972906 1.2091881 1.2091880
32.972907 1.2091881 1.2091881
32.972908 1.2091881 1.2091881
32.972909 1.2091881 1.2091881
32.972910 1.2091881 1.2091881
32.972911 1.2091881 1.2091882
The points are at latitudes 32.9729085 north and south (where the squares vertices touch the surface of the globe). The two squares are twisted 45 degrees relative to each other in longitude, and the chord distances are 1.2091881 units (in our case earth radii).
Edited on February 3, 2008, 12:55 pm
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Posted by Charlie
on 2008-02-02 01:53:25 |
re(2): Solution
