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Newsgroups: rec.puzzles,news.answers,rec.answers Path: senator-bedfellow.mit.edu!bloom-beacon.mit.edu!gatech!europa.eng.gtefsd.com!uunet!questrel!chris From: chris@questrel.com (Chris Cole) Subject: rec.puzzles Archive (geometry), part 14 of 35 Message-ID: <puzzles/archive/geometry/part2_745653851@questrel.com> Followup-To: rec.puzzles Summary: This is part of an archive of questions and answers that may be of interest to puzzle enthusiasts. Part 1 contains the index to the archive. Read the rec.puzzles FAQ for more information. Sender: chris@questrel.com (Chris Cole) Reply-To: archive-comment@questrel.com Organization: Questrel, Inc. References: <puzzles/archive/Instructions_745653851@questrel.com> Date: Wed, 18 Aug 1993 06:05:22 GMT Approved: news-answers-request@MIT.Edu Expires: Thu, 1 Sep 1994 06:04:11 GMT Lines: 280 Xref: senator-bedfellow.mit.edu rec.puzzles:25003 news.answers:11523 rec.answers:1923 Archive-name: puzzles/archive/geometry/part2 Last-modified: 17 Aug 1993 Version: 4 ==> geometry/tiling/rectangles.with.squares.p <== Given two sorts of squares, (axa) and (bxb), what rectangles can be tiled? ==> geometry/tiling/rectangles.with.squares.s <== A rectangle can be tiled with (axa) and (bxb) squares, iff (i) gcd(a,b)=1 , and any of the following hold: either: both sides of the rectangle are multiples of a; or: both sides of the rectangle are multiples of b; or: one side is a multiple of (ab), and the other is any length EXCEPT one of a finite number of "bad" lengths: those numbers which are NOT positive integer combinations of a & b. { By Sylvester's theorem there are (a-1)(b-1)/2 of these, the largest being (a-1)(b-1)-1. } (ii) gcd(a,b) = d . Then merely apply (i) to the problem with a,b replaced by a/d, b/d and the rectangle lengths also divided by d. i.e. all cells must appear in (dxd) subsquares. ------ PROOF It is clear that (ii) follows from (i), and that simple constructions give the "if" part of (i). For the "only if" part, we prove that... (S) If one side of the rectangle is not divisible by a, and the other is not divisible by b, then the tiling is impossible. The results in (i) follow immediately from (S). To prove (S): ( Chakraborty-Hoey style ). ~~~~~~~~~~~~~~~~ Let the width of the rectangle be a NON-(a-multiple). Then the number of bxb squares starting (i.e. top edge) at row 1 must be a NON-a-multiple. Thus the number of bxb starting at row 2 must BE an a-multiple. Similarly for the number starting at rows 3,4,...,b . Then the number starting at row (b+1) must be a NON-a-multiple again. Similarly the number starting at rows (2b+1), (3b+1), (4b+1),... must all be non-a-multiples. So if the number of rows is NOT a multiple of b, (call it bx+r), then row (bx+1) must have a NON-a-multiple of bxb squares starting there, i.e. at least one, and there is no room left to squeeze it in. [QED] ---- A Rickard-style proof of (S) is ..BBB....BBWWW...WBBB....BBWWW...W(..etc) ~~~~~~~ also possible, by ..BBB....BBWWW...WBBB....BBWWW...W coloring the rectangle in ..BBB....BBWWW...WBBB....BBWWW...W vertical strips as shown here: <- a ->< b-a ><- a ->< b-a > Every square tile covers an a-multiple of black squares. But if the width is a NON-b-multiple, and the number of rows is a NON-a-multiple, then there are a NON-a-multiple of black squares in total. [QED] (Note: the coloring must have 1 column of blacks on the right, and any ==== spare columns of whites on the left.) =================== Bill Taylor. wft@math.canterbury.ac.nz >A Rickard-style proof of (S) is ..BBB....BBWWW...WBBB....BBWWW...W(..etc) > ~~~~~~~ also possible, by ..BBB....BBWWW...WBBB....BBWWW...W >coloring the rectangle in ..BBB....BBWWW...WBBB....BBWWW...W >vertical strips as shown here: <- a ->< b-a ><- a ->< b-a > > >Every square tile covers an a-multiple of black squares. But if the width >is a NON-b-multiple, and the number of rows is a NON-a-multiple, then there >are a NON-a-multiple of black squares in total. [QED] > >(Note: the coloring must have 1 column of blacks on the right, and any > ==== spare columns of whites on the left.) This statement of how to position the colouring isn't good enough, I'm afraid. Take a=4, b=7 and consider e.g. a 19x10 rectangle. Coloured your way, you get: BWWWBBBBWWWBBBBWWWB BWWWBBBBWWWBBBBWWWB ::::::::::::::::::: BWWWBBBBWWWBBBBWWWB BWWWBBBBWWWBBBBWWWB The result has 10*10=100 black squares in it, which *is* a multiple of a=4, despite the fact that 19 is not a multiple of 7 and 10 is not a multiple of 4. Of course, there is an alternative offset for the pattern that does give you the result you want: WWBBBBWWWBBBBWWWBBB WWBBBBWWWBBBBWWWBBB ::::::::::::::::::: WWBBBBWWWBBBBWWWBBB WWBBBBWWWBBBBWWWBBB To show this happens in general: because the width of the rectangle is a non-multiple of b, it is possible to position it on the pattern so that the leftmost column in the rectangle is white and the column just right of the right edge of the rectangle is black. Suppose N columns are black with this positioning. Then the rectangle contains N*H black cells, where H is the height of the rectangle. If we then shift the rectangle right by one, the number of black columns increases by 1 and it contains (N+1)*H black cells. The difference between these two numbers of black cells is H, which is not a multiple of a. Therefore N*H and (N+1)*H cannot both be multiples of a, and so one of these two positionings of the pattern will suit your purposes. David Seal dseal@armltd.co.uk ==> geometry/tiling/scaling.p <== A given rectangle can be entirely covered (i.e. concealed) by an appropriate arrangement of 25 disks of unit radius. Can the same rectangle be covered by 100 disks of 1/2 unit radius? ==> geometry/tiling/scaling.s <== Yes. The same configuration of circles, when every distance is reduced by half (including the diameters), will cover a similar rectangle whose sides are one half of the original one. The original rectangle is the union of four such rectangles. ==> geometry/tiling/seven.cubes.p <== Consider 7 cubes of equal size arranged as follows. Place 5 cubes so that they form a Swiss cross or a + (plus) (4 cubes on the sides and 1 in the middle). Now place one cube on top of the middle cube and the seventh below the middle cube, to effectively form a 3-dimensional Swiss cross. Can a number of such blocks (of 7 cubes each) be arranged so that they are able to completely fill up a big cube (say 10 times the size of the small cubes)? It is all right if these blocks project out of the big cube, but there should be no holes or gaps. ==> geometry/tiling/seven.cubes.s <== Let n be a positive integer. Define the function f from Z^n to Z by f(x) = x_1+2x_2+3x_3+...+nx_n. For x in Z^n, say y is a neighbor of x if y and x differ by one in exactly one coordinate. Let S(x) be the set consisting of x and its 2n neighbors. It is easy to check that the values of f(y) for y in S(x) are congruent to 0,1,2,...,2n+1 (mod 2n+1) in some order. Using this, it is easy to check that every y in Z^n is a neighbor of one and only one x in Z^n such that f(x) is congruent to 0 (mod 2n+1). So Z^n can be tiled by clusters of the form S(x), where f(x) is congruent to 0 mod 2n+1. ==> geometry/topology/fixed.point.p <== A man hikes up a mountain, and starts hiking at 2:00 in the afternoon on a Friday. He does not hike at the same speed (a constant rate), and stops every once in a while to look at the view. He reaches the top in 4 hours. After spending the night at the top, he leaves the next day on the same trail at 2:00 in the afternoon. Coming down, he doesn't hike at a constant rate, and stops every once in a while to look at the view. It takes him 3 hours to get down the mountain. Q: What is the probability that there exists a point along the trail that the hiker was at on the same time Friday as Saturday? You can assume that the hiker never backtracked. ==> geometry/topology/fixed.point.s <== 100%. Superimpose the days: Friday starts walking up at 2:00, Saturday starts walking down at 2:00. Since they are on the same path, they must meet. ==> geometry/touching.blocks.p <== Can six 1x2x4 blocks be arranged so that each block touches n others, for all n? ==> geometry/touching.blocks.s <== n=0: 6 separate blocks n=1: 3 pairs n=2: 2 threesomes n=3: a 3x3 grid n=4: a box (each sides touches the four adjoining sides, but not the opposite) n=5: Crude ascii: Front view: Side view: /\ /\ ----- / \/ \ | | | / /\ \ | | | / / \ \ | | | \ /----\ / ---|.|.|--- \/| |\/ | | | | | ----------- ----------- | | | | ----------- ----- -- stein.kulseth@tf.tele.no [X.400] stein.kulseth@nta.no [internet] Place block A onto the x-y plane so that four of its corners are at (0,0), (0,1), (4,0), (4,1) (I give x and y coordinates only because the z coordinate will always be obvious). Place block B so four of its corners are at (2,1), (2,2), (6,1), (6,2). Now place block C with one 4x1 face on the x-y plane with one corner at (0,1) (tangent to block A) and tangent to block B at (2,1). Note that the angle between block A and block C is arctan(1/2), and a corner of block C will be at a point with approximate coordinates (3.5777, 2.7888). Call this point P. Now place an identical configuration of blocks on top of the first three as follows: block D with corners at (3.4,0.4), (4.4,0.4), (3.4,4.4), (4.4,4.4), block E with corners at (2.4,2.4), (3.4,2.4), (2.4,-1.6), (3.4,-1.6), and block F with one corner tangent to D at (3.4,4.4) and one side tangent to E at (2.4,2.4). If you have been plotting this on graph paper, then the following will be clear: Every block touches every other in its own layer. And A and B each touch D and E, and block C touches F. Point P falls under block D, so blocks C and D touch, and by symmetry so do blocks F and A. And the edge of block C intersects the edge of block E at (2.4,2.2) so blocks C and E touch, and by symmetry so do blocks F and B. Done! -- David Karr (karr@cs.cornell.edu) All the blocks are placed with their 2x4 face UP, although any face up would have worked, as it turns out. The blocks are called AAAA BBBB CCCC, etc. AAAA AAAA /_______ BBCC \ BBCC BBCC BBCC /\ || The two arrows point to the intersection of AC and BC. Now take block "D" and place the top edge along the diagonal (between the two arrows) so that the block extends SOUTH EAST of the line. Likely now the block does not touch either A or B. So slide the block towards the NORTH WEST so that it just touches A and B. You can now easily place blocks E and F perpendicular to block "D" so that they both touch all of ABC.....QED -- Guy Cousineau ==> geometry/trigonometry/euclidean.numbers.p <== For what numbers x is sin(x) expressible using only integers, +, -, *, / and square root? ==> geometry/trigonometry/euclidean.numbers.s <== Numbers generated by +, -, *, /, and sqrt from the integers are the Euclidean numbers, so called because they are those for which line segments can be constructed by use of straightedge and compass the ratio of whose lengths has that value. Using degrees, sin (360*M/N) (where (M,N)=1) is Euclidean if and only if the regular polygon with N sides can be constructed by straightedge and compass. This is true if (Gauss) and only if (easier) N is a power of 2 times the product of different Fermat primes (3, 5, 17, 257, 65537 and probably no more). So sin (3/17) = sin (360/(2^3*3*5*17)) is Euclidean, for example. Some particular values: sin(54) = (1 + sqrt(5))/4 sin(3) = sqrt(8 - sqrt(3) - sqrt(15) - sqrt(10 - 2*sqrt(5)))/4 ==> geometry/trigonometry/inequality.p <== Show that (sin x)^(sin x) < (cos x)^(cos x) when 0 < x < pi/4. ==> geometry/trigonometry/inequality.s <== The function f(x) = x^(1/sqrt(1-x^2)) is monotonically increasing for 0 < x < 1, easily verified by taking the derivative. Since 0 < sin x < cos x < 1 for 0 < x < pi/4, f(sin x) < f(cos x). But f(sin x) = (sin x)^(1/cos x) and f(cos x) = (cos x)^(1/sin x). Raising both sides to the power (cos x.sin x), we get the desired result.