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Newsgroups: rec.puzzles,news.answers,rec.answers Path: senator-bedfellow.mit.edu!bloom-beacon.mit.edu!spool.mu.edu!howland.reston.ans.net!europa.eng.gtefsd.com!uunet!questrel!chris From: chris@questrel.com (Chris Cole) Subject: rec.puzzles Archive (geometry), part 13 of 35 Message-ID: <puzzles/archive/geometry/part1_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: 1475 Xref: senator-bedfellow.mit.edu rec.puzzles:24997 news.answers:11517 rec.answers:1917 Archive-name: puzzles/archive/geometry/part1 Last-modified: 17 Aug 1993 Version: 4 ==> geometry/K3,3.p <== Can three houses be connected to three utilities without the pipes crossing? _______ _______ _______ | oil | |water| | gas | |_____| |_____| |_____| _______ _______ _______ |HOUSE| |HOUSE| |HOUSE| | one | | two | |three| ==> geometry/K3,3.s <== The problem you describe is to draw a bipartite graph of 3 nodes connected in all ways to 3 nodes, all embedded in the plane. The graph is called K3,3. A famous theorem of Kuratowsky says that all graphs can be embedded in the plane, EXCEPT those containing a subgraph that is topologically equivalent to K3,3 or K5 (the complete graph on 5 vertices, i.e., the graph with 5 nodes and 10 edges). So your problem is a minimal example of a graph that cannot be embedded in the plane. The proofs that K5 and K3,3 are non-planar are really quite easy, and only depend on Euler's Theorem that F-E+V=2 for a planar graph. For K3,3 V is 6 and E is 9, so F would have to be 5. But each face has at least 4 edges, so E >= (F*4)/2 = 10, contradiction. For K5 V is 5 and E is 10, so F = 7. In this case each face has at least 3 edges, so E >= (F*3)/2 = 10.5, contradiction. The difficult part of Kuratowsky is the proof in the other direction! A quick, informal proof by contradiction without assuming Euler's Theorem: Using a map in which the houses are 1, 2, and 3 and the utilities are A, B, and C, there must be continuous lines that connect the buildings and divide the area into three sections bounded by the loops A-1-B-2-A, A-1-B-3-A, and A-2-B-3-A. (One of the areas is the infinite plane *around* whichever loop is the outer edge of the network.) C must be in one of these three areas; whichever area it is in, either 1, or 2, or 3, is *not* part of the loop that rings its area and hence is inaccessible to C. The usual quibble is to solve the puzzle by running one of the pipes underneath one of houses on its way to another house; the puzzle's instructions forbid crossing other *pipes*, but not crossing other *houses*. ==> geometry/bear.p <== If a hunter goes out his front door, goes 50 miles south, then goes 50 miles west, shoots a bear, goes 50 miles north and ends up in front of his house. What color was the bear? ==> geometry/bear.s <== The hunter's door is in one of two locations. One is a foot or so from the North Pole, facing north, such that his position in front of the door is precisely upon the North Pole. Since that's a ridiculous place to build a house and since bears do not roam within fifty miles of the pole, the bear is either imaginary or imported, and there is no telling what color it is. There is another place (actually a whole set) on earth from which one can go fifty miles south, fifty miles west, and fifty miles north and end up where one started. Consider the parallel of latitude close enough to the South Pole that its length is 50/n miles, for some integer n. Take any point on that parallel of latitude and pick the point fifty miles north of it. Situate the hunter's front porch there. The hunter goes fifty miles south from his porch and is at a point we'll call A. He travels fifty miles west, circling the South Pole n times, and is at A again, where he shoots the bear. Fifty miles north from A he is back home. Since bears are not indigenous to the Antarctic, again the bear is either imaginary or imported and there is no telling what color it might be. ==> geometry/bisector.p <== Prove if two angle bisectors of a triangle are equal, then the triangle is isosceles (more specifically, the sides opposite to the two angles being bisected are equal). ==> geometry/bisector.s <== PROVE: <ABC = <BCA (i.e. triangle ABC is an isosceles triangle) A / \ / \ D E XP normal to AB / \ / \ XQ normal to AC P /----X----\ Q / / \ \ / / \ \ / / \ \ B/_______________\C PROOF : Let XP and XQ be normals to AC and AB. Since the three angle bisectors are concurrent, AX bisects angle A also and therefore XP = XQ. Let's assume XD > XE. Then ang(PDX) < ang(QEX) Now considering triangles BXD and CXE, the last condition requires that ang(DBX) > ang(ECX) OR ang(XBC) > ang(XCB) OR XC > XB Thus our assumption leads to : XC + XD > XE + XB OR CD > BE which is a contradiction. Similarly, one can show that XD < XE leads to a contradiction too. Hence XD = XE => CX = BX From which it is easy to prove that the triangle is isosceles. -- Manish S Prabhu (mprabhu@magnus.acs.ohio-state.edu) ==> geometry/calendar.p <== Build a calendar from two sets of cubes. On the first set, spell the months with a letter on each face of three cubes. Use lowercase three-letter English abbreviations for the names of all twelve months (e.g., "jan", "feb", "mar"). On the second set, number the days with a digit on each face of two cubes (e.g., "01", "02", etc.). ==> geometry/calendar.s <== First note that there are *nineteen* different letters in the month abbreviations (abcdef gjlmno prstuv y) so to get them all on the eighteen faces of 3 cubes, you know right away you're going to have to resort to trickery. So I wrote them all down and looked at which ones could be reversed to make another letter in the set. The only pair that jumped out at me was the d/p pair. Now I knew that it was at least feasible, as long as it wasn't necessary to duplicate any letters. Then I scanned the abbreviations to find ones that had a lot of common letters. The jan-jun-jul series looked like a good place to start: j a n u l was a good beginning but I realized right away that I had no room for duplicate letters and the second cube had both a and u so aug was going to be impossible. In fact I almost posted that answer. Then I realized that if Martin Gardner wrote about it, it must have a solution. :-) So I went back to the letter list. I don't put tails on my u's so it didn't strike me the first time through that n and u could be combined. Cube 1 Cube 2 Cube 3 j a n/u n/u l would let me get away with putting the g on the first cube to get aug, so I did. j a n/u g n/u l (1) Now came the fun part. The a was placed so I had to work around it for the other months that had an a in them (mar, apr, may). m a r d/p y (2) Now the d/p was placed so I had to work around that for sep and dec. This one was easy since they shared an e as well. d/p e s c (3) Now the e was placed so feb had to be worked in. f e b (4) The two months left (oct, nov) were far more complex. Not only did they have two "set" letters (c, n/u), there were two possible n/u's to be set with. That's why I left them for last. o t c n/u v (5) So now I had five pieces to fit together, so that no set would have more than six letters in it. Trial and error provided: j a n/u a b e g n/u l or, c d/p g r s m alphabetically: f l j y c d/p n/u m o e v t s n/u r o f b v t y Without some gimmick the days cannot be done. Because of the dates 11 and 22, there must be a 1 and a 2 on each cube. Thus there are 8 remaining spaces for the 8 remaining numbers, and because of 30, we put 3 and 0 on different cubes. I don't think the way you allocate the others matter. Now 6 numbers on each cube can produce at most 36 distinct pairs, and we need 31 distinct pairs to represent all possible dates. But since 3 each of {4,5,6,7,8,9} are on each cube, there are at least 9 representable numbers which can't be dates. Therefore there are at most 27 distinct numbers which are dates on the two cubes, and it can't be done. In particular, not all of {04,05,06,07,08,09} can be represented. The gimmick solution would be to represent the numbers in a stylised format (like say, on a digital clock or on a computer screen) such that the 6 can be turned upside down to be a 9. Then you can have 012 on both cubes, and three each of {3,4,5,6,7,8} on the other faces. Done. Example: 012468 012357 ==> geometry/circles.and.triangles.p <== Find the radius of the inscribed and circumscribed circles for a triangle. ==> geometry/circles.and.triangles.s <== Let a, b, and c be the sides of the triangle. Let s be the semiperimeter, i.e. s = (a + b + c) / 2. Let A be the area of the triangle, and let x be the radius of the incircle. Divide the triangle into three smaller triangles by drawing a line segment from each vertex to the incenter. The areas of the smaller triangles are ax/2, bx/2, and cx/2. Thus, A = ax/2 + bx/2 + cx/2, or A = sx. We use Heron's formula, which is A = sqrt(s(s-a)(s-b)(s-c)). This gives us x = sqrt((s-a)(s-b)(s-c)/s). The radius of the circumscribed circle is given by R = abc/4A. ==> geometry/coloring/cheese.cube.p <== A cube of cheese is divided into 27 subcubes. A mouse starts at one corner and eats each subcube, one at a time. Can it finish in the middle? ==> geometry/coloring/cheese.cube.s <== Give the subcubes a checkerboard-like coloring so that no two adjacent subcubes have the same color. If the corner subcubes are black, the cube will have 14 black subcubes and 13 white ones. The mouse always alternates colors and so must end in a black subcube. But the center subcube is white, so the mouse can't end there. ==> geometry/coloring/triominoes.p <== There is a chess board (of course with 64 squares). You are given 21 "triominoes" of size 3-by-1 (the size of an individual square on a chess board is 1-by-1). Which square on the chess board can you cut out so that the 21 triominoes exactly cover the remaining 63 squares? Or is it impossible? ==> geometry/coloring/triominoes.s <== |||||||| |||||||| |||||||| ---***+* ---...+* ---*+O+* ---*+... ---*+*** There is only one way to remove a square, aside from rotations and reflections. To see that there is at most one way, do this: Label all the squares of the chessboard with A, B or C in sequence by rows starting from the top: ABCABCAB CABCABCA BCABCABC ABCABCAB CABCABCA BCABCABC ABCABCAB CABCABCA Every triomino must cover one A, one B and one C. There is one extra A square, so an A must be removed. Now label the board again by rows starting from the bottom: CABCABCA ABCABCAB BCABCABC CABCABCA ABCABCAB BCABCABC CABCABCA ABCABCAB The square removed must still be an A. The only squares that got marked with A both times are these: ........ ........ ..A..A.. ........ ........ ..A..A.. ........ ........ ==> geometry/construction/4.triangles.6.lines.p <== Can you construct 4 equilateral triangles with 6 toothpicks? ==> geometry/construction/4.triangles.6.lines.s <== Use the toothpicks as the edges of a tetrahedron. ==> geometry/construction/5.lines.with.4.points.p <== Arrange 10 points so that they form 5 rows of 4 each. ==> geometry/construction/5.lines.with.4.points.s <== Draw a 5 pointed star, put a point where any two lines meet. ==> geometry/construction/square.with.compass.p <== Construct a square with only a compass and a straight edge. ==> geometry/construction/square.with.compass.s <== Draw a circle (C1 at P1). Now draw a diameter D1 (intersects at P2 and P3). Set the compass larger than before. From each of points P2 and P3 draw another larger circle (C2 and C3). Where these two circles cross, draw a line (D2). This line should go through the center of circle C1 at a right angle to the original diameter line. This line should cross circle C1 at P4 and P5. Points P2, P4, P3, P5 form a square. For extra credit: Reset the compass to its original size. From P2 and P4 draw a circle (C4 and C5). These circles intersect at P6 and P1. Connect P6, P2, P1, P4 for a square of the same size as the original compass setting. ==> geometry/corner.p <== A hallway of width A turns through 90 degrees into a hallway of width B. A ladder is to be passed around the corner. If the movement is within the horizontal plane, what is the maximum length of the ladder? ==> geometry/corner.s <== ------\--------- B / \.......|..B/sin(theta) theta\ | ---|-----X | |\ | | \...|..A/cos(theta) | \ | | \ | | A \| Theta is the angle off horizontal. Minimize length = A/cos(theta) + B/sin(theta) d(length)/d(theta) = A*sin(theta)/cos(theta)^2 - B*cos(theta)/sin(theta)^2 (?) = 0 A*sin(theta)/cos(theta)^2 = B*cos(theta)/sin(theta)^2 B/A = sin(theta)^3/cos(theta()^3 = tan(theta)^3 theta = inverse_tan(cube_root(B/A)) If you use the trigonometric formulas cos^2 x = 1/(1 + tan^2 x) and sin x = tan x cos x, and plug through the algebra, I believe that the formula for the length reduces to (A^(2/3) + B^(2/3))^(3/2) At any rate this is symmetric in A and B as one would expect, and has the right values at A=B and as either A-->0 or B-->0. ==> geometry/cover.earth.p <== A thin membrane covers the surface of the (spherical) earth. One square meter is added to the area of this membrane to form a larger sphere. How much is added to the radius and volume of this membrane? ==> geometry/cover.earth.s <== We know that V = (4/3)*pi*r^3 and A = 4*pi*r^2. We need to find out how much V increases if A increases by 1 m^2. dV / dr = 4 * pi * r^2 dA / dr = 8 * pi * r dV / dA = (dV / dr) / (dA / dr) = (4 * pi * r^2) / (8 * pi * r) = r/2 = 3,250,000 m If the area of the cover is increased by 1 square meter, then the volume it contains is increased by about 3.25 million cubic meters. We seem to be getting a lot of mileage out of such a small square of cotton. However, the new cover would not be very high above the surface of the planet -- about 6 nanometers (calculate dr/dA). ==> geometry/cycle.polynomial.p <== What are the cycle polynomials for the Platonic solids? ==> geometry/cycle.polynomial.s <== For future reference, here are the cycle polynomials for the five platonic solids (and I threw in the tesseract for good measure). Most combinatoric coloring problems are simple plug-ins to these polynomials. For details, see any book on combinatorics that presents Polya counting theory. tetrahedron: (x1^4+3x2^2+8x1*x3)/12 cube: (x1^6+6x2^3+3x1^2*x2^2+8x3^2+6x1^2*x4)/24 octahedron: (x1^8+9x2^4+8x1^2*x3^2+6x4^2)/24 dodecahedron: (x1^12+15x2^6+20x3^4+24x1^2*x5^2)/60 icosahedron: (x1^20+15x2^10+20x1^2*x3^6+24x5^4)/60 tesseract: (32x6^4+x2^12+48x8^3+x1^24+24x1^2*x2^11+12x2^2*x4^5+32x3^8+12x4^6 +18x1^4*x2^10+12x1^4*x4^5)/192 ==> geometry/dissections/disk.p <== Can a disk be cut into similar pieces without point symmetry about the midpoint? Can it be done with a finite number of pieces? ==> geometry/dissections/disk.s <== Yes. Draw a circle inside the circumference of the disk, sharing a common point on the right. Now draw another circle inside the second, sharing a point at the left. Now draw another inside the third, sharing a point at the right. Continue in this way, coloring in every other region thus generated. Now, all the colored regions touch, so count this as one piece and the uncolored regions as a second piece. So the circle has been divided into two similar pieces and there is no point symmetry about the midpoint. Maybe it is cheating to call these single pieces, though. ==> geometry/dissections/hexagon.p <== Divide the hexagon into: 1) 3 identical rhombuses. 2) 6 identical kites(?). 3) 4 identical trapezoids (trapeziums in Britain). 4) 8 identical shapes (any shape). 5) 12 identical shapes (any shape). ==> geometry/dissections/hexagon.s <== What is considered 'identical' for these questions? If mirror-image shapes are allowed, these are all pretty trivial. If not, the problems are rather more difficult... 1. Connect the center to every second vertex. 2. Connect the center to the midpoint of each side. 3. This is the hard one. If you allow mirror images, it's trivial: bisect the hexagon from vertex to vertex, then bisect with a perpendicular to that, from midpoint of side to midpoint of side. 4. This one's neat. Let the side length of the hexagon be 2 (WLOG). We can easily partition the hexagon into equilateral triangles with side 2 (6 of them), which can in turn be quartered into equilateral triangles with side 1. Thus, our original hexagon is partitioned into 24 unit equilateral triangles. Take the trapezoid formed by 3 of these little triangles. Place one such trapezoid on the inside of each face of the original hexagon, so that the long side of the trapezoid coincides with the side of the hexagon. This uses 6 trapezoids, and leaves a unit hexagon in the center as yet uncovered. Cover this little hexagon with two of the trapezoids. Voila. An 8-identical-trapezoid partition. 5. Easy. Do the rhombus partition in #1. Quarter each rhombus by connecting midpoints of opposite sides. This produces 12 small rhombi, each of which is equivalent to two adjacent small triangles as in #4. Except for #3, all of these partitions can be achieved by breaking up the hexagon into unit equilateral triangles, and then building these into the shapes desired. For #3, though, this would require (since there are 24 small triangles) trapezoids formed from 6 triangles each. The only trapezoid that can be built from 6 identical triangles is a parallelogram; I assume that the poster wouldn't have asked for a trapezoid if you could do it with a special case of trapezoid. At any rate, that parallelogram doesn't work. ==> geometry/dissections/largest.circle.p <== What is the largest circle that can be assembled from two semicircles cut from a rectangle with edges a and b? ==> geometry/dissections/largest.circle.s <== There are two methods: Method M1: The diameters of the semicircles have to be on the longer sides, starting at an endpoint of the rectangle. The two semicircles touch each other in the middle M of the rectangle. a D._______________________.C | | | | b | . M | | | | | |_______.___.___________| A R X B R should be the center of the semicircle, and because of RA = RM, it holds that: r^2 = (a/2 - r)^2 + (b/2)^2 Solving for r gives: r = min[b,(a^2+b^2)/(4a)], where a >= b. Method M2: We'll cut on the line y = c x, where c will turn out to be slightly less than d, the slope of the diagonal. We describe the semicircle lying above the line y = c x, having this line as the straight part of the semi-circle. The center P of the semicircle will be taken on the line y = d - x, and will be tangent to the left and top of the rectangle. Clearly the lower down P is on this line, the better. The naive solution is not optimal because the upper place where the semicircle meets the diagonal is interior to the rectangle. So we try to determine c in such a way that this latter point actually lies slightly down from the top, on the right side of the rectangle. This involves solving the quartic: 4r^4 - (4a+16b)r^3 + (16b^2+a^2+8ab)r^2 - (6b^3+4ab^2+2ba^2)r + b^4+(ab)^2 = 0, where r < b, the details of which will be left to the reader. The other semicircle is the reflection of the first through the origin. After a few calculations, we find that the value of r given by M2 is greater than the one given by M1 only if 1 < a/b < 2.472434. ==> geometry/dissections/square.70.p <== Since 1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2, can a 70x70 square be dissected into 24 squares of size 1x1, 2x2, 3x3, etc.? ==> geometry/dissections/square.70.s <== Martin Gardner asked this in his Mathematical Games column in the September 1966 issue of Scientific American. William Cutler was the first of 24 readers who reduced the uncovered area to 49, using all but the 7x7 square. All the patterns were the same except for interchanging the squares of orders 17 and 18 and rearranging the squares of orders 1, ..., 6, 8, 9, and 10. Nobody proved that the solution is minimal. +----------------+-------------+----------------------+---------------------+ | | | | | | | | | | | | 11 | | | | | | | | | 16 | | | | | +-----+--+----+ 22 | 21 | | | | 2| | | | | | 5 +--+----+ | | | | | | | | +----------------+--+--+ 6 | | | | | 3| | | | | ++-+-------+ | | | || | ++--------------------+ | || 8 +----------------------++ | | 18 || | | | | || | | | | ++---------+ | | | | | | 20 | | | 9 | | | +------------------++ | 23 | | | || | | | | ++----------+ | | | | | +---++---------------+ | | | | || | | 17 | 10 | | 4 || | | | +---------------+-------+---++ | | +-+---------+---------------+ | 15 | | | | | | | | | | | 12 | | +------------------+-+ | +-+-------------+ | | | |1| | | | +------------+-+ | | | 24 | | | | | | | | | 19 | | 13 | 14 | | | | | | | | | | | | | | | | +--------------------+-------------------------+--------------+-------------+ ==> geometry/dissections/square.five.p <== Can you dissect a square into 5 parts of equal area with just a straight edge? ==> geometry/dissections/square.five.s <== 1. Prove you can reflect points which lie on the sides of the square about the diagonals. 2. Construct two different rectangles whose vertices lie on the square and whose sides are parallel to the diagonals. 3. Construct points A, A', B, B' on one (extended) side of the square such that A/A' and B/B' are mirror image pairs with respect to another side of the square. 4. Construct the mirror image of the center of the square in one of the sides. 5. Divide the original square into 4 equal squares whose sides are parallel to the sides of the original square. 6. Divide one side of the square into 8 equal segments. 7. Construct a trapezoid in which one base is a square side and one base is 5/8 of the opposite square side. 8. Divide one side of the square into 5 equal segments. 9. Divide the square into 5 equal rectangles. ==> geometry/dissections/tesseract.p <== If you suspend a cube by one corner and slice it in half with a horizontal plane through its centre of gravity, the section face is a hexagon. Now suspend a tesseract (a four dimensional hypercube) by one corner and slice it in half with a hyper-horizontal hyperplane through its centre of hypergravity. What is the shape of the section hyper-face? ==> geometry/dissections/tesseract.s <== The 4-cube is the set of all points in [-1,1]^4 . The hyperplane { (x,y,z,w) : x + y + z + w = 0 } cuts the 4-cube in the desired manner. Now, { (.5,.5,-.5,-.5), (.5,-.5,.5,-.5), (.5,-.5,-.5,.5) } is an orthonormal basis for the hyperplane. Let (a,b,c) be a point on the hyperplane with respect to this basis. (a,b,c) is in the 4-cube if and only if |a| + |b| + |c| <= 2. The shape of the intersection is a regular octahedron. ==> geometry/duck.and.fox.p <== A duck is swimming about in a circular pond. A ravenous fox (who cannot swim) is roaming the edges of the pond, waiting for the duck to come close. The fox can run faster than the duck can swim. In order to escape, the duck must swim to the edge of the pond before flying away. Assume that the duck can't fly until it has reached the edge of the pond. How much faster must the fox run that the duck swims in order to be always able to catch the duck? ==> geometry/duck.and.fox.s <== Assume the ratio of the fox's speed to the duck's is a, and the radius of the pond is r. The duck's best strategy is: 1. Swim around a circle of radius (r/a - delta) concentric with the pond until you are diametrically opposite the fox (you, the fox, and the center of the pond are colinear). 2. Swim a distance delta along a radial line toward the bank opposite the fox. 3. Observe which way the fox has started to run around the circle. Turn at a RIGHT ANGLE in the opposite direction (i.e. if you started swimming due south in step 2 and the fox started running to the east, i.e. clockwise around the pond, then start swimming due west). (Note: If at the beginning of step 3 the fox is still in the same location as at the start of step 2, i.e. directly opposite you, repeat step 2 instead of turning.) 4. While on your new course, keep track of the fox. If the fox slows down or reverses direction, so that you again become diametrically opposite the fox, go back to step 2. Otherwise continue in a straight line until you reach the bank. 5. Fly away. The duck should make delta as small as necessary in order to be able to escape the fox. The key to this strategy is that the duck initially follows a radial path away from the fox until the fox commits to running either clockwise or counterclockwise around the pond. The duck then turns onto a new course that intersects the circle at a point MORE than halfway around the circle from the fox's starting position. In fact, the duck swims along a tangent of the circle of radius r/a. Let theta = arc cos (1/a) then the duck swims a path of length r sin theta + delta but the fox has to run a path of length r*(pi + theta) - a*delta around the circle. In the limit as delta goes to 0, the duck will escape as long as r*(pi + theta) < a*r sin theta that is, pi + arc cos (1/a) - a * sqrt(a^2 - 1) < 0 Maximize a in the above: a = 4.6033388487517003525565820291030165130674... The fox can catch the duck as long as he can run about 4.6 times as fast as the duck can swim. "But wait," I hear you cry, "When the duck heads off to that spot 'more than halfway' around the circle, why doesn't the fox just double back? That way he'll reach that spot much quicker." That is why the duck's strategy has instructions to repeat step 2 under certain circumstances. Note that at the end of step 2, if the fox has started to run to head off the duck, say in a clockwise direction, he and the duck are now on the same side of some diameter of the circle. This continues to be true as long as both travel along their chosen paths at full speed. But if the fox were now to try to reach the duck's destination in a counterclockwise direction, then at some instant he and the duck must be on a diameter of the pond. At that instant, they have exactly returned to the situation that existed at the end of step 1, except that the duck is a little closer to the edge than she was before. That's why the duck always repeats step 2 if the fox is ever diametrically opposite her. Then the fox must commit again to go one way or the other. Every time the fox fails to commit, or reverses his commitment, the duck gets a distance delta closer to the edge. This is a losing strategy for the fox. The limiting ratio of velocities that this strategy works against cannot be improved by any other strategy, i.e., if the ratio of the duck's speed to the fox's speed is less than a then the duck cannot escape given the best fox strategy. Given a ratio R of speeds less than the above a, the fox is sure to catch the duck (or keep it in water indefinitely) by pursuing the following strategy: Do nothing so long as the duck is in a radius of R around the centre. As soon as it emerges from this circle, run at top speed around the circumference. If the duck is foolish enough not to position itself across from the center when it comes out of this circle, run "the short way around", otherwise run in either direction. To see this it is enough to verify that at the circumference of the circle of radius R, all straight lines connecting the duck to points on the circumference (in the smaller segment of the circle cut out by the tangent to the smaller circle) bear a ratio greater than R with the corresponding arc the fox must follow. That this is enough follows from the observation that the shortest curve from a point on a circle to a point on a larger concentric circle (shortest among all curves that don't intersect the interior of the smaller circle) is either a straight line or an arc of the smaller circle followed by a tangential straight line. ==> geometry/earth.band.p <== How much will a band around the equator rise above the surface if it is made one meter longer? Assume the equator is a circle. ==> geometry/earth.band.s <== The formula for the circumference of a circle is 2 * pi * radius. Therefore, if you increase the circumference by 1 meter, you increase the radius by 1/(2 * pi) meters, or about 0.16 meters. ==> geometry/fence.p <== A farmer wishes to enclose the maximum possible area with 100 meters of fence. The pasture is bordered by a straight cliff, which may be used as part of the fence. What is the maximum area that can be enclosed? ==> geometry/fence.s <== A circle is the plane figure with highest ratio of area to perimeter. The cliff can be used to bisect a circle of radius 100/pi meters. By symmetry, this will form the pen of largest area. The resulting pen will contain 5000/pi meters squared. ==> geometry/ham.sandwich.p <== Consider a ham sandwich, consisting of two pieces of bread and one of ham. Suppose the sandwich was dropped into a machine and spindled, torn and mutilated. Is it still possible to divide the ham sandwich with a straight knife cut such that both the ham and each slice of bread are divided in two parts of equal volume? ==> geometry/ham.sandwich.s <== Yes. There is a theorem in topology called the Ham Sandwich Theorem, which says: Given 3 (finite) volumes (each may be of any shape, and in several pieces), there is a plane that cuts each volume in half. You would learn about it typically in a first course in algebraic topology, or maybe in a course on introductory topology (if you studied the fundamental group). ==> geometry/hike.p <== You are hiking in a half-planar woods, exactly 1 mile from the edge, when you suddenly trip and lose your sense of direction. What's the shortest path that's guaranteed to take you out of the woods? Assume that you can navigate perfectly relative to your current location and (unknown) heading. ==> geometry/hike.s <== Go 2/sqrt(3) away from the starting point, turn 120 degrees and head 1/sqrt(3) along a tangent to the unit circle, then traverse an arc of length 7*pi/6 along this circle, then head off on a tangent 1 mile. This gives a minimum of sqrt(3) + 7*pi/6 + 1 = 6.397... It remains to prove this is the optimal answer. ==> geometry/hole.in.sphere.p <== Old Boniface he took his cheer, Then he bored a hole through a solid sphere, Clear through the center, straight and strong, And the hole was just six inches long. Now tell me, when the end was gained, What volume in the sphere remained? Sounds like I haven't told enough, But I have, and the answer isn't tough! ==> geometry/hole.in.sphere.s <== The volume of the leftover material is equal to the volume of a 6" sphere. First, lets look at the 2 dimensional equivalent of this problem. Two concentric circles where the chord of the outer circle that is tangent to the inner circle has length D. What is the annular area between the circles? It is pi * (D/2)^2. The same area as a circle with that diameter. Proof: big circle radius is R little circle radius is r 2 2 area of donut = pi * R - pi * r 2 2 = pi * (R - r ) Draw a right triangle and apply the Pythagorean Theorem to see that 2 2 2 R - r = (D/2) so the area is 2 = pi * (D/2) Start with a sphere of radius R (where R > 6"), drill out the 6" high hole. We will now place this large "ring" on a plane. Next to it place a 6" high sphere. By Archemedes' theorem, it suffices to show that for any plane parallel to the base plane, the cross- sectional area of these two solids is the same. Take a general plane at height h above (or below) the center of the solids. The radius of the circle of intersection on the sphere is radius = srqt(3^2 - h^2) so the area is pi * ( 3^2 - h^2 ) For the ring, once again we are looking at the area between two concentric circles. The outer circle has radius sqrt(R^2 - h^2), The area of the outer circle is therefore pi (R^2 - h^2) The inner circle has radius sqrt(R^2 - 3^2). So the area of the inner circle is pi * ( R^2 - 3^2 ) the area of the doughnut is therefore pi(R^2 - h^2) - pi( R^2 - 3^2 ) = pi (R^2 - h^2 - R^2 + 3^2) = pi (3^2 - h^2) Therefore the areas are the same for every plane intersecting the solids. Therefore their volumes are the same. QED There also is a meta-theoretic answer to this puzzle. Assume the puzzle can be solved. Then it must be solvable with a hole of any diameter, even zero. But if you drill a hole of zero diameter that is six inches long, you leave behind the volume of a six inch diameter sphere. ==> geometry/hypercube.p <== How many vertices, edges, faces, etc. does a hypercube have? ==> geometry/hypercube.s <== Take any vertex of the hypercube, and ask how many k-V's it participates in. To make a k-V it needs to combine with k adjacent and orthogonal vertices, and there are (nCk) distinct ways of doing this [that is, choose k directions out of n possible ones]. Then multiply by 2^n, the total number of vertices. But this involves multiple counting, since each k-V is shared by 2^k vertices. So divide by 2^k, and this yields the answer: (nCk)*2^{n-k}. For example, 12d hypercube: 0-v: 4,096 1-v: 24,576 2-v: 67,584 3-v: 112,640 4-v: 126,720 5-v: 101,376 6-v: 59,136 7-v: 25,344 8-v: 7,920 9-v: 1,760 10-v: 264 11-v: 24 12-v: 1 ==> geometry/kissing.number.p <== How many n-dimensional unit spheres can be packed around one unit sphere? ==> geometry/kissing.number.s <== From the Feb. 1992 issue of Scientific American: Kissing Numbers Dimension Lower limit Upper limit 1* 2 2 2* 6 6 3* 12 12 4 24 25 5 40 46 6 72 82 7 126 140 8* 240 240 9 306 380 10 500 595 11 582 915 12 840 1,416 13 1,130 2,233 14 1,582 3,492 15 2,564 5,431 16 4,320 8,313 17 5,346 12,215 18 7,398 17,877 19 10,688 25,901 20 17,400 37,974 21 27,720 56,852 22 49,896 86,537 23 93,150 128,096 24* 196,560 196,560 * = dimensions for which the answer is known. REFERENCES (from the Sci. Am. article) The Problem of the Thirteen Spheres. John Leech in Mathematical Gazette, Vol. 40, No. 331, pages 22-23; February 1956 Sphere Packings, Lattics and Groups. John Horton Conway and Neil J. A. Sloane. Springer-Verlag, 1988. Sphere Packings and Spherical Geometry--Kepler's Conjecture and Beyond, preprint. Wu-Yi Hsiang. Center for Pure and Applied Mathematics, University of California, Berkeley, July 1991. -- David Radcliffe radcliff@csd4.csd.uwm.edu ==> geometry/konigsberg.p <== Can you draw a line through each edge on the diagram below without crossing any edge twice and without lifting your pencil from the paper? +---+---+---+ | | | | +---+-+-+---+ | | | +-----+-----+ ==> geometry/konigsberg.s <== This is solved in the same way as the famous "Seven Bridges of Konigsberg" problem first solved by Euler. In that problem, the task was to find a closed path that crossed each of the seven bridges of Konigsberg (now Kaliningrad, Russia) exactly once. For reasons given below, no such path existed. In this version, you cannot draw such a line without cheating by: (1) drawing a line along one of the edges, or (2) inscribing the diagram on a torus, or (3) defining a line segment as the entire length of each straight line, or (4) adding a vertex on one of the line segemnts, or (5) defining "crossing" as touching the endpoint of a segment. The method for determining if paths exist in all similar problems is given below. Turn each "room" into a point. Turn each line segment into a line connecting the two points representing the rooms it abuts. You should be able to see that drawing one continuous line across all segments in your drawing is equivalent to traversing all the edges in the resulting graph. Euler's Theorem states that for a graph to be traversable, the number of vertices with an odd number of edges proceeding from them must be either zero or two. For this graph, that number is four, and it cannot be traversed. +---+---+---+ | 1 | 2 | 3 | +---+-+-+---+ 6 (outside) | 4 | 5 | +-----+-----+ Number of edges proceeding from each vertex: 1: 4 2: 5 (*odd*) 3: 4 4: 5 (*odd*) 5: 5 (*odd*) 6: 9 (*odd*) To prove Euler's Theorem, think of walking along the graph from vertex to vertex. Each vertex must be entered as many times as it is exited, except for where you start and where you end. So, each vertex must have an even number of edges, except possibly for two vertices. And if there are two vertices with an odd number of edges, the path must start at one and end at the other. ==> geometry/ladders.p <== Two ladders form a rough X in an alley. The ladders are 11 and 13 meters long and they cross 4 meters off the ground. How wide is the alley? ==> geometry/ladders.s <== Ladders 1 and 2, denoted L1 and L2, respectively, will rest along two walls (taken to be perpendicular to the ground), and they will intersect at a point O = (a,s), a height s from the ground. Find the largest s such that this is possible. Then find the width of the alley, w = a+b, in terms of L1, L2, and s. This diagram is not to scale. B D |\ L1 L2 /| | \ / | BC = length of L1 | \ / | AD = length of L2 | \ O / | s = height of intersection x| \ / |y A = (0,0) | /|\ | AE = a | m / | \ n | EC = b | / |s \ | AO = m | / | \ | CO = n |/________|________\| (0,0) = A a E b C ----------------------------------------------------------------------------- Without loss of generality, let L2 >= L1. Observe that triangles AOB and DOC are similar. Let r be the ratio of similitude, so that x=ry. Consider right triangles CAB and ACD. By the Pythagorean theorem, L1^2 - x^2 = L2^2 - y^2. Substituting x=ry, this becomes y^2(1-r^2) = L2^2 - L1^2. Letting L= L2^2 -L1^2 (L>=0), and factoring, this becomes (*) y^2 (1+r)(1-r) = L Now, because parallel lines cut L1 (a transversal) in proportion, r = x/y = (L1-n)/n, and so L1/n = r+1. Now, x/s = L1/n = r+1, so ry = x = s(r+1). Solving for r, one obtains the formula r = s/(y-s). Substitute this into (*) to get (**) y^2 (y) (y-2s) = L (y-s)^2 NOTE: Observe that, since L>=0, it must be true that y-2s>=0. Now, (**) defines a fourth degree polynomial in y. It can be written in the form (by simply expanding (**)) (***) y^4 - 2s_y^3 - L_y^2 + 2sL_y - Ls^2 = 0 L1 and L2 are given, and so L is a constant. How large can s be? Given L, the value s=k is possible if and only if there exists a real solution, y', to (***), such that 2k <= y' < L2. Now that s has been chosen, L and s are constants, and (***) gives the desired value of y. (Make sure to choose the value satisfying 2s <= y' < L2. If the value of s is "admissible" (i.e., feasible), then there will exist exactly one such solution.) Now, w = sqrt(L2^2 - y^2), so this concludes the solution. L1 = 11, L2 = 13, s = 4. L = 13^2-11^2 = 48, so (***) becomes y^4 - 8_y^3 - 48_y^2 + 384_y - 768 = 0 Numerically find root y ~= 9.70940555, which yields w ~= 8.644504. ==> geometry/lattice/area.p <== Prove that the area of a triangle formed by three lattice points is integer/2. ==> geometry/lattice/area.s <== The formula for the area is A = | x1*y2 + x2*y3 + x3*y1 - x1*y3 - x2*y1 - x3*y2 | / 2 If the xi and yi are integers, A is of the form (integer/2) ==> geometry/lattice/equilateral.p <== Can an equlateral triangle have vertices at integer lattice points? ==> geometry/lattice/equilateral.s <== No. Suppose 2 of the vertices are (a,b) and (c,d), where a,b,c,d are integers. Then the 3rd vertex lies on the line defined by (x,y) = 1/2 (a+c,b+d) + t ((d-b)/(c-a),-1) (t any real number) and since the triangle is equilateral, we must have ||t ((d-b)/(c-a),-1)|| = sqrt(3)/2 ||(c,d)-(a,b)|| which yields t = +/- sqrt(3)/2 (c-a). Thus the 3rd vertex is 1/2 (a+c,b+d) +/- sqrt(3)/2 (d-b,a-c) which must be irrational in at least one coordinate. ==> geometry/manhole.cover.p <== Why is a manhole cover round? ==> geometry/manhole.cover.s <== It will not fall into the hole, even if rotated, tipped, etc. It gives maximal area for a given amount of material. It does not have to be carried, but can be rolled. Human beings are roughly round in horizontal cross section. Orientation of the cover with the access hole is not of concern. Orientation of the access hole with the ladder in the pipe below is not of concern. ==> geometry/pentomino.p <== Arrange pentominos in 3x20, 4x15, 5x12, 6x10, 2x3x10, 2x5x6 and 3x4x5 forms. ==> geometry/pentomino.s <== I've seen several different naming schemes used for pentominoes. This is the system I'm using (I think only F & N require a bit of imagination): FF I L N PP TTT U U V V W W W X X Y ZZ FF I L NN PP T UUU V V W W X YY Z F I L N P T V X X Y ZZ I LL N Y A 3x20 solution (the other solution is easily obtained by a rotation of the section from the Z to the L inclusive): UUXPPPZYYYYWTFNNNVVV UXXXPPZZZYWWTFFFNNLV UUXIIIIIZWWTTTFLLLLV A 4x15 solution: IIIIINNLLLLTVVV UUXNNNFZZWLTTTV UXXXYFFFZWWTPPV UUXYYYYFZZWWPPP 2 5x6 rectangles. Joined side-to-side, end-to-end, or stacked, these enable construction of the 6x10 & 5x12 rectangles, and the 2x5x6 prism: NFVVV YYYYI NFFFV LLYZI NNFXV LZZZI PNXXX LZWTI PPUXU LWWTI PPUUU WWTTT The 2x3x10 and 3x4x5 solutions are tricky to show - I hope these diagrams make sense: A 2x3x10 solution (shown as 2 layers; Y and L are shared between the 2 layers): VVVZIIIIIF UUXTTTWWPP VZZZNNNFFF UXXXTWWPPP VZYYYYNNFL UUXYTWLLLL A 3x4x5 solution (3 layers, V F W & L shared between 2 or more layers): VUUXF VZFFF VNYFW VUXXX TZZZW NNYPW VUUXW TTTZW NYYPP IIIII TLLLL NLYPP -- +------------------- pete@bignode.equinox.gen.nz -------------------+ | The effort to understand the universe is one of the very few things | | that lifts human life above the level of farce, and gives it some | | of the grace of tragedy - Steven Weinberg | +---------------------------------------------------------------------+ ==> geometry/points.in.sphere.p <== What is the expected distance between two random points inside a sphere? Assume the points are uniformly and independently distributed. ==> geometry/points.in.sphere.s <== Use spherical polar coordinates, and w.l.o.g. choose the polar axis through one of the points. Now the distance between the two points is sqrt ( r1^2 + r2^2 - 2 r1 r2 cos(theta)) and cos(theta) is (conveniently) uniformly distributed between -1 and +1, while r1 and r2 have densities 3 r1^2 d(r1) and 3 r2^2 d(r2). Split the total integral into two (equal) parts with r1 < r2 and r1 > r2, and it all comes down to integrating polynomials. More generally, the expectation of the n'th power of the distance between the two points is 2^n . 72 / ((n+3)(n+4)(n+6)) So the various means are: the (arithmetic) mean distance is 36/35 = 1.028571... the root mean square distance is sqrt(6/5) = 1.095445... the geometric mean distance is 2exp(-3/4) = 0.944733... the harmonic mean distance is 5/6 = 0.833333... the inverse root mean inverse square distance is 2/3 = 0.666666... ==> geometry/points.on.sphere.p <== What are the odds that n random points on a sphere lie in the same hemisphere? ==> geometry/points.on.sphere.s <== 1 - [1-(1/2)^(n-2)]^n where n is the # of points on the sphere. The question will become a lot easier if you restate it as the following: What is the probability in finding at least one point such that all the other points on the sphere are on one side of the great circle going through this point. When n=2, the probability= 1 , when n=infinity, it becomes 0. In his Scientific American column which was titled "Curious Maps", Martin Gardner ponders the fact that most of the land mass of the Earth is in one hemisphere and refers to a paper which models continents by small circular caps. He gives the above result. See "The Probability of Covering a Sphere With N Circular Caps" by E. N. Gilbert in Biometrika 52, 1965, p323. ==> geometry/revolutions.p <== A circle with radius 1 rolls without slipping once around a circle with radius 3. How many revolutions does the smaller circle make? ==> geometry/revolutions.s <== 4 if the smaller circle rolls on the outside of the larger circle; 2 if it rolls on the inside. Imagine you are rolling a wheel by pushing it along the equator of the earth. Suppose the circumference of the wheel is one third of that of the earth. By the time you return to your starting point, the wheel finishes 3 revolutions relative to you. But do not forget you yourself also finishes 1 revolution in the same direction. As a result, the number of absolute revolutions is 3+1=4. But if the small circle is rolling inside the large circle, the answer is then 3-1=2, because in this case the wheel makes a counter-revolution as you walk once around. ==> geometry/rotation.p <== What is the smallest rotation that returns an object to its original state? ==> geometry/rotation.s <== 720 degrees. Objects are made of bosons (integer-spin particles) and fermions (half-odd-integer spin particles), and the wave function of a fermion changes sign upon being rotated by 360 degrees. To get it back to its original state you must rotate by another 360 degrees, for a total of 720 degrees. This fact is the basis of Fermi-Dirac statistics, the Pauli Exclusion Principle, electron orbits, chemistry, and life. Mathematically, this is due to the continuous double cover of SO(2) by SO(3), where SO(2) is the internal symmetry group of fermions and SO(3) is the group of rotations in three dimensional space. A fermion can be modeled by a sphere with strings attached. It is possible to see that a 360 degree rotation will entangle the strings, which another 360 degree rotation will disentangle. You can also demonstrate this with a tray, which you hold in your right hand with the arm lowered, then rotate twice as you raise your arm and end up with the tray above your head, rotated twice about its vertical axis, but without having twisted your arm. Hospitals have machines which take out your blood, centrifuge it to take out certain parts, then return it to your veins. Because of AIDS they must never let your blood touch the inside of the machine which has touched others' blood. So the inside is lined with a single piece of disposable branched plastic tubing. This tube must rotate rapidly in the centrifuge where several branches come out. Thus the tube should twist and tangle up the branches. But the machine untwists the branches as in the above discussion. At several hundred rounds per minute! References R. Penrose and W. Rindler Spinors and Space-time, vol. 1, p. 43 Cambridge University Press, 1984 R. Feynman and S. Weinberg Elementary Particles and the Laws of Physics, p. 29 Cambridge University Press, 1987 M. Gardner The New Ambidextrous Universe, Revised (Third) Edition, pp. 329-332 W. H. Freeman, 1990 ==> geometry/shephard.piano.p <== What's the maximum area shape that will fit around a right-angle corner? ==> geometry/shephard.piano.s <== This problem is unsolved. A simple solution called the "Shephard piano" has area 2.2074+, but this can be improved upon with local modifications. A solution exists with area 2.215649+. It is known that a maximum area exists, but not whether the shape achieving it is symmetric, smooth, or even unique. See Problem G5 in Croft, Falconer, and Guy, _Unsolved Porblems in Geometry_, Springer-Verlag, 1991. ==> geometry/smuggler.p <== Somewhere on the high sees smuggler S is attempting, without much luck, to outspeed coast guard G, whose boat can go faster than S's. G is one mile east of S when a heavy fog descends. It's so heavy that nobody can see or hear anything further than a few feet. Immediately after the fog descends, S changes course and attempts to escape at constant speed under a new, fixed course. Meanwhile, G has lost track of S. But G happens to know S's speed, that it is constant, and that S is sticking to some fixed heading, unknown to G. How does G catch S? G may change course and speed at will. He knows his own speed and course at all times. There is no wind, G does not have radio or radar, there is enough space for maneuvering, etc. ==> geometry/smuggler.s <== One way G can catch S is as follows (it is not the fastest way). G waits until he knows that S has traveled for one mile. At that time, both S and G are somewhere on a circle with radius one mile, and with its center at the original position of S. G then begins to travel with a velocity that has a radially outward component equal to that of S, and with a tangential component as large as possible, given G's own limitation of total speed. By doing so, G and S will always both be on an identical circle having its center at the original position of S. Because G has a tangential component whereas S does not, G will always catch S (actually, this is not proven until you solve the o.d.e. associated with the problem). If G can go at 40 mph and S goes at 20 mph, you can work out that it will take G at most 1h 49m 52s to catch S. On average, G will catch S in: ( -2pi + sqrt(3) ( exp(2pi/sqrt(3)) - 1 )) / 40pi hours, which is, 27 min and 17 sec. ==> geometry/spiral.p <== How far must one travel to reach the North Pole if one starts from the equator and always heads northwest? ==> geometry/spiral.s <== One can't reach the North Pole by going northwest. If one could, then one could leave the North Pole by going southeast. But from the Pole all directions are due south. If one heads northwest continuously, one will spiral closer and closer to the North Pole, until finally one can't turn that sharply. ==> geometry/table.in.corner.p <== Put a round table into a (perpendicular) corner so that the table top touches both walls and the feet are firmly on the ground. If there is a point on the perimeter of the table, in the quarter circle between the two points of contact, which is 10 cm from one wall and 5 cm from the other, what's the diameter of the table? ==> geometry/table.in.corner.s <== Consider the +X axis and the +Y axis to be the corner. The table has radius r which puts the center of the circle at (r,r) and makes the circle tangent to both axis. The equation of the circle (table's perimeter) is (x-r)^2 + (y-r)^2 = r^2 . This leads to r^2 - 2(x+y) + x^2 + y^2 = 0 Using x = 10, y = 5 we get the solutions 25 and 5. The former is the radius of the table. It's diameter is 50 cm. The latter number is the radius of a table that has a point which satisfies the conditions but is not on the quarter circle nearest the corner. ==> geometry/tetrahedron.p <== Suppose you have a sphere of radius R and you have four planes that are all tangent to the sphere such that they form an arbitrary tetrahedron (it can be irregular). What is the ratio of the surface area of the tetrahedron to its volume? ==> geometry/tetrahedron.s <== For each face of the tetrahedron, construct a new tetrahedron with that face as the base and the center of the sphere as the fourth vertex. Now the original tetrahedron is divided into four smaller ones, each of height R. The volume of a tetrahedron is Ah/3 where A is the area of the base and h the height; in this case h=R. Combine the four tetrahedra algebraically to find that the volume of the original tetrahedron is R/3 times its surface area. ==> geometry/tiling/count.1x2.p <== Count the ways to tile an MxN rectangle with 1x2 dominos. ==> geometry/tiling/count.1x2.s <== The number of ways to tile an MxN rectangle with 1x2 dominos is 2^(M*N/2) times the product of { cos^2(m*pi/(M+1)) + cos^2(n*pi/(N+1)) } ^ (1/4) over all m,n in the range 0<m<M+1, 0<n<N+1. [Exercises: 0) Why does this work for M*N odd? 1) When M<3 the count can be determined directly; check that it agrees with the above formula. 2) Prove directly this formula gives an integer for all M,N, and further show that if M=N it is a perfect square when 4|N and twice a square otherwise. ] Where does this come from? For starters note that, with the usual checker- board coloring, each domino must cover one light and one dark square. Assume that M*N is even (but as it happens our formula will work also when both M,N are odd --- see exercise 0 above). Form a square matrix of size M*N/2 whose rows and columns are indexed by the light and dark squares, and whose (j,k) entry is 1 if the j-th light and k-th dark square are adjacent and zero otherwise. There are now three key ideas: First, the number of tilings is the number of ways to match each light square with an adjacent dark square; thus it is the _permanent_ of our matrix (recall that the permanent of a rxr matrix is a sum of the same r! terms that occur in its determinant, except without the usual +1/-1 sign factors). Second, that by modifying this matrix slightly we can convert the permanent to a determinant; this is nice because determinants are generally much easier to evaluate than permanents. One way to do this is to replace all the 1's that correspond to vertical adjacency to i's, and multiply the whole thing by a suitable power of i (which will disappear when we raise it to a fourth power). [Exercise 3: check that this transformation actually works as advertised!] Third, that we can diagonalize the resulting matrix A --- or, more conveniently, the square matrix of A' order M*N whose order-(M*N/2) blocks are 0,A;A-transpose,0 , whence det(A') = +-(det(A))^2. Then the rows and columns of A' are indexed by squares of either hue on our generalized checkerboard, and its entries are 1 for horizontally adjacent squares, i for vertically adjacent ones, and 0 for nonadjacent (including coincident) squares. This A' can be diagonalized by using the trigonometric basis of vectors v_ab (a,b as in the formula above) whose coordinate at the (m,n)-th square is sin(a*m*pi/(M+1)) * sin(b*n*pi/(N+1)). Exercise 4: verify that these are in fact orthogonal eigenvectors of A', determine their eigenvalues, and complete the proof of the above formula. (None of this is new, but it does not seem to be well-known: indeed each of the above steps seems to have been discovered independently several times, and I'm not sure whom to credit with the first discovery of this particular application of the method. For different approaches to exactly solvable problems involving the enumeration of domino tilings, see the two papers of G.Kuperberg, Larsen, Propp and myself on "Alternating-Sign Matrices and Domino Tilings" in the first volume of the _Journal of Algebraic Combinatorics_.) --Noam D. Elkies (elkies@zariski.harvard.edu) Dept. of Mathematics, Harvard University ==> geometry/tiling/rational.sides.p <== A rectangular region R is divided into rectangular areas. Show that if each of the rectangles in the region has at least one side with rational length then the same can be said of R. ==> geometry/tiling/rational.sides.s <== "Fourteen proofs of a result about tiling a rectangle" (Stan Wagon) _The American Mathematical Monthly_, Aug-Sep 1987, Vol 94 #7. There was also a fifteenth proof published a few issues later, attributed to a (University of Kentucky?) student.