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Frequently Asked Questions (FAQS);faqs.469 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 through every subcube. 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/dominoes.p <== There is a chess board (of course with 64 squares). You are given 21 dominoes 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 dominoes exactly cover the remaining 63 squares? Or is it impossible? ==> geometry/coloring/dominoes.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 trimino 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 points P2 and P3 draw another larger circle (C2 and C3). Where these two circles cross, draw a line (D2). This line should go the center of circle C1 at a rt angle to the original diameter line. This line should cross circle C1 at P4 and P5 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. ==> geometry/cover.earth.p <== A thin membrane covers the surface of the earth. One square meter is added to the area of this membrane. 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/dissections/circle.p <== Can a circle be cut into similar pieces without point symmetry about the midpoint? Can it be done with a finite number of pieces? ==> geometry/dissections/circle.s <== Yes. Draw a circle inside the original circle, 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 indentical rhombuses. 2) 6 indentical kites(?). 3) 4 indentical trapezoids. 4) 8 indentcal 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/square.70.p <== Since 1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2, can a 70x70 sqaure 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/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? ==> 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/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 mutiliated. Is it still possible to divide the ham sandwich with a straight knife cut such that both the ham and the 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. One 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 area of the "doughnut" 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 ==> 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/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.