The nth locker is opened or closed by student number k precisely when k divides n. So if student k changes locker n, so does student n/k. They cancel each other out. This always holds unless students k and n/k are precisely the same person. That is, k = n/k. The lockers that are exact squares will remain open. These are lockers 1, 4, 9, 16, 25, etc. How many of these are there in a row of 1,000? You can go all the way up to 31×31 = 961, hence there are 31 lockers open.
Each domino covers one black square and one white square. All 31 dominoes must cover 31 black squares and 31 white squares. However, the modified checkerboard contains 32 black squares and 30 white squares, or vice versa. It cannot be covered with 31 dominoes.
Let's try to construct a party that does not have three mutual friends or three mutual strangers. Suppose you are one of the six people at the party. Let's say you know three other people. We'll call them Alice, Bob, and Carol. If Alice and Bob also know each other, then we have three people, you Alice and Bob, who know each other. We're trying to construct a party where this doesn't happen, so Alice and Bob don't know each other. The same holds for Alice and Carol, and Bob and Carol. Thus the three people that you know, don't know each other, giving a triangle of strangers.
Now if you only know two people at the party, or one person, or perhaps nobody, then there are three people at the party who are strangers to you. Once again we'll call them Alice, Bob, and Carol. If Alice and Bob don't know each other, we have a triangle of strangers. So Alice and Bob do know each other, along with Alice and Carol, and Bob and Carol. These three form a triangle of friends. No matter how we arrange things, there will be three guests who know each other or three guests who are mutual strangers.
First we need to combine the two equations into one. Let h be the number of horses, p the number of pigs, and r the number of rabbits. Double the prices, just for a moment, so we can write 20h + 6p + r = 200. At the same time, h + p + r = 100. Subtract the second from the first to get 19h + 5p = 100. Since h and p are integers, this is a Diophantine equation. There are ways to solve these in general, but in this case we can just try various values of h, to see if p winds up being a whole number. For instance, if h = 0, p = 20, and that works. The farmer buys 20 pigs and 80 rabbits. The only other combination is 5 horses, 1 pig, and 94 rabbits.
Imagine there are two of you. One runs forward and one runs backward. The backwards runner reaches the beginning of the bridge just as the train reaches the bridge. He has covered 3/8 of a mile. At the same time the forward runner has covered 3/8 of a mile. He is now 3/8 + 3/8 = 3/4 of the way across the bridge. He runs the last quarter mile while the train moves across the entire one mile bridge. Thus the train is moving four times as fast as the runner. The train is traveling at 40 mph.
A cannibal and missionary go across, and the missionary comes back.
Two cannibals go across and one comes back.
Two missionaries go across and a missionary and cannibal come back.
Two missionaries go across and the cannibal comes back.
Two cannibals go across and one comes back.
Finally the last two cannibals go across.
In the beginning, the rock caused the water level to rise by an amount that exactly compensates for the weight of the rock. In the end the water level is increased by an amount equal to the volume of the rock. The rock is heavier than water, so its weight displaces more water than its volume. The water level is lower after you throw the rock overboard.
At the end, each glass has the same amount of liquid. If there were more water in the wine than wine in the water, there would be more water, in the two glasses taken together, than wine. Yet we started out with equal amounts of water and wine. Both ratios must be the same.
Here is another way to visualize it. Put the water (clear) on the left, and the wine (red) on the right. After the experiment, suppose the right glass is 8% water, and the left glass is 6% wine. The total water is94 + 8 or 102%. But it has to be 100%. So if 8% of the water has moved to the glass on the right, then 8% of the wine must have moved from right to left, so that both glasses have the same amount of liquid.
and it doesn't really matter if you thoroughly mix the spoonful of water into the wine before carrying a spoonful back. Stir it a lot, or don't stir it at all. If you happen to scoop the same spoonful of water out of the wine (completely unmixed), and bring it back to the water, the ratios are still the same.
First figure out what the total is. If each row adds up to t, then all three rows add up to 3t. Yet this is all the numbers 1 through 9, hence 3t = 1+2+3+4+5+6+7+8+9 = 45, and t is 15.
Next consider the for lines passing through the middle square. Taken together, these numbers add up to 4×15 = 60. Yet this is just the numbers 1 through 9, where the middle entry is used another 3 times over. Thus 3 times the middle is 60 - 45, and the middle square is 5. Some trial and error reveals the pattern.
| 2 | 9 | 4 |
| 7 | 5 | 3 |
| 6 | 1 | 8 |
suppose you are looking directly at the spider, i.e. at his end of the box. The spider must travel down and to the right, then cut across the bottom of the box until he reaches the right edge. He takes a long diagonal along the right side of the box, until he reaches the top. He then cuts across the top at an angle, and crawls down the far side until he reaches the fly. This seems like a crazy crooked path until you unfold the box. flip the spider's end down and the fly's end up, cut open the left side, and lay the whole thing flat. Now the spider's path becomes a straight line. It is the hypotenuse of a right triangle with base 24 inches and height 50 inches. The path length is 55.4 inches, well within the spider's range.
Take 1 coin from the first sack, 2 coins from the second, 3 coins from the third, and so on up to 10 coins from the tenth sack. Place them all on the scale and determine the weight of all 55 coins. If they were all counterfeit, the weight would be 55 ounces. If the first sack holds gold, one coin will be pure gold, and the weight will be 55 + 1 = 56 ounces. If the two coins from the second sack are gold, the weight will be 55 + 2 = 57 ounces. Similar reasoning holds for the other sacks. Subtract 55 from the total weight; This gives the number of the sack that holds the gold coins.
Add the numbers up in a different order: 1+100 + 2+99 + 3+98 + ... + 50+51. This produces 50 copies of 101, hence 50×101, or 5,050.
Place the light bulb at the center of your picture and draw a circle of radius 1 around it. Then draw another circle of radius 2 around the light bulb. Beams of light leave the bulb, bounce straight back from the inner circle, and return to the bulb. The outer room, the anulus between the two circles, remains dark, but that's because it is closed off. Cut a small door in the inner circle. Light now streams out through this door, bounces off the outer circle, and returns to the light bulb. As long as you're not standing near the doorway, the outer room remains dark.
This is a simple construction in theory, but it will not work in practice. If a person or object is placed in the inner room, some of the light bounces off that object and leaves the inner room at a different angle. This light does not bounce straight back into the inner room. Instead it bounces away and spreads throughout the anulus. Click here for a better solution.
You can place any of 6 books in the leftmost position. No matter which book you select, there will be 5 books to choose from as you make your next selection. Once the second book is in position, there are four books to choose from. This pattern continues down to the last book -- no choice there. Thus the answer is 6×5×4×3×2×1 = 720.
The simplest answer to this question is the north pole. The other solutions are all in Antarctica. Start approximately 1.16 miles north of the south pole, and walk south for one mile. You are now about 1/6 of a mile from the south pole. As you walk east, you trace an exact circle around the south pole, returning to your original longitude. Walk north a mile and you're back where you started from. If you move a little closer to the south pole, your eastward trek takes you twice around the south pole. A solution exists for every positive integer n. Your starting location is 1 + 1/(2πn) miles north of the south pole.
The base cell is a regular hexagon, ½ inch on a side. Note that the diameter of this hexagon is 1, hence it can be colored with one color. Tile the entire plane with these hexagons. As we move along one row of hexagons, color the cells red orange yellow green blue pink purple, and repeat. The row just above this one is colored the same way, but shifted ober, so that the red hexagon rests above yellow and green. The row above this is colored the same way, with the same phase shift. This pattern repeats as successive rows cover the entire plane. Verify that the closes red hexagons are sqrt(7)/2 = 1.32 inches apart. Therefore points that are 1 unit apart always have different colors.
The 7 color hexagon tiling is regional, i.e. small regions of the plane are assigned individual colors and it all works out. Any regional solution requires at least 6 colors. I have developed a proof of this; I may post it online at some point.
Make a graph of height verses time. Let the x axis run from 6 AM to 6 PM, and let the y axis run from base to summit. Using a blue pen, graph height, as a function of time, for Wednesday. This curve generally rises, though there are flat sections, when you stopped to eat lunch, and there may even be a dip here and there, when you step down into a small valley.
Next, using a red pen, draw Thursday's graph. This starts at the upper left and slopes down to the lower right. No matter how strange these curves are, they must cross somewhere. This is the point where your elevation was the same on both days.
There are 1,000 perfect squares between 1 and 1 million. These are the squares of the first thousand numbers. Similarly, there are 100 perfect cubes, the cubes of the numbers from 1 to 100. Subtract squares and cubes from a million to get 998,900. However, we have subtracted some numbers twice. Every number that is a perfect sixth power has been subtracted twice, so add these back in, giving 998,910.
The best possible score in a given frame is 19, 10 for the spare and 9 from the following frame. This is even true for the tenth frame, which is a bit weird. Thus the answer is 10×19 = 190.
Consider how much the two hands have moved since 3:00. Let t be the angle between 12:00 (straight up) and the two hands. The minute hand moves 12 times as fast as the hour hand, but the hour hand had a 90° head start. Thus we write t = 90 + t/12. This sets t = 1080/11. Now convert from degrees to minutes. The minute hand moves 6 degrees per minute, hence it has been traveling for 180/11 minutes. The time is 03:16:21.818181…, or 03:16:22 rounded to the nearest second.
You may also notice that the hands coincide 11 times in a 12 hour period. Everything is linear, so the time we seek is 12 hours times 3/11, which gives the time shown above.
We already discovered there are 2 cubes with 3 red and 3 blue sides. Let's see how many cubes have more red sides than blue. The entire cube could be red, or one side could be blue (that's 2). If two sides are blue, they could touch or they could be opposite (that's 2 more). Now we have 4 cubes with more red than blue, and that means there are 4 cubes with more blue than red. Add these together and bring in the 2 cubes with equal amounts of red and blue. There are 10 different cubes.
If we have 3 colors to choose from there are 3 cubes that are all one color, and 6 cubes that have five faces painted one color and the sixth face another. If a cube has four faces one color and two faces another, those two faces could be together or opposite, and there are 6 ways to assign the colors, giving another 12 cubes. If the cube has three of one color and three of another, the same-color faces could meet at a corner or wrap around the cube, and there are 3 ways to assign colors, giving another 6 cubes. Those are all the cubes with two colors; we must now consider the threee-color cubes. If four faces are one color, the two odd colors could be together or apart, and there are 3 ways to assign the colors, giving another 6 cubes. Oh well - I'll leave the rest to you. If you got 57, you did it right.
If there are k colors, there are (k6 + 3k4 + 12k3 + 8k2) / 24 cubes, as per the Burnside Counting Theorem.
Draw an inner circle of radius 1 and an outer circle of radius 2. Then draw 3 lines connecting the inner circle to the outer circle. This creates four regions, and each touches the other three. Such a map requires 4 colors.
Actually this can be generalized. Draw 5 lines connecting the inner and outer circles, or 7 lines, or 9, or any odd number of lines. As you walk all the way around the anulus between the two circles, you step through an odd number of countries, forming a ring. If you try to color them with just 2 colors, the countries alternate red blue red blue red blue ..., but when you get back to start, two red countries run into each other. Thus you need 3 colors for this ring of countries, and the country in the middle, the inner circle, requires a fourth color.
In the map of the United States, Nevada is surrounded by a ring of 5 states, hence the map requires 4 colors.
suppose there is such a fraction. We may as well reduce it, just as you would turn 6/8 into 3/4. After this is done, x and y cannot both be even. Now if x/y is the square root of 2, Then x×x divided by y×y is exactly 2. Put another way, x×x = 2×y×y. This means x is even, so y is odd. Now x×x is divisible by 4, but 2×y×y is not (since y is odd). Therefore there is no fraction that is exactly the square root of 2.
A string encloses the most area against one wall when it forms a semicircle. The proof is simple and beautiful! Suppose the curve was anything else. It touches the wall at points a and b. Reflect this curve on the other side of the wall, then remove the wall. Now the two curves together enclose an area larger than would be contained by a perfect circle. That is a contradiction, therefore a half circle holds the most area against one wall.
Let two walls meet at a corner. Let the string have length 1. If you only use one wall, the string is a half circle whose radius is 1/π. The area of the whole circle is 1/π, and the area of the half circle is 1/2π. That's about 1/6 which isn't great. You could draw a hypotenuse against the corner, of length 1, making a right triangle with legs of length sqrt(½), and that triangle has an area of 1/4, which is larger than 1/6. So you have to use the corner, not just one wall. Let the curve meet one wall at point a and the other wall at point b. Reflect the curve through the first wall, meeting the wall at a. Reflect through the second wall, meeting the wall at b. And rotate the curve around the corner 180 degrees to complete the picture. The result has to be a circle, hence the largest area enclosed by a string and a corner is a quarter circle.
If two walls meet at 60 degrees, a sixth of 360 degrees, you can do exactly the same thing. Prove that you can get a bigger area by using both walls, then reflect 6 times around, and the result must be a circle, thus the string forms a sixth of a circle with the two walls.
The odd case, pun intended, is when the walls are an odd quotient of 360. Let the walls meet at an angle of 120 degrees. Again, start by showing you need to use both walls. The curve meets one wall at a and one wall at b. Reflect through the first wall, so that both curves touch at a. Now there are two curves, whose end points are at a distance b from the origin. We can't put the third curve in place to complete the picture, because our curve runs from a to b. We need a and b to be equidistant from the origin. Here is a trick. Draw a line segment l from a to b. The enclosed area is in two parts, between l and the curve, and between the walls and l. Slide l along the two walls, keeping the curve intact. The area between l and the curve doesn't change, but the area between l and the walls does. Intuition, or calculus, tells us the area is maximized when a = b. Now we can rotate the curve three times around, and we must find a circle. The same holds if the walls meet at 72 degrees (a fifth of 360), and so on. Maximum area is always a slice of pizza.
I skipped past an important detail. As you slide l along the two walls, how do you know the curve won't collide with one of the walls? The curve would have to bend inward, at a or b, if it collides with the wall when l moves. If a curve bends inward at a wall, you could always run straight into the wall, at 90 degrees, and use less string, and enclose more area. The extra string then increases the area further. Therefore the curve touches a and b perpendicular to the walls, or outward, away from the corner. Sliding l will not cause the curve to collide with a wall.
There is probably a way to generalize this to any rational angle, and then any irrational angle by continuity. It's always a slice of pizza.
Apply the angle addition formula. Sin(t-120°) is -½ (sin(t) + sqrt(3)×cos(t)). To simplify notation, let x be the first term and y the second, giving -½ (x+y). Now sin(t-240°) expands into a similar expression, -½ (x-y) to be exact. We have to square these expressions and add them together, giving ¼×2×(x2+y2). Substitute for x and y to obtain ½ + cos(t)2. Bring in sin(t)2 and the result is ½+1.
The power extracted from the generator is constant, perfectly balanced between the three cities. If it weren't, the time varying megawatts of power would literally shake the generator to pieces. This is surely the most useful trig identity of them all.
Large factories are sometimes given "three phase power", so that the generator remains in balance as they turn on and off their machines.
Break each edge in half, at its midpoint, whence each vertex accounts for 3 half edges. Similarly, each vertex accounts for 1 sixth of a hexagon plus a sixth of a hexagon plus a fifth of a pentagon. In euler's formula, v-e+f = 2, where v is the number of vertices, e the number of edges, and f the number of faces. Write e = 3v/2, and f = 8v/15. solve and get v = 60, e = 90, and f = 32. The pentagons consume all the vertices, and consume them without overlap, hence there are 12 pentagons, leaving 20 hexagons.
If you don't feel like building your own paper model, just pick up a soccer ball. Every vertex joins two hexagons and a pentagon, and there are 32 faces.
The king of the Archemedian solids, with 92 faces, Has a pentagon and four triangles at each vertex.
If you try to place two pentagons and a hexagon at each vertex, just walk around the perimeter of a pentagon. The shapes alternate pentagon, hexagon, pentagon, hexagon, pentagon, oops - there's a vertex with three pentagons. This generalizes to a rule about using odd-numberd gons at vertices.
Yes. Standard currency is quite thin. Toss a nickel 100 times; does it ever wind up standing on edge? Only on the Twilight Zone. (remember that episode?) Now imagine stretching that nickel into a long rod with heads at one end and tails at the other. If you toss this rod high into the air, it certainly won't land on one of its ends (heads or tails) and stay there, just as a dropped pencil won't stand up on its eraser. Somewhere between these two extremes, there is an ideal thickness, where heads, tails, and edge are equally likely.
In fact you can make a fair coin with equally likely outcomes for any modest value of n. Start with a regular polygon with n-2 sides, then stretch it to an appropriate thickness, so that heads and tails are just as likely as any of the n-2 sides around the edge. Determining these thicknesses is quite difficult in practice. Of course when n = 6 you obtain the standard die (cube).
At first you are tempted to run up the powers of 2. Weights of 1, 2, 4, and 8 pounds, in various combinations, can weigh everything from 1 to 15 pounds, but that's nowhere near 40. The trick is, some of the weights can sit on the left pan, with the box. Each weight can be on the left, the right, or not used at all. This suggests powers of 3.
Using 1 and 3 pound weights, you see immediately how to measure 1, 3, and 4 pounds. If the box weighs 2 pounds, put the 1 pound weight on the left pan with the box, and the 3 pound weight on the right, and everything is in balance. Next, bring in a 9 pound weight and put it on the right. We already showed we can add a net weight of 1 through 4 to the box, so if the box weighs anywhere from 5 to 8 pounds, we'll be able to balance it against the 9 pound weight. Of course there is no trouble if the box weighs 9 pounds, and if it weighs anywhere from 10 to 13 pounds, we achieve balance by adding (net) 1 to 4 pounds to the right. Now we've covered everything from 1 to 13 pounds, and the last weight to bring in is 27 pounds. This lets us weigh everything from 27-13 pounds up to 27+13 pounds, or 40 pounds. An inductive proof shows successive weights should always be powers of 3.
To weigh a box from 1 to 81 pounds, double the weights. Thus the weights are 2, 6, 18, and 54 pounds. If the weight of the box is even, you can determine it using the procedure outlined above. If the weight is odd, the balance will always tip one way or the other. Determine that the box is heavier than n and lighter than n+2. For instance, the box might be more than 30 pounds and less than 32 pounds, whence it is 31 pounds.
Overlay a black and white checkerboard onto the 10 by 10 square and note that there are just as many black squares as white. Each tetromino, no matter how it is placed, introduces either an excess or a deficiency of 2 black squares. The "excess" tetrominoes must equal the "deficient" tetrominoes, hence there must be an even number of pieces in the puzzle. We can't tile any rectangle using an odd number of pieces. The 10 by 10 board is out, as is the 6 by 14 board, and so on.
Similarly, you can't tile the 10 by 10 board with the 1 by 4 tetrominno (a rectangle of length 4 and width 1). This time the overlaid checkerboard has larger squares; each square is 2 pixels on a side. There are 25 such squares. If the lower left square is white, then there is an extra white square. However, each piece, no matter how placed, consumes equal amounts of black and white. It is not possible to cover more white than black; it is not possible to tile the rectangle. This is just the beginning of the wonderful world of polyominoes.
You'd think it didn't matter, one box or the other, but actually you should keep the one you started with. I could draw you a series of probability trees, but that really doesn't help explain it. Instead, let me illustrate by jumping to the extreme.
Monty shows you 500 boxes, you pick one, and he opens 498, all with beautiful prizes. That leaves only the one you selected and one other box. Surely you didn't pick the joke box from the start. but if you switch, you're probably going to get the chewing gum. Keep the one you have!
The math is not as extreme when there are only three boxes, but the answer is the same. If you stick with your first choice you'll win 2/3 of the time. If you switch you'll only win 1/3 of the time.
Note that the math is reversed if there is only one prize and two gag gifts. In that case you should switch. again the math is clearer when there are 500 boxes and just one prize. You select a box that is almost surely a joke, Monty shows you 498 worthless boxes, and then you should switch to get the prize.
If the buckets are labeled A B and C, holding 10 7 and 3 quarts respectively, here are the transfers.
A→B B→C C→A B→C C→A B→C A→B B→C
Here are the coordinates of the jumping pegs, in order.
G5 G4 E5 F3 H3 H5 I3 G3 E3 G2 G1 E1 I2 I1 G1 I5 J3 H3 F3 D3
It is impossible to get a peg up to row A, even if the rows are infinite left to right, and even if there are infinitely many rows below row E. You have an entire half plane of pegs to work with, and yet you still can't get to row A. The proof is a bit technical; you can see it here.
Given the list, write the decimal expansion of a real number r as follows. The first digit after the decimal point is anything other than 6, because 6 is the first digit in the first number on the list. The second digit of r is anything other than 3, because 3 is the second digit in the second number on the list. The third digit of r is anything other than 2, and so on. If r is in position j on the list, then the jth digit of r must be something other than the jth digit of r. This is a contradiction.
Actually we need to be a bit careful here, because some real numbers have two decimal representations. Suppose you construct r = .3699999999[forever]. That same real number might be on the list, but in the form of .3700000000[forever]. This is easy to fix. When you select the digits of r, never choose 0 or 9. Thus r cannot be represented in any other way.
You can construct such an r for any list at all. Thus there is no list that includes all the real numbers from 0 to 1. This diagonalization argument shows there are more real numbers than integers. Both sets are infinite, but one is a larger flavor of infinity than the other. There are infinitely many flavors of infinity, and even more beyond those.
Slide the left and right points of the diamond towards the origin, shrinking the diamond along its horizontal dimension. Stop when the left and right points have coordinates +2 and -2, just like the top and bottom points. Now our diamond has become a square, standing on one corner. Any ellipse that was drawn inside the diamond becomes an ellipse inside the square, and its area has been reduced by half. The largest ellipse that fits inside a square is a perfect circle. In this case its radius is the square root of 2, hence the area of the circle is 2π. Now stretch the square back out into a diamond, and the circle becomes the largest possible inscribed ellipse, having an area of 4π.
Arrange the puzzle so that its lower left corner is at the origin. For each of the 21 squares, I will give its length and the coordinates of its lower left corner. The big square, 50 units on a side, is at the lower left, hence its lower left corner is at 0,0.
50 @ 0,0
29 @ 0,50
33 @ 0,79
25 @ 29,50
4 @ 29,75
37 @ 33,75
35 @ 50,0
15 @ 50,35
9 @ 54,50
16 @ 54,59
2 @ 63,50
7 @ 63,52.
17 @ 65,35
18 @ 70,52
42 @ 70,70
11 @ 82,35
6 @ 82,46
27 @ 85,0
8 @ 85,27
24 @ 88,46
19 @ 93,27
If all this is a bit algebraic and abstract, you can see some pictures here.
Consider the smallest square in a square of squares. If this smallest square is by an edge, three larger squares surround it - and one of these larger squares extends past the edge of the puzzle. Similarly the smallest square cannot be in a corner, else one of the two squares surrounding it extends beyond the puzzle. The smallest square has to be in the interior.
If a large cube consists of smaller cubes, all of them different sizes, then the floor of the cube is a square made up of smaller squares, all of them different. Focus on the smallest square on the floor - the smallest cube in the bottom layer. It must be in the interior, not next to a wall. It is completely surrounded by larger cubes - cubes that rise higher than the smallest cube. The top of this cube is an enclosed square that must be tiled with a layer of even smaller cubes. The smallest cube in this layer is somewhere in the interior, and the top of this smallest cube must be tiled with yet another layer of even smaller cubes. This continues forever, hence a cube cannot be built from a finite number of smaller distinct cubes.
Your first impulse is to place a connect point in the center, and connect all four houses to that. However that consumes 2 diagonals, or 2 times the square root of 2, or 2.828 miles of wire. This is still too much.
Connection points that reduce edge lengths are called steiner points, and they always have three lines coming into them (not four), and the angles are always 120°. Place two steiner points halfway up the square, approximately 0.288 miles from the left and right sides. Connect the left point to the left houses and the right point to the right houses; then connect them to each other. The angles are all 120°, and you only need 1 + sqrt(3), or 2.732 miles of wire.
It's tempting to solve this by constructing a geometric series, but there's an easier way. The trains collide in one hour, and in an hour the fly travels 70 miles. That's it.
"Would you have to disobey orders, if you told me I could pass through your exit unharmed?"
Try all the combinations, you'll see that it works.
There is no need to add $27 and $2, since the $2 is part of the $27. In fact, the clerk has $2 and the store owner has $25, which makes $27, so it all works out.
Place the queens in the following squares, using algebraic notation.
A4 B7 C5 D2 E6 F1 G3 H8
Look at any one of the 9 points. Its integer coordinates are even or odd. So there are 8 combinations, as x y and z are even or odd. Since there are 9 points, and 8 combinations, one of the combinations repeats. Perhaps x is even and y and z are odd for the points A and B. The midpoint of A and B is (A+B)/2, and for all three coordinates, the result is a whole integer.
6 centimeters on a side. It's an easy calculation; I include it because it's interesting that the sum of 3 consecutive cubes is another cube. There are formulas out there for the sum of consecutive cubes being another cube.
Zero. If 2 ment have their hats back, then the third must have his hat back as well.
Leave all the switches off for an hour. Turn the first switch on and wait an hour. Turn the second switch on and run over to the bedroom. The first switch controls the hot lamp, the second switch controls the lamp that is on but still cool, and the third switch controls the lamp that is dark.
This was a great problem 20 years ago, before cfl bulbs, which don't really get that hot. I'm not sure if a modern bulb gets warm enough for this approach to work. Probably it does, but not as pronounced.
Four lamps and switches are easily accommodated. Leave switches 1 and 2 on for an hour, turn switch 2 off and turn switch 3 on, then run to the bedroom. The lamps will be on and hot, off and hot, on and cold, and off and cold, respectively.
This is one of those problems that is really hard if you try to use calculus, but is quite easy with some high school geometry. Draw the triangle determined by the centers of the pennies. This has area sqrt(3). Within the triangle, the pennies consume three wedges, each 60 degrees of arc. This is an area of π/2, hence the space between has area sqrt(3)-π/2, or 0.161.
Don't use V W X Y or Z as your model, use the letter C. Make it a perfect semicircle with radius 1/π, and the enclosed area is 1/(2π). I believe this is best, but a formal proof is beyond the scope of this article.
What about 3 dimensions? The convex hull is created by stretching shrink wrap around the shape. What is the largest volume?
Connect 3 segments of length 1/3 at the corner of a cube. The shape is a right simplex with volume 1/(27×6).
Connect 4 sticks at a central point, like a methane molecule. Use some geometry to find the volume of this shape. For the moment, let the edge of a tetrahedron have length 1. The base is an equilateral triangle. The distance from its corner to its center is sqrt(1/3). The altitude has height sqrt(2/3). The volume is bh/3 = sqrt(2)/12. The distance from the center of the tetrahedron to any of its 4 corners is sqrt(6)/4. The total length of the 4 sticks, from center to corners, is sqrt(6). Scale the shape so that this length is 1. This divides volume by sqrt(6) cubed, giving sqrt(3)/216 = 0.00801. this is clearly better than 1/162.
Draw a semicircle in the plane and stand a goal post up at one end. for completeness, draw in the other half of the circle. The convex hull is determined by lines joining the top of the post to all the points around the circle. I want to convince you that this is really the same as the cone with the post in the middle. Let the top of the post be the origin, pointing up the z axis. A linear transformation fixes the xy plane, and slides each plane above and below to the left and right proportionally, according to the z coordinate. Our circle slides over, and the goal post stands on the edge where it belongs. This linear transformation maps lines to lines, and the transformed cone becomes our convex hull. Thus the volume is the area of our semicircle times height divided by 3. This is maximized when the semicircle has length 2/3 and the post has height 1/3. The volume is 2/(81×π), or 0.00785. This is surprising; I really thought it would come out better than the tetrahedron. The same cone argument applies if we put the goal post anywhere along the semicircle.
Erect two goal posts at the ends of the semicircle. At each level, the cross section starts at one post, travels a quarter circle, moves in a straight line towards the other post, and runs a quarter circle to that post. Combine the quarter circles and find the volume of the cone shown above. This is πr2h/6. Then add the wedge, which is r2h/3. The best solution has r = 2/(3π) and h = 1/6, giving a sad0.00643.
Join 3 quarter circles together at a point, with angles of 120 degrees, so that they cradle a hemisphere. Of course the convex hull is not a hemisphere; each cross section is an inscribed equilateral triangle, rather than a circle. Adjust the volume of the hemisphere by this ratio and get 4×sqrt(3) over 27π3 = 0.00827.
This is the best I have come up with so far. I haven't had time to pursue this problem further, but it is fascinating.
He can't. Such a path is called an Eulerian trail, after Leonhard Euler, an eighteenth century mathematician. When the spider passes through a node, or in this case a corner, he consumes two edges, the incoming edge and the outgoing edge. If he passes through this point again another two edges are consumed. Therefore each point has to have an even number of edges, except perhaps his start and end points. If he wants to return to start, then every point has to have an even number of edges. However, each corner of the cube has three edges incident to it; that's no good.
The octahedron, with four edges at each corner, 12 edges and 6 corners altogether, works perfectly. Find a path such that the spider starts at the top, travels along every edge exactly once, and returns to the top.
Let the pool table be the unit square in the xy plane, with x running from 0 to 1 and y running from 0 to 1. As a warmup, launch the ball out of the origin with slope 1/3. It hits the right side of the table at the coordinates x = 1 and y = 1/3. Then the ball bounces back, traveling to the left along a line whose slope is -1/3, but instead, let the ball break through the right side and into a reflected copy of the pool table. It is a mirror image of the original, reflected through the vertical line x = 1. Now the ball can continue on its original trajectory with slope 1/3. When x = 2, the ball hits the right side of the reflected pool table, which is the left side of the original pool table. Instead of bouncing back, push the ball through into a third copy of the pool table, which is a reflection of the second, and a direct copy of the first. The ball continues on with slope 1/3 until it reaches the point x = 3 and y = 1. This is a pocket in our pool table. Returning to the original square, the ball launches out of the lower left corner, bounces off the right wall, bounces off the left wall, and lands in the upper right pocket.
Cover the plane with direct and reflected copies of the pool table, forming a checkerboard; then launch the ball with slope p/q, where p and q are relatively prime. Let the ball travle on in a straight line in the first quadrant, corresponding to the ball bouncing around in the original pool table. It lands in a pocket with integer coordinate x iff xp/q is an integer, iff x is a multiple of q. If x is kq then the coordinates are kq and kp. But the ball falls into the pocket at the first opportunity, so set k = 1 and the point is q,p.
Since q and p can be as large as you like, the ball can travel for as long as you like. In the simplest example, the slope is 1/q, almost horizontal, and the ball bounces back and forth across the table, slowly moving up to the top, until it finally falls into the upper left pocket (q even), or the upper right pocket (q odd).
the pocket depends on the parity of p and q. If both are odd then the ball lands in the upper right pocket. If both are even then the ball lands in the lower left pocket from which it was launched. However, p and q are relatively prime, and cannot both be even. The ball can only land in one of the other three pockets.
If the slope is irrational then the x and y coordinates can never be integers at the same time. that would produce a rational slope. thus the ball bounces around the table forever.
Here is the grid again; imagine the lower left corner is at the origin. This is where your pencil will start.
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Draw a line from the origin up to 0,3, then down to 3,0, then back to the origin, then up to 2,2. At the outset, people are reluctant to leave the confines of the 3 by 3 grid, but this constraint is not part of the problem description. If the 3 by 3 grid lives in an imaginary box, then you must "think outside the box" to solve the problem. This is generally accepted as the origin of the phrase "think outside the box", which has become metaphorical for creative solutions beyond the current paradigm.
Note however that there is an "inside the box" solution as well, requiring only 4 lines, if you define a line as a continuous path that forms a line segment in the plane. (And there's no reason not to, given my problem description.) Pause for a moment and see if you can find it.
In topology, a path is any continuous function from a line segment into the plane. A circle for instance, or the letter S, or the figure 8. A path can intersect itself, or even draw back over itself. Start at the origin and draw a line up to 0,2 and back down to the origin, (this is technically one line), then across to 2,0 and back to the origin, then up to 2,2 and back to 1.5,1.5, then up to 1,2 and down to 2,1. Anticipating this mathematically correct solution, a modern description of the 9 dots problem now states that a line cannot retrace itself. Thus you are forced, once again, to think outside the box. For me, the "inside the box" solution is trickier, and more interesting.