bonanova
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A really simple connect-the-dots problem
bonanova replied to bonanova's question in New Logic/Math Puzzles
Very clever. I like this solution so much that I submitted it to the combined Town Boards of the ABCD Municipality for approval. They texted me back and said they were willing to proceed, even to bear the cost of all that additional asphalt, provided that -
Nice. But you can do better on part 1.
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Correct. Each play leaves another knight on the board.
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I can't do spoilers on iPad, but this answer is a bit of hoax, I judge. So showing it doesn't disclose the real (npi) solution.
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A classic puzzle notes that 3 miles of road is sufficient to connect four cities situated on the corners of a square one mile on a side. Yes, the cities are really small; point-like, for our purposes. The question is then asked: what length of road(s) is necessary to connect them? I.e. the shortest distance of road(s) required.
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This puzzle reportedly dates from the 13th century. So some of you older (ahem) Denizens may have encountered it already. Anticipating this possibility, I've added a competitive wrinkle to the mix. On a 3x3 chessboard you are to place a knight on an unoccupied square and then move it to another unoccupied square using a legal knight's move. If you alternate turns with an opponent, the board eventually fills up to a point where there are not sufficient unoccupied squares for one of the players to complete a turn. The player who has that turn then loses the game. I.e., the last player to successfully complete a turn is the winner. If first player is A and second player is B, which player wins the game if 1, Players cooperate to maximize the number of turns. 2. Players compete to win the game? If you wish to describe a sequence of moves in your answer, it may be helpful to number the squares row-wise left to right starting with the top row 1 2 3 4 5 6 7 8 9 A legal move would then be 1 - 6, for example, provided that quares 1 and 6 are unoccupied prior to the move. Square 6 would remain occupied. Assume there are enough knights available to each player.
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That's it. Nice explanation. And the star?
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It should. I've made that edit. Thanks for making the point.
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Waving across the forum to Y-san. Welcome back! LTNS. I have no answers yet.
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Construct a regular 5-pointed star, and cut it into acute triangles. Or sketch the star and draw lines. The triangles need not be congruent. What is the smallest number of cuts (lines) needed to accomplish this? A Greek cross is the union of five squares: one each above, below and either side of a central square. Ignoring the lines joining the squares and taking only the outside perimeter, or by constructing the shape, divide a Greek cross into the smallest number of acute triangles. How many?
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Yup. Very nice.
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Precisely. Well done.
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Yeah, you have it. Nice.
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Yes. Assume complete mixing.
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Of course! Here it is [thanks to Plainglazed for a tip].
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A familiar problem concerns two glasses, one containing water, the other wine [edit: in equal amounts]. A certain amount of water is transferred to the wine, and an equal amount of the mixture is transferred back to the water. Is there now more wine in the water glass than there is water in the wine glass? A moment's thought shows the two amounts must be the same. Is it possible, by thus transferring the same amount of fluid, any number of times, to reach a state where the percentage of wine in each mixture is the same?
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A Sam Lloyd classic puzzle has two ferry boats starting at the same instant from opposite sides of a river, traveling across the water at right angles to the shores. They travel at constant, but possibly different speeds. The boats pass at a point 720 yards from the nearest shore. Both boats dock for 15 minutes, to discharge and receive passengers, before heading back. On their return trips they meet again, 400 yards from the other shore. How wide is the river?
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Four cards are taken from a standard deck, one from each suit. They are shuffled and dealt face down on a table. Coins are then placed at random on two of the cards. What is the probability that the two chosen cards are the same color?
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Nicely done ... and the interesting property of the third circle is that it's in all cases centered above the left end of the triangle's base. I was looking around for a geometrical proof of that property.
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Not at all. I agree that's how it looks, but that was chance at work. I intended for r to be any value between 0 and R. It might be interesting to get a solution first with r=R/2 and then generalize. Or even simpler, the extreme cases of r equalling 0 or R.
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I mean really ... magnified ... 2915 times. Got five minutes? The ending is worth the wait. Or just scrub to 4:30 and watch from there.
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Welcome back, Denizens. Sharpen your pencils and brush up on your sines and cosines. Here's a simple geometry / trig puzzle to start the new BD site off. Draw two circles with radii R > r that are centered on the positive x-axis and tangent at the origin. Draw an isosceles triangle, in the upper half-plane, whose base is the portion of the x-axis between the circles -- that is, outside the smaller circle and inside the larger circle -- and whose apex touches the larger circle. Draw a third circle, also in the upper half-plane, that does not intersect any of the three existing shapes but is tangent to all of them. For the third circle, determine Its radius ro The x-coordinate of its center Have fun!
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Classical treatment of friction says it's independent of surface area. I'm betting your experiment shows otherwise, and that greater surface area gives [slight but measurably] greater friction. The simple coefficient of friction formula is practical in use, but nonetheless approximate.
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Since the light path can't be re-entrant, that is, loop back on itself, and then leave the grid, there must be a theoretical upper limit to the path length - the total length of all possible diagonals. Wondering what percent of the theoretical maximum the best solution is ...
