bonanova
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Clarify: . Since the table is round, between any two persons there are two sets of intervening people. Do we always take the smaller number? Doh! Forget that one ... but, . The King sits after all the spacings have been set. So the King does not apply to the separation count?
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Bingo. Yeah, the things we remember sometimes.. .
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Correct for the outer circle; close but not quite for the inner circle.
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Yes. In the words of the OP, "... unique, single points ...." i.e. tangents, not intersections.
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For the second part Edit: added second part
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and Yes. There is a cute way to arrive at the answer without performing a calculation and to phrase it without using numbers or symbols.
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A moment's reflection will verify that in the plane a set of circles, each of which touches all the others at unique, single points, numbers at most four. Let three circles have radii 1, 2 and 3, respectively, and arrange them so each touches the other two externally. That is, the center of each circle is outside both of the other circles. Like a dime, nickel and quarter might look. There are now two ways a 4th circle can be in contact with each of these circles. Make a sketch of each: . It may lie in the space between the first three.It may enclose the first three, so that their outer edges each touch the inside of the 4th circle at different points.. Determine the radius of the 4th circle in each case.
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Sketch four circles, two side by side and two more directly above them. All four circles have the same diameter and touch their two nearest neighbors. Quickly determine the area of the space between the circles. Extra credit: Sketch three circles, two side by side and a third resting above and between them. All three circles have the same diameter and touch the other two. Quickly determine the area of the space between the circles. Enjoy, and please use spoilers.
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Yup, the first of those is in my post 26.
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Nice job on Challenge 2. Very counter-intuitive result.
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Nice work all. Note that nobody's incorrect assumption leads nevertheless to the correct answer. And the pretty relationship among the four distances, in both 2-D and 3-D. We're waiting for word of a drill on G-pa's wheat field.
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Hats on a death row!! One of my favorites puzzles!
bonanova replied to roolstar's question in New Logic/Math Puzzles
You haven't missed anything. The OP does not specify the relative numbers of red and black hats. Nothing about their relative numbers is taken for granted. You are correct in stating there could be 6 black and 14 red hats. The solution must apply for any numbers of red and black hats. -
The consensus has it.
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Yes. Check the OP - I added a clue
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A couple turns of the century ago my grandfather homesteaded a few hundred acres of wheat land in NW North Dakota. Today that land is famous not for its wheat but for its oil, buried in shale, miles under the ND prairies. This gives rise to a geometry challenge, and this time it's 3-dimensional. An oil drill [unfortunately not on Grandpa's land] found oil-bearing shale at an underground location that was determined to be exactly 28,000 feet from one corner of a [rectangular] wheat field, 24,000 feet from the opposite corner and 36,000 feet from a third corner. How far is the underground location from the fourth corner of the field? Note: we're not looking for the location on the surface where the drilling began.
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Yeah ... you all got this one.
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Here are two related challenges; one simple, one more difficult Challenge 1: A magician produces a deck of four cards, 2 red and 2 black, and deals them face down [the backs of the cards are indistinguishable.] He bets that you cannot point to two cards of the same color, and offers even odds. A friend urges you to take the bet: "There are only three cases: 2 red, 2 black and one each. You win 2 out of 3 times." .Another friend says you will neither win nor lose: "There are four cases: red-red, red-black, black-red, and black black. You win 2 out of 4 times, and that just matches the odds." Quickly decide which friend is right. Challenge 2: A deck with equal numbers of red and black cards gives a probability p that two selected cards will have the same color. If a card is removed the color symmetry is broken, and one color is more likely to be represented in the two chosen cards. Calculate the increase in p resulting from the removal of the card. Enjoy, and please use spoilers
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Sketch a circle, and another circle whose center lies on the perimeter of the first. At one intersection of the circles center a third circle - all three of the same radius. The center of each circle lies on an intersection of the other two. See that funny little curved triangle in the center? Quickly compare its area to 1/4 of the area of one of the circles. . Less than?Equal to?Greater than?.
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Draw a sketch of a square with three-inch sides. On one side, mark off 1-inch distances - i.e., trisect one of the sides. Draw a four-inch line segment from the center of the square through one of these 1-inch marks. Use that line as a side of a four-inch square that has one of its corners at the center of the three-inch square. Complete the sketch of the four-inch square, and shade in the region of overlap. This is much simpler than I've probably made it sound. Anyway, the challenge is to quickly determine the area of the overlapping region. Please use spoilers
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Good point. That thought came to me halfway through. But since I thoroughly enjoyed the process ...
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.. I am led to believe that I may have mis-stated the OP in a way that could give rise to these answers. There was an error in calculating time of flight of the bullet. Here are the new results.
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That works. Can we do it without extinguishing a fuse? Some types are difficult to stop.
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I'll allow the interpretation that would make this solution work: Equal time fuses are also equal length, and burn rate - inches per second, say, all the same and constant. Still, there's a method that emulates the hourglass solution given above with no waiting time.