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Posts posted by bonanova


My freshman physics prof drew these on the board one morning.
He asked, if the thing on the left is a centimeter, what is the thing on the right?
Your turn (spoilers please.)

At a busy airport a conveyor belt stretches from the runway, where all the planes land, to the baggage claim area inside the terminal. At any given time it may contain hundreds of pieces of luggage, placed there at what we may consider to be random time intervals. Each bag has two neighbors, one of which is nearer to it than the other. Each segment of the belt is bounded by two bags, which may or may not be near neighbors (to each other.)
On average, what fraction of the conveyor belt is not bounded by nearneighbors?
Example:
 belt segment bounded by near neighbors
===== belt segment not bounded by near neighbors... AB==========CDE=============FGHI ...

At 5second intervals, 1024 automobiles enter a straight (and otherwise empty) singlelane highway traveling at initial speeds chosen at random from the interval [50, 70] miles per hour. Cars may not pass nor collide with other cars. When a slower car is encountered, a car must simply reduce its speed, and for the purposes of this puzzle we may consider the cars in such a case become permanently attached, traveling at the slower car's speed.
Eventually there will be N clusters of cars. What is the expected value of N? (Equivalently, what is the expected cluster size?)

On 2/8/2018 at 1:48 AM, Donald Cartmill said:I have it ...the 6th solution
1) a square with 4 sides and two diagonals =6 lines
2) would be a diamond shape created by two equilateral triangles; 5 sides equal and one long diagonal
3) An equilateral triangle ,3 equal sides ; The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter
4) an equilateral triangle is 3 lines connecting points "A" ,"B" and "C", Assume "A" is the vertex and "B" and "C" are the base points. A 4th point "X"gives us a line the same length as a sides of the triangle ,and directly off of vertex point "A" away from the triangle , but perpendicular to the opposing side "B"C"; This arrangement gives us 4 lines of equal length . Now the lines connecting the 2 base points "B"and "C", and point "X" ,are of equal length and longer than the other 4 lines
5) an equilateral triangle is 3 lines connecting points "A" ,"B" and "C", Assume "A" is the vertex and "B" and "C" are the base points. A 4th point "X"gives us a line the same length as a sides of the triangle ,and directly off of vertex point "A",but downward bisecting line BC = 4 lines of equal length; 2 short lines BX and CX
6) 4 points A,B,C,D, lay on two intersecting circles. point A is the center of one circle C and D lay on this circle. D is the center of the 2nd circle with A and B laying on the circle. The 4 points create a trapezoid in which the distance BC , is the same length as chords AB and CD (that is 3 short lengths ) The diagonals AC ,AD and BD are equal all being equal to the radius.
That should be all of them ????
6) 4 points A,B,C,D, lay on two intersecting circles having the same radius. point A is the center of one circle C and D lay on this circle. D is the center of the 2nd circle with A and B laying on the circle. The 4 points create a trapezoid in which the distance BC , is the same length as chords AB and CD (that is 3 short lengths ) The diagonals AC ,AD and BD are equal all being equal to the radius.
I'm not sure I recognize your solution 6 from its description of how it's constructed, but your final description of it seems right. Nice solve.
SpoilerAnother way to describe it is to take a 5pointed star (which fulfills the 2 length criterion) and remove one of the points.

One might start by constraining the points to be on the circle's perimeter.

Another approach:
SpoilerLet the small angle in the diagram be theta. Total length then is 1 + 2 sec(theta)  tan(theta), whose derivative is (2 sin(theta)  1)/cos^{2}(theta) which is zero for theta=30^{o}. The three lines intersect, nicely enough, at 120^{o}. Using 30^{o} for theta, the length is 1 + (41)/sqrt(3) = 1 + sqrt(3), which is ThunderCloud's result

Did OP leave out the part about starting from a random distribution of the (total number of) b = 3x2^{n} beans on the plates? Apologies.



Ten years ago I called attention to a number that when divided by a single integer p it left a remainder of p1. (Help, a remainder is chasing me) Here is a chance to construct a ninedigit number, a permutation of { 1 2 3 4 5 6 7 8 9 } that has no remainders, sort of. The task is to permute { 1 2 3 4 5 6 7 8 9 } to create a number whose first n digits is a multiple of n for any singledigit n.
For example, consider 123654987. Its first 2 digits (12) are divisible by 2. It's first 5 digits (12365) are divisible by 5.
However this is not a solution, since 1236549 is not a multiple of 7.
 1

Arrange the Jacks, Queens, Kings and Aces of the four suits { Spades, Hearts, Diamonds, Clubs }
in a 4x4 array, in such a manner that: Each row has exactly one card of each rank and one card of each suit.
 Each column has exactly one card of each rank and one card of each suit.
 Both major diagonals have exactly one card of each rank and one card of each suit.
Euler, the great mathematician, proposed the task of constructing a similar 6x6 array, but instead it was proven to be impossible.
Does a 5x5 array exist?


Four towns, A, B, C and D, are located such that their centers form the vertices of a square 1 mile on a side. Town planners want to build a set of roads that connect the four town centers while minimizing the cost, which can be considered to increase linearly with road length.
What set of roads minimizes that cost?
A B
C D

Clue:
SpoilerWhat property of the painted squares (collectively) remains constant as more squares are painted?

Nice work. Bonus solution is the one I had in mind.

Bumping this puzzle and offering a clue  unless someone's actively searching for the final case.

In a previous puzzle @plasmid found that by successively doubling the jelly beans on a plate by transferring from one of the other of three plates, it's possible to empty one of the plates.
Suppose the starting number of jelly beans distributed among three plates is a sufficiently nice multiple of 3, namely b = 3x2^{n}. By making successive doubling moves, as in the first puzzle, is it always possible to end up with an equal number ( 2^{n} ) of jelly beans on the three plates?

Hi Cygnet, and welcome to the Den. And nice pic.
SpoilerYour and plasmid's result (.639) obtains if [0, 1] is taken to be the the longest of the sides. While p=.821 (similar to ThunderCloud's result) obtains if it is taken as the middle side. Finally, if it's the shortest side, so that the other two sides are unconstrained in length, plasmid's first result (p~1) obtains. Which makes putting all three points anywhere inside a circle a case that gives an unambiguous and pleasing result.

What fraction of triangles in a circle are obtuse?

This puzzle is an ancient one that doesn't have a definitive answer so far as I can find. Points given to ThunderCloud and plasmid for proofs of possible answers, but will leave the puzzle open for further comments.
I have a criticism of this puzzle, closely related to Pickett's comments, that I haven't seen raised elsewhere: There is no such thing as a random triangle in the plane. How do you pick random points in the plane? We can impose a coordinate system that makes (0, 0) a reference point, and we can add (1, 0) to provides a scale factor and orient the axes, but that's it. What we can't do is pick three arbitrary points in the plane. The origin can be the first point, WOLOG, but the other two, if truly chosen at random, are both points at infinity. It's like asking the average value of the integers. A finite value would, by any measure be disproportionately "close" to the origin. I haven't found any reference to this objection in other discussions of this puzzle. The issue plagues any attempt at a solution.
In the analyses plasmid gave us, the two divergent answers are equally correct  or equally incorrect. They assume one of the sides of the triangle has finite length. But any random line segment in the plane must have infinite length. And if so, then the analysis compares areas that are both of infinite extent. The analysis ThunderCloud gave us includes the premise that "scale doesn't matter" so let b=1, and then let a and c be anything. Same issue: if b is constrained, we can't let a and c be infinite. Or, if b=1, then it's not random over an infinite space, where things cannot be "scaled".
Other approaches that I found let the triangle, instead, be randomly chosen in a circle. This preserves angular randomness and permits comparison of side lengths, by eliminating the infinity problem. As a bonus, it gives a unique answer.
Followon puzzle: What fraction of triangles in a circle are obtuse?
SpoilerIf we revisit the puzzle I recently posted that asked the fraction of random triangles in a circle that cover its center, there are at least three ways to convert that solution to an unambiguous solution of this puzzle.

On 2/2/2018 at 9:39 AM, Pickett said:My brain isn't exactly firing on all cylinders today, but...

However, if you are asking for total number of obtuse triangles divided by total number of triangles, I'd say it's indeterminate.
 It's similar to asking "what fraction of points are inside of a circle on a plane?"
 Technically, since there are an infinite number of points inside the circle AND outside of the circle, you could map every single point on the inside of a circle to some point outside of the circle and vice versa...however, conceptually, it's obvious that there should be more points outside of the circle...but that defies the idea that there's an infinite number inside...
In some cases we can get around this point. We could for example divide the plane into increasingly small squares. If we picked a point at random we could say it lies in all squares with equal probability and then count squares. Since a finite circle includes a finite number of squares but excludes an (uncountably) infinite number, the fraction of squares inside the circle is no longer indeterminate  it's zero. That would allow us to reach the reasonable conclusion say that the probability of hitting a finite circle embedded within an infinite dartboard is unambiguously zero. So in some cases where we're picking points at random we can start out picking very small areas, getting an answer, then take the limit of that answer as the areas go to zero. In cases where we're dealing with two infinite areas, however, this approach does not work. (Unless perhaps if the infinities are of different cardinalities.)
Using these "geometric probabilities" is something like saying that points have equal "density" everywhere. It's kind of a reasonable approach, but it's contradicted by the point that you make, namely that there is a surjection between the interior and exterior of a circle.

However, if you are asking for total number of obtuse triangles divided by total number of triangles, I'd say it's indeterminate.

On 2/2/2018 at 10:08 AM, plasmid said:The four laws of card shifting
1. If you see [2] [Empty] [1], then move card 1 onto card 2.
2. If you see any empty slot(s), move the card to the left of the empty slot into the empty slot, unless doing so would conflict with law 1.
3. If there are no empty slots and card 3 is in the far left slot, then move card 2 onto card 3.
4. If there are no empty slots and card 3 in NOT in the far left slot, then move card 3 one slot to the right.Start with this
Spoiler[ 2 ] [  ] [ 1/3 ]. Invoke Rule 1.
[ 1/2 ] [  ] [ 3 ]. Invoke Rule 2.
[ 2 ] [ 1 ] [ 3 ]. Invoke Rule 4.
[ 3/2 ] [ 1 ] [  ]. Invoke Rule 2.
[ 3/2 ] [  ] [ 1 ]. Invoke Rule 2.
[ 2 ] [ 3 ] [ 1 ]. Invoke Rule 4.
[ 2 ] [  ] [ 3/1 ]. Invoke Rule 2.
[  ] [ 2 ] [ 3/1 ]. Invoke Rule 2.
[ 3 ] [ 2 ] [ 1 ]. Invoke Rule 3.
[ 2/3 ] [  ] [ 1 ]. Invoke Rule 1.
[ 1/2/3 ] [  ] [  ]. FTWNice.

You can fit all the answers on a sheet of paper. Nice thinking tho.

On 2/2/2018 at 2:37 AM, rocdocmac said:SpoilerThere are eight equally likely sets of marbles in the bag:
 WWWW 6 ways to pick (WW) each with p(W_{next}) = 1  6x1=6 desired outcomes
 WWWB 3 ways to pick (WW) each with p(W_{next}) = .5  3x1=1.5 "
 WWBW 3 ways to pick (WW) each with p(W_{next}) = .5  3x1=1.5 "
 WBWW 3 ways to pick (WW) each with p(W_{next}) = .5  3x1=1.5 "
 WWBB 1 way to pick (WW) each with p(W_{next}) = 0  1x0=0 "
 WBWB 1 way to pick (WW) each with p(W_{next}) = 0  1x0=0 "
 WBBW 1 way to pick (WW) each with p(W_{next}) = 0  1x0=0 "
 WBBB 0 ways to pick (WW)
Of the 18 ways (WW) could have been picked, there are 10.5 desired outcomes.
p(W_{next}) = 10.5/18 = 21/36
Pairing the points
in New Logic/Math Puzzles
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This puzzle has a nice "Aha!" proof.