On randomly searching for physics puzzles I have come across QuantumBoffin‘s site.
The puzzle is about how to determine the center of mass of a body using the plumb line method, given there is some uncertainty due to experimental errors:
You have found three plumb lines not intersecting in a single point; thus the three intersections of two lines each form a triangle. The question is:
What is the probability that the center of mass is actually located within the triangle?
The following assumption should be used:
The probability of finding the center of mass on either side of each line is 50%
If you want to entertain yourself with trying to solve the puzzle on your own, do not scroll down. I did not find a published solution, so the following is just my proposal. Thus there is a chance that I make a fool of myself. But I enjoy those deceptively simple science puzzles, and there is always a good chance so-called intuition might lead you astray.
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I have tackled the problem as follows:
If there is a single line, it divides the body into two pieces, the center of mass (COM) is found with 50% probability in either piece.
If there are two lines, they would intersect in a point, and the body would be divided into four quadrants. The COM is located in one of these quadrants with a probability of 25%. I’d like to stress that the reason is that the body is divided into 4 equivalent pieces, each of them is located either to the left or to the right of each line. In this case the result is the same as 0,5 times 0,5.
However, if there are three lines, the probability is not simply equal to 0,53 = 1/8 (as confirmed by QuantumBoffin)
I propose: The solution is 1/7.[*]
The three lines divide the piece into 7 (not 8!) parts – the triangle in the center and 6 sections adjacent to the triangle: Each of these parts is located either ‘to the left’ or ‘to the right’ of each line, I call these the ‘+’ and ‘-‘ parts (halves) of the body:
The seven parts are equivalent. This may sound a bit awkward as you might expect the probability to find the COM on the periphery lower than in the middle. But this is due to the assumption that should be made – the assumption (50%) should be replaced by a probability distribution. So in a sense, this is more a math puzzle than a physics puzzle.
[*] Edit as per Nov. 2013: I think I have found a flaw in my argument as the sum of probabilities for areas on either side of a line would not add up to 0,5 if I assign the same probability to each of the 7 areas depicted below (It is: 4/7 on one side versus 3/7 on the other side). I am now searching for a system of equations that let me determining the probabilities for the three different types of areas: the triangle, the corner areas (wedges) and the open trapezoids.
You can find my updated proposal here. I keep the rest of the post unchanged as I consider my random musings about curved spaces (on the bottom) correct.
I did not use the number of 0,5 in a calculation, but as a justification of the equivalence of the 7 pieces.
Now there is a remaining – mathematical – puzzle that still baffles me: There are 8 (23) possibilities of combining + and -: Which is missing and why?
-+- is the missing combination.
The picture exhibits some nice symmetry, but note that the definition of + and – as such is arbitrary. So we could have ended up with other missing combinations.
- +-+ in the center. The missing combination is the ‘negation’ of the combination reflecting the triangle.
- +++ and — opposite to each other (opposite = one attached to a corner of the triangle and the other to the side of the triangle not connected to this corner)
- Other pairs / opponents:
–+ and ++-
-++ and +–
Imagine the negative of the triangle now: To the left of the red line, to the right of the blue and above the green line. Obviously there is no intersection between these, and this is the only combination that does not result in any intersection at all. I suppose there is a better way to state that in stricter mathematical terms. And if we put the figure on a sphere (Non-Euclidian geometry), there would be an intersection – actually the figure would collapse onto 4 equivalent triangles (a ‘bulged tetrahedron’).
In case of two lines only, all combinations can be ‘realized’ in flat space, therefore the solution is simply equal to 1 over the total number of combinations (4).
This stuff is addictive – finally I understand why Feynman was fascinated by flexagons.