# Physics / Math Puzzle: Where Is the Center of Mass?

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.

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.

1. random8042 says:

You don’t have enough information. A little algebraic manipulation proves that all the corners have equal probability, and that all the trapezoids have equal probability. We can then label the triangle a, the trapezoids b and the corners c. However, there are three variables and only two equations, so there must be infinitely many solutions. A trivial example is a=0, b=c=1/6.
If you want a=1/8, then b=1/12 and c=5/24.

1. elkement says:

Thanks for your comment – I agree. I have come to exactly the same conclusion (only two equations) in my follow-up post published yesterday (Nov. 16) – and I picked the same trivial solution finally (Triangle –> zero):
https://elkement.wordpress.com/2013/11/15/revisiting-the-enigma-of-the-intersecting-lines-and-that-pesky-triangle/

However, I was still wondering, if there could be a third equation that might make more sense than others. Finally I think the main issue is that the 50% assumption is unphysical. If you want to make a statement about the probability to find a point on an infinite plane you’d need to have a probability distribution – and it had better go to zero at infinity to keep the total probability finite. Thus based on the assumption given (50%) you could pick any probability density function peaked at the line, then multiplying the three functions.

2. Matthew Rave says:

The reason that the answer is not 1/8 is that the three events (left/right of line 1, left/right of line 2, and left/right of line 3) cannot be independent. As you point out, one of the 8 combinations is impossible. If the three events were truly independent, then nothing would prevent this (impossible) combination.

1/7 is a good guess, but I would argue that without more information about the initial assumption(s) we can’t say for sure. In particular, what if the triangle were very small or very large?

Imagine that the plumb lines are wildly inaccurate. The triangle formed by the lines is huge, covering most of the object, but I would be very, very confident that the COM was within the triangle, since it can’t exist outside of a convex shape. Conversely, if the area of the triangle were 10^−34 m^2 (i.e. 1 outhouse) I would seriously doubt that the COM was within the triangle.

Maybe we have to use Bayesian statistics: given that the area of the triangle is a certain fraction of the object’s area, what is the probability that the COM falls within the triangle?

1. elkement says:

I fully agree – the initial assumptions seem to be incomplete. They would imply that the COM could be located either very close to a line or far away with the same probability. It seems that a more “physical” assumption needs to be added. Now we only know that the probability distribution is symmetrical to each line, but we do not know its shape. If you would e.g. say there is a Gaussian distribution centered around each line, the probability to find the COM within the triangle would be equivalent to the product of the three distribution functions integrated over the area – so the size of the area is taken into account.

3. The “children’s” book The Phantom Tollbooth proposes a land where two kingdoms, one of words and one of numbers wage eternal war with each other because the harmony imposed by the imprisoned nymphs Rhyme and Reason is lacking. Your thinking and more importantly your enthusiasm for thinking reflects what the world should be, though alas it is so lacking in Rhyme and Reason across all borders.

1. elkement says:

Thanks a lot for your comment – much appreciated! “Enthusiasm for thinking” … I have never thought about it this way, but yes – I guess, you are right!
I have noticed you have also liked my post on spam poems – so I am really glad to see that the probably weird mixture of “genres” my blog comprises makes sense to some readers! If my ventures into creating “poems” from search terms and spam would count as Rhyme, it’s probably really Rhyme and Reason!
The book http://en.wikipedia.org/wiki/The_Phantom_Tollbooth seems to be a true classic – thanks for the pointer!
Now I am going to click Post Comment and return to my Castle in the Air!

4. So, basically you’re saying that the probalbility for the center of mass lying within the triangle is 25%. Sounds about right.

1. elkement says:

No, I am saying it is 1/7, this is 14% because there are seven distinct areas associated with equal probabilities: the triangle and 6 areas adjacent to the triangle and extending to the edge of the body. (0,25% is the probability per area in case there are only 2 lines. In this case there would be 4 quadrants only.)

1. I probably got confused reading your elaborations. I just had baked beans – Boston style – with hot dogs for dinner, so, understandably, I am a bit preoccupied with digesting :)

1. elkement says:

OK, that’s a valid explanation of course ;-) Actually, these “simple” puzzle are usually not that simple anyway. I had experienced that “experts” also do not get the answers right immediately. I guess this is way science shows are so successful: The experiments and physics involved is usually “simple”, but nevertheless intricate. My favorite puzzle is: Imagine balloons filled with Helium in a car. When the car accelerates – in which direction do they move and why?

1. That is really a good one. I couldn’t say for sure, If I were to conduct the experiment, I’d expect them to move to the back of the car because of their inertia, if they move relatively to the car at all. Then, because of the aerodynamics working on very obects with a big surface, they would begin to swirl around, that is, if they moved in the first place. Since cars don’t accelerate linear, the balloons would do so until the acceleration ceased and the movement of the balloons matched that of the vehicle – so, in a real car on a real street, only if the driver were forced to stop the car

1. elkement says: