Bing Says We Are Weird. I Prove It. Using Search Term Poetry.

Bing has done so repeatedly:

Bing Places asks for this every few months.

In order to learn more about this fundamental confusion I investigated my Bing search terms. [This blog has now entered the phase of traditionally light summer entertainment.]

Rules:

  • Raw material: Search terms shown in Bing Web Master Tools for any of my / our websites.
  • Each line is a search term of a snippet of a search term – snippets must not be edited but truncation of phrases at the beginning or end is allowed.
  • Images: Random pick from the media library of elkement.blog or our German blog (‘Professional Tinkerers – Restless Settlers‘)

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you never know
my life and my work

is life just about working
please try to give a substantial answer

A substantial answer.

element not found
map random elkement

Random elkement, confused by maps.

so as many of you may know (though few of you may care)
my dedication to science

this can happen if
the internet is always a little bit broken

The internet is always a little broken

google translate repetetive glitch poetry

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So let’s hear what Google has to say!
Same rules, only for Google Search Console:

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so called art
solar energy poem

ploughing through
proof of carnot theorem
thermodynamics in a nutshell

Thermodynamics in a nutshell.

mr confused
why him?

Mr Confused

plastic cellar
sublime attic
in three sentences or fewer, explain the difference

Plastic tank - 'ice storage'

Sublime attic.

name the three common sources of heat for heat pumps
magic gyroscope
frozen herbs
mice in oven

Had contained frozen herbs. Isomorphic to folding of ice/water tank pond liner.

shapeshift vs kraken
tinkers construct slimey
you found that planet z should not have seasons

Slimey

the best we can hope for is that
tv is dangerous

what is the main function of the mulling phase?
you connect a packet sniffer to a switch

Connect a sniffer - to your heat pump

how to keep plastic water tank cool in summer
throwing boiling water into freezing air

Freezing air.

just elke
self employed physicist
the force is strong in me

The force is strong in me.

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Revisiting the Enigma of the Intersecting Lines and That Pesky Triangle

Chances are I made a fool of myself when trying to solve an intriguing math/physics puzzle described in this post.

I wanted to create a German version but found it needs a revision. I will just give you my stream of consciousness as I cannot make it worse anyway.

The puzzle is presented as a ‘physics puzzle’ but I think its enigmatic nature is described better if stated in purely mathematical terms:

Consider three lines in a flat plane, not parallel to each other and not intersecting in a single point. Their mutual intersection points are the corners of a triangle.

Assuming that the probability to find an arbitrary point on either side of each line is 50% – what is the probability to find a point within the triangle?

I had proposed a solution 1/7. My earlier line of reasoning was this:

The three lines divide the full area into 7 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, called the ‘+’ and ‘-’ parts.

Center of mass, physics puzzle.

Proposed solution in post published in Feb. 2013: The body is divided into 7 parts; the center of mass being located in either with equal probability. (Image Credits: Mine)

There are 8 possible combinations of + and – signs, but note that the inverse of the symbols assigned to the triangle is missing: (-+-)  Digression: It would be there if we painted these lines on a ball instead of a flat plane – then each line would close on itself in a circle and there would be 8 equivalent triangles. The combination missing here would correspond to the triangle opposite to the distinguished and singular triangle in our flat plane.

I had assumed that the 7 areas are equivalent based on ‘symmetry’ – each area being positioned on either side of one of three planes – and assuming that the condition given (50%) is not physical anyway. A physical probability would vary with distance from the line – imagine something like a Gaussian symmetrical distribution function centered around each line. Than the triangle would approximately correspond to the area of highest probability (where the peaks of the three Gaussians overlap most).

Do you spot the flaw?

Intersecting lines, two halves

The lower half contains 4 different parts (1x triangle, 2x open trapezoid, 1x open wedge), and the lower half contains two open wedges and one open trapezoid. Probabilities should add up to 50% in each half though.

We can do two cross-checks:

1) the sum of the probabilities of all parts should add up to 1 – OK as 7 x 1/7 is 1. But :

2) the sum of probabilities of all pieces on either side of a line should add up to 0,5! This was the assumption after all.

Probabilities don’t add up correctly if I assign the same probability to each of the 7 pieces – it is 4/7 for the lower half and 3/7 for the upper half.

So I need to amend my theory and rethink the probability assigned to different kinds of areas (I guess mathematicians have a better term for ‘kinds of areas’ – more like ‘topologically equivalent’ or something.).

We spot three distinct shapes:

  • A triangle formed by the three lines.
  • Three ‘open wedges’ formed by two lines – e.g. part (- – -) in the lower half.
  • Three ‘open trapezoids’ formed by three lines, e.g. part (+++) in the upper half.

I am assuming now that probabilities assigned to all wedges are the same and those assigned to trapezoids are the same. I am aware of the fact that this will not work out if we consider a limiting case: Assume the angle between two of the three lines gets smaller and smaller – this will result in one very small wedge (between the red and the blue line) and two ‘wedges’ which are nearly equivalent to a quarter of the total area:

Intersecting lines, narrow wedges

In the limiting case of the blue and red lines coalescing we would end up with four quarters, and you would find an arbitrary point with a probability of 25% in either quarter.

In the video Quantum Boffin has asked for the probability of the triangle – which can be any triangle, of any arbitrary shape and size, and he states that there is a definitive answer. Therefore I think also the details of size and shape of the other areas does not matter, and the 50% assumption is somewhat unphysical.

As there are three distinct types of shapes – I need three equations to calculate them all.

Notation in the following: p… probability. T…triangle, W…wedge, Z…trapezoid. p(T) denotes the probability to find a point in the triangle.

The sum of all probalities to meet the point in either of the 7 pieces must be 1, and we have 3 wedges and 3 trapezoids:
i) p(T) + 3p(W) +3p(Z) = 1

We need 50% on either side of a line.

There is one Z and 2W on one side…
ii) p(Z) + 2p(W) = 0,5

…and T and 2W on the other side:
iii) p(T) + 2p(Z) + p(W) = 0,5

Now the sum ii) and iii) is just i) that these equations are not independent. We need one more information to solve for p(T), p(W), and p(Z)!

And here is my great educated guess: You have to make another assumption and this has to be based on a limiting case. How else could we make an assumption for an arbitrary shape?

I played with different ones, such as letting iv) p(W) = 0,25 motivated by the limiting case of a nearly right angle. Interestingly, you obtain a self-consistent solution. Just plugging in and solving you get: p(T)=0,25 and p(Z)=0. Cross-checking you see immediately that this is consistent with the assumptions – probabilities sum of to 50%: You either have two Ws or one W and the T in one half of the plane.

Assigning 0 to the trapezoid does not seem physical though. We can do better.
So what about assigning equal probabilities to Z and W? iv) p(Z) = p(W)?

I don’t need to do the algebra to see that p(T) has to be zero as you would have 3 equivalent pieces on each side, but the triangle can only be located one one side.

This assumption is in line with the limiting case of a really infinite plane. The triangle has finite size compared to 6 other infinite areas.

I change my proposal to: The probability to find an arbitrary point in the triangle is zero – given the probability to find it on either side of each line is 50% and given that the area is infinite.

Again I’d like to stress that I consider this a math puzzle as the 50% assumption does not make sense without considering a spatial variation of probability (probability density, actually).

Addition as per November 21:

Based on the ingenious proposal by Jacques Pienaar in the comments, I am adding a sketch highlighting his idea.

Theoretically, the center of mass would correspond to the intersection of the 3 “perfect” solid lines. Now allow for some “measurement error” and add an additional line denoting the deviations. I depicted the “left” and “right” lines as dashed and dotted, respectively.

Now take a break, get a coffee, and look at the position of the true center of mass with respect to the triangles made up by the dashed and dotted lines:

Intersecting-Lines-Proposal-Jacques-PienaarSince we have 3 colors and either a dashed or a dotted lines, there are 8 distinct triangles. I tried to make the angles and distances as random as possible, so I think Jacques’ proof does not depend on the details of the configuration or the probability distribution function (yet beware the limiting cases such as parallel lines). The intersection of the solid lines is within 2 of 8 triangles – hence a probability of 2/8 = 1/4.

I was intrigued by an odd coincidence as I had played with o,25, too (see above), but based on the assumption that of a probability of 0,25 for the wedges/corners – which by cranking the algebra or just cross-checking the 50% criterion results in p(Triangle)=0,25, too, and in p(Trapezoid)=0.

Looking hard at this new figure introduced by Jacques I see something closely related, but unfortunately a new puzzle as well: The true center of mass is in exactly two of eight trapezoids built from dashed or dotted lines. So I am tempted to state:

p(Trapezoid)=0,25.

But it is difficult to make a statement on the corners or wedges as any intersecting two lines cut the plane in 4 parts and any point is found in one of them. I was tempted to pick p(W) = 0 though, but this would result in p(Triangle)= -0,25.

So this was probably not the last update or the last post related to the enigma of the intersecting lines.

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.

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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:

Center of mass, physics puzzle.

Proposed solution: The body is divided into 7 parts; the center of mass being located in either with equal probability. (Image Credits: Mine)

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.