Triangle puzzle

[ceptimus]

This one is an old classic. Work out the angle, theta ( θ ) in the attached diagram. The big triangle is isosceles.

[seebs]

Oooh, I love that.

I once spent an hour or so trying to figure out the inside angles of a pentagram. (No, I didn't remember the inside angle of a pentagon, that would have been too easy.)

[seebs]

Hmm. I missed the isosceles bit, which made it seem very hard.

I have, however, now gotten stuck. I have four unknown angles (theta, the one above it, the other corner of the top triangle, and the one just below that), and I seem to be able to find multiple solutions for them which work out just fine. I am probably missing something.

[seebs]

Hmm. It seems obvious to casual inspection that, in fact, some proposed values for theta cannot be correct, but I can't find a good way to prove them, except to observe that I can't draw the triangle that way. I must have forgotten some triangle rule. The angles always add up to 180, even with ludicrous values for theta, and I don't end up with any contradictions...

[cappuccino]

I've been working for a hour on the puzzle and I'm currently stuck, I just need to figure out one more equation and I'll have my system of equations solution for theta.

Gotta go and take a shower now.

[viscousmemories]

I solved the puzzle in 2 minutes, but I'm going to give someone else a chance before I reveal the answer.

[seebs]

Did you have to do anything with all the sin and cosine rules? They're the tangled mess I don't think I want to get involved with.

[ceptimus]

It's possible to solve the puzzle with just basic geometry (but you have to use some elegant moves). If you use trigonometry to find the answer, that's fine, but there will be extra kudos for a simple geometric solution. :yup:

[seebs]

Knowing that it's even possible certainly helps motivate me.

I love this kind of thing.

Huh! I have an answer. My solution seems convincing, but I am not sure whether it's correct. It's one of those things where I have some worry that the construction I used to say "ahh, X = Y!!" may actually not be valid. (I once saw a beautiful proof that all angles are equal to the right angle that relies on confusing the reader over which side of a line a given construction ends up on.)

[slimshady2357]

It took me about 10 mins to solve this, in a purely geometric way.

I'll try to attach a pic and put it in a spoiler....

Edit:

Oops, I cocked it up. I still haven't solved it.

[viscousmemories]

Did you have to do anything with all the sin and cosine rules? They're the tangled mess I don't think I want to get involved with.
I was only kidding, seebs. I couldn't solve a math puzzle if my life depended on it.

[seebs]

This one's actually not bad, unless I did it wrong. Which I might have. I have an assumption I'm not comfortable with, that I may not have thought through correctly.

[Joshua Adams]

http://img.photobucket.com/albums/v406/jaganshihieiyyh/triangle.png

theta = 50?

[ceptimus]

No Josh. That right angle you marked doesn't have to be a right angle - you could make it any angle you want and so get different answers for θ

[Joshua Adams]

I would not be able to make that right angle something else without losing the 40 degree angle in there as well. I don't see any problem with my proof. I constructed a triangle with 2 known angles and one that is definitely theta. :shrug:

Edit: Actually, I think I get why it is wrong now... in order to deduce that 40 degree angle, I would need to make sure the 'vertical' line is parallel to the big triangle's side instead...

[IRON MAN]

:eager: Ooh, a bunch of geometry geniuses! Just what I always wanted!

If you are done with that one - try this on for size ...


1. Equal lateral Triangle Problem:

http://img124.echo.cx/img124/5486/triangulationproblem2vp.jpg

DO NOT tell me what the value of x actually is, unless you can show me the process by which you arrived at it.

You would be amazed how many times someone has claimed to know the answer to this problem, but can't, (or according to their plethora of excuses, wont), tell me how they worked it out. Yes, I can draw a diagram on a piece of graph paper and measure the distance too. Big Fucking Deal!

So be warned, anyone who tells me the answer is 7.143 or something like that, and doesn't have a formula that backs it up - I will find out where you live and come kill you, I swear.

What is important here is the formula: As in, x = L^2-((x+2)^2+(x+4)^2) or some shit like that. To really get the point home your answer should be a formula that starts with, "x=", so no matter where I put the centre point your formula is still correct, (it should also work if that point lies outside the triangle).

As you can see, x describes the shortest distance to any vertex of the triangle. The remaining distances will therefore be x+something, (known values), and x will have a different value when the other variables are changed.

For example:

If the central point where the blue lines meet was in the centre of the triangle, then each blue line would be x long, 2/3 the height of the triangle.

If that intersection point lies on one of the vertices, then x = 0 and the other lengths = x + 10.

These are simple points that you can use to test your formula for accuracy.

The values are added for ease of understanding of the problem, but here is the actual problem with all variable terms added:

http://img124.echo.cx/img124/236/triangulationproblem29sh.jpg


2. Scalene Triangle: The triangle is equal lateral for simplicity. If you are feeling particularly clever, change the triangle to scalene, and change L to L1, L2, and L3.

3. Tetrahedron: If you want to really show your genius, make the triangle a tetrahedron and make a three dimensional calculation for x, (you will have to add another value to account for the extra vertex of course).

4. Unequal Tetrahedron: If you really think that you're up for a Nobel Prize, vary the sides of the tetrahedron so that they are not equal and do it again.


ETA: Not only is the actual answer for x not relevant - neither is my crappy spelling of the word "relevant" in that jpg. :)

[seebs]

I came to the same value for theta; my solution, which I am unsure of, was to just extend the line above theta through the right side of the big triangle to the same distance, and then draw a line down to the bottom-right of the big triangle. However, I am not at all convinced of my assumption that the resulting big triangle (with theta in the middle of one side) is actually isosceles. In fact, I'm not sure of it at all.

It is worth pointing out that I'm quite sure the original construction is inaccurate. The triangle made of the top point of the big triangle, the point defining theta, and the bottom left point of the big triangle should be isosceles, but the two sides are drawn as very different lengths.

[ceptimus]

Yes, I deliberately drew it not to scale. :P

If you have a CAD program, such as Autocad or similar, one way to find out θ is to just construct the diagram and have the CAD measure and display the angle for you. Of course, that doesn't give a satisfying proof.

[seebs]

Yes, I deliberately drew it not to scale. :P

If you have a CAD program, such as Autocad or similar, one way to find out θ is to just construct the diagram and have the CAD measure and display the angle for you. Of course, that doesn't give a satisfying proof.

Yeah, so at this point, I am not convinced that I have a valid solution, and I'm a little sleepy-headed to nail it down. Just looking at the angles within that triangle, there's a pretty broad range of numbers that add up to the right values, but obviously they aren't all workable.

The not-to-scale makes it a little tricky to figure out which constructions are actually possible.

[slimshady2357]

Ok, I tried again on this one this morning and it took me about 15 mins to solve it. But I had to resort to trigonometry.

Theta = 30 degrees.

I still can't see a nice smooth geometric solution.

To solve this using trig, just asign some value to the equal sides of the big triangle, then use the sin and cosine laws to fill in all the rest of the details needed. I ended up with 0.49999993 = sin(theta) and figured it was just rounding error for the very small difference.

[Fahrenheit]

I got theta as 60 degrees, by some weird calculations purely geometrical.

[seebs]

I'm convinced of 30, but I can't find an elegant proof.

[ceptimus]

If you put copies of the original triangle together, like slices of a pie, you will find some very interesting results. The apex angle of the big triangle is 20:degrees: so there is room for 18 of them, if you construct a complete pie.

Once you've seen how they go together, you only need three or four such pie slices to come up with an elegant proof. :wink:

[ceptimus]

Here's the way I do it (Warning! Diagrams and explanation in spoiler below).


We start by drawing an equilateral triangle and dividing it in half. We know that the angles of an equilateral triangle are 60:degrees: so we can mark in the 30:degrees: angle.

http://ceptimus.co.uk/freethought/images/triangle/triangle1.gif

Now we trisect the apex of the equilateral triangle into three 20:degrees: angles and construct three isosceles triangles with 20:degrees: apexes, and two sides the same length as the sides of the equilateral triangle. We add a line over on the right, thus making a small isosceles triangle that has two 20:degrees: angles, We can also mark in the 20:degrees: angle over on the left - we see that this is 20:degrees: as the left hand isosceles triangle is congruent with the centre one, but rotated 20:degrees: clockwise.

http://ceptimus.co.uk/freethought/images/triangle/triangle2.gif

Now we can rub out the base of the original equilateral triangle and add the 30:degrees: and 20:degrees: together, to mark in the 50:degrees: angle. As we know that the base angles of an isosceles triangle with a 20:degrees: apex must both be 80:degrees: we can also mark in the 60:degrees: angle over on the right.

http://ceptimus.co.uk/freethought/images/triangle/triangle3.gif

Now we add two lines, and using symmetry we can label the angles as 50:degrees: and 60:degrees: from the ones we already know.

http://ceptimus.co.uk/freethought/images/triangle/triangle4.gif

Now we recognize the centre isosceles triangle as our old, puzzling friend so we can mark in θ. We also can mark in an 80:degrees: angle (the base angle of one of the isosceles triangles) and a 20:degrees: angle (this is an 80:degrees: isosceles base angle, less the 60:degrees: we already know).

http://ceptimus.co.uk/freethought/images/triangle/triangle5.gif

Now we have θ trapped in a triangle where we know the other two angles! So we can see that θ is 180:degrees: - ( 20:degrees: + 80:degrees: + 50:degrees: ) and hence arrive at the answer: θ = 30:degrees:

I figured out this method about three years ago, while flying on a plane back from Sudan. I had nothing better to do!

[Dragar]

Skillfully done, ceptimus!

[ceptimus]

I made some progress with IRON MAN's problem, though I don't have a solution yet...

To make the notation simpler, I renamed y and z so that IRON MAN's x+y is my y and IRON MAN's x+z is my z. Thus my x, y and z are simply the distances of the point from the three corners of the triangle.

I get this formula for the length of one side, L, given the distances: x, y, z of a point from the three corners:

L4 - (x2 + y2 + z2)L2 + (x4 + y4 + z4 - x2y2 - x2z2 - y2z2) = 0

As this is a quartic, I don't fancy my chances of rearranging it to get a formula for x. :P

[IRON MAN]

It's an interesting problem isn't it?

I have asked this of several people with varying mathematics ability and nobody can seem to derive an answer. It also seems to be beyond my personal mathematical abilities.

While it may be discouraging to tell you that I do not know the answer, let me offer you some encouragement.

It is conspicuously obvious that there can always only be one answer for x for any given values of, (my), y and z.

Nice idea redefining y and z to simplify the process, I never thought of that, (and you should see some of the monster formulae I've got). But of course, one would have to split them up again to simplify for x.

Since you display an interest in this problem, and have the ability to derive that first step, I'll tell you how I arrived at this problem in the first place.

When I was in high school science class in about 1986, the teacher demonstrated how earthquake epicentres are triangulated by working out the distance the shock wave travelled before arriving at each seismic station.

Plotted on a common time scale you would get three sets of shock waves like this:

http://img45.imageshack.us/img45/5643/triangulationps8.jpg
Earthquakes contain both primary waves, (P), and secondary waves, (S), that travel at different speeds, so the distance a set of shock waves have travelled can be directly derived from the delay between both those waves, taking into account the composition of the material they are travelling through.

Thus, it is a simple matter to plot a circle around each station at the distance each detected, and by plotting three stations triangulate the epicentre, sometimes resulting in a "triangle of error". The epicentre need not be inside the triangle formed by the stations to be triangulated.

This got me thinking. Why do you need the S-waves at all? There should be sufficient available information to derive the epicentre without the added benefit of this information, and the problem boils down to the geometric problem I presented.

http://img45.imageshack.us/img45/5442/triangulation2ay7.jpg

There is indeed only one value for x when y and z have specific values. The information we seek appears lost to us initially, because the first inkling we get of an earthquake is at time=0 when it hits the first station and we have no idea how long it took to get there. But knowing that the missing time/distance, (x), is a component of the other two readings, (x+y), and (x+z), gives us the information we need in order to leave x with a single unique value for each location, and value of y and z.

Obviously we can instantly figure out that if the earthquake occurred directly underneath station A then we would get:

y = 10
z = 10

and therefore x=0.

Or if the earthquake occurred directly in the middle of an equal lateral triangle, it would hit all stations at the same time, and therefore:

y=0
z=0

Therefore x must equal the distance from the apex to the centre-point.

Now obviously A wont always be the first station to detect the earthquake, so you might get:

x=14
y=27
z=0

In this case the answer you would be after is z, and the answer in this case lies outside the equal lateral triangle also. But obviously the formula would be the same except you would change the x and z terms around.

Now obviously we have a couple of options:

1. We can plug random values of x into the formula and try to home in on the answer that way.

2. We can do the calculations prior and put them in a look up table.

3. We can work out the formula and derive the exact answer to a potentially infinite number of decimal places for any point extending off to infinity anywhere on that plane.

Having access to fast computers might make one opt for the lazy option 1 or 2, but that still doesn't solve the problem.

And here's the best bit of all. This formula has endless practical value.

Imagine a security system with 3 microphones or seismic detectors that can instantly triangulate your position in a building by the noises you make and your footsteps.

Better yet imagine putting this concept in 3 dimensions, (a tetrahedron of sensors), and being able to instantly work out the location of any object emitting a particular type of energy/radiation, at any range, (with decreasing accuracy over distance in the real world).

http://img158.imageshack.us/img158/4239/tricorder1le0.jpg

Now I don't want to get all geeky on you, but can you say, "Tricorder" boys and girls?

Better yet, you could add a different type of sensor in another tetrahedron within the same space, say you have seismic, and radiation.

http://img87.imageshack.us/img87/2867/tricorder2wu2.jpg

Going straight to the logical conclusion you could have an entire sphere of tetrahedrons within each other all in the same space, each slightly offset, (the individual orientation of each one obviously doesn't matter).

The size of those tetrahedrons, (distance sensors are apart), would have a bearing on the accuracy also.

The accuracy with which they were placed in their positions would also have a bearing if you were relying on the tetrahedron being perfect.

However, if a formula could be devised for a "non-equal lateral" tetrahedron, then that is simply a matter of calibration, perhaps even self-calibration if additional sensors were added and/or mathematical derivation used so the device could triangulate the location of it's own sensors in relation to one another.

http://img158.imageshack.us/img158/6394/tricordersqg1.jpg

Ta-da!

Time for the Mark II (http://en.wikipedia.org/wiki/Tricorder)?

[IRON MAN]

Coincidently, I made up an excel document to assist me with verifying any formulae I came up with against seven known points.

I ended up just mucking around putting in any old nonsense that seemed about right and at one stage wound up with 5 verified points and the other values as multiples of the input factors or numbers like SQR(2) and shit like that.

It's just circumstantial of course, but seems to suggest that there may well be a solution. So tantalisingly close I can almost taste it.

http://img133.imageshack.us/img133/5040/triangulationxlld2.th.jpg (http://img133.imageshack.us/my.php?image=triangulationxlld2.jpg)

[IRON MAN]

Another application is robotic vision. With this a robot could actually "see" in accurate and genuine 3D - passively and/or actively - on a wide spectrum with minimum processing delay.

It would sure as shit beat trying to replicate our own vision with all it's flaws giving only the impression of 3D from a couple of 2D images.

If the robot could see in 3D there is no necessity to do additional processing to convert the image it sees - as we do. Also it would be immune to the optical illusions and other visual errors that we suffer from.

Human:

3D world => Sensed by our eyes in => 2 x 2-Dimensional images => Converted by our brains to => Fake 3D impression/approximation.

I sure don't envy anyone trying to replicate that system in robotics. That's a lot of processing requiring, (a lot of gap filling and data reconstruction).

Wouldn't it be better to go this route:

3D world => Sensed with enough information to process it quickly into => True and accurate 3D information.

Of course you don't need this formula just for that. You could probably use an algorithm to narrow down the answer to quite a few decimal places pretty quickly, (as pointed out by a friend of mine), and there is practical value in that.

But it sure would be cool to get the answer in one hit.

You would basically only have to identify similar waveform characteristics from each sensor, which is easier if they are clustered, rather than far away from each other, since the intensity will be about the same. But like I say the distance between them is a trade-off with accuracy.

Add active sensors, say like a bat's sonic pulse and radar system and you have a pile of incoming data on four 2-dimensional channels within a very short space of time.

There is also another precedent for this in nature. There is a desert scorpion that senses not only the angle, but distance to it's prey by the differential between seismic vibrations it receives through it's legs - all planted in the sand at different points.

Since the damn thing is practically blind, it's pretty impressive to see it jump exactly the right distance onto it's prey with pinpoin accuracy, (although I also figure the intensity of the vibration is an additional factor in the "calculation" of distance - there's no penalty for cheating in natural selection). Especially considering it can't use it's eyes to calibrate that sense, (like I'm assuming we do with our directional sense for sound).

It is also interesting to note, that by simply recording the incoming data trains on the 4 sensors on a common timescale, (only twice as taxing as recording a couple of high fidelity stereo channels). You would then be able to extract any available information later, even if you didn't notice it, (process it out), at the time of recording. You effectively have a recording of your 3-dimensional environment and it just depends on the resolution, sensitivity, (at the time), and the processing power, (either applied at the time or afterward), as to how much information you can actually extract.

Really gives new meaning to the term, "Tricorder", doesn't it? Maybe it should be, "Quadcorder", to be more precise. :)

Examples:

1. You might be able to distinguish the sound and location of a minor metal deformation in a machine that was not large enough to appear to be a problem at the time, but you discern it from the, "black box", recording after a major failure.

2. You might be able to filter out an unnoticed individual background conversation in a busy public place, based on specifying the location of the speakers afterward.

That last one is an interesting example of kinda working this idea backwards too. Which makes you wonder if there is a useful application for a device that emits a pulse from 4 points at a specific time, in order to get them to intersect at an exact point in 3-Dimensional space.

If you figure any of this stuff out, by all means go make a million dollars with it. I wont lose any sleep as long as I can buy the resultant consumer products, and I'll be the first in line.

I just want my goddamn tricorder! :beam:

[Dragar]

I had a brief crack at that puzzle, based on the original 'seismic waves' idea, presented by Iron Man.

So you treat each point of his triangle as the centre of a circle. The one at the bottom left has radius r+y, the bottom right r+z, and the top point has radius r. (Basically, realise I have replaced his x with r).

You know the equations for each of these circles (they will be related to L). But for that you need a coordinate system. So I placed the origin at the bottom left corner. As I needed x and y for my coordinates, I renamed the quantity y as a, and z as b. These are constants, so it makes more sense to use a conventional notation.

So when I wrote them down I got:


x2 + y2 = (r+a)2
(x - Lcos(pi/3))2 + (y-Lsin(pi/3))2 = r2
(x-L)2 + y2 = (r+b)2

I may have done this wrong; I did it very rushed. But these are three simultaneous equations, with three unknowns (r, x and y). (x,y) will be with the coordinates of the place where all three circles meet. r will be the unknown quantity in the question (Iron Man's 'x').

With three unknowns and three equations, this should be solvable for r, provided the constants a, b and L are supplied.

It might not be particularly nice to solve, but it is (in principle) doable. Personally, I'd use a computer to solve them.

[ceptimus]

I've got complicated expressions for x, and the coordinates of the intersection point. The one for x has just "x =" on the left hand side, but also includes various functions of x on the right hand side. :sadcheer: Looking at the various power terms for x, it looks as though the expression (given enough algebraic manipulation) will simplify to a quartic.

However, quartics are not particularly nice things to deal with, so I agree with Dragar that a numerical approach using a computer program is probably the most practical way to solve the problem...

...but that's not what IRON MAN asked for, so I'll keep plugging away at the algebra, hoping that the x4 and x3 terms will magically cancel out, leaving some nice tractable equations. :crossed:

[ceptimus]

I made a utility to draw IRON MAN's triangle. You have to specify the width (pixels) of the diagram, and the x and y coordinates of the point (taking the bottom left corner as the origin and scaling to a side length of 1.0). Here I'm just testing whether the triangle will render in a forum post. Quote this post to see how it works.
This example is asking for a triangle 600 pixels wide, and I worked out the coordinates so that the three lines to the centre have lengths: 3, 4, 5 (and the side would therefore be 6.76643).

http://ceptimus.co.uk/freethought/triangle/tridiagram.php?w=600&x=0.402&y=0.434

I'll post my numerical solver when I've tidied it up a bit. That solver uses this diagram utility to do display the results.

[Shake]

I drew in a couple of lines parallel to the base and extended some other lines outside of the original triangle, but eventually came to the answer of theta being 30º, also. Too bad I can't draw it out here.