Others probably did, but I never realized this actually makes a perfect fit. This means that you can construct the angle of an angled liftarm - which suites the Pythagorean triple (3-4-5) - with a right-angled 3x5 liftarm. Might be useful.

On 1/10/2017 at 2:04 PM, Didumos69 said:

Some may ask themselves, what's the practical use of all this? And even when you see the use of it, it turns out very hard to tell when you're in a situation where you can actually use this. So that got me thinking and after applying this 'perfect fit' several times, I came to the following rule of thumb:

Take the shorter end of any angled (or bend) liftarm. This end is not long enough to span a 5 stud length required for a full (3, 4, 5) triangle; it is either 2 studs long or 3 studs long (measured between pinhole centers). So, when working on a 1-stud grid, the outer pinhole is out-of-grid. However, the outer pinhole is always one stud distant from the grid:

• Outer end 3 studs long (measured between pinhole centers): To get 'on-grid', move one stud away from the outer pinhole - orthogonal to the outer end - towards the other end of the liftarm.
• Outer end 2 studs long (measured between pinhole centers): To get 'on-grid', move one stud away from the outer pinhole - orthogonal to the outer end - away from the other end of the liftarm.

 

Edited November 20, 20177 yr by Didumos69

Cool!

There are many more perfect fits, actually, if anyone is looking: http://www.brickshelf.com/cgi-bin/gallery.cgi?f=193780

Edited December 28, 20168 yr by Sariel

7 minutes ago, Sariel said:

There are many more perfect fits, actually, if anyone is looking: http://www.brickshelf.com/cgi-bin/gallery.cgi?f=193780

Very nice overview! Although some of them aren't perfect.

Very nice find indeed, I like the compactness of this

Taking this a bit further, one can get a visual proof of (the 3-4-5 triangle instance of) Pythagoras' theorem:

6 hours ago, Jeroen Ottens said:

Very nice find indeed, I like the compactness of this

There is an even more compact way to get the same angle. The trick is that you need to construct a kite-shape from two right triangles with one leg twice the length of the other leg:

4 hours ago, aeh5040 said:

Taking this a bit further, one can get a visual proof of (the 3-4-5 triangle instance of) Pythagoras' theorem:

I'm afraid you're still showcasing and not proving the theorem. This is the easiest visual proof I know. The green and orange triangles sum up to equal areas in both the left and right situation. Hence the gray squares must also sum up to equal areas. In other words, c^2 must be a^2 + b^2.

Edited December 29, 20168 yr by Didumos69

If one leg is not 2, but 3 times as long as the other one (on both sides of the "kite", then you get the other angle of the 3-4-5- triangle. See the image below, where every two connected dots have a distance of 1.

30 minutes ago, Erik Leppen said:

If one leg is not 2, but 3 times as long as the other one (on both sides of the "kite", then you get the other angle of the 3-4-5- triangle. See the image below, where every two connected dots have a distance of 1.

Which can be easily constructed using 2x4L liftarms. This closes the circle - or should I say 'triangle'  - very nicely, thanks!

It would be nice to see a simple proof of the thing, it's a shame that my math is so eroded that I can't prove it myself any more.

Anyways, I also "discovered" this fit and utilized in my Telehandler.

24 minutes ago, Lipko said:

It would be nice to see a simple proof of the thing, it's a shame that my math is so eroded that I can't prove it myself any more.

Anyways, I also "discovered" this fit and utilized in my Telehandler.

It must be related to the incircle of a triangle, which divides each triangle in 3 perfect kites (a square is also a perfect kite). The radius of the incircle equals the area of the triangle divided by half the sum of the side lengths. In case of the (3, 4, 5) triangle this radius is 1. This is illustrated by Erik's dotted triangle a few posts back.

My girlfriend prooved it in a few minutes (but also Didumos proof is valid, though it includes a statement that is not trivial: radius = area / half_of_circumference)

So a right triangle:

indeed, the triangle has an inner circle with radius x. The lines designated with x are perpendicular to the sides (since the inner circle is touching the triangle).Plus we have three equations:

1) A=z+x

2) B=x+y

3) C=y+z  which is the same as -C=-y-z

Summing the three equations: 1) + 2) + 3) gives

A+B-C = 2x. Which in our case is 4+3-5 = 2x -> 2=2x ->1=x.

EDIT2: My girlfriend has a very good visual method for calculating x (which leads to the same equation as I posted above):

Edited December 29, 20168 yr by Lipko

5 hours ago, Lipko said:

My girlfriend prooved it in a few minutes (but also Didumos proof is valid, though it includes a statement that is not trivial: radius = area / half_of_circumference)

So a right triangle:

indeed, the triangle has an inner circle with radius x. The lines designated with x are perpendicular to the sides (since the inner circle is touching the triangle).Plus we have three equations:

1) A=z+x

2) B=x+y

3) C=y+z  which is the same as -C=-y-z

Summing the three equations: 1) + 2) + 3) gives

A+B-C = 2x. Which in our case is 4+3-5 = 2x -> 2=2x ->1=x.

EDIT2: My girlfriend has a very good visual method for calculating x (which leads to the same equation as I posted above):

So there can be no doubt we are talking about a mathematically sound fit!

On 12/29/2016 at 8:30 AM, Didumos69 said:

On 12/29/2016 at 3:36 AM, aeh5040 said:

Taking this a bit further, one can get a visual proof of (the 3-4-5 triangle instance of) Pythagoras' theorem:

I'm afraid you're still showcasing and not proving the theorem.

There is a common misconception that pictures cannot yield proofs, but in this case your fear is misplaced.  The construction definitely gives a fully rigorous proof, and in fact it even extends to the general case of Pythagoras' theorem.  Filling in some details:

Consider two congruent right-triangles ABC and BCD with shorter sides 1 and 2 and sharing a hypotenuse.  Add two further right-triangles CDE and CEF, congruent to each other, with shorter sides 2 and 4 and sharing a hypotenuse, as shown.  Since all four triangles are similar, ACF are collinear.  Finally, drop the perpendicular BG.  Then triangle EGB is right-angled and has sides 4-1=3, 2+2=4, and 4+1=5.

Similarly, starting with AB=x, AC=xy, DE=x y^2 gives the general Pythagoras theorem after a bit of algebra.

It's interesting that the same construction also proves the tangent double-angle formula (again in a special case, which is easily generalized):

tan <FEC = 1/2;

tan (2 <FEC) = tan <FED = 4/3.

5 hours ago, aeh5040 said:

There is a common misconception that pictures cannot yield proofs, but in this case your fear is misplaced.  The construction definitely gives a fully rigorous proof, and in fact it even extends to the general case of Pythagoras' theorem.  Filling in some details:

Consider two congruent right-triangles ABC and BCD with shorter sides 1 and 2 and sharing a hypotenuse.  Add two further right-triangles CDE and CEF, congruent to each other, with shorter sides 2 and 4 and sharing a hypotenuse, as shown.  Since all four triangles are similar, ACF are collinear.  Finally, drop the perpendicular BG.  Then triangle EGB is right-angled and has sides 4-1=3, 2+2=4, and 4+1=5.

Similarly, starting with AB=x, AC=xy, DE=x y^2 gives the general Pythagoras theorem after a bit of algebra.

It's interesting that the same construction also proves the tangent double-angle formula (again in a special case, which is easily generalized):

tan <FEC = 1/2;

tan (2 <FEC) = tan <FED = 4/3.

I get it now. I couldn't read this from the construction only, but I agree a construction or a simple picture can serve as proof. They sometimes even provide immediate insight without the need to deduce anything, at least not consciously.

Edited December 31, 20168 yr by Didumos69

12 hours ago, Didumos69 said:

I get it now. I couldn't read this from the construction only, but I agree a construction or a simple picture can serve as proof. They sometimes even provide immediate insight without the need to deduce anything, at least not consciously.

Cool!  To me it really is a slightly new way of looking at why Pythagoras works (although doubtless many people have been there before).  As it happens I did notice essentially the same phenomenon a few months ago by looking at two trans-clear base plates, as shown below, so I was primed to notice the significance of your construction.  So this really is an example of a (minor) mathematical insight from Lego!

 

Edited January 1, 20178 yr by aeh5040

Some may ask themselves, what's the practical use of all this? And even when you see the use of it, it turns out very hard to tell when you're in a situation where you can actually use this. So that got me thinking and after applying this 'perfect fit' several times, I came to the following rule of thumb:

Take the shorter end of any angled (or bend) liftarm. This end is not long enough to span a 5 stud length required for a full (3, 4, 5) triangle; it is either 2 studs long or 3 studs long (measured between pinhole centers). So, when working on a 1-stud grid, the outer pinhole is out-of-grid. However, the outer pinhole is always one stud distant from the grid:

• Outer end 3 studs long (measured between pinhole centers): To get 'on-grid', move one stud away from the outer pinhole - orthogonal to the outer end - towards the other end of the liftarm.
• Outer end 2 studs long (measured between pinhole centers): To get 'on-grid', move one stud away from the outer pinhole - orthogonal to the outer end - away from the other end of the liftarm.

Edited January 10, 20178 yr by Didumos69

This is very useful indeed. Obvious in hindsight, nice to be pointed out. Thanks for taking the effort to make the render. 

I used it on my F40, it took me a while to understand that it was legal, everything fit too well...  .

@Didumos69  First of all, thank you for bringing that topic up. For me, there is no question about practical use of such connections. Triangles are very stiff structures, so every large construction will benefit from them. You can see angled liftarms in some official TLG sets, used to reinforce the structure. If we are talking more specifically about these smaller ' Pythagoras ' triangles, the perfect illustration is your Rugged Supercar (it is brilliant, b.t.w.). And I have an example too. Right now I am working on a Unimog. Unimogs have that famous double-S shaped frame. If you make this frame using single layer of liftarms, the chassis will be very soft ( and you won't need a suspension in that case ). So I am using triangles to reinforce the frame around the bends:

It is old WIP picture . Right now I've switched to the legal ' Pythagoras ' triangle and the frame looks better. Can post image later. Updated frame sketch:

But I would agree, that applications for connections other than this simple  'Pythagoras ' triangle could be not so easy to come up with.

Edited January 11, 20178 yr by proran

I agree - these are very useful and interesting constructions.  This very interesting article is somewhat related (it is about Meccano, but it works just as well with Lego liftarms).  Among other things, it shows how to perfectly brace a regular pentagon.

http://www.staff.science.uu.nl/~hooft101/lectures/meccano.pdf

That's awesome, Aeh!

I used it in the telehandler, I also scratched my head with it that time, but LDD implied it's good enough fit.

12 hours ago, aeh5040 said:

I agree - these are very useful and interesting constructions.  This very interesting article is somewhat related (it is about Meccano, but it works just as well with Lego liftarms).  Among other things, it shows how to perfectly brace a regular pentagon.

http://www.staff.science.uu.nl/~hooft101/lectures/meccano.pdf

Indeed an awesome article! Now I wonder if we could go 3D with this. I mean regular bodies like the dodecahedron. A bit off topic and not of practical use for LEGO, but all these triangles and regularities remind me of a study I ever did - and a paper I wrote - a decade ago into equal-area map projections for polyhedral globes (http://bricksafe.com/files/Didumos/miscellaneous/Cagis.pdf). I remember that in the process of developing my method I also gained immediate insight from a few drawings.

3 hours ago, Lipko said:

That's awesome, Aeh!

I used it in the telehandler, I also scratched my head with it that time, but LDD implied it's good enough fit.

And now we know it is a perfect fit .

Edited January 12, 20178 yr by Didumos69

On 29-12-2016 at 7:41 PM, Lipko said:

My girlfriend prooved it in a few minutes (but also Didumos proof is valid, though it includes a statement that is not trivial: radius = area / half_of_circumference)

So a right triangle:

indeed, the triangle has an inner circle with radius x. The lines designated with x are perpendicular to the sides (since the inner circle is touching the triangle).Plus we have three equations:

1) A=z+x

2) B=x+y

3) C=y+z  which is the same as -C=-y-z

Summing the three equations: 1) + 2) + 3) gives

A+B-C = 2x. Which in our case is 4+3-5 = 2x -> 2=2x ->1=x.

EDIT2: My girlfriend has a very good visual method for calculating x (which leads to the same equation as I posted above):

So the Pythagorean triple (5, 12, 13) has an incircle with radius (5 + 12 - 13) / 2 = 2. Only realized this just now .

I had to study this one a little, but it's also a perfect fit. Can anyone tell me why? LXF-file here.

Edited March 23, 20177 yr by Didumos69