I Geometry problem of interest with a 3-4-5 triangle

[bob012345]

@Gavran On the video of post 28, they missed the boat I think when they tried to do the regular octagon. I sketched it myself at Starbucks today, (with the lines spaced in the ratio of 1,1, and 1), and I got what they did=no octagon, as I observed after looking at the video again.

I then calculated that the lines need to be separated in a ratio of 1,root 2, and 1 , both across and up and down, and then you get a regular octagon in the center. I then sketched it this way, and I was successful at getting the regular octagon in the center.

They did ok with the one that has the 3-4-5 triangles, where they connected corners to midpoints of the other sides.

I was looking at how to make a regular octagon in the spirit of the OP but it’s difficult to construct the ratio you mentioned without invoking a compass or just measuring which I’m trying to avoid.

 

[Charles Link]

I was looking at how to make a regular octagon in the spirit of the OP but it’s difficult to construct the ratio you mentioned without invoking a compass or just measuring which I’m trying to avoid.

Unless someone can come up with a better procedure, I think it is ok to measure the spacing of 1,root 2, and 1 to generate the regular octagon.

In any case, I thought the video of post 28 for the regular octagon was misleading. I thought they had generated a regular octagon, but when I went back to it after unsuccessfully trying to sketch it myself at Starbucks, I saw I had done exactly what they did and that the 1,1,1 spacing doesn't give you any octagon, but instead you simply get a square in the center.

The video of post 28 displays their result for perhaps a half second or less, so you really got to look at it carefully and watch it a couple of times to see that they don't give you what they claim they are giving you.

 

[Gavran]

Both octagrams in the post #28 are irregular octagrams and the first one has an irregular octagon in the centre while the second one has no octagon in the centre, as you mentioned in the post #30. The two inner regular squares form the octagon in the first case and they exist in the second case too, but they do not form the octagon because their sides only touch each other.

I then calculated that the lines need to be separated in a ratio of 1,root 2, and 1 , both across and up and down, and then you get a regular octagon in the center. I then sketched it this way, and I was successful at getting the regular octagon in the center.

What you get here is a regular octagram with a regular octagon in the centre. Now, you have eight equal interior angles and eight equal sides. In this case all existing triangles have sides in the same ratio as the segments of every side of the initial square, 1:the square root of 2:1.

 

[Charles Link]

@Gavran It helps to go back to the diagrams of post 23 to see this, (for the 3-4-5 case), because the results are not displayed nearly long enough in the video of post 28. Thank you again @DaveC426913 :)

We could use a diagram with the spacing of 1,root 2, and 1 (for the regular octagon case), if someone could draw it up.

 

[bob012345]

Here is a simple drawing of the regular octagon within the square. If the square has corners at (±1,±1), the outside points are root(2) from the center. The angle of the lines to the sides is 22.5 degrees.

 

[Charles Link]

@bob012345 ok, but try making the ratio of horizontal and vertical lines with spacing 1,root 2,1 and connecting the points on the sides where the slope of the lines is +/- 1. That will get you the regular octagon the way the video of post 28 was doing, (i.e. trying to do), where they incorrectly used a 1,1,1 spacing.

Edit: as mentioned in post 34, maybe someone could draw this case up of the regular octagon. @DaveC426913 You did a good job in post 18 of the irregular case.

 

[Charles Link]

It may be of interest that in computing the ratio of the radius of the circle inscribed in the octagon to the length of a side of the square for these two cases, I get $\frac{1}{ 2 \sqrt{5}} \approx .22$ for the irregular case, and $\frac{\sqrt{2}-1}{2} \approx .21$ for the regular case.

It might need a very close inspection to tell these two apart. The regular one is horizontal at the top and bottom though, and vertical on the sides.

 

[bob012345]

It may be of interest that in computing the ratio of the radius of the circle inscribed in the octagon to the length of a side of the square for these two cases, I get $\frac{1}{ 2 \sqrt{5}} \approx .22$ for the irregular case, and $\frac{\sqrt{2}-1}{2} \approx .21$ for the regular case.

It might need a very close inspection to tell these two apart. The regular one is horizontal at the top and bottom though, and vertical on the sides.

For the irregular octagon, there are two squares inside it but they are not equal. The ratio of areas is 1.125.

 

[Charles Link]

If you look at the diagram by @DaveC426913 in post 23, I think by symmetry that both of those squares have the same area. I'll need to take a second look, but I think they are identical.

Edit: and I think I figured out the problem: They look like squares, but they are not squares. They are rectangles. and on further review, no, I think they are indeed squares, but I think they are identical. :)
The distance across those parallel lines is the same in all cases.

 

[bob012345]

@bob012345 ok, but try making the ratio of horizontal and vertical lines with spacing 1,root 2,1 and connecting the points on the sides where the slope of the lines is +/- 1. That will get you the regular octagon the way the video of post 28 was doing, (i.e. trying to do), where they incorrectly used a 1,1,1 spacing.

Edit: as mentioned in post 34, maybe someone could draw this case up of the regular octagon. @DaveC426913 You did a good job in post 18 of the irregular case.

 

[Charles Link]

@bob012345 Read my post 36 again please. I originally had +/- 1/2 as the slope, but changed it to +/- 1 about an hour later when I saw I had made a mistake.

I wish I could draw it, but I don't even have easy access to even post photos.
The lower point on the left side connects to the right point on the upper side for a slope of +1 , etc. You need both of the horizontal and both of the vertical lines to complete the octagon.

 

[bob012345]

If you look at the diagram by @DaveC426913 in post 23, I think by symmetry that both of those squares have the same area. I'll need to take a second look, but I think they are identical.

Edit: and I think I figured out the problem: They look like squares, but they are not squares. They are rectangles. and on further review, no, I think they are indeed squares, but I think they are identical. :)

I thought you meant the squares inscribed in the irregular octagon.

 

[Charles Link]

@bob012345 Please see the additions to post 39. :) and yes, I'm referring to the diagrams of post 23. :)

I think post 40 is a regular octagon, but I have a different way to construct it. (post 36) :)

 

[bob012345]

@bob012345 Read my post 36 again please. I originally had +/- 1/2 as the slope, but changed it to +/- 1 about an hour later when I saw I had made a mistake.

I wish I could draw it, but I don't even have easy access to even post photos.
The lower point on the left side connects to the right point on the upper side for a slope of +1 , etc. You need both of the horizontal and both of the vertical lines to complete the octagon.

 

[Charles Link]

I need to learn to work the computer graphics package. :) If you follow my post 36 carefully, you will get a small regular octagon in the center of the figure, nearly the same in size as the one of the irregular case of post 23. :)

It will be oriented so that the top and bottom are both horizontal, and the right and left sides are both vertical lines.

 

[bob012345]

I need to learn to work the computer graphics package. :) If you follow my post 36 carefully, you will get a small regular octagon in the center of the figure, nearly the same in size as the one of the irregular case of post 23. :)

It will be oriented so that the top and bottom are both horizontal, and the right and left sides are both vertical lines.

Finally, I think you meant this.

 

[Charles Link]

@bob012345 Yes, the octagon in the very center is what I was looking for. Excellent. :)

That's what I think they were trying to do in the second part of the video of post 28, but they didn't get it right.

 

[bob012345]

It looks off but I computed the coordinates and checked the edge lengths which are $2-√2$.

 

[DaveC426913]

Edit: as mentioned in post 34, maybe someone could draw this case up of the regular octagon. @DaveC426913 You did a good job in post 18 of the irregular case.

This, by the way, is the actual fold template for the origami diamond.

You can see that the pseudo-diagonals are now bisections (22.5°) of the diagonals because they're folded to be equal:

 

[Charles Link]

It looks off but I computed the coordinates and checked the edge lengths which are $2-√2$.

I believe you have it completely correct, (post 46). You didn't get the black marker right on top of the fine line in all cases, but very good. Excellent. :) and I agree with your calculation for the length of each side of the octagon. :)

Very close inspection shows the root 2 in your diagram is 1.375=11(.125), instead of 1.414, but it's more than accurate enough to illustrate the concept. :)

 

[bob012345]

I believe you have it completely correct, (post 46). You didn't get the black marker right on top of the fine line in all cases, but very good. Excellent. :) and I agree with your calculation for the length of each side of the octagon. :)

Very close inspection shows the root 2 in your diagram is 1.375=11(.125), instead of 1.414, but it's more than accurate enough to illustrate the concept. :)

Yes, I rounded it to fit the grids of the graph paper from 2.828 to 2.75 since the grids were quarter inch. Made the drawing a lot easier!

 

[bob012345]

Apparently, this construction, the original square with semi-diagonals as @DaveC426913 drew in post #18, goes back thousands of years and has numerous properties of interest. There is a Wooden book called The Diagram all about it. This construction is also called the Helicon. Wooden books allows all their books to be read online if one isn’t bothered by the ‘sample’ watermark. Charles, you may be interested in page 10 which discusses 3-4-5 triangles.

https://woodenbooks.com/index.php?id_cms=8&controller=cms#!DIA

 

[SammyS]

@bob012345 ok, but try making the ratio of horizontal and vertical lines with spacing 1,root 2,1 and connecting the points on the sides where the slope of the lines is +/- 1. That will get you the regular octagon the way the video of post 28 was doing, (i.e. trying to do), where they incorrectly used a 1,1,1 spacing.

Edit: as mentioned in post 34, maybe someone could draw this case up of the regular octagon. @DaveC426913 You did a good job in post 18 of the irregular case.

Here I have quoted Post#36, where you refer to @bob012345 's figure in Post #35.

Rotate bob0.., 's figure by $22.5^\circ$, ignoring the larger square. Now the resulting horizontal lines are cut by the vertical lines in the ratios; $1:\sqrt 2:1$ . Similarly, the resulting vertical lines are cut by the horizontal lines in the ratios; $1:\sqrt 2:1$ .

Indeed, bob012345 has drawn a regular octagram which, of course, has a regular octagon at its center.

It looks to me that your drawing is more accurate than what's claimed. As drawn, the upper most vertex should be $12\cdot\cot(22.5^\circ ) \approx 28.97$ grid spacings above the lower edge of the auxiliary square. That's pretty close to the 29 grid spacings as I count them.

 

[Charles Link]

@SammyS I agree in post 35 that @bob012345 has a regular octagon. It is in post 46 though that he finally drew it the way I was looking for (see post 36) which draws it how the video of post 28 should have drawn it, and it seems they attempted to draw a regular octagon in the second half of the video, but they were unsuccessful. They labeled it an isogonal octagon in the heading, and it wasn't.

Edit: Looking at it more carefully, the keyword may be octagram instead of octagon. I was looking for a regular octagon to match the irregular one made from the 3-4-5 triangles that we have been discussing. :) @bob012345 drawing of post 46 is exactly what I was looking for. :)

(In post 35 @bob012345 goes outside the square to make the construction).

 

[bob012345]

It might be interesting to consider a square inscribed in a 3-4-5 triangle. If the square must touch all three sides, what is the largest, the smallest possible?

 

[Charles Link]

It might be interesting to consider a square inscribed in a 3-4-5 triangle. If the square must touch all three sides, what is the largest, the smallest possible?

I think I have a solution to this, but I need to double-check my calculations. It was a little tricky computing the size of the square for the case of some arbitrary slope m of the top side of the square. I get for the side of the square $s=3 \sqrt{m^2+1}/(7/4+m)$. This works for $0 \leq m \leq 4/3$ and has a minimum for $m=4/7$ if I computed the derivative correctly that I set to zero.

I have yet to compute it for $m>4/3$ or $m<0$.

Edit: I now also worked it for $4/3 \leq m<+\infty$. I get $s= 3 \sqrt{m^2+1}/(7m/4+3/4)$, with a minimum at $m=7/3$. This now means that the orientation of the square has now been considered over the complete 90 degree span of rotation.

The absolute maximum turns out to be for $m=0$, with $s=12/7$, and the absolute minimum is at $m=4/7$ with $s=12/\sqrt{65}$. There is also a relative maximum for $m=4/3$ with $s=60/37$, and a relative minimum at $m=7/3$ with $s=12/\sqrt{58}$.

Note: In the above calculations, the 3,4,5 triangle has the 3 on the y axis, and the 4 on the x axis, with the hypotenuse having a slope of -3/4.

 

[bob012345]

I got the maximum and one relative maximum but missed the minimum. I had the diagonal of the square vertical which gives s=12/11 root(2).

 

[Charles Link]

I got the maximum and one relative maximum but missed the minimum. I had the diagonal of the square vertical which gives s=12/11 root(2).

When the diagonal is vertical, that has $m=1$, and my formula for $s$ agrees with that result, s=12/11 root(2), but it is not the minimum, and it really is necessary to take the derivative and set it to zero to find the minimum. There doesn't seem to be any way of deducing or guessing the minimum by other means.

There seemed to be a possibility that having the side lying on the hypotenuse ($m=4/3$), (note the slope of the side in this formula is perpendicular to the hypotenuse which has slope of -3/4), might give a minimum, but that is instead a relative maximum. (Presumably that is the relative maximum that you are referring to in the above).

One other item that may be worth mentioning: The point $m=4/3$ is the endpoint for both of my formulas, (post 56), and it looks like you might get two different answers for $s$ when you put $m=4/3$ into them. It turns out though, upon doing the arithmetic, in both cases you do get $s=60/37$.

Notice also that $m=0$ in the first formula is the same case as $m=+\infty$ of the second formula, (rotating a square by 90 degrees gives the same square), and for both the result is $s=12/7$.

 

[Gavran]

I can recognize a lot of squares with three corners touching the three triangle’s sides while the fourth square’s corner does not touch any of the three triangle’s sides. The two triangle legs will always be touched with the same two square’s corners A and B, while the triangle’s hypotenuse will be touched with the one of the remaining two square’s corners C or D. There are two cases depending on which corner of the remaining two square’s corners has been used for touching the triangle’s hypotenuse. These two cases are described by two different formulas for s in the post #56 and they have two minimums, the first one with the side s=12/(65)^1/2 and the second one with the side s=12/(58)^1/2, the absolute minimum and the relative minimum as @Charles Link said in the post #56.

Also, I can recognize only two different squares with all four corners touching the three triangle’s sides, the first one with the side s=12/7 and the second one with the side s=60/37, the absolute maximum and the relative maximum as @Charles Link said in the post #56. These two squares are two maximums of two different formulas for s in the post #56.

 

[bob012345]

The two solutions with square edges along the triangle sides cut the triangle into either two or three smaller 3-4-5 triangles. The case where only three points of the square touch the sides of the triangle cuts it into two triangles different from the 3-4-5 and one four sided shape.