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

[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.