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

[Charles Link]

TL;DR Summary

Rather interesting video that shows that a 3-4-5 right triangle results when you draw lines connecting the corners to the midpoints of a square

It was a week or two ago that a friend of mine showed me a video of interest. The video is somewhat incomplete mathematically, but still is a good one. The author should state that when the product of the slopes of the two lines is minus one, (slopes are +2 and -1/2), that the lines are perpendicular.

In addition, using the $\tan(\theta-\phi)$ identity, it is easy to show that the triangle is a 3-4-5 because the tangent of the apex angle, (slopes of +2 and -2 for the lines forming the angle), is computed to be 4/3. For the hypotenuse to be 5, the square needs to have sides of 2 root 5, which the author also omits.

I worked it a little further and computed the radius of a circle that is inscribed in a (right) 3-4-5 triangle. I found somewhat surprisingly that it has a radius of one unit. Following that, I was also able to show that the center of the circle coincides with the center of the square, so that all four 3-4-5 triangles have the same inscribed circle.

See

I see the video didn't load, but you may be able to follow the discussion, even without the video.

 

[Charles Link]

In a related calculation, I just computed the other day, (I think for the first time for me), that the 5-12-13 triangle has an inscribed circle of radius of 2. I did a google which confirms I got it correct.

Edit: and the 8-15-17 triangle has a radius of 3 for its inscribed circle if my calculations are correct.

Edit 2: In a correction to the above post, there are in fact a total of eight 3-4-5 triangles. The same inscribed circle of unit radius sits inside all of these with an octagon around it with all sides of the same length, but the sides come in 4 pairs, and calculations show that it is not a regular octagon. Instead of 45 degree increments, the angles alternate at 53 degrees (approximately) and 37 degrees.

Edit 3: In drawing all of the lines connecting corners to opposite midpoints, you get 8 larger 3-4-5 triangles, but upon further review you get many smaller 3-4-5 triangles as well. The 53 degrees and 37 degrees found in the calculations of Edit 2 are to be more precise the angles of smaller 3-4-5 triangles that appear when you draw all of the lines as just mentioned.

 

[DaveE]

I got to the video on Facebook by clicking "Video on Facebook".

 

[Charles Link]

See my Edits in post 2. I was a little surprised the other day when I computed the tangent of the angle between adjacent sides of the octagon that forms in the center of this thing around the inscribed circle. The sides are all the same length, but it comes in four pairs. The tangent of the angle alternates from 4/3 to 3/4, (it's based upon 3-4-5 triangles), and thereby going around two sides of the octagon, the direction changes by 90 degrees, but it is not 45 degrees for each increment, (draw arrows along the sides as you go counterclockwise around the octagon. The direction changes by 53 degrees, and then 37 degrees, etc.), even though all the sides have the same length. I found it to be an interesting application of trigonometry to solve this thing. When I saw all sides to be the same length, I was erroneously expecting a regular octagon, when it is only semi-regular.

 

[Gavran]

I worked it a little further and computed the radius of a circle that is inscribed in a (right) 3-4-5 triangle. I found somewhat surprisingly that it has a radius of one unit. Following that, I was also able to show that the center of the circle coincides with the center of the square, so that all four 3-4-5 triangles have the same inscribed circle.

Just to add that a triangle with sides in the ratio 3:4:5 is the unique right triangle whose sides are in arithmetic progression.

 

[bob012345]

In a related calculation, I just computed the other day, (I think for the first time for me), that the 5-12-13 triangle has an inscribed circle of radius of 2. I did a google which confirms I got it correct.

Edit: and the 8-15-17 triangle has a radius of 3 for its inscribed circle if my calculations are correct.

Edit 2: In a correction to the above post, there are in fact a total of eight 3-4-5 triangles. The same inscribed circle of unit radius sits inside all of these with an octagon around it with all sides of the same length, but the sides come in 4 pairs, and calculations show that it is not a regular octagon. Instead of 45 degree increments, the angles alternate at 53 degrees (approximately) and 37 degrees.

Edit 3: In drawing all of the lines connecting corners to opposite midpoints, you get 8 larger 3-4-5 triangles, but upon further review you get many smaller 3-4-5 triangles as well. The 53 degrees and 37 degrees found in the calculations of Edit 2 are to be more precise the angles of smaller 3-4-5 triangles that appear when you draw all of the lines as just mentioned.

There appears to be a simple formula for the radius of the inscribed circle of any right triangle that goes as the area of the triangle divided by the semi-perimeter of the triangle. That is just half the perimeter. It works for all these cases you mention but I’m not certain it is really general yet.

 

[SammyS]

There appears to be a simple formula for the radius of the inscribed circle of any right triangle that goes as the area of the triangle divided by the semi-perimeter of the triangle. That is just half the perimeter. It works for all these cases you mention but I’m not certain it is really general yet.

Not only is this true for all right triangles, it's true for all triangles in general.

 

[Charles Link]

Just to add that a triangle with sides in the ratio 3:4:5 is the unique right triangle whose sides are in arithmetic progression.

Very good. $x^2+(x+d)^2=(x+2d)^2$ has two solutions when you solve for $x$ in terms of $d$: $x=3d$ and $x=-d$, with the second one basically being extraneous.

 

[Charles Link]

Not only is this true for all right triangles, it's true for all triangles in general.

Yes, last night I verified this in general, using a triangle with one slope of zero, with vertices at $(0,0)$ and $(a,0)$, and with the other two sides of slope $m_1$ and $m_2$. The distance from a point $(x_o,y_o)$ to the line $Ax+By+C=0$ is $|Ax_o+By_o+C|/(A^2+B^2)^{1/2}=r$. Solving this for both lines gave the general formula for $r$ posted by @bob012345 . :)

Note: It was also necessary to compute the point where the two lines intersect in order to compute the lengths of the other two sides to compute the perimeter, and now having the height, we then could also calculate the area.
Edit: and note also the center of the circle is at $(x_o,r)$. We solve for $r$ eliminating $x_o$ from the two equations.

 

[bob012345]

Have you tried a 13,14,15 sided triangle? You can guess the radius.

 

[Charles Link]

Have you tried a 13,14,15 sided triangle? You can guess the radius.

Very clever. I see for that one you have a base of 14, that can be split into 5 and 9, so that the height of the triangle is 12, making for right triangles (side by side) of 5,12,13, and 9, 12, 15 (3x 3,4,5). The radius from our formula will be 4. :)

Edit: It is even possible to guess where the center of the circle is. I guessed it to be around (6,4), and I believe I got the calculations to confirm that.

 

[Charles Link]

It may be stating the obvious, (for the inscribed circle of the triangle of post 10), but with the base of 14, where the center of the circle is 6 from the left and 4 up, with the base of 13, the center is 7 from the left and 4 up, and with the base of 15, the center of the circle is 8 from the left and 4 up. (The angle bisectors of the three angles of the triangle meet at the center of the inscribed circle).

 

[DaveC426913]

... there are in fact a total of eight 3-4-5 triangles. ...

How is there more than one 3-4-5 triangle?

(I cannot see the video).

 

[Charles Link]

How is there more than one 3-4-5 triangle?

(I cannot see the video).

The video just shows only one, but if you draw all of the lines connecting the four corners to the midpoints of the other sides, you get 8 larger 3-4-5 triangles and many smaller 3-4-5 triangles.

 

[SammyS]

The video just shows only one, but if you draw all of the lines connecting the four corners to the midpoints of the other sides, you get 8 larger 3-4-5 triangles and many smaller 3-4-5 triangles.

I couldn't see the video either, but following your early posts, especially Posts 2 & 4, I was able understand how this procedure produced multiple 3-4-5 right triangles. Some crude sketches also helped me see that many smaller triangles also resulted., but, the crudeness of these sketches was also somewhat misleading.

More accurate drawings led me to the following conclusions:
There are 8 large 3-4-5 triangles, all the same size.
There are also eight 3-4-5 triangles, each of which is 1/2 the size of the large triangles.
There are eight more 3-4-5 triangles, each of which is 1/3 the size of the large triangles.
And finally, there are also eight 3-4-5 triangles, each of which is 1/6 the size of the large triangles. These smallest tringles are arranged in such a way that the 8 hypotenuses form the octagon which you have described.

 

[DaveC426913]

The video just shows only one, but if you draw all of the lines connecting the four corners to the midpoints of the other sides, you get 8 larger 3-4-5 triangles and many smaller 3-4-5 triangles.

OK, since I can't see the video, I can't follow what you're saying.

I thought when you said "there were eight 3-4-5- triangles" you meant there are eight types of 3-4-5 triangles that exist - like "there are five types platonic solids".

But it sounds like you are saying "in this particular illustrated scenario, there are eight instances of a 3-4-5 triangle rendered". (all of which are similar, by definition.)

I will step back, since I can't contribute meaningfully.

 

[Charles Link]

@DaveC426913 Start with a square and draw lines from all four corners to the midpoints of the two other sides of the square. I don't know how these graphics programs work, but it should be simple for you to sketch by hand. If you are proficient with these graphic packages, perhaps you could even supply the diagram for all of us. :)

 

[DaveC426913]

That looks like an origami 'diamond' base.

Almost; not quite.

 

[Charles Link]

Thank you @DaveC426913 <3 for the above diagram (post 18). The video on this is very incomplete, claiming that the triangles are of type 3-4-5, but failed to show it mathematically. It really is some interesting trigonometry, where the lines of slope 2 and -1/2 are perpendicular because the product is minus one. The triangles can then be shown to be of type 3-4-5, (there are also some of type 1 ,2, root 5), as mentioned in the OP by computing the tangent of the angle between a pair of lines.

It is also of interest to inscribe a circle in the center which turns out to have unit radius if the dimensions of the larger triangles are 3,4,5.

 

[Charles Link]

For the octagon in the center, all sides are the same length, but the angles alternate as mentioned previously, (post 4), so it is not a regular octagon. Be sure and see the diagram in post 18 above, supplied by @DaveC426913 .

 

[DaveC426913]

For the octagon in the center, all sides are the same length, but the angles alternate as mentioned previously, (post 4), so it is not a regular octagon. Be sure and see the diagram in post 18 above, supplied by @DaveC426913 .

I don't follow. The figure is perfectly symmetical. How can the central octagon not have eight 135 degree inner angles?
Sunuvagun.

So it's a 127/143° octagon.

 

[Charles Link]

I don't follow. The figure is perfectly symmetical. How can the central octagon not have eight 135 degree inner angles?

It surprised me at first when I computed the angles between two of the lines. I thought I would get the result that the tangent of the angle is 1, making for 45 degrees, but instead the angles alternate between having 4/3 for their tangent and 3/4 for their tangent. The result is that in going across two sides, you turn 90 degrees, but a single turn is not 45 degrees. It really is very interesting. :)

Note: Draw arrows on the sides of the octagon, each going in the counterclockwise direction. In the above, I'm computing the angle change from one arrow to the next.

If you place a protractor over your diagram, I think you might be able to see the angles are not identical, but in fact alternate between about 53 degrees and 37 degrees.

 

[DaveC426913]

It surprised me at first when I computed the angles between two of the lines. I thought I would get the result that the tangent of the angle is 1, making for 45 degrees, but instead the angles alternate between having 4/3 for their tangent and 3/4 for their tangent. The result is that in going across two sides, you turn 90 degrees, but a single turn is not 45 degrees. It really is very interesting. :)

Note: Draw arrows on the sides of the octagon, each going in the counterclockwise direction. In the above, I'm computing the angle change from one arrow to the next.

If you place a protractor over your diagram, I think you might be able to see the angles are not identical, but in fact alternate between about 53 degrees and 37 degrees.

Oh I see. The diagram is perfectly four-fold symmetrical - but it is not eight-fold symmetrical.

And the 37/53 is reflected in the two different 'X' shapes:

 

[Charles Link]

@DaveC426913 I see your additions to post 21. Very good. It surprised me also when I first computed it. I was at Starbucks at the time, and I really wondered, how could it be possible to have four pairs of sides all the same, and still not be a regular octagon? :-)

Edit: You can also see that the 53 and 37 degree angles do make for the many 3-4-5 type triangles that appear in the diagram.

 

[SammyS]

@DaveC426913 I see your additions to post 21. Very good. It surprised me also when I first computed it. I was at Starbucks at the time, and I really wondered, how could it be possible to have four pairs of sides all the same, and still not be a regular octagon? :-)

Edit: You can also see that the 53 and 37 degree angles do make for the many 3-4-5 type triangles that appear in the diagram.

Of course, those angles are in degrees and are rounded to the nearest degree.

 

[bob012345]

@DaveC426913 I see your additions to post 21. Very good. It surprised me also when I first computed it. I was at Starbucks at the time, and I really wondered, how could it be possible to have four pairs of sides all the same, and still not be a regular octagon? :-)

Edit: You can also see that the 53 and 37 degree angles do make for the many 3-4-5 type triangles that appear in the diagram.

The edges are the same lengths but parallel edges are skewed wrt each other, they are parallelograms not rectangles.

 

[Charles Link]

The edges are the same lengths but parallel edges are skewed wrt each other, they are parallelograms not rectangles.

From symmetry, it is readily seen that the sides come in 4 identical pairs. The angle between the two sides in each pair is not 135 degrees though. If it were, then it would be a regular octagon.

 

[Gavran]

Be sure and see the diagram in post 18 above, supplied by @DaveC426913 .

The polygon is an isotoxal octagram.

The source: https://en.wikipedia.org/wiki/Octagram.

There is an interesting video which shows a difference in drawing an isotoxal octagram and an isogonal octagram.

 

[Charles Link]

as mentioned in the OP, $\tan(\theta-\phi)=\frac{\tan(\theta)-\tan(\phi)}{1+\tan(\theta) \tan(\phi)}$ makes it very easy to compute the tangent of the angle $\Phi$ between two lines of slopes $m_1$ and $m_2$ :
$\tan(\Phi)=\frac{m_1-m_2}{1+m_1 m_2 }$.

In the above, $\Phi=\theta-\phi$.

Very good videos, :) but the mathematics of this identity, which I didn't spell out completely in the OP, is for me the most interesting part of the construction of both the triangles and the octagon. IMO every physics student should make sure they can do this calculation routinely, without needing to dig out a textbook for the formula, etc.

 

[Charles Link]

@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.

 

[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. :)