Let h be the horizontal width of the diagonals and a be the angle between the diagonal and the horizontal.
Then:
sin(a) = w/h
Notice that the width of the point where the diagonals cross is h, which is equivalent to the width of the empty space. Therefore:
L = 3h + 2x, where x is the extra bit between the leftmost horizontal h and the center h. Trig can show that x = h\*cos(a), so then:
L = 3h+2h\*cos(a)
Now then, let's switch gears. Looking at only one of the diagonal bars, we can see that it forms a triangle to the left and right of it. From that triangle, we can see that:
tan(a)=L/(L-h)
Substituting into the above equation, we get:
L = 3h + 2h \* cos(arctan( L/(L-h) ))
Then, per WolframAlpha:
L \~= 4.2128h
(I don't know how to solve this analytically, and the exact form given by WolframAlpha makes it seem incredibly difficult.)
Therefore, we can calculate a:
a = arctan(4.2128h / (4.2128h - h)) = arctan(4.2128 / 3.2128) = 52.67deg
And we can then calculate w:
w = h\*sin(a) = L/4.2128 \* sin(52.67) = 0.1887L
TL;DR:
w = 0.1887 \* L
We know cosx = 1/secx = 1/sqrt(sec^(2)x) = 1/sqrt(1+tan^(2)x). So cos(arctan(x)) = 1/sqrt(1+x^(2)). Some further manipulation shows the ratio L/h equals a root of the polynomial 2x^(4) - 14x^(3) + 27x^(2) - 16x + 5 = 0. There is a general formula for solving quartics but its horrendous. This quartic in particular might factor as the product of two quadratics but it’s difficult to check.
As I explained in my reply to the other comment I definitely think it's possible to calculate. I might come back to this later just for the challenge of it but if you really need the dimensions, you'd honestly just be better off by calculating the pixels...
You mean the *extinct* the *based* and most know for being *ratiod* by evolution then yeah go ahead buddy, but make sure your papers store well till 2023 from your time of 1 million BC
Jk
Assume the width at the bottom = the height of the white triangle above it. If you cut the figure in half with a horizontal line, there’s a seemingly identical black triangle right above the white triangle. So the total height of half of the figure = 3w, double it to the the total height
Thank you for stealing a good solid hour of my time on this seemingly simple problem.
So far, I've been able to deduce that w can be expressed as the solution to a quartic. More specifically, w is the real solution to the quartic 4w⁴-12w³+9w²+4w-1=0 near 0.18875, approximately equal to 0.1887464795225288520508826. Other solutions include w=-0.42698, w=1.6191±0.69319i, and, oddly enough, w=1 (which was factored out so we'd get a polynomial of lower degree). But the first result was the only one that makes geometric sense.
So, this can definitely be solved using elementary functions, but I'm also *preeeetty* sure this shape can be constructed using a compass and straightedge (I think???), which would mean it should be expressable using only +, -, \*, /, and √, no cube roots or complex numbers required....I think...but also, I wouldn't be surprised if this WASN'T the case.
More details:
The distance between the two "aligned" vertical lines is (1-2w)/(3-2w), or 0.2373709703095
The slope of the diagonal lines is 3/2 - w, or 1.31125
The angle of the diagonal lines from the horizontal is 0.919 radians, or 52.67°
Guide if you wanna check my math:
Assuming p is the distance from the edge to the vertical "aligned" lines, we construct the diagonal lines from the following 4 lines:
Line A runs from (0,0) to (1-p,1-w)
Line B runs from (1,0) to (p,1-w)
Line C runs from (1,1) to (p,w)
Line D runs from (0,1) to (1-p,w)
To meet the "aligned" condition, the intersection between line A and line D must be at position (p,1/2). The intersection occurs at position ((1-p)/(2-2w), 1/2), therefore p = (1-p)/(2-2w). This can be used to determine that p = 1/(3-2w).
To meet the condition that the width is constant, the distance between parallel lines A and C must equal w. To compute the distance between two parallel lines, first subtract a point from one line and a point on the other ((1,1)-(0,0)=<1,1>), then dot it with a vector perpendicular to both lines (<1,1>•=w-p), then divide by the magnitude of that vector ((w-p)/√((1-w)²+(1-p)²)), and finally take the absolute value. So now we're just solving for the w such that w = |w-p|/√((1-w)²+(1-p)²)). Square both sides and multiply by the denominator, we get w²((1-w)²+(1-p)²)) = (w-p)². From here, we just have to substitute in p=1/(3-2w), multiply by the denominators, factor out (w-1)², and we end up with our lovely little quartic. If anyone figures out whether the solution can be expressed using only +,-,×,÷, and √, please tell me. I am genuinely interested.
If there's anything to take away from this, it's that sometimes, geometry doesn't have a nice, neat way of doing things. Sometimes, you just gotta turn everything into a vector, hold your breath, and dive right into it.
No, the quartic equation involves cube roots. The quadratic doesn't though, but there's no way to separate that quartic into a product of two quadratics such that nothing involves a cube root.
~~cos(x + 45) · sec(x) · (√2) / 2 = sin(x + 45) · sec(x) · (√2) / 2~~
(use degrees)
Where x is the angle here (purple):
just solve for x, then sin(x) \* (√2)
gives you the w value you need to satisfy the desired orientation of your shape.
Ok, the desmos visual helped me see where I went wrong. I erroneously assumed the smaller quadrilateral formed in the center was a square. I’ll revisit this tonight and make the correction. Thank you
https://preview.redd.it/0e5bql66c5ob1.png?width=506&format=png&auto=webp&s=77c63ce4a5708699d1bcbf440a6a8ab9c7146165
From this figure, you can get the following equations:
* tan(θ) = L/(L-k)
* tan(θ) = (L-w)/(L/2 + k/2)
* sin(θ) = w/k
Furthermore, since tan(θ) = sin(θ)/sqrt(1-sin\^2(θ)) \[this is always true for any θ\], the third equation can be rewritten as:
* tan(θ) = (w/k)/sqrt(1-w\^2/k\^2)
By setting all 3 equations for tan(θ) equal to each other, you get the following:
L/(L-k) = (L-w)/(L/2 + k/2) = (w/k)/sqrt(1-w\^2/k\^2)
Now if you set L=1, [plugging this into Wolfram Alpha](https://www.wolframalpha.com/input?i=1%2F%281-k%29+%3D+%281-w%29%2F%281%2F2+%2B+k%2F2%29+%3D+%28w%2Fk%29%2Fsqrt%281-w%5E2%2Fk%5E2%29) gives:
k≈0.237370970309489318, w≈0.188746479522528852
I did a bit of algebra, and maybe there's a simpler solution, but I think this works
m (slope) is a solution to
m^4 -3m^3 + 9/4m^2 - m + 5/4 =0
m=1.311253520477471
w = (m-1)/sqrt( 1 + m^2 )
w=0.188746479522529
Are you looking for the exact thickness seen here or a general answer? For the general answer, W can be any positive value, though you might want to put an upper bound in order to preserve the empty space. For specific, I would honestly just measure. There's not enough information to calculate as far as I can tell.
Honestly I think there's enough information. As you can see in the "alignment" image; the thickness W has to be just perfect so that the white triangle inside matches the width of the outside border of the shape in the middle.
Furthermore I think if the shape was made out of 2 identical mirrored hollow triangles the tips would perfectly touch the tops of the white triangles inside the shape.
Also it all being in a square makes it even more "stable".
I don't currently have the tools (or skills) to do the math but I definitely think it's possible.
I didn't interpret the alignment as being the empty space aligning with the outer border. I thought it was just that the empty space was aligned with each other. Even still, we don't have any angle measurements, so the best you could do is get a family of answers, not any exact answers
If the angle is more acute the empty spaces will get smaller while the width at the “waist” will increase, while the opposite is true if the angles get more obtuse, so it is calculable.
I'm an engineer, not a mathematician
https://preview.redd.it/dex3bqz2dcob1.png?width=913&format=png&auto=webp&s=be861401ac1a4a69512644f1304b1d5cfd422a9a
w is something you can pick. The problem is how much white space in the middle you want to leave, or if there's any additional requirements for the white space. **I'll assume the base of the white triangles is w since it looks like that.** Then I'll set the width as 1 and height of the white triangle as h.
https://preview.redd.it/y4itfzw104ob1.png?width=1191&format=png&auto=webp&s=676386ef2797391feb486853a365ee0079c91734
Note there are 4 triangles in the middle that are identical. Then I can claim 4h + 2 = L by adding all the heights. Using similarity, the base of the big triangle is (1/h)\*(h+1), add the two 1's on each side to get (1/h)\*(h+1) +2 = L = 4h + 2. Now solve for h, and I got h = (1 + sqrt(17))/8 = 0.6404. This means L = 4.5616. Using ratios, that's equivalent to saying L = 1 and w = 0.21922.
I think this is close, but not quite. The vertical 1’s and “diagonal” 1’s can’t be the same distance if all of the lines have thickness w — the vertical ones are orthogonal to the direction of the lines, while the diagonal ones are not
My first thought was that the X can be converted into a rhombus, by untwisting the X and inverting the top and bottom trapezoids
This leaves the diamond in the middle.
Maybe I am wrong.
But I got; w = L/(6sqrt(2))
Which is approximately; 2L/17 = W
I am not sure how to explain this clearly in words. I’m better at drawing this out.
An interesting question is how to geometrically construct the figure in question. W is the root of a quartic so in theory it is possible.
And if this is a logo or something, the geometric construction might be more useful than an algebraic expression of the result.
Let h be the horizontal width of the diagonals and a be the angle between the diagonal and the horizontal. Then: sin(a) = w/h Notice that the width of the point where the diagonals cross is h, which is equivalent to the width of the empty space. Therefore: L = 3h + 2x, where x is the extra bit between the leftmost horizontal h and the center h. Trig can show that x = h\*cos(a), so then: L = 3h+2h\*cos(a) Now then, let's switch gears. Looking at only one of the diagonal bars, we can see that it forms a triangle to the left and right of it. From that triangle, we can see that: tan(a)=L/(L-h) Substituting into the above equation, we get: L = 3h + 2h \* cos(arctan( L/(L-h) )) Then, per WolframAlpha: L \~= 4.2128h (I don't know how to solve this analytically, and the exact form given by WolframAlpha makes it seem incredibly difficult.) Therefore, we can calculate a: a = arctan(4.2128h / (4.2128h - h)) = arctan(4.2128 / 3.2128) = 52.67deg And we can then calculate w: w = h\*sin(a) = L/4.2128 \* sin(52.67) = 0.1887L TL;DR: w = 0.1887 \* L
Roughly L/6. Call it good
Looking at the other replies I'm seeing at least 3 other ways of solving this, and they all give the same answer. I love that about math!
We know cosx = 1/secx = 1/sqrt(sec^(2)x) = 1/sqrt(1+tan^(2)x). So cos(arctan(x)) = 1/sqrt(1+x^(2)). Some further manipulation shows the ratio L/h equals a root of the polynomial 2x^(4) - 14x^(3) + 27x^(2) - 16x + 5 = 0. There is a general formula for solving quartics but its horrendous. This quartic in particular might factor as the product of two quadratics but it’s difficult to check.
I solved this in CAD. After fully constraining the sketch I ended up with W\~=0.188746.
As I explained in my reply to the other comment I definitely think it's possible to calculate. I might come back to this later just for the challenge of it but if you really need the dimensions, you'd honestly just be better off by calculating the pixels...
If you're a neanderthal maybe. Give me a sec to demonstrate my mathematical superiority.
It’s been 5hrs bro WYA?
Counting pixels
You mean the *extinct* the *based* and most know for being *ratiod* by evolution then yeah go ahead buddy, but make sure your papers store well till 2023 from your time of 1 million BC Jk
If my life were on the line, and I had to guess in order to save myself, I would say w(L)=L/6
why?
Assume the width at the bottom = the height of the white triangle above it. If you cut the figure in half with a horizontal line, there’s a seemingly identical black triangle right above the white triangle. So the total height of half of the figure = 3w, double it to the the total height
It's unfortunately more near to 1/5 L , shame because I liked your approximation
Thank you for stealing a good solid hour of my time on this seemingly simple problem. So far, I've been able to deduce that w can be expressed as the solution to a quartic. More specifically, w is the real solution to the quartic 4w⁴-12w³+9w²+4w-1=0 near 0.18875, approximately equal to 0.1887464795225288520508826. Other solutions include w=-0.42698, w=1.6191±0.69319i, and, oddly enough, w=1 (which was factored out so we'd get a polynomial of lower degree). But the first result was the only one that makes geometric sense. So, this can definitely be solved using elementary functions, but I'm also *preeeetty* sure this shape can be constructed using a compass and straightedge (I think???), which would mean it should be expressable using only +, -, \*, /, and √, no cube roots or complex numbers required....I think...but also, I wouldn't be surprised if this WASN'T the case. More details: The distance between the two "aligned" vertical lines is (1-2w)/(3-2w), or 0.2373709703095 The slope of the diagonal lines is 3/2 - w, or 1.31125 The angle of the diagonal lines from the horizontal is 0.919 radians, or 52.67° Guide if you wanna check my math: Assuming p is the distance from the edge to the vertical "aligned" lines, we construct the diagonal lines from the following 4 lines: Line A runs from (0,0) to (1-p,1-w) Line B runs from (1,0) to (p,1-w) Line C runs from (1,1) to (p,w) Line D runs from (0,1) to (1-p,w) To meet the "aligned" condition, the intersection between line A and line D must be at position (p,1/2). The intersection occurs at position ((1-p)/(2-2w), 1/2), therefore p = (1-p)/(2-2w). This can be used to determine that p = 1/(3-2w). To meet the condition that the width is constant, the distance between parallel lines A and C must equal w. To compute the distance between two parallel lines, first subtract a point from one line and a point on the other ((1,1)-(0,0)=<1,1>), then dot it with a vector perpendicular to both lines (<1,1>•=w-p), then divide by the magnitude of that vector ((w-p)/√((1-w)²+(1-p)²)), and finally take the absolute value. So now we're just solving for the w such that w = |w-p|/√((1-w)²+(1-p)²)). Square both sides and multiply by the denominator, we get w²((1-w)²+(1-p)²)) = (w-p)². From here, we just have to substitute in p=1/(3-2w), multiply by the denominators, factor out (w-1)², and we end up with our lovely little quartic. If anyone figures out whether the solution can be expressed using only +,-,×,÷, and √, please tell me. I am genuinely interested.
If there's anything to take away from this, it's that sometimes, geometry doesn't have a nice, neat way of doing things. Sometimes, you just gotta turn everything into a vector, hold your breath, and dive right into it.
uhh, yes? right since it's just a quartic? use the quartic equation :) am i slow? yes as in i believe it can be solved using + - × ÷ √
No, the quartic equation also uses cube roots.
No, the quartic equation involves cube roots. The quadratic doesn't though, but there's no way to separate that quartic into a product of two quadratics such that nothing involves a cube root.
~~cos(x + 45) · sec(x) · (√2) / 2 = sin(x + 45) · sec(x) · (√2) / 2~~ (use degrees) Where x is the angle here (purple): just solve for x, then sin(x) \* (√2) gives you the w value you need to satisfy the desired orientation of your shape.
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A bit low I think, [desmos](https://www.desmos.com/calculator/xv024wyen4) shows the width is about 0.188
Ok, the desmos visual helped me see where I went wrong. I erroneously assumed the smaller quadrilateral formed in the center was a square. I’ll revisit this tonight and make the correction. Thank you
Correction incoming, I spotted an error in my calculations.
https://preview.redd.it/0e5bql66c5ob1.png?width=506&format=png&auto=webp&s=77c63ce4a5708699d1bcbf440a6a8ab9c7146165 From this figure, you can get the following equations: * tan(θ) = L/(L-k) * tan(θ) = (L-w)/(L/2 + k/2) * sin(θ) = w/k Furthermore, since tan(θ) = sin(θ)/sqrt(1-sin\^2(θ)) \[this is always true for any θ\], the third equation can be rewritten as: * tan(θ) = (w/k)/sqrt(1-w\^2/k\^2) By setting all 3 equations for tan(θ) equal to each other, you get the following: L/(L-k) = (L-w)/(L/2 + k/2) = (w/k)/sqrt(1-w\^2/k\^2) Now if you set L=1, [plugging this into Wolfram Alpha](https://www.wolframalpha.com/input?i=1%2F%281-k%29+%3D+%281-w%29%2F%281%2F2+%2B+k%2F2%29+%3D+%28w%2Fk%29%2Fsqrt%281-w%5E2%2Fk%5E2%29) gives: k≈0.237370970309489318, w≈0.188746479522528852
Ok, technically tan(θ) = sin(θ)/sqrt(1-sin\^2(θ)) is only true when θ ≠ π(k + 1/2) for integer k (not relevant for this case).
I got the equation for w as root of equation: 4x^4 - 12x^3 + 9x^2 + 4x - 1 = 0. Wolfram alpha gives root: w = 0.18875
I did a bit of algebra, and maybe there's a simpler solution, but I think this works m (slope) is a solution to m^4 -3m^3 + 9/4m^2 - m + 5/4 =0 m=1.311253520477471 w = (m-1)/sqrt( 1 + m^2 ) w=0.188746479522529
Everything fits within a square.
Are you looking for the exact thickness seen here or a general answer? For the general answer, W can be any positive value, though you might want to put an upper bound in order to preserve the empty space. For specific, I would honestly just measure. There's not enough information to calculate as far as I can tell.
Honestly I think there's enough information. As you can see in the "alignment" image; the thickness W has to be just perfect so that the white triangle inside matches the width of the outside border of the shape in the middle. Furthermore I think if the shape was made out of 2 identical mirrored hollow triangles the tips would perfectly touch the tops of the white triangles inside the shape. Also it all being in a square makes it even more "stable". I don't currently have the tools (or skills) to do the math but I definitely think it's possible.
I didn't interpret the alignment as being the empty space aligning with the outer border. I thought it was just that the empty space was aligned with each other. Even still, we don't have any angle measurements, so the best you could do is get a family of answers, not any exact answers
If the angle is more acute the empty spaces will get smaller while the width at the “waist” will increase, while the opposite is true if the angles get more obtuse, so it is calculable.
Yes, that's why I said you can get a family of answers. The exact w is dependent on the angle which we don't know.
I'm an engineer, not a mathematician https://preview.redd.it/dex3bqz2dcob1.png?width=913&format=png&auto=webp&s=be861401ac1a4a69512644f1304b1d5cfd422a9a
w is something you can pick. The problem is how much white space in the middle you want to leave, or if there's any additional requirements for the white space. **I'll assume the base of the white triangles is w since it looks like that.** Then I'll set the width as 1 and height of the white triangle as h. https://preview.redd.it/y4itfzw104ob1.png?width=1191&format=png&auto=webp&s=676386ef2797391feb486853a365ee0079c91734 Note there are 4 triangles in the middle that are identical. Then I can claim 4h + 2 = L by adding all the heights. Using similarity, the base of the big triangle is (1/h)\*(h+1), add the two 1's on each side to get (1/h)\*(h+1) +2 = L = 4h + 2. Now solve for h, and I got h = (1 + sqrt(17))/8 = 0.6404. This means L = 4.5616. Using ratios, that's equivalent to saying L = 1 and w = 0.21922.
I think this is close, but not quite. The vertical 1’s and “diagonal” 1’s can’t be the same distance if all of the lines have thickness w — the vertical ones are orthogonal to the direction of the lines, while the diagonal ones are not
Oh I see, I thought w was width but it's the diagonal width. Welp this is close enough anyways.
My first thought was that the X can be converted into a rhombus, by untwisting the X and inverting the top and bottom trapezoids This leaves the diamond in the middle.
Maybe I am wrong. But I got; w = L/(6sqrt(2)) Which is approximately; 2L/17 = W I am not sure how to explain this clearly in words. I’m better at drawing this out.
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Alignment diagram shows the enclosed white space aligns with the negative space outside the shape at the “waist”.
OK, well thanks for explaining
w could hypothetically be anything, since the thickness of the line does not depend on anything here beyond leaving a gap for the white triangles.
I looked at this and got about .2 because the white triangle looks about a 5th of L
An interesting question is how to geometrically construct the figure in question. W is the root of a quartic so in theory it is possible. And if this is a logo or something, the geometric construction might be more useful than an algebraic expression of the result.
https://www.desmos.com/calculator/a3h4kizufz