Is there an exact solution for tan(36) using origami from a unit square?
This question is more for chagrins and curiosity than anything else: Is there a way to use origami to construct the tangent of 36 degrees (~0.7265425)?
I've come up with the image below, which is accurate to about 5 decimal places (the exact decimal eludes me at the moment), but I'd like something closer if it exists.
The vertical red line on the right is at x = ~0.7265403.
As I said, the question is mostly for chagrins (as the method above is close enough in practice), but I'd appreciate any more creative solutions.
trigonometry origami
edited Nov 24 at 0:13
Shaun
8,507113580
asked Nov 23 at 23:59
Dirge of Dreams
1184
add a comment |
This question is more for chagrins and curiosity than anything else: Is there a way to use origami to construct the tangent of 36 degrees (~0.7265425)?
I've come up with the image below, which is accurate to about 5 decimal places (the exact decimal eludes me at the moment), but I'd like something closer if it exists.
The vertical red line on the right is at x = ~0.7265403.
As I said, the question is mostly for chagrins (as the method above is close enough in practice), but I'd appreciate any more creative solutions.
trigonometry origami
edited Nov 24 at 0:13
Shaun
8,507113580
asked Nov 23 at 23:59
Dirge of Dreams
1184
add a comment |
This question is more for chagrins and curiosity than anything else: Is there a way to use origami to construct the tangent of 36 degrees (~0.7265425)?
I've come up with the image below, which is accurate to about 5 decimal places (the exact decimal eludes me at the moment), but I'd like something closer if it exists.
The vertical red line on the right is at x = ~0.7265403.
As I said, the question is mostly for chagrins (as the method above is close enough in practice), but I'd appreciate any more creative solutions.
trigonometry origami
edited Nov 24 at 0:13
Shaun
8,507113580
asked Nov 23 at 23:59
Dirge of Dreams
1184
This question is more for chagrins and curiosity than anything else: Is there a way to use origami to construct the tangent of 36 degrees (~0.7265425)?
I've come up with the image below, which is accurate to about 5 decimal places (the exact decimal eludes me at the moment), but I'd like something closer if it exists.
The vertical red line on the right is at x = ~0.7265403.
As I said, the question is mostly for chagrins (as the method above is close enough in practice), but I'd appreciate any more creative solutions.
trigonometry origami
trigonometry origami
edited Nov 24 at 0:13
Shaun
8,507113580
asked Nov 23 at 23:59
Dirge of Dreams
1184
edited Nov 24 at 0:13
Shaun
8,507113580
asked Nov 23 at 23:59
Dirge of Dreams
1184
edited Nov 24 at 0:13
Shaun
8,507113580
edited Nov 24 at 0:13
Shaun
8,507113580
edited Nov 24 at 0:13
Shaun
8,507113580
8,507113580
asked Nov 23 at 23:59
Dirge of Dreams
1184
asked Nov 23 at 23:59
Dirge of Dreams
1184
asked Nov 23 at 23:59
Dirge of Dreams
1184
1184
add a comment |
add a comment |
1 Answer
1
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votes
Form a pentagon base. All ten resulting central angles are 36 degrees. Taking the short radial fold to have length $1$, the half-edge has length $tan(36 ^circ)$. (That is, you have ten right triangles with adjacent $1$ and opposite $tan(36 ^circ)$.)
So start with this step of the pentagon base.
Now unfold to the beginning. Observe that there is a central vertex with four 36 degree angles to its right. The two right triangles sharing a leg along the first bisection fold are what we want to project to an edge. Extend either hypotenuse all the way across the paper. (This extension almost auto-folds for me. You have the two long legs of the adjacent right triangles to guide this fold.) Taking the extended hypotenuse as a unaxial base, construct a hinge passing through a (either) corner. (The extended hypotenuse, the axis for this hinge, is indicated with pink arrows in the following diagram.)
The hinge meets the edge opposite the corner at the desired point (indicated by the light blue arrow).
Note that the foreground flap is a $36^circ text{-} 54^circ text{-} 90^circ$ triangle, as desired. Why? The right triangle we started with has legs parallel to the edges of the paper. Any perpendicular to its hypotenuse meets an edge at either $36^circ$ or $54^circ$. We just arranged for the hinge to hit one of the two sides forming a $54^circ$ angle.
answered Nov 24 at 0:14
Eric Towers
31.7k22265
The question asks to start from a unit square, though, so I don't think you're allowed to just set the length of the short radial fold to have length $1$.
– YiFan
Nov 24 at 0:20
@YiFan : As is common, the question does not ask this. The title does. So much time is wasted on this site due to this variety of mismatch. As per usual, we will have to wait for the OP to resolve which is normative.
– Eric Towers
Nov 24 at 0:23
Sorry if I was unclear. I'm looking for a way to find tan(36) in relation to the edge of a unit square--in other words, find the point (0.7265425, 0) with the bottom left corner of the square at the origin.
– Dirge of Dreams
Nov 25 at 1:44
@YiFan : ... and ... done.
– Eric Towers
Nov 25 at 3:33
add a comment |
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1 Answer
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Form a pentagon base. All ten resulting central angles are 36 degrees. Taking the short radial fold to have length $1$, the half-edge has length $tan(36 ^circ)$. (That is, you have ten right triangles with adjacent $1$ and opposite $tan(36 ^circ)$.)
So start with this step of the pentagon base.
Now unfold to the beginning. Observe that there is a central vertex with four 36 degree angles to its right. The two right triangles sharing a leg along the first bisection fold are what we want to project to an edge. Extend either hypotenuse all the way across the paper. (This extension almost auto-folds for me. You have the two long legs of the adjacent right triangles to guide this fold.) Taking the extended hypotenuse as a unaxial base, construct a hinge passing through a (either) corner. (The extended hypotenuse, the axis for this hinge, is indicated with pink arrows in the following diagram.)
The hinge meets the edge opposite the corner at the desired point (indicated by the light blue arrow).
Note that the foreground flap is a $36^circ text{-} 54^circ text{-} 90^circ$ triangle, as desired. Why? The right triangle we started with has legs parallel to the edges of the paper. Any perpendicular to its hypotenuse meets an edge at either $36^circ$ or $54^circ$. We just arranged for the hinge to hit one of the two sides forming a $54^circ$ angle.
answered Nov 24 at 0:14
Eric Towers
31.7k22265
The question asks to start from a unit square, though, so I don't think you're allowed to just set the length of the short radial fold to have length $1$.
– YiFan
Nov 24 at 0:20
@YiFan : As is common, the question does not ask this. The title does. So much time is wasted on this site due to this variety of mismatch. As per usual, we will have to wait for the OP to resolve which is normative.
– Eric Towers
Nov 24 at 0:23
Sorry if I was unclear. I'm looking for a way to find tan(36) in relation to the edge of a unit square--in other words, find the point (0.7265425, 0) with the bottom left corner of the square at the origin.
– Dirge of Dreams
Nov 25 at 1:44
@YiFan : ... and ... done.
– Eric Towers
Nov 25 at 3:33
add a comment |
Form a pentagon base. All ten resulting central angles are 36 degrees. Taking the short radial fold to have length $1$, the half-edge has length $tan(36 ^circ)$. (That is, you have ten right triangles with adjacent $1$ and opposite $tan(36 ^circ)$.)
So start with this step of the pentagon base.
Now unfold to the beginning. Observe that there is a central vertex with four 36 degree angles to its right. The two right triangles sharing a leg along the first bisection fold are what we want to project to an edge. Extend either hypotenuse all the way across the paper. (This extension almost auto-folds for me. You have the two long legs of the adjacent right triangles to guide this fold.) Taking the extended hypotenuse as a unaxial base, construct a hinge passing through a (either) corner. (The extended hypotenuse, the axis for this hinge, is indicated with pink arrows in the following diagram.)
The hinge meets the edge opposite the corner at the desired point (indicated by the light blue arrow).
Note that the foreground flap is a $36^circ text{-} 54^circ text{-} 90^circ$ triangle, as desired. Why? The right triangle we started with has legs parallel to the edges of the paper. Any perpendicular to its hypotenuse meets an edge at either $36^circ$ or $54^circ$. We just arranged for the hinge to hit one of the two sides forming a $54^circ$ angle.
answered Nov 24 at 0:14
Eric Towers
31.7k22265
The question asks to start from a unit square, though, so I don't think you're allowed to just set the length of the short radial fold to have length $1$.
– YiFan
Nov 24 at 0:20
@YiFan : As is common, the question does not ask this. The title does. So much time is wasted on this site due to this variety of mismatch. As per usual, we will have to wait for the OP to resolve which is normative.
– Eric Towers
Nov 24 at 0:23
Sorry if I was unclear. I'm looking for a way to find tan(36) in relation to the edge of a unit square--in other words, find the point (0.7265425, 0) with the bottom left corner of the square at the origin.
– Dirge of Dreams
Nov 25 at 1:44
@YiFan : ... and ... done.
– Eric Towers
Nov 25 at 3:33
add a comment |
Form a pentagon base. All ten resulting central angles are 36 degrees. Taking the short radial fold to have length $1$, the half-edge has length $tan(36 ^circ)$. (That is, you have ten right triangles with adjacent $1$ and opposite $tan(36 ^circ)$.)
So start with this step of the pentagon base.
Now unfold to the beginning. Observe that there is a central vertex with four 36 degree angles to its right. The two right triangles sharing a leg along the first bisection fold are what we want to project to an edge. Extend either hypotenuse all the way across the paper. (This extension almost auto-folds for me. You have the two long legs of the adjacent right triangles to guide this fold.) Taking the extended hypotenuse as a unaxial base, construct a hinge passing through a (either) corner. (The extended hypotenuse, the axis for this hinge, is indicated with pink arrows in the following diagram.)
The hinge meets the edge opposite the corner at the desired point (indicated by the light blue arrow).
Note that the foreground flap is a $36^circ text{-} 54^circ text{-} 90^circ$ triangle, as desired. Why? The right triangle we started with has legs parallel to the edges of the paper. Any perpendicular to its hypotenuse meets an edge at either $36^circ$ or $54^circ$. We just arranged for the hinge to hit one of the two sides forming a $54^circ$ angle.
answered Nov 24 at 0:14
Eric Towers
31.7k22265
Form a pentagon base. All ten resulting central angles are 36 degrees. Taking the short radial fold to have length $1$, the half-edge has length $tan(36 ^circ)$. (That is, you have ten right triangles with adjacent $1$ and opposite $tan(36 ^circ)$.)
So start with this step of the pentagon base.
Now unfold to the beginning. Observe that there is a central vertex with four 36 degree angles to its right. The two right triangles sharing a leg along the first bisection fold are what we want to project to an edge. Extend either hypotenuse all the way across the paper. (This extension almost auto-folds for me. You have the two long legs of the adjacent right triangles to guide this fold.) Taking the extended hypotenuse as a unaxial base, construct a hinge passing through a (either) corner. (The extended hypotenuse, the axis for this hinge, is indicated with pink arrows in the following diagram.)
The hinge meets the edge opposite the corner at the desired point (indicated by the light blue arrow).
Note that the foreground flap is a $36^circ text{-} 54^circ text{-} 90^circ$ triangle, as desired. Why? The right triangle we started with has legs parallel to the edges of the paper. Any perpendicular to its hypotenuse meets an edge at either $36^circ$ or $54^circ$. We just arranged for the hinge to hit one of the two sides forming a $54^circ$ angle.
answered Nov 24 at 0:14
Eric Towers
31.7k22265
edited Nov 25 at 3:31
answered Nov 24 at 0:14
Eric Towers
31.7k22265
answered Nov 24 at 0:14
Eric Towers
31.7k22265
answered Nov 24 at 0:14
Eric Towers
31.7k22265
31.7k22265
The question asks to start from a unit square, though, so I don't think you're allowed to just set the length of the short radial fold to have length $1$.
– YiFan
Nov 24 at 0:20
@YiFan : As is common, the question does not ask this. The title does. So much time is wasted on this site due to this variety of mismatch. As per usual, we will have to wait for the OP to resolve which is normative.
– Eric Towers
Nov 24 at 0:23
Sorry if I was unclear. I'm looking for a way to find tan(36) in relation to the edge of a unit square--in other words, find the point (0.7265425, 0) with the bottom left corner of the square at the origin.
– Dirge of Dreams
Nov 25 at 1:44
@YiFan : ... and ... done.
– Eric Towers
Nov 25 at 3:33
add a comment |
The question asks to start from a unit square, though, so I don't think you're allowed to just set the length of the short radial fold to have length $1$.
– YiFan
Nov 24 at 0:20
@YiFan : As is common, the question does not ask this. The title does. So much time is wasted on this site due to this variety of mismatch. As per usual, we will have to wait for the OP to resolve which is normative.
– Eric Towers
Nov 24 at 0:23
Sorry if I was unclear. I'm looking for a way to find tan(36) in relation to the edge of a unit square--in other words, find the point (0.7265425, 0) with the bottom left corner of the square at the origin.
– Dirge of Dreams
Nov 25 at 1:44
@YiFan : ... and ... done.
– Eric Towers
Nov 25 at 3:33
The question asks to start from a unit square, though, so I don't think you're allowed to just set the length of the short radial fold to have length $1$.
– YiFan
Nov 24 at 0:20
The question asks to start from a unit square, though, so I don't think you're allowed to just set the length of the short radial fold to have length $1$.
– YiFan
Nov 24 at 0:20
@YiFan : As is common, the question does not ask this. The title does. So much time is wasted on this site due to this variety of mismatch. As per usual, we will have to wait for the OP to resolve which is normative.
– Eric Towers
Nov 24 at 0:23
@YiFan : As is common, the question does not ask this. The title does. So much time is wasted on this site due to this variety of mismatch. As per usual, we will have to wait for the OP to resolve which is normative.
– Eric Towers
Nov 24 at 0:23
Sorry if I was unclear. I'm looking for a way to find tan(36) in relation to the edge of a unit square--in other words, find the point (0.7265425, 0) with the bottom left corner of the square at the origin.
– Dirge of Dreams
Nov 25 at 1:44
Sorry if I was unclear. I'm looking for a way to find tan(36) in relation to the edge of a unit square--in other words, find the point (0.7265425, 0) with the bottom left corner of the square at the origin.
– Dirge of Dreams
Nov 25 at 1:44
@YiFan : ... and ... done.
– Eric Towers
Nov 25 at 3:33
@YiFan : ... and ... done.
– Eric Towers
Nov 25 at 3:33
add a comment |
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