Special triangles in convex polygons
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Given equal-sided 30-60-90 triangles, what is the convex polygon with the highest number of sides that I can build from them?
This seems a very easy task by first look, but IâÂÂm totally stuck right now. The only thing IâÂÂm pretty sure of is that the polygon can at most have 12 sides, because a 150 degrees angle is the maximum convex angle I can find combining triangles. Obviously this does not guarantee that a decagon can be filled with 30-60-90 triangles.
Any suggestion? I was tempted to post it in Math, so if you think it is not pertinent I will shut the question down.
mathematics geometry triangle
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add a comment |Â
up vote
4
down vote
favorite
Given equal-sided 30-60-90 triangles, what is the convex polygon with the highest number of sides that I can build from them?
This seems a very easy task by first look, but IâÂÂm totally stuck right now. The only thing IâÂÂm pretty sure of is that the polygon can at most have 12 sides, because a 150 degrees angle is the maximum convex angle I can find combining triangles. Obviously this does not guarantee that a decagon can be filled with 30-60-90 triangles.
Any suggestion? I was tempted to post it in Math, so if you think it is not pertinent I will shut the question down.
mathematics geometry triangle
New contributor
Did you mean "regular" polygon? It would be trivial to assemble an irregular polygon
â Hugh
5 hours ago
1
The regular polygon with the highest number of sides that I can build is an hexagon indeed. I mean a general convex (I specified it in the title but didnâÂÂt add it in the text, just edited) polygon.
â Francesco Arnaudo
5 hours ago
If say six â I can't think of anything better.
â Hugh
4 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Given equal-sided 30-60-90 triangles, what is the convex polygon with the highest number of sides that I can build from them?
This seems a very easy task by first look, but IâÂÂm totally stuck right now. The only thing IâÂÂm pretty sure of is that the polygon can at most have 12 sides, because a 150 degrees angle is the maximum convex angle I can find combining triangles. Obviously this does not guarantee that a decagon can be filled with 30-60-90 triangles.
Any suggestion? I was tempted to post it in Math, so if you think it is not pertinent I will shut the question down.
mathematics geometry triangle
New contributor
Given equal-sided 30-60-90 triangles, what is the convex polygon with the highest number of sides that I can build from them?
This seems a very easy task by first look, but IâÂÂm totally stuck right now. The only thing IâÂÂm pretty sure of is that the polygon can at most have 12 sides, because a 150 degrees angle is the maximum convex angle I can find combining triangles. Obviously this does not guarantee that a decagon can be filled with 30-60-90 triangles.
Any suggestion? I was tempted to post it in Math, so if you think it is not pertinent I will shut the question down.
mathematics geometry triangle
mathematics geometry triangle
New contributor
New contributor
edited 5 hours ago
New contributor
asked 5 hours ago
Francesco Arnaudo
212
212
New contributor
New contributor
Did you mean "regular" polygon? It would be trivial to assemble an irregular polygon
â Hugh
5 hours ago
1
The regular polygon with the highest number of sides that I can build is an hexagon indeed. I mean a general convex (I specified it in the title but didnâÂÂt add it in the text, just edited) polygon.
â Francesco Arnaudo
5 hours ago
If say six â I can't think of anything better.
â Hugh
4 hours ago
add a comment |Â
Did you mean "regular" polygon? It would be trivial to assemble an irregular polygon
â Hugh
5 hours ago
1
The regular polygon with the highest number of sides that I can build is an hexagon indeed. I mean a general convex (I specified it in the title but didnâÂÂt add it in the text, just edited) polygon.
â Francesco Arnaudo
5 hours ago
If say six â I can't think of anything better.
â Hugh
4 hours ago
Did you mean "regular" polygon? It would be trivial to assemble an irregular polygon
â Hugh
5 hours ago
Did you mean "regular" polygon? It would be trivial to assemble an irregular polygon
â Hugh
5 hours ago
1
1
The regular polygon with the highest number of sides that I can build is an hexagon indeed. I mean a general convex (I specified it in the title but didnâÂÂt add it in the text, just edited) polygon.
â Francesco Arnaudo
5 hours ago
The regular polygon with the highest number of sides that I can build is an hexagon indeed. I mean a general convex (I specified it in the title but didnâÂÂt add it in the text, just edited) polygon.
â Francesco Arnaudo
5 hours ago
If say six â I can't think of anything better.
â Hugh
4 hours ago
If say six â I can't think of anything better.
â Hugh
4 hours ago
add a comment |Â
3 Answers
3
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oldest
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up vote
2
down vote
It is possible to do better than a hexagon, if an irregular polygon is acceptable.
One can construct an irregular dodecagon with $180$ identical 30-60-90 triangles as shown below. The dodecagon would have sides of alternating lengths, with ratio $4:sqrt12$; as a specific size for the 30-60-90 triangle is not given.
In the diagram below, the larger white triangles are not unit 30-60-90 triangles â they are actually made up of four, smaller 30-60-90 triangles, which is demonstrated in the top-left (shaded in pink). The image also includes two regular hexagons (of different sizes), a bunch of equilateral triangles, and some rectangles (all constructed from 30-60-90 triangles). It is not possible to get more than 12 sides because the minimum external angle that can be created from 30-60-90 triangles is 210 degrees.
It is also possible to construct an equilateral triangle or a hexagon. On reflection (and thanks to @Hugh's comment) you probably can't make a square â but can get relatively close to a square. See below for a rectangle with side length ratio $~1.0785:1$. By taking $sqrt3$ by $1$ squares, each made from two triangles, you could stack them $n$ by $fracnsqrt3$ which, for sufficiently large $n$, would approach $1:1$.
1
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong â regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
â Hugh
46 mins ago
Yep - you are right - I sketched to soon :)
â Penguino
42 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
â Hugh
16 mins ago
1
Feel free to edit Hugh.
â Penguino
11 mins ago
add a comment |Â
up vote
0
down vote
The answer is that
a regular dodecagon can indeed be made out of 30-60-90 tiles; you can see one rather complicated construction here.
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
â Jaap Scherphuis
34 mins ago
add a comment |Â
up vote
0
down vote
Here is a convex dodecahedron made of $76$ of those triangles.
Can it be done with fewer?
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It is possible to do better than a hexagon, if an irregular polygon is acceptable.
One can construct an irregular dodecagon with $180$ identical 30-60-90 triangles as shown below. The dodecagon would have sides of alternating lengths, with ratio $4:sqrt12$; as a specific size for the 30-60-90 triangle is not given.
In the diagram below, the larger white triangles are not unit 30-60-90 triangles â they are actually made up of four, smaller 30-60-90 triangles, which is demonstrated in the top-left (shaded in pink). The image also includes two regular hexagons (of different sizes), a bunch of equilateral triangles, and some rectangles (all constructed from 30-60-90 triangles). It is not possible to get more than 12 sides because the minimum external angle that can be created from 30-60-90 triangles is 210 degrees.
It is also possible to construct an equilateral triangle or a hexagon. On reflection (and thanks to @Hugh's comment) you probably can't make a square â but can get relatively close to a square. See below for a rectangle with side length ratio $~1.0785:1$. By taking $sqrt3$ by $1$ squares, each made from two triangles, you could stack them $n$ by $fracnsqrt3$ which, for sufficiently large $n$, would approach $1:1$.
1
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong â regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
â Hugh
46 mins ago
Yep - you are right - I sketched to soon :)
â Penguino
42 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
â Hugh
16 mins ago
1
Feel free to edit Hugh.
â Penguino
11 mins ago
add a comment |Â
up vote
2
down vote
It is possible to do better than a hexagon, if an irregular polygon is acceptable.
One can construct an irregular dodecagon with $180$ identical 30-60-90 triangles as shown below. The dodecagon would have sides of alternating lengths, with ratio $4:sqrt12$; as a specific size for the 30-60-90 triangle is not given.
In the diagram below, the larger white triangles are not unit 30-60-90 triangles â they are actually made up of four, smaller 30-60-90 triangles, which is demonstrated in the top-left (shaded in pink). The image also includes two regular hexagons (of different sizes), a bunch of equilateral triangles, and some rectangles (all constructed from 30-60-90 triangles). It is not possible to get more than 12 sides because the minimum external angle that can be created from 30-60-90 triangles is 210 degrees.
It is also possible to construct an equilateral triangle or a hexagon. On reflection (and thanks to @Hugh's comment) you probably can't make a square â but can get relatively close to a square. See below for a rectangle with side length ratio $~1.0785:1$. By taking $sqrt3$ by $1$ squares, each made from two triangles, you could stack them $n$ by $fracnsqrt3$ which, for sufficiently large $n$, would approach $1:1$.
1
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong â regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
â Hugh
46 mins ago
Yep - you are right - I sketched to soon :)
â Penguino
42 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
â Hugh
16 mins ago
1
Feel free to edit Hugh.
â Penguino
11 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It is possible to do better than a hexagon, if an irregular polygon is acceptable.
One can construct an irregular dodecagon with $180$ identical 30-60-90 triangles as shown below. The dodecagon would have sides of alternating lengths, with ratio $4:sqrt12$; as a specific size for the 30-60-90 triangle is not given.
In the diagram below, the larger white triangles are not unit 30-60-90 triangles â they are actually made up of four, smaller 30-60-90 triangles, which is demonstrated in the top-left (shaded in pink). The image also includes two regular hexagons (of different sizes), a bunch of equilateral triangles, and some rectangles (all constructed from 30-60-90 triangles). It is not possible to get more than 12 sides because the minimum external angle that can be created from 30-60-90 triangles is 210 degrees.
It is also possible to construct an equilateral triangle or a hexagon. On reflection (and thanks to @Hugh's comment) you probably can't make a square â but can get relatively close to a square. See below for a rectangle with side length ratio $~1.0785:1$. By taking $sqrt3$ by $1$ squares, each made from two triangles, you could stack them $n$ by $fracnsqrt3$ which, for sufficiently large $n$, would approach $1:1$.
It is possible to do better than a hexagon, if an irregular polygon is acceptable.
One can construct an irregular dodecagon with $180$ identical 30-60-90 triangles as shown below. The dodecagon would have sides of alternating lengths, with ratio $4:sqrt12$; as a specific size for the 30-60-90 triangle is not given.
In the diagram below, the larger white triangles are not unit 30-60-90 triangles â they are actually made up of four, smaller 30-60-90 triangles, which is demonstrated in the top-left (shaded in pink). The image also includes two regular hexagons (of different sizes), a bunch of equilateral triangles, and some rectangles (all constructed from 30-60-90 triangles). It is not possible to get more than 12 sides because the minimum external angle that can be created from 30-60-90 triangles is 210 degrees.
It is also possible to construct an equilateral triangle or a hexagon. On reflection (and thanks to @Hugh's comment) you probably can't make a square â but can get relatively close to a square. See below for a rectangle with side length ratio $~1.0785:1$. By taking $sqrt3$ by $1$ squares, each made from two triangles, you could stack them $n$ by $fracnsqrt3$ which, for sufficiently large $n$, would approach $1:1$.
edited 6 mins ago
answered 1 hour ago
Penguino
6,8121866
6,8121866
1
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong â regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
â Hugh
46 mins ago
Yep - you are right - I sketched to soon :)
â Penguino
42 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
â Hugh
16 mins ago
1
Feel free to edit Hugh.
â Penguino
11 mins ago
add a comment |Â
1
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong â regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
â Hugh
46 mins ago
Yep - you are right - I sketched to soon :)
â Penguino
42 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
â Hugh
16 mins ago
1
Feel free to edit Hugh.
â Penguino
11 mins ago
1
1
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong â regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
â Hugh
46 mins ago
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong â regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
â Hugh
46 mins ago
Yep - you are right - I sketched to soon :)
â Penguino
42 mins ago
Yep - you are right - I sketched to soon :)
â Penguino
42 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
â Hugh
16 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
â Hugh
16 mins ago
1
1
Feel free to edit Hugh.
â Penguino
11 mins ago
Feel free to edit Hugh.
â Penguino
11 mins ago
add a comment |Â
up vote
0
down vote
The answer is that
a regular dodecagon can indeed be made out of 30-60-90 tiles; you can see one rather complicated construction here.
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
â Jaap Scherphuis
34 mins ago
add a comment |Â
up vote
0
down vote
The answer is that
a regular dodecagon can indeed be made out of 30-60-90 tiles; you can see one rather complicated construction here.
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
â Jaap Scherphuis
34 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The answer is that
a regular dodecagon can indeed be made out of 30-60-90 tiles; you can see one rather complicated construction here.
The answer is that
a regular dodecagon can indeed be made out of 30-60-90 tiles; you can see one rather complicated construction here.
answered 2 hours ago
Gareth McCaughanâ¦
57.8k3144221
57.8k3144221
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
â Jaap Scherphuis
34 mins ago
add a comment |Â
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
â Jaap Scherphuis
34 mins ago
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
â Jaap Scherphuis
34 mins ago
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
â Jaap Scherphuis
34 mins ago
add a comment |Â
up vote
0
down vote
Here is a convex dodecahedron made of $76$ of those triangles.
Can it be done with fewer?
add a comment |Â
up vote
0
down vote
Here is a convex dodecahedron made of $76$ of those triangles.
Can it be done with fewer?
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here is a convex dodecahedron made of $76$ of those triangles.
Can it be done with fewer?
Here is a convex dodecahedron made of $76$ of those triangles.
Can it be done with fewer?
answered 8 mins ago
Jaap Scherphuis
13.1k12259
13.1k12259
add a comment |Â
add a comment |Â
Francesco Arnaudo is a new contributor. Be nice, and check out our Code of Conduct.
Francesco Arnaudo is a new contributor. Be nice, and check out our Code of Conduct.
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Did you mean "regular" polygon? It would be trivial to assemble an irregular polygon
â Hugh
5 hours ago
1
The regular polygon with the highest number of sides that I can build is an hexagon indeed. I mean a general convex (I specified it in the title but didnâÂÂt add it in the text, just edited) polygon.
â Francesco Arnaudo
5 hours ago
If say six â I can't think of anything better.
â Hugh
4 hours ago