Overlapping inscribed triangles.
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Question: Let $T_1$ be the equilateral triangle inscribed by the unit circle centered at the origin of $mathbb{R^2}.$ Now let $T_2$ be the triangle induced by clockwise rotating each vertex of $T_1$ by $1^circ.$ What is the area of $T_1 cap T_2$ ?
The not-to-scale-drawing below illustrates $T_1 cap T_2.$ It is a convex hexagon. Careful $T_1 cap T_2$ is not cyclic - none of its vertices are incident to the boundary of the unit circle. For short handedness I write $muleft(T_iright)$ for the area of $T_i;$ where $1leq ileq 2.$ I know that $muleft(T_iright)={3sqrt{3} above 1.5pt 4}$ and so I am certain $mu(T_1 cap T_2) < {3sqrt{3} above 1.5pt 4}.$
geometry triangles area
asked Dec 20 '18 at 19:03
Antonio Hernandez MaquivarAntonio Hernandez Maquivar
1,452623
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add a comment |
$begingroup$
Question: Let $T_1$ be the equilateral triangle inscribed by the unit circle centered at the origin of $mathbb{R^2}.$ Now let $T_2$ be the triangle induced by clockwise rotating each vertex of $T_1$ by $1^circ.$ What is the area of $T_1 cap T_2$ ?
The not-to-scale-drawing below illustrates $T_1 cap T_2.$ It is a convex hexagon. Careful $T_1 cap T_2$ is not cyclic - none of its vertices are incident to the boundary of the unit circle. For short handedness I write $muleft(T_iright)$ for the area of $T_i;$ where $1leq ileq 2.$ I know that $muleft(T_iright)={3sqrt{3} above 1.5pt 4}$ and so I am certain $mu(T_1 cap T_2) < {3sqrt{3} above 1.5pt 4}.$
geometry triangles area
asked Dec 20 '18 at 19:03
Antonio Hernandez MaquivarAntonio Hernandez Maquivar
1,452623
$endgroup$
add a comment |
$begingroup$
Question: Let $T_1$ be the equilateral triangle inscribed by the unit circle centered at the origin of $mathbb{R^2}.$ Now let $T_2$ be the triangle induced by clockwise rotating each vertex of $T_1$ by $1^circ.$ What is the area of $T_1 cap T_2$ ?
The not-to-scale-drawing below illustrates $T_1 cap T_2.$ It is a convex hexagon. Careful $T_1 cap T_2$ is not cyclic - none of its vertices are incident to the boundary of the unit circle. For short handedness I write $muleft(T_iright)$ for the area of $T_i;$ where $1leq ileq 2.$ I know that $muleft(T_iright)={3sqrt{3} above 1.5pt 4}$ and so I am certain $mu(T_1 cap T_2) < {3sqrt{3} above 1.5pt 4}.$
geometry triangles area
asked Dec 20 '18 at 19:03
Antonio Hernandez MaquivarAntonio Hernandez Maquivar
1,452623
$endgroup$
Question: Let $T_1$ be the equilateral triangle inscribed by the unit circle centered at the origin of $mathbb{R^2}.$ Now let $T_2$ be the triangle induced by clockwise rotating each vertex of $T_1$ by $1^circ.$ What is the area of $T_1 cap T_2$ ?
The not-to-scale-drawing below illustrates $T_1 cap T_2.$ It is a convex hexagon. Careful $T_1 cap T_2$ is not cyclic - none of its vertices are incident to the boundary of the unit circle. For short handedness I write $muleft(T_iright)$ for the area of $T_i;$ where $1leq ileq 2.$ I know that $muleft(T_iright)={3sqrt{3} above 1.5pt 4}$ and so I am certain $mu(T_1 cap T_2) < {3sqrt{3} above 1.5pt 4}.$
geometry triangles area
geometry triangles area
asked Dec 20 '18 at 19:03
Antonio Hernandez MaquivarAntonio Hernandez Maquivar
1,452623
asked Dec 20 '18 at 19:03
Antonio Hernandez MaquivarAntonio Hernandez Maquivar
1,452623
edited Dec 21 '18 at 15:07
Antonio Hernandez Maquivar
asked Dec 20 '18 at 19:03
Antonio Hernandez MaquivarAntonio Hernandez Maquivar
1,452623
asked Dec 20 '18 at 19:03
Antonio Hernandez MaquivarAntonio Hernandez Maquivar
1,452623
asked Dec 20 '18 at 19:03
Antonio Hernandez MaquivarAntonio Hernandez Maquivar
1,452623
1,452623
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2 Answers
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In a figure like the one in the question, in my experience people would say the unit circle is circumscribed about each of the triangles $T_1$ and $T_2,$
and they would say $T_1$ and $T_2$ are inscribed triangles,
specifically, triangles inscribed in the unit circle.
The entire figure has lines of reflection symmetry through the center of the circle and any of the vertices of the resulting figure.
There are three such lines of symmetry; each of the lines goes through two vertices.
So the segments from the center to any pair of adjacent vertices make an angle of $60$ degrees.
If you find the distance from the center to each of the vertices, you can use the angle between these radial segments to find the area of the figure.
answered Dec 21 '18 at 14:40
David KDavid K
55.6k345121
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Perhaps a better way to write this is to say that $T_1$ is centered at the origin of $mathbb{R^2}$ and has a circumradius equal to $1$ ?
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– Antonio Hernandez Maquivar
Dec 21 '18 at 14:51
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The question was clear enough. I was just being pedantic.
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– David K
Dec 21 '18 at 22:37
add a comment |
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Find the area of one of the triangles that is not part of the overlapping region, then subtract three times this area from the area of the original triangle. Hint: Start by solving the triangle with vertices at the center of the circle, a vertex of $T_1$, and the corresponding vertex of $T_2$.
answered Dec 23 '18 at 0:57
Daniel MathiasDaniel Mathias
1,40518
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
In a figure like the one in the question, in my experience people would say the unit circle is circumscribed about each of the triangles $T_1$ and $T_2,$
and they would say $T_1$ and $T_2$ are inscribed triangles,
specifically, triangles inscribed in the unit circle.
The entire figure has lines of reflection symmetry through the center of the circle and any of the vertices of the resulting figure.
There are three such lines of symmetry; each of the lines goes through two vertices.
So the segments from the center to any pair of adjacent vertices make an angle of $60$ degrees.
If you find the distance from the center to each of the vertices, you can use the angle between these radial segments to find the area of the figure.
answered Dec 21 '18 at 14:40
David KDavid K
55.6k345121
$endgroup$
$begingroup$
Perhaps a better way to write this is to say that $T_1$ is centered at the origin of $mathbb{R^2}$ and has a circumradius equal to $1$ ?
$endgroup$
– Antonio Hernandez Maquivar
Dec 21 '18 at 14:51
$begingroup$
The question was clear enough. I was just being pedantic.
$endgroup$
– David K
Dec 21 '18 at 22:37
add a comment |
$begingroup$
In a figure like the one in the question, in my experience people would say the unit circle is circumscribed about each of the triangles $T_1$ and $T_2,$
and they would say $T_1$ and $T_2$ are inscribed triangles,
specifically, triangles inscribed in the unit circle.
The entire figure has lines of reflection symmetry through the center of the circle and any of the vertices of the resulting figure.
There are three such lines of symmetry; each of the lines goes through two vertices.
So the segments from the center to any pair of adjacent vertices make an angle of $60$ degrees.
If you find the distance from the center to each of the vertices, you can use the angle between these radial segments to find the area of the figure.
answered Dec 21 '18 at 14:40
David KDavid K
55.6k345121
$endgroup$
$begingroup$
Perhaps a better way to write this is to say that $T_1$ is centered at the origin of $mathbb{R^2}$ and has a circumradius equal to $1$ ?
$endgroup$
– Antonio Hernandez Maquivar
Dec 21 '18 at 14:51
$begingroup$
The question was clear enough. I was just being pedantic.
$endgroup$
– David K
Dec 21 '18 at 22:37
add a comment |
$begingroup$
In a figure like the one in the question, in my experience people would say the unit circle is circumscribed about each of the triangles $T_1$ and $T_2,$
and they would say $T_1$ and $T_2$ are inscribed triangles,
specifically, triangles inscribed in the unit circle.
The entire figure has lines of reflection symmetry through the center of the circle and any of the vertices of the resulting figure.
There are three such lines of symmetry; each of the lines goes through two vertices.
So the segments from the center to any pair of adjacent vertices make an angle of $60$ degrees.
If you find the distance from the center to each of the vertices, you can use the angle between these radial segments to find the area of the figure.
answered Dec 21 '18 at 14:40
David KDavid K
55.6k345121
$endgroup$
In a figure like the one in the question, in my experience people would say the unit circle is circumscribed about each of the triangles $T_1$ and $T_2,$
and they would say $T_1$ and $T_2$ are inscribed triangles,
specifically, triangles inscribed in the unit circle.
The entire figure has lines of reflection symmetry through the center of the circle and any of the vertices of the resulting figure.
There are three such lines of symmetry; each of the lines goes through two vertices.
So the segments from the center to any pair of adjacent vertices make an angle of $60$ degrees.
If you find the distance from the center to each of the vertices, you can use the angle between these radial segments to find the area of the figure.
answered Dec 21 '18 at 14:40
David KDavid K
55.6k345121
answered Dec 21 '18 at 14:40
David KDavid K
55.6k345121
answered Dec 21 '18 at 14:40
David KDavid K
55.6k345121
answered Dec 21 '18 at 14:40
David KDavid K
55.6k345121
55.6k345121
$begingroup$
Perhaps a better way to write this is to say that $T_1$ is centered at the origin of $mathbb{R^2}$ and has a circumradius equal to $1$ ?
$endgroup$
– Antonio Hernandez Maquivar
Dec 21 '18 at 14:51
$begingroup$
The question was clear enough. I was just being pedantic.
$endgroup$
– David K
Dec 21 '18 at 22:37
add a comment |
$begingroup$
Perhaps a better way to write this is to say that $T_1$ is centered at the origin of $mathbb{R^2}$ and has a circumradius equal to $1$ ?
$endgroup$
– Antonio Hernandez Maquivar
Dec 21 '18 at 14:51
$begingroup$
The question was clear enough. I was just being pedantic.
$endgroup$
– David K
Dec 21 '18 at 22:37
$begingroup$
Perhaps a better way to write this is to say that $T_1$ is centered at the origin of $mathbb{R^2}$ and has a circumradius equal to $1$ ?
$endgroup$
– Antonio Hernandez Maquivar
Dec 21 '18 at 14:51
$begingroup$
Perhaps a better way to write this is to say that $T_1$ is centered at the origin of $mathbb{R^2}$ and has a circumradius equal to $1$ ?
$endgroup$
– Antonio Hernandez Maquivar
Dec 21 '18 at 14:51
$begingroup$
The question was clear enough. I was just being pedantic.
$endgroup$
– David K
Dec 21 '18 at 22:37
$begingroup$
The question was clear enough. I was just being pedantic.
$endgroup$
– David K
Dec 21 '18 at 22:37
add a comment |
$begingroup$
Find the area of one of the triangles that is not part of the overlapping region, then subtract three times this area from the area of the original triangle. Hint: Start by solving the triangle with vertices at the center of the circle, a vertex of $T_1$, and the corresponding vertex of $T_2$.
answered Dec 23 '18 at 0:57
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
add a comment |
$begingroup$
Find the area of one of the triangles that is not part of the overlapping region, then subtract three times this area from the area of the original triangle. Hint: Start by solving the triangle with vertices at the center of the circle, a vertex of $T_1$, and the corresponding vertex of $T_2$.
answered Dec 23 '18 at 0:57
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
add a comment |
$begingroup$
Find the area of one of the triangles that is not part of the overlapping region, then subtract three times this area from the area of the original triangle. Hint: Start by solving the triangle with vertices at the center of the circle, a vertex of $T_1$, and the corresponding vertex of $T_2$.
answered Dec 23 '18 at 0:57
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
Find the area of one of the triangles that is not part of the overlapping region, then subtract three times this area from the area of the original triangle. Hint: Start by solving the triangle with vertices at the center of the circle, a vertex of $T_1$, and the corresponding vertex of $T_2$.
answered Dec 23 '18 at 0:57
Daniel MathiasDaniel Mathias
1,40518
answered Dec 23 '18 at 0:57
Daniel MathiasDaniel Mathias
1,40518
answered Dec 23 '18 at 0:57
Daniel MathiasDaniel Mathias
1,40518
answered Dec 23 '18 at 0:57
Daniel MathiasDaniel Mathias
1,40518
1,40518
add a comment |
add a comment |
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