Inscribe an equilateral triangle inside a triangle
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Given a triangle ΔABC, how to draw all possible inscribed equilateral triangles with given side whose vertices lie on different sides of ΔABC?
geometry triangle problem-solving geometric-construction
asked Nov 23 at 15:34
Stepii
106
add a comment |
up vote
1
down vote
favorite
Given a triangle ΔABC, how to draw all possible inscribed equilateral triangles with given side whose vertices lie on different sides of ΔABC?
geometry triangle problem-solving geometric-construction
asked Nov 23 at 15:34
Stepii
106
Are you looking for the maximum possible equilateral triangle, or an arbitrary one? Would you consider any equilateral triangle which shares part of a side with the bounding triangle?
– abiessu
Nov 23 at 15:37
@abiessu I am looking for all equilateral triangles with given side, not maximum nor sharing side, just inscribed inside another arbitrary triangle
– Stepii
Nov 23 at 15:44
That’s a lot of triangles, or one, or none. Do you have anything else to go on?
– abiessu
Nov 23 at 16:46
@abiessu Well, my teacher said that there's maximum 2 solutions.
– Stepii
Nov 23 at 17:36
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a triangle ΔABC, how to draw all possible inscribed equilateral triangles with given side whose vertices lie on different sides of ΔABC?
geometry triangle problem-solving geometric-construction
asked Nov 23 at 15:34
Stepii
106
Given a triangle ΔABC, how to draw all possible inscribed equilateral triangles with given side whose vertices lie on different sides of ΔABC?
geometry triangle problem-solving geometric-construction
geometry triangle problem-solving geometric-construction
asked Nov 23 at 15:34
Stepii
106
asked Nov 23 at 15:34
Stepii
106
edited 12 hours ago
asked Nov 23 at 15:34
Stepii
106
asked Nov 23 at 15:34
Stepii
106
asked Nov 23 at 15:34
Stepii
106
106
Are you looking for the maximum possible equilateral triangle, or an arbitrary one? Would you consider any equilateral triangle which shares part of a side with the bounding triangle?
– abiessu
Nov 23 at 15:37
@abiessu I am looking for all equilateral triangles with given side, not maximum nor sharing side, just inscribed inside another arbitrary triangle
– Stepii
Nov 23 at 15:44
That’s a lot of triangles, or one, or none. Do you have anything else to go on?
– abiessu
Nov 23 at 16:46
@abiessu Well, my teacher said that there's maximum 2 solutions.
– Stepii
Nov 23 at 17:36
add a comment |
Are you looking for the maximum possible equilateral triangle, or an arbitrary one? Would you consider any equilateral triangle which shares part of a side with the bounding triangle?
– abiessu
Nov 23 at 15:37
@abiessu I am looking for all equilateral triangles with given side, not maximum nor sharing side, just inscribed inside another arbitrary triangle
– Stepii
Nov 23 at 15:44
That’s a lot of triangles, or one, or none. Do you have anything else to go on?
– abiessu
Nov 23 at 16:46
@abiessu Well, my teacher said that there's maximum 2 solutions.
– Stepii
Nov 23 at 17:36
Are you looking for the maximum possible equilateral triangle, or an arbitrary one? Would you consider any equilateral triangle which shares part of a side with the bounding triangle?
– abiessu
Nov 23 at 15:37
Are you looking for the maximum possible equilateral triangle, or an arbitrary one? Would you consider any equilateral triangle which shares part of a side with the bounding triangle?
– abiessu
Nov 23 at 15:37
@abiessu I am looking for all equilateral triangles with given side, not maximum nor sharing side, just inscribed inside another arbitrary triangle
– Stepii
Nov 23 at 15:44
@abiessu I am looking for all equilateral triangles with given side, not maximum nor sharing side, just inscribed inside another arbitrary triangle
– Stepii
Nov 23 at 15:44
That’s a lot of triangles, or one, or none. Do you have anything else to go on?
– abiessu
Nov 23 at 16:46
That’s a lot of triangles, or one, or none. Do you have anything else to go on?
– abiessu
Nov 23 at 16:46
@abiessu Well, my teacher said that there's maximum 2 solutions.
– Stepii
Nov 23 at 17:36
@abiessu Well, my teacher said that there's maximum 2 solutions.
– Stepii
Nov 23 at 17:36
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
Take a point $P$ on a side of $ABC$ and rotate $ABC$ around $P$ by $60^circ$ clockwise/counterclockwise. The intersections between the sides of the rotated triangle and the original triangle provide two points $Q,R$ such that $PQR$ is equilateral. In follows that there are infinite equilateral triangles inscribed in a given triangle.
On the other hand, if the sides length is fixed, there are at most two solutions. I am going to provide a proof almost-without words:
The red locus is convex (since it is the arc of an ellipse centered at the upper vertex), hence it cannot meet the horizontal side at more than two points.
answered Nov 23 at 18:46
Jack D'Aurizio
283k33275653
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
– abiessu
Nov 24 at 3:41
Can you explain how to find theese two solutions?
– Stepii
12 hours ago
The axis of the involved ellipse are given by an internal/external angle bisector and its vertices are simple to find. To find the solutions is equivalent to intersecting an ellipse and a line, or, up to affine maps, to intersecting a circle and a line.
– Jack D'Aurizio
12 hours ago
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
– Jack D'Aurizio
11 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Take a point $P$ on a side of $ABC$ and rotate $ABC$ around $P$ by $60^circ$ clockwise/counterclockwise. The intersections between the sides of the rotated triangle and the original triangle provide two points $Q,R$ such that $PQR$ is equilateral. In follows that there are infinite equilateral triangles inscribed in a given triangle.
On the other hand, if the sides length is fixed, there are at most two solutions. I am going to provide a proof almost-without words:
The red locus is convex (since it is the arc of an ellipse centered at the upper vertex), hence it cannot meet the horizontal side at more than two points.
answered Nov 23 at 18:46
Jack D'Aurizio
283k33275653
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
– abiessu
Nov 24 at 3:41
Can you explain how to find theese two solutions?
– Stepii
12 hours ago
The axis of the involved ellipse are given by an internal/external angle bisector and its vertices are simple to find. To find the solutions is equivalent to intersecting an ellipse and a line, or, up to affine maps, to intersecting a circle and a line.
– Jack D'Aurizio
12 hours ago
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
– Jack D'Aurizio
11 hours ago
add a comment |
up vote
2
down vote
Take a point $P$ on a side of $ABC$ and rotate $ABC$ around $P$ by $60^circ$ clockwise/counterclockwise. The intersections between the sides of the rotated triangle and the original triangle provide two points $Q,R$ such that $PQR$ is equilateral. In follows that there are infinite equilateral triangles inscribed in a given triangle.
On the other hand, if the sides length is fixed, there are at most two solutions. I am going to provide a proof almost-without words:
The red locus is convex (since it is the arc of an ellipse centered at the upper vertex), hence it cannot meet the horizontal side at more than two points.
answered Nov 23 at 18:46
Jack D'Aurizio
283k33275653
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
– abiessu
Nov 24 at 3:41
Can you explain how to find theese two solutions?
– Stepii
12 hours ago
The axis of the involved ellipse are given by an internal/external angle bisector and its vertices are simple to find. To find the solutions is equivalent to intersecting an ellipse and a line, or, up to affine maps, to intersecting a circle and a line.
– Jack D'Aurizio
12 hours ago
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
– Jack D'Aurizio
11 hours ago
add a comment |
up vote
2
down vote
up vote
2
down vote
Take a point $P$ on a side of $ABC$ and rotate $ABC$ around $P$ by $60^circ$ clockwise/counterclockwise. The intersections between the sides of the rotated triangle and the original triangle provide two points $Q,R$ such that $PQR$ is equilateral. In follows that there are infinite equilateral triangles inscribed in a given triangle.
On the other hand, if the sides length is fixed, there are at most two solutions. I am going to provide a proof almost-without words:
The red locus is convex (since it is the arc of an ellipse centered at the upper vertex), hence it cannot meet the horizontal side at more than two points.
answered Nov 23 at 18:46
Jack D'Aurizio
283k33275653
Take a point $P$ on a side of $ABC$ and rotate $ABC$ around $P$ by $60^circ$ clockwise/counterclockwise. The intersections between the sides of the rotated triangle and the original triangle provide two points $Q,R$ such that $PQR$ is equilateral. In follows that there are infinite equilateral triangles inscribed in a given triangle.
On the other hand, if the sides length is fixed, there are at most two solutions. I am going to provide a proof almost-without words:
The red locus is convex (since it is the arc of an ellipse centered at the upper vertex), hence it cannot meet the horizontal side at more than two points.
answered Nov 23 at 18:46
Jack D'Aurizio
283k33275653
edited Nov 23 at 19:34
answered Nov 23 at 18:46
Jack D'Aurizio
283k33275653
answered Nov 23 at 18:46
Jack D'Aurizio
283k33275653
answered Nov 23 at 18:46
Jack D'Aurizio
283k33275653
283k33275653
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
– abiessu
Nov 24 at 3:41
Can you explain how to find theese two solutions?
– Stepii
12 hours ago
The axis of the involved ellipse are given by an internal/external angle bisector and its vertices are simple to find. To find the solutions is equivalent to intersecting an ellipse and a line, or, up to affine maps, to intersecting a circle and a line.
– Jack D'Aurizio
12 hours ago
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
– Jack D'Aurizio
11 hours ago
add a comment |
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
– abiessu
Nov 24 at 3:41
Can you explain how to find theese two solutions?
– Stepii
12 hours ago
The axis of the involved ellipse are given by an internal/external angle bisector and its vertices are simple to find. To find the solutions is equivalent to intersecting an ellipse and a line, or, up to affine maps, to intersecting a circle and a line.
– Jack D'Aurizio
12 hours ago
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
– Jack D'Aurizio
11 hours ago
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
– abiessu
Nov 24 at 3:41
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
– abiessu
Nov 24 at 3:41
Can you explain how to find theese two solutions?
– Stepii
12 hours ago
Can you explain how to find theese two solutions?
– Stepii
12 hours ago
The axis of the involved ellipse are given by an internal/external angle bisector and its vertices are simple to find. To find the solutions is equivalent to intersecting an ellipse and a line, or, up to affine maps, to intersecting a circle and a line.
– Jack D'Aurizio
12 hours ago
The axis of the involved ellipse are given by an internal/external angle bisector and its vertices are simple to find. To find the solutions is equivalent to intersecting an ellipse and a line, or, up to affine maps, to intersecting a circle and a line.
– Jack D'Aurizio
12 hours ago
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
– Jack D'Aurizio
11 hours ago
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
– Jack D'Aurizio
11 hours ago
add a comment |
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Are you looking for the maximum possible equilateral triangle, or an arbitrary one? Would you consider any equilateral triangle which shares part of a side with the bounding triangle?
– abiessu
Nov 23 at 15:37
@abiessu I am looking for all equilateral triangles with given side, not maximum nor sharing side, just inscribed inside another arbitrary triangle
– Stepii
Nov 23 at 15:44
That’s a lot of triangles, or one, or none. Do you have anything else to go on?
– abiessu
Nov 23 at 16:46
@abiessu Well, my teacher said that there's maximum 2 solutions.
– Stepii
Nov 23 at 17:36