Inscribe an equilateral triangle inside a triangle
$begingroup$
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 '18 at 15:34
StepiiStepii
106
$endgroup$
add a comment |
$begingroup$
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 '18 at 15:34
StepiiStepii
106
$endgroup$
$begingroup$
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?
$endgroup$
– abiessu
Nov 23 '18 at 15:37
$begingroup$
@abiessu I am looking for all equilateral triangles with given side, not maximum nor sharing side, just inscribed inside another arbitrary triangle
$endgroup$
– Stepii
Nov 23 '18 at 15:44
$begingroup$
That’s a lot of triangles, or one, or none. Do you have anything else to go on?
$endgroup$
– abiessu
Nov 23 '18 at 16:46
$begingroup$
@abiessu Well, my teacher said that there's maximum 2 solutions.
$endgroup$
– Stepii
Nov 23 '18 at 17:36
add a comment |
$begingroup$
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 '18 at 15:34
StepiiStepii
106
$endgroup$
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 '18 at 15:34
StepiiStepii
106
asked Nov 23 '18 at 15:34
StepiiStepii
106
edited Nov 30 '18 at 17:21
Stepii
asked Nov 23 '18 at 15:34
StepiiStepii
106
asked Nov 23 '18 at 15:34
StepiiStepii
106
asked Nov 23 '18 at 15:34
StepiiStepii
106
106
$begingroup$
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?
$endgroup$
– abiessu
Nov 23 '18 at 15:37
$begingroup$
@abiessu I am looking for all equilateral triangles with given side, not maximum nor sharing side, just inscribed inside another arbitrary triangle
$endgroup$
– Stepii
Nov 23 '18 at 15:44
$begingroup$
That’s a lot of triangles, or one, or none. Do you have anything else to go on?
$endgroup$
– abiessu
Nov 23 '18 at 16:46
$begingroup$
@abiessu Well, my teacher said that there's maximum 2 solutions.
$endgroup$
– Stepii
Nov 23 '18 at 17:36
add a comment |
$begingroup$
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?
$endgroup$
– abiessu
Nov 23 '18 at 15:37
$begingroup$
@abiessu I am looking for all equilateral triangles with given side, not maximum nor sharing side, just inscribed inside another arbitrary triangle
$endgroup$
– Stepii
Nov 23 '18 at 15:44
$begingroup$
That’s a lot of triangles, or one, or none. Do you have anything else to go on?
$endgroup$
– abiessu
Nov 23 '18 at 16:46
$begingroup$
@abiessu Well, my teacher said that there's maximum 2 solutions.
$endgroup$
– Stepii
Nov 23 '18 at 17:36
$begingroup$
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?
$endgroup$
– abiessu
Nov 23 '18 at 15:37
$begingroup$
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?
$endgroup$
– abiessu
Nov 23 '18 at 15:37
$begingroup$
@abiessu I am looking for all equilateral triangles with given side, not maximum nor sharing side, just inscribed inside another arbitrary triangle
$endgroup$
– Stepii
Nov 23 '18 at 15:44
$begingroup$
@abiessu I am looking for all equilateral triangles with given side, not maximum nor sharing side, just inscribed inside another arbitrary triangle
$endgroup$
– Stepii
Nov 23 '18 at 15:44
$begingroup$
That’s a lot of triangles, or one, or none. Do you have anything else to go on?
$endgroup$
– abiessu
Nov 23 '18 at 16:46
$begingroup$
That’s a lot of triangles, or one, or none. Do you have anything else to go on?
$endgroup$
– abiessu
Nov 23 '18 at 16:46
$begingroup$
@abiessu Well, my teacher said that there's maximum 2 solutions.
$endgroup$
– Stepii
Nov 23 '18 at 17:36
$begingroup$
@abiessu Well, my teacher said that there's maximum 2 solutions.
$endgroup$
– Stepii
Nov 23 '18 at 17:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 '18 at 18:46
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
$begingroup$
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
$endgroup$
– abiessu
Nov 24 '18 at 3:41
$begingroup$
Can you explain how to find theese two solutions?
$endgroup$
– Stepii
Nov 30 '18 at 17:16
$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 17:53
$begingroup$
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 18:17
$begingroup$
@JackD'Aurizio I don't see it so simple. Can you give more details on finding the two solutions?
$endgroup$
– Stepii
Dec 2 '18 at 13:15
|
show 1 more comment
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 '18 at 18:46
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
$begingroup$
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
$endgroup$
– abiessu
Nov 24 '18 at 3:41
$begingroup$
Can you explain how to find theese two solutions?
$endgroup$
– Stepii
Nov 30 '18 at 17:16
$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 17:53
$begingroup$
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 18:17
$begingroup$
@JackD'Aurizio I don't see it so simple. Can you give more details on finding the two solutions?
$endgroup$
– Stepii
Dec 2 '18 at 13:15
|
show 1 more comment
$begingroup$
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 '18 at 18:46
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
$begingroup$
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
$endgroup$
– abiessu
Nov 24 '18 at 3:41
$begingroup$
Can you explain how to find theese two solutions?
$endgroup$
– Stepii
Nov 30 '18 at 17:16
$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 17:53
$begingroup$
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 18:17
$begingroup$
@JackD'Aurizio I don't see it so simple. Can you give more details on finding the two solutions?
$endgroup$
– Stepii
Dec 2 '18 at 13:15
|
show 1 more comment
$begingroup$
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 '18 at 18:46
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
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 '18 at 18:46
Jack D'AurizioJack D'Aurizio
288k33280659
edited Nov 23 '18 at 19:34
answered Nov 23 '18 at 18:46
Jack D'AurizioJack D'Aurizio
288k33280659
answered Nov 23 '18 at 18:46
Jack D'AurizioJack D'Aurizio
288k33280659
answered Nov 23 '18 at 18:46
Jack D'AurizioJack D'Aurizio
288k33280659
288k33280659
$begingroup$
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
$endgroup$
– abiessu
Nov 24 '18 at 3:41
$begingroup$
Can you explain how to find theese two solutions?
$endgroup$
– Stepii
Nov 30 '18 at 17:16
$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 17:53
$begingroup$
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 18:17
$begingroup$
@JackD'Aurizio I don't see it so simple. Can you give more details on finding the two solutions?
$endgroup$
– Stepii
Dec 2 '18 at 13:15
|
show 1 more comment
$begingroup$
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
$endgroup$
– abiessu
Nov 24 '18 at 3:41
$begingroup$
Can you explain how to find theese two solutions?
$endgroup$
– Stepii
Nov 30 '18 at 17:16
$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 17:53
$begingroup$
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 18:17
$begingroup$
@JackD'Aurizio I don't see it so simple. Can you give more details on finding the two solutions?
$endgroup$
– Stepii
Dec 2 '18 at 13:15
$begingroup$
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
$endgroup$
– abiessu
Nov 24 '18 at 3:41
$begingroup$
Is it safe to assume that with a perfectly-chosen given length, there will be exactly one solution?
$endgroup$
– abiessu
Nov 24 '18 at 3:41
$begingroup$
Can you explain how to find theese two solutions?
$endgroup$
– Stepii
Nov 30 '18 at 17:16
$begingroup$
Can you explain how to find theese two solutions?
$endgroup$
– Stepii
Nov 30 '18 at 17:16
$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 17:53
$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 17:53
$begingroup$
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 18:17
$begingroup$
@abiessu: the critical length is clearly given by the smallest inscribed equilateral triangle, which is related to the Napoleon triangle of $ABC$.
$endgroup$
– Jack D'Aurizio
Nov 30 '18 at 18:17
$begingroup$
@JackD'Aurizio I don't see it so simple. Can you give more details on finding the two solutions?
$endgroup$
– Stepii
Dec 2 '18 at 13:15
$begingroup$
@JackD'Aurizio I don't see it so simple. Can you give more details on finding the two solutions?
$endgroup$
– Stepii
Dec 2 '18 at 13:15
|
show 1 more comment
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$begingroup$
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?
$endgroup$
– abiessu
Nov 23 '18 at 15:37
$begingroup$
@abiessu I am looking for all equilateral triangles with given side, not maximum nor sharing side, just inscribed inside another arbitrary triangle
$endgroup$
– Stepii
Nov 23 '18 at 15:44
$begingroup$
That’s a lot of triangles, or one, or none. Do you have anything else to go on?
$endgroup$
– abiessu
Nov 23 '18 at 16:46
$begingroup$
@abiessu Well, my teacher said that there's maximum 2 solutions.
$endgroup$
– Stepii
Nov 23 '18 at 17:36