Given three lines that intersect at a point, construct an equilateral triangle of given length so that its vertices lie on these lines. [closed] The 2019 Stack Overflow Developer Survey Results Are InTriangle given its three perpendicular bisectors and a point of an edgeConstruct a triangle given one side, its height and inradiusGiven a line segment. Construct an equilateral triangle with one side the given line segment.construct an equilateral triangle with out knowing its scaleHow to construct the circumcenter of a triangle using a compass ONLY.Construct an equilateral triangle with area equal to a given triangleThree lines are given. Find three points on these lines, one point on each line, that are vertices of an equilateral triangleGiven two points calculate a third point in a way that three point construct an equilateral triangle.Constructing an equilateral triangle with a given segment as a side, using a compass whose radius is less than the length of the segmentHow do you calculate the weighting of a point inside of an equilateral triangle compared to its vertices?
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Given three lines that intersect at a point, construct an equilateral triangle of given length so that its vertices lie on these lines. [closed]
The 2019 Stack Overflow Developer Survey Results Are InTriangle given its three perpendicular bisectors and a point of an edgeConstruct a triangle given one side, its height and inradiusGiven a line segment. Construct an equilateral triangle with one side the given line segment.construct an equilateral triangle with out knowing its scaleHow to construct the circumcenter of a triangle using a compass ONLY.Construct an equilateral triangle with area equal to a given triangleThree lines are given. Find three points on these lines, one point on each line, that are vertices of an equilateral triangleGiven two points calculate a third point in a way that three point construct an equilateral triangle.Constructing an equilateral triangle with a given segment as a side, using a compass whose radius is less than the length of the segmentHow do you calculate the weighting of a point inside of an equilateral triangle compared to its vertices?
$begingroup$
This problem has to be solved with a compass and a straight edge. I haven't found any solutions online.
geometry triangles
edited Mar 23 at 19:28
Aretino
25.8k31545
asked Mar 23 at 17:53
Sebastián CórdobaSebastián Córdoba
82
$endgroup$
closed as off-topic by Travis, John Douma, abiessu, Maria Mazur, Cesareo Mar 24 at 0:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, John Douma, abiessu, Maria Mazur, Cesareo
add a comment |
$begingroup$
This problem has to be solved with a compass and a straight edge. I haven't found any solutions online.
geometry triangles
edited Mar 23 at 19:28
Aretino
25.8k31545
asked Mar 23 at 17:53
Sebastián CórdobaSebastián Córdoba
82
$endgroup$
closed as off-topic by Travis, John Douma, abiessu, Maria Mazur, Cesareo Mar 24 at 0:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, John Douma, abiessu, Maria Mazur, Cesareo
4
$begingroup$
What have you tried?
$endgroup$
– weareallin
Mar 23 at 18:18
add a comment |
$begingroup$
This problem has to be solved with a compass and a straight edge. I haven't found any solutions online.
geometry triangles
edited Mar 23 at 19:28
Aretino
25.8k31545
asked Mar 23 at 17:53
Sebastián CórdobaSebastián Córdoba
82
$endgroup$
This problem has to be solved with a compass and a straight edge. I haven't found any solutions online.
geometry triangles
geometry triangles
edited Mar 23 at 19:28
Aretino
25.8k31545
asked Mar 23 at 17:53
Sebastián CórdobaSebastián Córdoba
82
edited Mar 23 at 19:28
Aretino
25.8k31545
asked Mar 23 at 17:53
Sebastián CórdobaSebastián Córdoba
82
edited Mar 23 at 19:28
Aretino
25.8k31545
edited Mar 23 at 19:28
Aretino
25.8k31545
edited Mar 23 at 19:28
Aretino
25.8k31545
25.8k31545
asked Mar 23 at 17:53
Sebastián CórdobaSebastián Córdoba
82
asked Mar 23 at 17:53
Sebastián CórdobaSebastián Córdoba
82
asked Mar 23 at 17:53
Sebastián CórdobaSebastián Córdoba
82
82
closed as off-topic by Travis, John Douma, abiessu, Maria Mazur, Cesareo Mar 24 at 0:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, John Douma, abiessu, Maria Mazur, Cesareo
closed as off-topic by Travis, John Douma, abiessu, Maria Mazur, Cesareo Mar 24 at 0:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, John Douma, abiessu, Maria Mazur, Cesareo
4
$begingroup$
What have you tried?
$endgroup$
– weareallin
Mar 23 at 18:18
add a comment |
4
$begingroup$
What have you tried?
$endgroup$
– weareallin
Mar 23 at 18:18
4
4
$begingroup$
What have you tried?
$endgroup$
– weareallin
Mar 23 at 18:18
$begingroup$
What have you tried?
$endgroup$
– weareallin
Mar 23 at 18:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We use the fact that a compass and a straight edge suffice for rotating any point $60^circ $ about an origin.
Let $O$ be the point of intersection of the three lines. Choose an arbitrary point $A $ on $t $ on the right from $O$ and rotate the line $r $ $60^circ $ clockwise about the point. Let the rotated line be $r'$. Let $B$ be the intersection point of $r'$ with $s$. Rotate the point $B $ $60^circ $ counterclockwise, name the rotated point $C $. Observe that it lies on the line $r $. By the construction the triangle $ABC $ is equilateral and its vertices lie on the lines $r,s,t $. It remains only to rescale the triangle using homothetic transformation with respect to $O$.
Can you take it from here?
Note that the first point ($A$) can be chosen lying on any of three lines and the first rotation can be done both clock- and counterclockwise. The latter can be required for example if the angle between the lines $r$ and $s$ is $60^circ $.
answered Mar 23 at 20:18
useruser
6,46311031
$endgroup$
$begingroup$
Thanks for the insight. This definitely helped.
$endgroup$
– Sebastián Córdoba
Mar 24 at 8:43
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We use the fact that a compass and a straight edge suffice for rotating any point $60^circ $ about an origin.
Let $O$ be the point of intersection of the three lines. Choose an arbitrary point $A $ on $t $ on the right from $O$ and rotate the line $r $ $60^circ $ clockwise about the point. Let the rotated line be $r'$. Let $B$ be the intersection point of $r'$ with $s$. Rotate the point $B $ $60^circ $ counterclockwise, name the rotated point $C $. Observe that it lies on the line $r $. By the construction the triangle $ABC $ is equilateral and its vertices lie on the lines $r,s,t $. It remains only to rescale the triangle using homothetic transformation with respect to $O$.
Can you take it from here?
Note that the first point ($A$) can be chosen lying on any of three lines and the first rotation can be done both clock- and counterclockwise. The latter can be required for example if the angle between the lines $r$ and $s$ is $60^circ $.
answered Mar 23 at 20:18
useruser
6,46311031
$endgroup$
$begingroup$
Thanks for the insight. This definitely helped.
$endgroup$
– Sebastián Córdoba
Mar 24 at 8:43
add a comment |
$begingroup$
We use the fact that a compass and a straight edge suffice for rotating any point $60^circ $ about an origin.
Let $O$ be the point of intersection of the three lines. Choose an arbitrary point $A $ on $t $ on the right from $O$ and rotate the line $r $ $60^circ $ clockwise about the point. Let the rotated line be $r'$. Let $B$ be the intersection point of $r'$ with $s$. Rotate the point $B $ $60^circ $ counterclockwise, name the rotated point $C $. Observe that it lies on the line $r $. By the construction the triangle $ABC $ is equilateral and its vertices lie on the lines $r,s,t $. It remains only to rescale the triangle using homothetic transformation with respect to $O$.
Can you take it from here?
Note that the first point ($A$) can be chosen lying on any of three lines and the first rotation can be done both clock- and counterclockwise. The latter can be required for example if the angle between the lines $r$ and $s$ is $60^circ $.
answered Mar 23 at 20:18
useruser
6,46311031
$endgroup$
$begingroup$
Thanks for the insight. This definitely helped.
$endgroup$
– Sebastián Córdoba
Mar 24 at 8:43
add a comment |
$begingroup$
We use the fact that a compass and a straight edge suffice for rotating any point $60^circ $ about an origin.
Let $O$ be the point of intersection of the three lines. Choose an arbitrary point $A $ on $t $ on the right from $O$ and rotate the line $r $ $60^circ $ clockwise about the point. Let the rotated line be $r'$. Let $B$ be the intersection point of $r'$ with $s$. Rotate the point $B $ $60^circ $ counterclockwise, name the rotated point $C $. Observe that it lies on the line $r $. By the construction the triangle $ABC $ is equilateral and its vertices lie on the lines $r,s,t $. It remains only to rescale the triangle using homothetic transformation with respect to $O$.
Can you take it from here?
Note that the first point ($A$) can be chosen lying on any of three lines and the first rotation can be done both clock- and counterclockwise. The latter can be required for example if the angle between the lines $r$ and $s$ is $60^circ $.
answered Mar 23 at 20:18
useruser
6,46311031
$endgroup$
We use the fact that a compass and a straight edge suffice for rotating any point $60^circ $ about an origin.
Let $O$ be the point of intersection of the three lines. Choose an arbitrary point $A $ on $t $ on the right from $O$ and rotate the line $r $ $60^circ $ clockwise about the point. Let the rotated line be $r'$. Let $B$ be the intersection point of $r'$ with $s$. Rotate the point $B $ $60^circ $ counterclockwise, name the rotated point $C $. Observe that it lies on the line $r $. By the construction the triangle $ABC $ is equilateral and its vertices lie on the lines $r,s,t $. It remains only to rescale the triangle using homothetic transformation with respect to $O$.
Can you take it from here?
Note that the first point ($A$) can be chosen lying on any of three lines and the first rotation can be done both clock- and counterclockwise. The latter can be required for example if the angle between the lines $r$ and $s$ is $60^circ $.
answered Mar 23 at 20:18
useruser
6,46311031
edited Mar 24 at 19:12
answered Mar 23 at 20:18
useruser
6,46311031
answered Mar 23 at 20:18
useruser
6,46311031
answered Mar 23 at 20:18
useruser
6,46311031
6,46311031
$begingroup$
Thanks for the insight. This definitely helped.
$endgroup$
– Sebastián Córdoba
Mar 24 at 8:43
add a comment |
$begingroup$
Thanks for the insight. This definitely helped.
$endgroup$
– Sebastián Córdoba
Mar 24 at 8:43
$begingroup$
Thanks for the insight. This definitely helped.
$endgroup$
– Sebastián Córdoba
Mar 24 at 8:43
$begingroup$
Thanks for the insight. This definitely helped.
$endgroup$
– Sebastián Córdoba
Mar 24 at 8:43
add a comment |
4
$begingroup$
What have you tried?
$endgroup$
– weareallin
Mar 23 at 18:18