Is the Torricelli/Fermat point unique?
$begingroup$
In a triangle with all angles less than 120 degrees can there be two such points?
I am wondering because in the geometry game Euclidea it gives two "v-stars" for this problem (theta 8.6). This usually means that there are two solutions possible.
I found one solution, basically by constructing equilateral triangles on the sides and connecting their external vertices.
However I am confused what the second solution is supposed to be.
geometry
asked May 4 '18 at 22:11
user1583209user1583209
1133
$endgroup$
add a comment |
$begingroup$
In a triangle with all angles less than 120 degrees can there be two such points?
I am wondering because in the geometry game Euclidea it gives two "v-stars" for this problem (theta 8.6). This usually means that there are two solutions possible.
I found one solution, basically by constructing equilateral triangles on the sides and connecting their external vertices.
However I am confused what the second solution is supposed to be.
geometry
asked May 4 '18 at 22:11
user1583209user1583209
1133
$endgroup$
add a comment |
$begingroup$
In a triangle with all angles less than 120 degrees can there be two such points?
I am wondering because in the geometry game Euclidea it gives two "v-stars" for this problem (theta 8.6). This usually means that there are two solutions possible.
I found one solution, basically by constructing equilateral triangles on the sides and connecting their external vertices.
However I am confused what the second solution is supposed to be.
geometry
asked May 4 '18 at 22:11
user1583209user1583209
1133
$endgroup$
In a triangle with all angles less than 120 degrees can there be two such points?
I am wondering because in the geometry game Euclidea it gives two "v-stars" for this problem (theta 8.6). This usually means that there are two solutions possible.
I found one solution, basically by constructing equilateral triangles on the sides and connecting their external vertices.
However I am confused what the second solution is supposed to be.
geometry
geometry
asked May 4 '18 at 22:11
user1583209user1583209
1133
asked May 4 '18 at 22:11
user1583209user1583209
1133
asked May 4 '18 at 22:11
user1583209user1583209
1133
asked May 4 '18 at 22:11
user1583209user1583209
1133
asked May 4 '18 at 22:11
user1583209user1583209
1133
1133
add a comment |
add a comment |
1 Answer
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There is one Fermat–Torricelli point, i.e. the point such that the total distance from the three vertices of the triangle to the point is the minimum possible
There are two isogonic centres of a triangle (except for special cases where one of the original angles is $120^circ$ or $60^circ$ such as an equilateral triangle) which form angles of $120^circ$ or $60^circ$ with each of the three pairs of vertices of the original triangle
When all the original angles are less than $120^circ$ then the Fermat–Torricelli point is one of the isogonic centres (the one inside the original triangle). When one of original angles is greater $120^circ$ then that vertex is the Fermat–Torricelli point while both isogonic centres are outside the original triangle
So perhaps if something suggests there are two solutions to this type of question, it is referring to isogonic centres rather than Fermat–Torricelli points
answered May 4 '18 at 22:26
HenryHenry
100k481167
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$begingroup$
Indeed they wanted the second isogonic centre.
$endgroup$
– user1583209
May 5 '18 at 9:55
$begingroup$
There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
$endgroup$
– Henry
May 5 '18 at 10:33
add a comment |
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1 Answer
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1 Answer
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$begingroup$
There is one Fermat–Torricelli point, i.e. the point such that the total distance from the three vertices of the triangle to the point is the minimum possible
There are two isogonic centres of a triangle (except for special cases where one of the original angles is $120^circ$ or $60^circ$ such as an equilateral triangle) which form angles of $120^circ$ or $60^circ$ with each of the three pairs of vertices of the original triangle
When all the original angles are less than $120^circ$ then the Fermat–Torricelli point is one of the isogonic centres (the one inside the original triangle). When one of original angles is greater $120^circ$ then that vertex is the Fermat–Torricelli point while both isogonic centres are outside the original triangle
So perhaps if something suggests there are two solutions to this type of question, it is referring to isogonic centres rather than Fermat–Torricelli points
answered May 4 '18 at 22:26
HenryHenry
100k481167
$endgroup$
$begingroup$
Indeed they wanted the second isogonic centre.
$endgroup$
– user1583209
May 5 '18 at 9:55
$begingroup$
There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
$endgroup$
– Henry
May 5 '18 at 10:33
add a comment |
$begingroup$
There is one Fermat–Torricelli point, i.e. the point such that the total distance from the three vertices of the triangle to the point is the minimum possible
There are two isogonic centres of a triangle (except for special cases where one of the original angles is $120^circ$ or $60^circ$ such as an equilateral triangle) which form angles of $120^circ$ or $60^circ$ with each of the three pairs of vertices of the original triangle
When all the original angles are less than $120^circ$ then the Fermat–Torricelli point is one of the isogonic centres (the one inside the original triangle). When one of original angles is greater $120^circ$ then that vertex is the Fermat–Torricelli point while both isogonic centres are outside the original triangle
So perhaps if something suggests there are two solutions to this type of question, it is referring to isogonic centres rather than Fermat–Torricelli points
answered May 4 '18 at 22:26
HenryHenry
100k481167
$endgroup$
$begingroup$
Indeed they wanted the second isogonic centre.
$endgroup$
– user1583209
May 5 '18 at 9:55
$begingroup$
There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
$endgroup$
– Henry
May 5 '18 at 10:33
add a comment |
$begingroup$
There is one Fermat–Torricelli point, i.e. the point such that the total distance from the three vertices of the triangle to the point is the minimum possible
There are two isogonic centres of a triangle (except for special cases where one of the original angles is $120^circ$ or $60^circ$ such as an equilateral triangle) which form angles of $120^circ$ or $60^circ$ with each of the three pairs of vertices of the original triangle
When all the original angles are less than $120^circ$ then the Fermat–Torricelli point is one of the isogonic centres (the one inside the original triangle). When one of original angles is greater $120^circ$ then that vertex is the Fermat–Torricelli point while both isogonic centres are outside the original triangle
So perhaps if something suggests there are two solutions to this type of question, it is referring to isogonic centres rather than Fermat–Torricelli points
answered May 4 '18 at 22:26
HenryHenry
100k481167
$endgroup$
There is one Fermat–Torricelli point, i.e. the point such that the total distance from the three vertices of the triangle to the point is the minimum possible
There are two isogonic centres of a triangle (except for special cases where one of the original angles is $120^circ$ or $60^circ$ such as an equilateral triangle) which form angles of $120^circ$ or $60^circ$ with each of the three pairs of vertices of the original triangle
When all the original angles are less than $120^circ$ then the Fermat–Torricelli point is one of the isogonic centres (the one inside the original triangle). When one of original angles is greater $120^circ$ then that vertex is the Fermat–Torricelli point while both isogonic centres are outside the original triangle
So perhaps if something suggests there are two solutions to this type of question, it is referring to isogonic centres rather than Fermat–Torricelli points
answered May 4 '18 at 22:26
HenryHenry
100k481167
edited May 4 '18 at 22:31
answered May 4 '18 at 22:26
HenryHenry
100k481167
answered May 4 '18 at 22:26
HenryHenry
100k481167
answered May 4 '18 at 22:26
HenryHenry
100k481167
100k481167
$begingroup$
Indeed they wanted the second isogonic centre.
$endgroup$
– user1583209
May 5 '18 at 9:55
$begingroup$
There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
$endgroup$
– Henry
May 5 '18 at 10:33
add a comment |
$begingroup$
Indeed they wanted the second isogonic centre.
$endgroup$
– user1583209
May 5 '18 at 9:55
$begingroup$
There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
$endgroup$
– Henry
May 5 '18 at 10:33
$begingroup$
Indeed they wanted the second isogonic centre.
$endgroup$
– user1583209
May 5 '18 at 9:55
$begingroup$
Indeed they wanted the second isogonic centre.
$endgroup$
– user1583209
May 5 '18 at 9:55
$begingroup$
There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
$endgroup$
– Henry
May 5 '18 at 10:33
$begingroup$
There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
$endgroup$
– Henry
May 5 '18 at 10:33
add a comment |
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