There are 2013 random points on the plain, prove exactly 8 lies within a circle
$begingroup$
This was posed as a homework problem to me and I have made progress, but I don't think I have a solution.
So there are a lot of random points on the plane and we need to find a circle that contains exactly eight of them. The way I would attack this is to list the distances of each point pairs. Now take a pair (among perhaps several) that has a minimal distance.
I think that even if the minimal distances are equal, the most you can do with equal minimal distances without having a circle covering them is seven points (suspiciously one short of that eight!) by picking the centre of a circle and then six equidistant points on the circle (belonging to a 60 degree central degree).
geometry
asked Oct 11 '13 at 3:25
chxchx
823720
$endgroup$
add a comment |
$begingroup$
This was posed as a homework problem to me and I have made progress, but I don't think I have a solution.
So there are a lot of random points on the plane and we need to find a circle that contains exactly eight of them. The way I would attack this is to list the distances of each point pairs. Now take a pair (among perhaps several) that has a minimal distance.
I think that even if the minimal distances are equal, the most you can do with equal minimal distances without having a circle covering them is seven points (suspiciously one short of that eight!) by picking the centre of a circle and then six equidistant points on the circle (belonging to a 60 degree central degree).
geometry
asked Oct 11 '13 at 3:25
chxchx
823720
$endgroup$
add a comment |
$begingroup$
This was posed as a homework problem to me and I have made progress, but I don't think I have a solution.
So there are a lot of random points on the plane and we need to find a circle that contains exactly eight of them. The way I would attack this is to list the distances of each point pairs. Now take a pair (among perhaps several) that has a minimal distance.
I think that even if the minimal distances are equal, the most you can do with equal minimal distances without having a circle covering them is seven points (suspiciously one short of that eight!) by picking the centre of a circle and then six equidistant points on the circle (belonging to a 60 degree central degree).
geometry
asked Oct 11 '13 at 3:25
chxchx
823720
$endgroup$
This was posed as a homework problem to me and I have made progress, but I don't think I have a solution.
So there are a lot of random points on the plane and we need to find a circle that contains exactly eight of them. The way I would attack this is to list the distances of each point pairs. Now take a pair (among perhaps several) that has a minimal distance.
I think that even if the minimal distances are equal, the most you can do with equal minimal distances without having a circle covering them is seven points (suspiciously one short of that eight!) by picking the centre of a circle and then six equidistant points on the circle (belonging to a 60 degree central degree).
geometry
geometry
asked Oct 11 '13 at 3:25
chxchx
823720
asked Oct 11 '13 at 3:25
chxchx
823720
asked Oct 11 '13 at 3:25
chxchx
823720
asked Oct 11 '13 at 3:25
chxchx
823720
asked Oct 11 '13 at 3:25
chxchx
823720
823720
add a comment |
add a comment |
2 Answers
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You can find such a circle with almost any center in the plane you want. Imagine picking some point and slowly expanding the radius of a circle until you have eight points inside the circle. The only time you fail is when there are at least two points equidistant from the center you pick, and as you expand the circle they take you from less than eight points within the circle to nine or more. Each pair of points has a line of points they are equidistant from. There are a little more than 2 million pairs of points. Pick any point not on any of these two million lines as a center and as the radius expands you will cross the points one at a time. As the lines have zero area, there are still plenty of centers to choose from. When you have crossed eight and not nine, you have your circle.
answered Oct 11 '13 at 3:37
Ross MillikanRoss Millikan
295k23198371
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$begingroup$
Pick an extremal line passing through two points $A$ and $B$ in this set.
We can do this by picking the two points with the least y coordinates say.
Now consider the angles formed by joining $A$ and $B$ to any point in the set.
Each of these points will form a different angle.
Sort these angles.
Let the 9th largest angle have vertex $P_9$.
Now the 8 angles greater than this angle must necessarily lie within the circumcircle of $ABP_9$.
Now, we pick the circle with its center the same as the circumcentre of $ABP_9$ but with a radius that is just infinitesimally smaller than that of $ABP_9$.
We have constructed a circle that contains exactly 8 points.
answered Jan 5 at 20:33
ArchimedesprincipleArchimedesprinciple
34417
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2 Answers
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2 Answers
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$begingroup$
You can find such a circle with almost any center in the plane you want. Imagine picking some point and slowly expanding the radius of a circle until you have eight points inside the circle. The only time you fail is when there are at least two points equidistant from the center you pick, and as you expand the circle they take you from less than eight points within the circle to nine or more. Each pair of points has a line of points they are equidistant from. There are a little more than 2 million pairs of points. Pick any point not on any of these two million lines as a center and as the radius expands you will cross the points one at a time. As the lines have zero area, there are still plenty of centers to choose from. When you have crossed eight and not nine, you have your circle.
answered Oct 11 '13 at 3:37
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
You can find such a circle with almost any center in the plane you want. Imagine picking some point and slowly expanding the radius of a circle until you have eight points inside the circle. The only time you fail is when there are at least two points equidistant from the center you pick, and as you expand the circle they take you from less than eight points within the circle to nine or more. Each pair of points has a line of points they are equidistant from. There are a little more than 2 million pairs of points. Pick any point not on any of these two million lines as a center and as the radius expands you will cross the points one at a time. As the lines have zero area, there are still plenty of centers to choose from. When you have crossed eight and not nine, you have your circle.
answered Oct 11 '13 at 3:37
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
You can find such a circle with almost any center in the plane you want. Imagine picking some point and slowly expanding the radius of a circle until you have eight points inside the circle. The only time you fail is when there are at least two points equidistant from the center you pick, and as you expand the circle they take you from less than eight points within the circle to nine or more. Each pair of points has a line of points they are equidistant from. There are a little more than 2 million pairs of points. Pick any point not on any of these two million lines as a center and as the radius expands you will cross the points one at a time. As the lines have zero area, there are still plenty of centers to choose from. When you have crossed eight and not nine, you have your circle.
answered Oct 11 '13 at 3:37
Ross MillikanRoss Millikan
295k23198371
$endgroup$
You can find such a circle with almost any center in the plane you want. Imagine picking some point and slowly expanding the radius of a circle until you have eight points inside the circle. The only time you fail is when there are at least two points equidistant from the center you pick, and as you expand the circle they take you from less than eight points within the circle to nine or more. Each pair of points has a line of points they are equidistant from. There are a little more than 2 million pairs of points. Pick any point not on any of these two million lines as a center and as the radius expands you will cross the points one at a time. As the lines have zero area, there are still plenty of centers to choose from. When you have crossed eight and not nine, you have your circle.
answered Oct 11 '13 at 3:37
Ross MillikanRoss Millikan
295k23198371
edited Oct 11 '13 at 3:53
answered Oct 11 '13 at 3:37
Ross MillikanRoss Millikan
295k23198371
answered Oct 11 '13 at 3:37
Ross MillikanRoss Millikan
295k23198371
answered Oct 11 '13 at 3:37
Ross MillikanRoss Millikan
295k23198371
295k23198371
add a comment |
add a comment |
$begingroup$
Pick an extremal line passing through two points $A$ and $B$ in this set.
We can do this by picking the two points with the least y coordinates say.
Now consider the angles formed by joining $A$ and $B$ to any point in the set.
Each of these points will form a different angle.
Sort these angles.
Let the 9th largest angle have vertex $P_9$.
Now the 8 angles greater than this angle must necessarily lie within the circumcircle of $ABP_9$.
Now, we pick the circle with its center the same as the circumcentre of $ABP_9$ but with a radius that is just infinitesimally smaller than that of $ABP_9$.
We have constructed a circle that contains exactly 8 points.
answered Jan 5 at 20:33
ArchimedesprincipleArchimedesprinciple
34417
$endgroup$
add a comment |
$begingroup$
Pick an extremal line passing through two points $A$ and $B$ in this set.
We can do this by picking the two points with the least y coordinates say.
Now consider the angles formed by joining $A$ and $B$ to any point in the set.
Each of these points will form a different angle.
Sort these angles.
Let the 9th largest angle have vertex $P_9$.
Now the 8 angles greater than this angle must necessarily lie within the circumcircle of $ABP_9$.
Now, we pick the circle with its center the same as the circumcentre of $ABP_9$ but with a radius that is just infinitesimally smaller than that of $ABP_9$.
We have constructed a circle that contains exactly 8 points.
answered Jan 5 at 20:33
ArchimedesprincipleArchimedesprinciple
34417
$endgroup$
add a comment |
$begingroup$
Pick an extremal line passing through two points $A$ and $B$ in this set.
We can do this by picking the two points with the least y coordinates say.
Now consider the angles formed by joining $A$ and $B$ to any point in the set.
Each of these points will form a different angle.
Sort these angles.
Let the 9th largest angle have vertex $P_9$.
Now the 8 angles greater than this angle must necessarily lie within the circumcircle of $ABP_9$.
Now, we pick the circle with its center the same as the circumcentre of $ABP_9$ but with a radius that is just infinitesimally smaller than that of $ABP_9$.
We have constructed a circle that contains exactly 8 points.
answered Jan 5 at 20:33
ArchimedesprincipleArchimedesprinciple
34417
$endgroup$
Pick an extremal line passing through two points $A$ and $B$ in this set.
We can do this by picking the two points with the least y coordinates say.
Now consider the angles formed by joining $A$ and $B$ to any point in the set.
Each of these points will form a different angle.
Sort these angles.
Let the 9th largest angle have vertex $P_9$.
Now the 8 angles greater than this angle must necessarily lie within the circumcircle of $ABP_9$.
Now, we pick the circle with its center the same as the circumcentre of $ABP_9$ but with a radius that is just infinitesimally smaller than that of $ABP_9$.
We have constructed a circle that contains exactly 8 points.
answered Jan 5 at 20:33
ArchimedesprincipleArchimedesprinciple
34417
answered Jan 5 at 20:33
ArchimedesprincipleArchimedesprinciple
34417
answered Jan 5 at 20:33
ArchimedesprincipleArchimedesprinciple
34417
answered Jan 5 at 20:33
ArchimedesprincipleArchimedesprinciple
34417
34417
add a comment |
add a comment |
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