Bob selects four points on a $10times 10$ square.
$begingroup$
Bob selects four points on a $10times 10$ square.
Is it true that two of them are less than $sqrt{101}$ units apart?
I know how to prove things like this for five points.
These seems intuitively true because the 'best' selection is the four corners.
Let's say I generalize this: Bob selects four points on a $10times10$ square.
If it is always true that two of them are less than $k$ units apart, what is the maximum value of $k$?
combinatorics pigeonhole-principle upper-lower-bounds
asked Dec 18 '18 at 12:12
user627514user627514
393
$endgroup$
add a comment |
$begingroup$
Bob selects four points on a $10times 10$ square.
Is it true that two of them are less than $sqrt{101}$ units apart?
I know how to prove things like this for five points.
These seems intuitively true because the 'best' selection is the four corners.
Let's say I generalize this: Bob selects four points on a $10times10$ square.
If it is always true that two of them are less than $k$ units apart, what is the maximum value of $k$?
combinatorics pigeonhole-principle upper-lower-bounds
asked Dec 18 '18 at 12:12
user627514user627514
393
$endgroup$
$begingroup$
If anyone could just show me a configuration where the points are all more than $10$ apart, that would be cool. Because I have no idea what that would look like.
$endgroup$
– Arthur
Dec 18 '18 at 12:15
add a comment |
$begingroup$
Bob selects four points on a $10times 10$ square.
Is it true that two of them are less than $sqrt{101}$ units apart?
I know how to prove things like this for five points.
These seems intuitively true because the 'best' selection is the four corners.
Let's say I generalize this: Bob selects four points on a $10times10$ square.
If it is always true that two of them are less than $k$ units apart, what is the maximum value of $k$?
combinatorics pigeonhole-principle upper-lower-bounds
asked Dec 18 '18 at 12:12
user627514user627514
393
$endgroup$
Bob selects four points on a $10times 10$ square.
Is it true that two of them are less than $sqrt{101}$ units apart?
I know how to prove things like this for five points.
These seems intuitively true because the 'best' selection is the four corners.
Let's say I generalize this: Bob selects four points on a $10times10$ square.
If it is always true that two of them are less than $k$ units apart, what is the maximum value of $k$?
combinatorics pigeonhole-principle upper-lower-bounds
combinatorics pigeonhole-principle upper-lower-bounds
asked Dec 18 '18 at 12:12
user627514user627514
393
asked Dec 18 '18 at 12:12
user627514user627514
393
asked Dec 18 '18 at 12:12
user627514user627514
393
asked Dec 18 '18 at 12:12
user627514user627514
393
asked Dec 18 '18 at 12:12
user627514user627514
393
393
$begingroup$
If anyone could just show me a configuration where the points are all more than $10$ apart, that would be cool. Because I have no idea what that would look like.
$endgroup$
– Arthur
Dec 18 '18 at 12:15
add a comment |
$begingroup$
If anyone could just show me a configuration where the points are all more than $10$ apart, that would be cool. Because I have no idea what that would look like.
$endgroup$
– Arthur
Dec 18 '18 at 12:15
$begingroup$
If anyone could just show me a configuration where the points are all more than $10$ apart, that would be cool. Because I have no idea what that would look like.
$endgroup$
– Arthur
Dec 18 '18 at 12:15
$begingroup$
If anyone could just show me a configuration where the points are all more than $10$ apart, that would be cool. Because I have no idea what that would look like.
$endgroup$
– Arthur
Dec 18 '18 at 12:15
add a comment |
1 Answer
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active
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$begingroup$
Suppose there were four points in a $10times10$ square such that each pair is more than $10$ units apart. We first argue that we can assume the four points are on the four sides of the square. We then show this can't happen.
There cannot be two (or more) leftmost points that are more than $10$ units apart, since only one such point could be inside the square. So if the leftmost point is not on the left side of the square, we can increase its distance from the other points by moving it leftward. Likewise we can move the rightmost point to the right (if it's not already on the right side of the square), the topmost point higher, and the bottommost point lower. So we can assume the four points are on the four sides of the square, with one point on each of the four sides.
One of the four points must be closest to a corner of the square. By rotation and/or reflection, we can assume it's the point on the bottom side and it's closest to the bottom right corner. But a simple picture and the Pythagorean theorem shows that the point on the right side of the square must be closer to the upper right corner of the square than the point on the bottom is to the bottom right corner, and that's a contradiction. Thus for any four points in a $10times10$ square, some pair of them are no more than $10$ units apart.
answered Dec 18 '18 at 13:37
Barry CipraBarry Cipra
60.5k655128
$endgroup$
add a comment |
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$begingroup$
Suppose there were four points in a $10times10$ square such that each pair is more than $10$ units apart. We first argue that we can assume the four points are on the four sides of the square. We then show this can't happen.
There cannot be two (or more) leftmost points that are more than $10$ units apart, since only one such point could be inside the square. So if the leftmost point is not on the left side of the square, we can increase its distance from the other points by moving it leftward. Likewise we can move the rightmost point to the right (if it's not already on the right side of the square), the topmost point higher, and the bottommost point lower. So we can assume the four points are on the four sides of the square, with one point on each of the four sides.
One of the four points must be closest to a corner of the square. By rotation and/or reflection, we can assume it's the point on the bottom side and it's closest to the bottom right corner. But a simple picture and the Pythagorean theorem shows that the point on the right side of the square must be closer to the upper right corner of the square than the point on the bottom is to the bottom right corner, and that's a contradiction. Thus for any four points in a $10times10$ square, some pair of them are no more than $10$ units apart.
answered Dec 18 '18 at 13:37
Barry CipraBarry Cipra
60.5k655128
$endgroup$
add a comment |
$begingroup$
Suppose there were four points in a $10times10$ square such that each pair is more than $10$ units apart. We first argue that we can assume the four points are on the four sides of the square. We then show this can't happen.
There cannot be two (or more) leftmost points that are more than $10$ units apart, since only one such point could be inside the square. So if the leftmost point is not on the left side of the square, we can increase its distance from the other points by moving it leftward. Likewise we can move the rightmost point to the right (if it's not already on the right side of the square), the topmost point higher, and the bottommost point lower. So we can assume the four points are on the four sides of the square, with one point on each of the four sides.
One of the four points must be closest to a corner of the square. By rotation and/or reflection, we can assume it's the point on the bottom side and it's closest to the bottom right corner. But a simple picture and the Pythagorean theorem shows that the point on the right side of the square must be closer to the upper right corner of the square than the point on the bottom is to the bottom right corner, and that's a contradiction. Thus for any four points in a $10times10$ square, some pair of them are no more than $10$ units apart.
answered Dec 18 '18 at 13:37
Barry CipraBarry Cipra
60.5k655128
$endgroup$
add a comment |
$begingroup$
Suppose there were four points in a $10times10$ square such that each pair is more than $10$ units apart. We first argue that we can assume the four points are on the four sides of the square. We then show this can't happen.
There cannot be two (or more) leftmost points that are more than $10$ units apart, since only one such point could be inside the square. So if the leftmost point is not on the left side of the square, we can increase its distance from the other points by moving it leftward. Likewise we can move the rightmost point to the right (if it's not already on the right side of the square), the topmost point higher, and the bottommost point lower. So we can assume the four points are on the four sides of the square, with one point on each of the four sides.
One of the four points must be closest to a corner of the square. By rotation and/or reflection, we can assume it's the point on the bottom side and it's closest to the bottom right corner. But a simple picture and the Pythagorean theorem shows that the point on the right side of the square must be closer to the upper right corner of the square than the point on the bottom is to the bottom right corner, and that's a contradiction. Thus for any four points in a $10times10$ square, some pair of them are no more than $10$ units apart.
answered Dec 18 '18 at 13:37
Barry CipraBarry Cipra
60.5k655128
$endgroup$
Suppose there were four points in a $10times10$ square such that each pair is more than $10$ units apart. We first argue that we can assume the four points are on the four sides of the square. We then show this can't happen.
There cannot be two (or more) leftmost points that are more than $10$ units apart, since only one such point could be inside the square. So if the leftmost point is not on the left side of the square, we can increase its distance from the other points by moving it leftward. Likewise we can move the rightmost point to the right (if it's not already on the right side of the square), the topmost point higher, and the bottommost point lower. So we can assume the four points are on the four sides of the square, with one point on each of the four sides.
One of the four points must be closest to a corner of the square. By rotation and/or reflection, we can assume it's the point on the bottom side and it's closest to the bottom right corner. But a simple picture and the Pythagorean theorem shows that the point on the right side of the square must be closer to the upper right corner of the square than the point on the bottom is to the bottom right corner, and that's a contradiction. Thus for any four points in a $10times10$ square, some pair of them are no more than $10$ units apart.
answered Dec 18 '18 at 13:37
Barry CipraBarry Cipra
60.5k655128
answered Dec 18 '18 at 13:37
Barry CipraBarry Cipra
60.5k655128
answered Dec 18 '18 at 13:37
Barry CipraBarry Cipra
60.5k655128
answered Dec 18 '18 at 13:37
Barry CipraBarry Cipra
60.5k655128
60.5k655128
add a comment |
add a comment |
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$begingroup$
If anyone could just show me a configuration where the points are all more than $10$ apart, that would be cool. Because I have no idea what that would look like.
$endgroup$
– Arthur
Dec 18 '18 at 12:15