Mixing
Given a line segment, a line parallel to it, and a straightedge, divide the segment into $n$ equal segments
$begingroup$
Given a line segment, a line parallel to it, and a straightedge, how to divide the segment into $n$ equal segments?
(With a straightedge, you are allowed only to draw straight lines. You are not allowed to mark off distances on the straightedge.)
geometry
edited Feb 1 at 20:20
Blue
49.6k870158
asked Feb 1 at 20:12
DHRUV JOSHIDHRUV JOSHI
11
$endgroup$
add a comment |
$begingroup$
Given a line segment, a line parallel to it, and a straightedge, how to divide the segment into $n$ equal segments?
(With a straightedge, you are allowed only to draw straight lines. You are not allowed to mark off distances on the straightedge.)
geometry
edited Feb 1 at 20:20
Blue
49.6k870158
asked Feb 1 at 20:12
DHRUV JOSHIDHRUV JOSHI
11
$endgroup$
1
$begingroup$
You can't. Not without a compass. And if you have a compass ... you don't need the line parallel to it. Where did you get this problem? And what exactly does the question ask?
$endgroup$
– fleablood
Feb 1 at 20:55
$begingroup$
If you have a compass then.... hint. You can extend the parallel line so that it in $n$ times as long and you can mark of $n$ tick marks on the long parallel line so that parallel line is divided evenly into $n$ equal parts (each part as long as the original line).
$endgroup$
– fleablood
Feb 1 at 21:00
add a comment |
$begingroup$
Given a line segment, a line parallel to it, and a straightedge, how to divide the segment into $n$ equal segments?
(With a straightedge, you are allowed only to draw straight lines. You are not allowed to mark off distances on the straightedge.)
geometry
edited Feb 1 at 20:20
Blue
49.6k870158
asked Feb 1 at 20:12
DHRUV JOSHIDHRUV JOSHI
11
$endgroup$
Given a line segment, a line parallel to it, and a straightedge, how to divide the segment into $n$ equal segments?
(With a straightedge, you are allowed only to draw straight lines. You are not allowed to mark off distances on the straightedge.)
geometry
geometry
edited Feb 1 at 20:20
Blue
49.6k870158
asked Feb 1 at 20:12
DHRUV JOSHIDHRUV JOSHI
11
edited Feb 1 at 20:20
Blue
49.6k870158
asked Feb 1 at 20:12
DHRUV JOSHIDHRUV JOSHI
11
edited Feb 1 at 20:20
Blue
49.6k870158
edited Feb 1 at 20:20
Blue
49.6k870158
edited Feb 1 at 20:20
Blue
49.6k870158
49.6k870158
asked Feb 1 at 20:12
DHRUV JOSHIDHRUV JOSHI
11
asked Feb 1 at 20:12
DHRUV JOSHIDHRUV JOSHI
11
asked Feb 1 at 20:12
DHRUV JOSHIDHRUV JOSHI
11
11
1
$begingroup$
You can't. Not without a compass. And if you have a compass ... you don't need the line parallel to it. Where did you get this problem? And what exactly does the question ask?
$endgroup$
– fleablood
Feb 1 at 20:55
$begingroup$
If you have a compass then.... hint. You can extend the parallel line so that it in $n$ times as long and you can mark of $n$ tick marks on the long parallel line so that parallel line is divided evenly into $n$ equal parts (each part as long as the original line).
$endgroup$
– fleablood
Feb 1 at 21:00
add a comment |
1
$begingroup$
You can't. Not without a compass. And if you have a compass ... you don't need the line parallel to it. Where did you get this problem? And what exactly does the question ask?
$endgroup$
– fleablood
Feb 1 at 20:55
$begingroup$
If you have a compass then.... hint. You can extend the parallel line so that it in $n$ times as long and you can mark of $n$ tick marks on the long parallel line so that parallel line is divided evenly into $n$ equal parts (each part as long as the original line).
$endgroup$
– fleablood
Feb 1 at 21:00
1
1
$begingroup$
You can't. Not without a compass. And if you have a compass ... you don't need the line parallel to it. Where did you get this problem? And what exactly does the question ask?
$endgroup$
– fleablood
Feb 1 at 20:55
$begingroup$
You can't. Not without a compass. And if you have a compass ... you don't need the line parallel to it. Where did you get this problem? And what exactly does the question ask?
$endgroup$
– fleablood
Feb 1 at 20:55
$begingroup$
If you have a compass then.... hint. You can extend the parallel line so that it in $n$ times as long and you can mark of $n$ tick marks on the long parallel line so that parallel line is divided evenly into $n$ equal parts (each part as long as the original line).
$endgroup$
– fleablood
Feb 1 at 21:00
$begingroup$
If you have a compass then.... hint. You can extend the parallel line so that it in $n$ times as long and you can mark of $n$ tick marks on the long parallel line so that parallel line is divided evenly into $n$ equal parts (each part as long as the original line).
$endgroup$
– fleablood
Feb 1 at 21:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Partial Answer
Here's a solution for the case $n = 2$, which might provide you some inspiration for the more general case. The picture tells the whole story. The red line $PQ$ is the initial segment, the blue line is the parallel one. We pick points $A$ and $B$ on the blue line at random. We join each of $A$ and $B$ to each end of the initial segment. The lines $PA$ and $QB$ intersect at the magenta point; the lines $PB$ and $QA$ intersect at the blue point, and joining the blue and magenta points and taking the intersection with the original line $PQ$ gives us the red point, which is the midpoint of $PQ$.
Note that I did need to assume I could pick two distinct points on a given line; I'm not certain that this is allowed in straightedge-only constructions.
Clearly this solution generalizes to handle all cases where $n$ is a power of $2$ (just apply recursively to sub-lines), but I don't see how to do $n = 3$; presumably once that's clear, the rest is downwind sailing.
answered Feb 1 at 22:04
John HughesJohn Hughes
65.4k24293
$endgroup$
add a comment |
$begingroup$
Hint. Assuming you are allowed a compass...
It might seem hard to divide the given line $n$ times but you can take the parallel line; extend it $n$ times to get a parallel line that is $n$ times longer than it was originally and you have successfully divided that big new line in $n$ equal parts.
Is there any way to take those $n$ parts an the parallel line to make $n$ equal parts of the original line?
What do you know about parallel lines, and equal angles, and similar triangles?
Time to brainstorm....
......
answered Feb 1 at 21:04
fleabloodfleablood
1
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Partial Answer
Here's a solution for the case $n = 2$, which might provide you some inspiration for the more general case. The picture tells the whole story. The red line $PQ$ is the initial segment, the blue line is the parallel one. We pick points $A$ and $B$ on the blue line at random. We join each of $A$ and $B$ to each end of the initial segment. The lines $PA$ and $QB$ intersect at the magenta point; the lines $PB$ and $QA$ intersect at the blue point, and joining the blue and magenta points and taking the intersection with the original line $PQ$ gives us the red point, which is the midpoint of $PQ$.
Note that I did need to assume I could pick two distinct points on a given line; I'm not certain that this is allowed in straightedge-only constructions.
Clearly this solution generalizes to handle all cases where $n$ is a power of $2$ (just apply recursively to sub-lines), but I don't see how to do $n = 3$; presumably once that's clear, the rest is downwind sailing.
answered Feb 1 at 22:04
John HughesJohn Hughes
65.4k24293
$endgroup$
add a comment |
$begingroup$
Partial Answer
Here's a solution for the case $n = 2$, which might provide you some inspiration for the more general case. The picture tells the whole story. The red line $PQ$ is the initial segment, the blue line is the parallel one. We pick points $A$ and $B$ on the blue line at random. We join each of $A$ and $B$ to each end of the initial segment. The lines $PA$ and $QB$ intersect at the magenta point; the lines $PB$ and $QA$ intersect at the blue point, and joining the blue and magenta points and taking the intersection with the original line $PQ$ gives us the red point, which is the midpoint of $PQ$.
Note that I did need to assume I could pick two distinct points on a given line; I'm not certain that this is allowed in straightedge-only constructions.
Clearly this solution generalizes to handle all cases where $n$ is a power of $2$ (just apply recursively to sub-lines), but I don't see how to do $n = 3$; presumably once that's clear, the rest is downwind sailing.
answered Feb 1 at 22:04
John HughesJohn Hughes
65.4k24293
$endgroup$
add a comment |
$begingroup$
Partial Answer
Here's a solution for the case $n = 2$, which might provide you some inspiration for the more general case. The picture tells the whole story. The red line $PQ$ is the initial segment, the blue line is the parallel one. We pick points $A$ and $B$ on the blue line at random. We join each of $A$ and $B$ to each end of the initial segment. The lines $PA$ and $QB$ intersect at the magenta point; the lines $PB$ and $QA$ intersect at the blue point, and joining the blue and magenta points and taking the intersection with the original line $PQ$ gives us the red point, which is the midpoint of $PQ$.
Note that I did need to assume I could pick two distinct points on a given line; I'm not certain that this is allowed in straightedge-only constructions.
Clearly this solution generalizes to handle all cases where $n$ is a power of $2$ (just apply recursively to sub-lines), but I don't see how to do $n = 3$; presumably once that's clear, the rest is downwind sailing.
answered Feb 1 at 22:04
John HughesJohn Hughes
65.4k24293
$endgroup$
Partial Answer
Here's a solution for the case $n = 2$, which might provide you some inspiration for the more general case. The picture tells the whole story. The red line $PQ$ is the initial segment, the blue line is the parallel one. We pick points $A$ and $B$ on the blue line at random. We join each of $A$ and $B$ to each end of the initial segment. The lines $PA$ and $QB$ intersect at the magenta point; the lines $PB$ and $QA$ intersect at the blue point, and joining the blue and magenta points and taking the intersection with the original line $PQ$ gives us the red point, which is the midpoint of $PQ$.
Note that I did need to assume I could pick two distinct points on a given line; I'm not certain that this is allowed in straightedge-only constructions.
Clearly this solution generalizes to handle all cases where $n$ is a power of $2$ (just apply recursively to sub-lines), but I don't see how to do $n = 3$; presumably once that's clear, the rest is downwind sailing.
answered Feb 1 at 22:04
John HughesJohn Hughes
65.4k24293
answered Feb 1 at 22:04
John HughesJohn Hughes
65.4k24293
answered Feb 1 at 22:04
John HughesJohn Hughes
65.4k24293
answered Feb 1 at 22:04
John HughesJohn Hughes
65.4k24293
65.4k24293
add a comment |
add a comment |
$begingroup$
Hint. Assuming you are allowed a compass...
It might seem hard to divide the given line $n$ times but you can take the parallel line; extend it $n$ times to get a parallel line that is $n$ times longer than it was originally and you have successfully divided that big new line in $n$ equal parts.
Is there any way to take those $n$ parts an the parallel line to make $n$ equal parts of the original line?
What do you know about parallel lines, and equal angles, and similar triangles?
Time to brainstorm....
......
answered Feb 1 at 21:04
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
Hint. Assuming you are allowed a compass...
It might seem hard to divide the given line $n$ times but you can take the parallel line; extend it $n$ times to get a parallel line that is $n$ times longer than it was originally and you have successfully divided that big new line in $n$ equal parts.
Is there any way to take those $n$ parts an the parallel line to make $n$ equal parts of the original line?
What do you know about parallel lines, and equal angles, and similar triangles?
Time to brainstorm....
......
answered Feb 1 at 21:04
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
Hint. Assuming you are allowed a compass...
It might seem hard to divide the given line $n$ times but you can take the parallel line; extend it $n$ times to get a parallel line that is $n$ times longer than it was originally and you have successfully divided that big new line in $n$ equal parts.
Is there any way to take those $n$ parts an the parallel line to make $n$ equal parts of the original line?
What do you know about parallel lines, and equal angles, and similar triangles?
Time to brainstorm....
......
answered Feb 1 at 21:04
fleabloodfleablood
1
$endgroup$
Hint. Assuming you are allowed a compass...
It might seem hard to divide the given line $n$ times but you can take the parallel line; extend it $n$ times to get a parallel line that is $n$ times longer than it was originally and you have successfully divided that big new line in $n$ equal parts.
Is there any way to take those $n$ parts an the parallel line to make $n$ equal parts of the original line?
What do you know about parallel lines, and equal angles, and similar triangles?
Time to brainstorm....
......
answered Feb 1 at 21:04
fleabloodfleablood
1
edited Feb 1 at 21:17
answered Feb 1 at 21:04
fleabloodfleablood
1
answered Feb 1 at 21:04
fleabloodfleablood
1
answered Feb 1 at 21:04
fleabloodfleablood
1
1
add a comment |
add a comment |
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$begingroup$
You can't. Not without a compass. And if you have a compass ... you don't need the line parallel to it. Where did you get this problem? And what exactly does the question ask?
$endgroup$
– fleablood
Feb 1 at 20:55
$begingroup$
If you have a compass then.... hint. You can extend the parallel line so that it in $n$ times as long and you can mark of $n$ tick marks on the long parallel line so that parallel line is divided evenly into $n$ equal parts (each part as long as the original line).
$endgroup$
– fleablood
Feb 1 at 21:00