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Equal areas of segments in the lazy caterer problem?
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraWhat's wrong with this solution of Tarski's circle-squaring problem?square cake with raisinscutting a cake without destroying the toppingsAbout cutting AlmondsCut the rectangleLazy Caterer's Problem: why a new line can cut all the othersTool for the partition problem with planar rectanglesThe reverse pizza problem .Is there a way to determine the cheapest way to cut a line if each cut costs the current length of a line?Cutting a pie into n equal pieces with fewest straight cuts
$begingroup$
In the book "Build Your Brain Power" by Wootton and Horne, they mention the lazy caterer's problem, asking for a way to cut a circular cake into 8 equally sized pieces with 3 cuts. Clearly since the maximum number of possible segments is 7 for the $n=3$ case, that is impossible.
But, I was nonetheless wondering about this: is it possible, for any $ngeq 3$ (3 cuts or more) case of the lazy caterer's problem, for each of the resulting areas to be equal? How would one prove or disprove this?
geometry coordinate-systems
asked Dec 19 '15 at 4:06
James HarrisonJames Harrison
834615
$endgroup$
add a comment |
$begingroup$
In the book "Build Your Brain Power" by Wootton and Horne, they mention the lazy caterer's problem, asking for a way to cut a circular cake into 8 equally sized pieces with 3 cuts. Clearly since the maximum number of possible segments is 7 for the $n=3$ case, that is impossible.
But, I was nonetheless wondering about this: is it possible, for any $ngeq 3$ (3 cuts or more) case of the lazy caterer's problem, for each of the resulting areas to be equal? How would one prove or disprove this?
geometry coordinate-systems
asked Dec 19 '15 at 4:06
James HarrisonJames Harrison
834615
$endgroup$
$begingroup$
Lazy, but fair.
$endgroup$
– Eric Tressler
Dec 19 '15 at 4:11
add a comment |
$begingroup$
In the book "Build Your Brain Power" by Wootton and Horne, they mention the lazy caterer's problem, asking for a way to cut a circular cake into 8 equally sized pieces with 3 cuts. Clearly since the maximum number of possible segments is 7 for the $n=3$ case, that is impossible.
But, I was nonetheless wondering about this: is it possible, for any $ngeq 3$ (3 cuts or more) case of the lazy caterer's problem, for each of the resulting areas to be equal? How would one prove or disprove this?
geometry coordinate-systems
asked Dec 19 '15 at 4:06
James HarrisonJames Harrison
834615
$endgroup$
In the book "Build Your Brain Power" by Wootton and Horne, they mention the lazy caterer's problem, asking for a way to cut a circular cake into 8 equally sized pieces with 3 cuts. Clearly since the maximum number of possible segments is 7 for the $n=3$ case, that is impossible.
But, I was nonetheless wondering about this: is it possible, for any $ngeq 3$ (3 cuts or more) case of the lazy caterer's problem, for each of the resulting areas to be equal? How would one prove or disprove this?
geometry coordinate-systems
geometry coordinate-systems
asked Dec 19 '15 at 4:06
James HarrisonJames Harrison
834615
asked Dec 19 '15 at 4:06
James HarrisonJames Harrison
834615
edited Dec 19 '15 at 5:55
James Harrison
asked Dec 19 '15 at 4:06
James HarrisonJames Harrison
834615
asked Dec 19 '15 at 4:06
James HarrisonJames Harrison
834615
asked Dec 19 '15 at 4:06
James HarrisonJames Harrison
834615
834615
$begingroup$
Lazy, but fair.
$endgroup$
– Eric Tressler
Dec 19 '15 at 4:11
add a comment |
$begingroup$
Lazy, but fair.
$endgroup$
– Eric Tressler
Dec 19 '15 at 4:11
$begingroup$
Lazy, but fair.
$endgroup$
– Eric Tressler
Dec 19 '15 at 4:11
$begingroup$
Lazy, but fair.
$endgroup$
– Eric Tressler
Dec 19 '15 at 4:11
add a comment |
1 Answer
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oldest
votes
$begingroup$
Well, it's not possible for three cuts. Each cut has 3 pieces on one side of it, and 4 on the other, so it must cut off 3/7 of the circle on the side away from the center. That allows you to solve for the distance of the cut from the center of the circle; it turns out to be about 1/9 of the radius. Since all three cuts must be the same distance from the center, and since the three pieces between the acute angle formed by a pair of cuts and the circle must all have the same area, the acute angle between any pair of cuts must be the same, and so must be $tau/6$ ($tau$ being the angle measure of a full circle). So the central piece must be an equilateral triangle centered on the center of the circle, with height roughly 1/3 the radius (3 times the distance from the center to each side). But unfortunately the area of such an equilateral triangle is much less than 1/7 the area of the circle.
Similarly, 4 cuts won't work: 2 of the lines cut off segments with area of 5/11 of the circle, and 2 cut off segments with area 4/11. Again, you can compute the distance each line must be from the center of the circle, and guarantee that there will be pieces near the center that will be too small.
Presumably, the situation just worsens from there, and there will always be too-small slivers near the center, but this presumption falls short of a proof that there is no equal-area lazy caterer dissection for any $n$. But I'd be astonished if there were.
answered Mar 31 at 6:59
Glen WhitneyGlen Whitney
471311
$endgroup$
add a comment |
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$begingroup$
Well, it's not possible for three cuts. Each cut has 3 pieces on one side of it, and 4 on the other, so it must cut off 3/7 of the circle on the side away from the center. That allows you to solve for the distance of the cut from the center of the circle; it turns out to be about 1/9 of the radius. Since all three cuts must be the same distance from the center, and since the three pieces between the acute angle formed by a pair of cuts and the circle must all have the same area, the acute angle between any pair of cuts must be the same, and so must be $tau/6$ ($tau$ being the angle measure of a full circle). So the central piece must be an equilateral triangle centered on the center of the circle, with height roughly 1/3 the radius (3 times the distance from the center to each side). But unfortunately the area of such an equilateral triangle is much less than 1/7 the area of the circle.
Similarly, 4 cuts won't work: 2 of the lines cut off segments with area of 5/11 of the circle, and 2 cut off segments with area 4/11. Again, you can compute the distance each line must be from the center of the circle, and guarantee that there will be pieces near the center that will be too small.
Presumably, the situation just worsens from there, and there will always be too-small slivers near the center, but this presumption falls short of a proof that there is no equal-area lazy caterer dissection for any $n$. But I'd be astonished if there were.
answered Mar 31 at 6:59
Glen WhitneyGlen Whitney
471311
$endgroup$
add a comment |
$begingroup$
Well, it's not possible for three cuts. Each cut has 3 pieces on one side of it, and 4 on the other, so it must cut off 3/7 of the circle on the side away from the center. That allows you to solve for the distance of the cut from the center of the circle; it turns out to be about 1/9 of the radius. Since all three cuts must be the same distance from the center, and since the three pieces between the acute angle formed by a pair of cuts and the circle must all have the same area, the acute angle between any pair of cuts must be the same, and so must be $tau/6$ ($tau$ being the angle measure of a full circle). So the central piece must be an equilateral triangle centered on the center of the circle, with height roughly 1/3 the radius (3 times the distance from the center to each side). But unfortunately the area of such an equilateral triangle is much less than 1/7 the area of the circle.
Similarly, 4 cuts won't work: 2 of the lines cut off segments with area of 5/11 of the circle, and 2 cut off segments with area 4/11. Again, you can compute the distance each line must be from the center of the circle, and guarantee that there will be pieces near the center that will be too small.
Presumably, the situation just worsens from there, and there will always be too-small slivers near the center, but this presumption falls short of a proof that there is no equal-area lazy caterer dissection for any $n$. But I'd be astonished if there were.
answered Mar 31 at 6:59
Glen WhitneyGlen Whitney
471311
$endgroup$
add a comment |
$begingroup$
Well, it's not possible for three cuts. Each cut has 3 pieces on one side of it, and 4 on the other, so it must cut off 3/7 of the circle on the side away from the center. That allows you to solve for the distance of the cut from the center of the circle; it turns out to be about 1/9 of the radius. Since all three cuts must be the same distance from the center, and since the three pieces between the acute angle formed by a pair of cuts and the circle must all have the same area, the acute angle between any pair of cuts must be the same, and so must be $tau/6$ ($tau$ being the angle measure of a full circle). So the central piece must be an equilateral triangle centered on the center of the circle, with height roughly 1/3 the radius (3 times the distance from the center to each side). But unfortunately the area of such an equilateral triangle is much less than 1/7 the area of the circle.
Similarly, 4 cuts won't work: 2 of the lines cut off segments with area of 5/11 of the circle, and 2 cut off segments with area 4/11. Again, you can compute the distance each line must be from the center of the circle, and guarantee that there will be pieces near the center that will be too small.
Presumably, the situation just worsens from there, and there will always be too-small slivers near the center, but this presumption falls short of a proof that there is no equal-area lazy caterer dissection for any $n$. But I'd be astonished if there were.
answered Mar 31 at 6:59
Glen WhitneyGlen Whitney
471311
$endgroup$
Well, it's not possible for three cuts. Each cut has 3 pieces on one side of it, and 4 on the other, so it must cut off 3/7 of the circle on the side away from the center. That allows you to solve for the distance of the cut from the center of the circle; it turns out to be about 1/9 of the radius. Since all three cuts must be the same distance from the center, and since the three pieces between the acute angle formed by a pair of cuts and the circle must all have the same area, the acute angle between any pair of cuts must be the same, and so must be $tau/6$ ($tau$ being the angle measure of a full circle). So the central piece must be an equilateral triangle centered on the center of the circle, with height roughly 1/3 the radius (3 times the distance from the center to each side). But unfortunately the area of such an equilateral triangle is much less than 1/7 the area of the circle.
Similarly, 4 cuts won't work: 2 of the lines cut off segments with area of 5/11 of the circle, and 2 cut off segments with area 4/11. Again, you can compute the distance each line must be from the center of the circle, and guarantee that there will be pieces near the center that will be too small.
Presumably, the situation just worsens from there, and there will always be too-small slivers near the center, but this presumption falls short of a proof that there is no equal-area lazy caterer dissection for any $n$. But I'd be astonished if there were.
answered Mar 31 at 6:59
Glen WhitneyGlen Whitney
471311
answered Mar 31 at 6:59
Glen WhitneyGlen Whitney
471311
answered Mar 31 at 6:59
Glen WhitneyGlen Whitney
471311
answered Mar 31 at 6:59
Glen WhitneyGlen Whitney
471311
471311
add a comment |
add a comment |
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$begingroup$
Lazy, but fair.
$endgroup$
– Eric Tressler
Dec 19 '15 at 4:11