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I am interested in the topic of simple polygons, such as triangles and rectangles, that can be drawn with all their verices on the perimeter of a smooth curve which does not intersect itself.
However, I do not know what this topic is called, and thus cannot find it on either Wikipedia or Google.
Your assistance is appreciated. Bh12 ( talk) 08:51, 31 October 2016 (UTC)
That's it - the Inscribed square problem is the key to what I am looking for. Thank you. Bh12 ( talk) 15:47, 31 October 2016 (UTC)
Convexity is not required. I am writing up the proof off-line, and hope to post it here before they finish the World Series. Bh12 ( talk) 15:47, 31 October 2016 (UTC)
Consider the smooth, continuous non-intersecting perimeter of a closed area.
I claim that every point on this perimeter is the vertex of at least one equilateral triangle, where the other two vertices are also on this perimeter.
Note: By “smooth” and “continuous” I mean to exclude perimeters that have cusps, that are tangential to themselves, that have abrupt changes in direction, or that extend to infinity, if they are incompatible with this theorem.
Consider a point A on the perimeter, where without loss of generality we have positioned the area such that, in the vicinity of point A, the perimeter is approximately horizontal and the inside of the area is above the perimeter.
Starting from point A, we consider a point B that initially moves to the right of point A along the perimeter. As point B traverses the perimeter, its movement relative to point A will of course change, but this is immaterial - the essential feature is that, along the perimeter, B continues to move away from A.
Consider a point C such that ABC describes an equilateral triangle, and such that we move from points A to B to C in a counterclockwise direction. Initially, as point B moves to the right of point A, this places point C “above” the perimeter and thus inside the area. Point C moves so as to maintain the equilateral triangle.
As point B finishes traversing the perimeter, it approaches point A from the left. Because the points ABC are counterclockwise, this places point C “below” the perimeter and thus outside the area.
So, initially point C is inside the area, but at the finish it is outside the area. Because the perimeter is smooth and continuous, this is possible only if somewhere along the perimeter, point C crosses it. Whereever point C touches or crosses the perimeter, this gives us an equilateral triangle all of whose vertices are on the perimeter. QED. Bh12 ( talk) 19:50, 31 October 2016 (UTC)
Some small changes were just made to the proof. Bh12 ( talk) 21:51, 31 October 2016 (UTC)
The perimeter has cusps, which is not allowed. Nonetheless, I believe that one can form an equilateral triangle by taking the specified point on the lower semicircle and either two correctly placed points on the upper semicircle or two correctly placed points on the left or right vertical lines.
Correct, by inscribed I mean only that the vertices are on the perimeter, not that the whole triangle is within the perimeter.
I have erased the part about inscribed squares, it leads to nowhere. Bh12 ( talk) 15:54, 1 November 2016 (UTC)
The proof also works if one considers only convex areas; that is, the entire triangle being inside the area. Bh12 ( talk) 21:15, 1 November 2016 (UTC)
Mathematics desk | ||
---|---|---|
< October 30 | << Sep | October | Nov >> | November 1 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
Help!
I am interested in the topic of simple polygons, such as triangles and rectangles, that can be drawn with all their verices on the perimeter of a smooth curve which does not intersect itself.
However, I do not know what this topic is called, and thus cannot find it on either Wikipedia or Google.
Your assistance is appreciated. Bh12 ( talk) 08:51, 31 October 2016 (UTC)
That's it - the Inscribed square problem is the key to what I am looking for. Thank you. Bh12 ( talk) 15:47, 31 October 2016 (UTC)
Convexity is not required. I am writing up the proof off-line, and hope to post it here before they finish the World Series. Bh12 ( talk) 15:47, 31 October 2016 (UTC)
Consider the smooth, continuous non-intersecting perimeter of a closed area.
I claim that every point on this perimeter is the vertex of at least one equilateral triangle, where the other two vertices are also on this perimeter.
Note: By “smooth” and “continuous” I mean to exclude perimeters that have cusps, that are tangential to themselves, that have abrupt changes in direction, or that extend to infinity, if they are incompatible with this theorem.
Consider a point A on the perimeter, where without loss of generality we have positioned the area such that, in the vicinity of point A, the perimeter is approximately horizontal and the inside of the area is above the perimeter.
Starting from point A, we consider a point B that initially moves to the right of point A along the perimeter. As point B traverses the perimeter, its movement relative to point A will of course change, but this is immaterial - the essential feature is that, along the perimeter, B continues to move away from A.
Consider a point C such that ABC describes an equilateral triangle, and such that we move from points A to B to C in a counterclockwise direction. Initially, as point B moves to the right of point A, this places point C “above” the perimeter and thus inside the area. Point C moves so as to maintain the equilateral triangle.
As point B finishes traversing the perimeter, it approaches point A from the left. Because the points ABC are counterclockwise, this places point C “below” the perimeter and thus outside the area.
So, initially point C is inside the area, but at the finish it is outside the area. Because the perimeter is smooth and continuous, this is possible only if somewhere along the perimeter, point C crosses it. Whereever point C touches or crosses the perimeter, this gives us an equilateral triangle all of whose vertices are on the perimeter. QED. Bh12 ( talk) 19:50, 31 October 2016 (UTC)
Some small changes were just made to the proof. Bh12 ( talk) 21:51, 31 October 2016 (UTC)
The perimeter has cusps, which is not allowed. Nonetheless, I believe that one can form an equilateral triangle by taking the specified point on the lower semicircle and either two correctly placed points on the upper semicircle or two correctly placed points on the left or right vertical lines.
Correct, by inscribed I mean only that the vertices are on the perimeter, not that the whole triangle is within the perimeter.
I have erased the part about inscribed squares, it leads to nowhere. Bh12 ( talk) 15:54, 1 November 2016 (UTC)
The proof also works if one considers only convex areas; that is, the entire triangle being inside the area. Bh12 ( talk) 21:15, 1 November 2016 (UTC)