Search: -meta:BP908
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| BP39 |
| Segments approximately parallel to each other vs. large angles between segments. |
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REFERENCE
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M. M. Bongard, Pattern Recognition, Spartan Books, 1970, p. 226. |
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CROSSREFS
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Adjacent-numbered pages:
BP34 BP35 BP36 BP37 BP38 * BP40 BP41 BP42 BP43 BP44
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KEYWORD
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finished, unorderedtriplet, traditional, bongard
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CONCEPT
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all (info | search),
line_slope (info | search),
same (info | search),
similar (info | search),
three (info | search),
parallel (info | search)
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WORLD
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three_segments [smaller | same | bigger]
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AUTHOR
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Mikhail M. Bongard
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| BP78 |
| Extensions of segments cross at one point vs. extensions of segments do not cross at one point. |
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REFERENCE
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M. M. Bongard, Pattern Recognition, Spartan Books, 1970, p. 239. |
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CROSSREFS
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Adjacent-numbered pages:
BP73 BP74 BP75 BP76 BP77 * BP79 BP80 BP81 BP82 BP83
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KEYWORD
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finished, unorderedtriplet, traditional, bongard
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CONCEPT
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all (info | search),
lines_coincide (info | search),
imagined_point (info | search),
imagined_line_or_curve (info | search),
imagined_entity (info | search)
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WORLD
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three_segments [smaller | same | bigger]
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AUTHOR
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Mikhail M. Bongard
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| BP161 |
| Midpoints are collinear vs. midpoints are not collinear. |
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| BP907 |
| One dot cluster is the product of the other two vs. not so. |
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| BP934 |
| If "distance" is taken to be the sum of horizontal and vertical distances between points, the 3 points are equidistant from each other vs. not so. |
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COMMENTS
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In other words, we take the distance between points (a,b) and (c,d) to be equal to |c-a| + |d-b|, or, in other words, the distance of the shortest path between points that travels along grid lines. In mathematics, this way of measuring distance is called the 'taxicab' or 'Manhattan' metric. The points on the left hand side form equilateral triangles in this metric.
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An alternate (albeit more convoluted) solution that someone may arrive at for this Problem is as follows: The triangles formed by the points on the left have some two points diagonal to each other (in the sense of bishops in chess), and considering the corresponding edge as their base, they also have an equal height. However, this was proven to be equivalent to the Manhattan distance answer by Sridhar Ramesh. Here is the proof:
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An equilateral triangle amounts to points A, B, and C such that B and C lie on a circle of some radius centered at A, and the chord from B to C is as long as this radius.
A Manhattan circle of radius R is a turned square, ♢, where the Manhattan distance between any two points on opposite sides is 2R, and the Manhattan distance between any two points on adjacent sides is the larger distance from one of those points to the corner connecting those sides. Thus, to get two of these points to have Manhattan distance R, one of them must be a midpoint of one side of the ♢ (thus, bishop-diagonal from its center) and the other can then be any point on an adjacent side of the ♢ making an acute triangle with the aforementioned midpoint and center. |
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CROSSREFS
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Adjacent-numbered pages:
BP929 BP930 BP931 BP932 BP933 * BP935 BP936 BP937 BP938 BP939
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KEYWORD
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hard, allsorted, solved, left-finite, right-finite, perfect, pixelperfect, unorderedtriplet, finishedexamples
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CONCEPT
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triangle (info | search)
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WORLD
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3_dots_on_square_grid [smaller | same | bigger]
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AUTHOR
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Leo Crabbe
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