Search: concept:triangle
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Displaying 31-39 of 39 results found.
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BP897 |
| Wide angles connected to narrow angles vs. not so. |
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BP898 |
| Can fold into tetragonal disphenoid ("isosceles tetrahedron") vs. cannot. |
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COMMENTS
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Which two sides are the long sides and which side is the short side, or equivalently which angles are the wider angles and which angle is the narrower angle, is the only relevant information to consider for each triangle. Triangles are all assumed isosceles and congruent to one another.
All examples in this Problem feature four of these triangles connected by corners and/or edges. |
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CROSSREFS
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BP897 was conceived as a false solution for this.
Adjacent-numbered pages:
BP893 BP894 BP895 BP896 BP897  *  BP899 BP900 BP901 BP902 BP903
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KEYWORD
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hard, precise, allsorted, notso, math, preciseworld
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CONCEPT
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triangle (info | search)
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WORLD
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[smaller | same | bigger]
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AUTHOR
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Molly C Klenzak
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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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BP969 |
| Triangle is smallest black shape vs. square is smallest black shape. |
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BP970 |
| Triangle is largest black shape vs. circle is largest black shape. |
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BP1201 |
| The only triangles are the individual regions and the whole vs. there is some other triangle made of triangles. |
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BP1208 |
| More triangles left than right vs. more triangles right than left. |
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BP1211 |
| Triangle vs. anything else. |
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BP1260 |
| Same transformation applied to circle, triangle, and square vs. different transformations applied. |
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CROSSREFS
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BP839 is about applying opposite transformations to a single object.
Adjacent-numbered pages:
BP1255 BP1256 BP1257 BP1258 BP1259  *  BP1261 BP1262 BP1263 BP1264 BP1265
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KEYWORD
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easy, nice, abstract, arbitrary, anticomputer, left-null, structure, orderedtriplet, traditional, rules
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CONCEPT
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circle (info | search), analogy (info | search), square (info | search), same (info | search), triangle (info | search), function (info | search)
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AUTHOR
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Aaron David Fairbanks
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