All the triangles can stick together. The triangles counted is the independent triangles, triangles made up of two shapes, a triangle made up from 3 shapes, or the outline of the shape consisting of 4, 5 or 6 lines.
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$\begingroup$ In Euclidean geometry? $\endgroup$– Peter TaylorCommented Jan 6, 2019 at 9:11
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1 Answer
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I seem to recall that the solution to this problem for $n$ lines is
$n \choose 3$, so for 4 lines it's 4, for 5 lines it's 10 and for 6 lines it's 20.
The idea is that
each combination of 3 lines generates exactly one triangle. They intersect in at most 3 points, which form the triangle. Actually, as long as you don't choose parallel lines, they will intersect in exactly 3 points. There are $n \choose 3$ of these combinations.
For example, here is the solution for 6 lines:
The triangles are formed by:
ABC, ADE, AGI, AHJ
FGH, FIJ, FLN, FMO
KLM, KNO, KBD, KCE
BIO, BJM, CGO, CHM
DIN, DJL, EGN, EHL
answered Jan 5, 2019 at 21:27
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$\begingroup$ Very nice! I suppose you could just ignore 1 or 2 lines in the above diagram to get the optimal solution for 5 and 4 lines. $\endgroup$ Commented Jan 6, 2019 at 4:43