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A091908 Number of interior intersection points made by the straight line segments connecting the edges of an equilateral triangle with the n-1 points resulting from a subdivision of the sides into n equal pieces, counting coinciding intersection points only once. 12
0, 1, 12, 13, 48, 49, 108, 109, 192, 193, 300, 301, 432, 433, 576, 589, 768, 769, 972, 961, 1200, 1201, 1452, 1405, 1728, 1729, 2028, 2029, 2352, 2341, 2700, 2701, 3072, 3073, 3444, 3469, 3888, 3889, 4332, 4297, 4800, 4777, 5292, 5293, 5724, 5809, 6348 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

In a drawing the distinction between simple and multiple intersection points may be difficult due to near-coincidences. E.g. there are no coincident intersections for n=7.

Note that 3 divides a(2k)-1 and a(2k+1). - T. D. Noe, Jun 29 2005

The interior intersection points only can be the result of the concurrency of 2 or 3 segments by construction. It is easy to see that the total number of 2-intersections N2 is 3*(n-1)^2 (which includes every 3-intersection as two 2-intersections) by symmetry. But we are interested in excluding the concurrency of more than 2. By Ceva's theorem necessary and sufficient condition for 3 concurrent segments that connect the edges with the opposite side, the number of 3-intersections N3 is the same as the number of (i,j,k) belonging to [1,n-1]x[1,n-1]x[1,n-1] such that (i/(n-i))*(j/(n-j))*(k/(n-k))=1. Thus the terms a(n)=N2-2*N3. - Herman Jamke (hermanjamke(AT)fastmail.fm), Oct 26 2006

If n is even then a(n) < 3*(n-1)^2; if n is odd then a(n) = 3*(n-1)^2 except for n in A332378. - N. J. A. Sloane, Feb 14 2020

REFERENCES

Mohammad K. Azarian, A Trigonometric Characterization of  Equilateral Triangle, Problem 336, Mathematics and Computer Education, Vol. 31, No. 1, Winter 1997, p. 96.  Solution published in Vol. 32, No. 1, Winter 1998, pp. 84-85.

Mohammad K. Azarian, Equating Distances and Altitude in an Equilateral Triangle, Problem 316, Mathematics and Computer Education, Vol. 28, No. 3, Fall 1994, p. 337.  Solution published in Vol. 29, No. 3, Fall 1995, pp. 324-325.

LINKS

Hugo Pfoertner, Table of n, a(n) for n = 1..1000

Hugo Pfoertner, Visualization of diagonal intersections in an equilateral triangle.

Sequences formed by drawing all diagonals in regular polygon

EXAMPLE

a(3)=12 because the 3*2 line segments intersect each other in 12 distinct points (see pictures given at link)

a(4)=13 because the 27 intersections form 6 two line intersection points and 7 three line intersection points.

PROG

(PARI) for(n=1, 70, conc=0; for(i=1, n-1, for(j=1, n-1, for(k=1, n-1, if(i*j*k/((n-i)*(n-j)*(n-k))==1, conc++)))); print1(3*(n-1)^2-2*conc, ", ")) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), Oct 26 2006

CROSSREFS

Cf. A091910 = radial locations of intersection points, A092098 = number of regions that the line segments cut the triangle into, A006561.

For the basic properties of the underlying graph, see A092098 (cells), A331782 (vertices), A331782 (vertices), A332376 & A332377 (edges). - N. J. A. Sloane, Feb 14 2020

Sequence in context: A041298 A041689 A041296 * A329597 A041300 A041302

Adjacent sequences:  A091905 A091906 A091907 * A091909 A091910 A091911

KEYWORD

nonn

AUTHOR

Hugo Pfoertner, Feb 19 2004

EXTENSIONS

More terms from T. D. Noe, Jun 29 2005

More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), Oct 26 2006

STATUS

approved

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Last modified April 1 04:15 EDT 2020. Contains 333155 sequences. (Running on oeis4.)