intersect one another at one point. Divide the number 8 25. Let us now suppose the lines to be divided into 3 sets. 1. The lines in each set parallel among themselves. In this case each line of one set can intersect each line of the other two sets. To find the greatest number of intersection points, the numbers which the sets contain must be multiplied together in pairs. If then the numbers in the 3 sets be 6, 2, and 2, we have (6×2)+(6×2)+(2×2)=28 points 2. The lines of the 1st and 2d sets parallel among themselves, and the lines of the 3d set intersecting one another at one point. In this case, the lines of the first two sets intersect one another, and the lines of the 3d set, besides intersecting all the lines of the other two, give one intersection point among themselves. If the numbers in the 3 sets be 8, 2, 2 we have (8×2)+[(8+2)×2]+1=37 points 3. The lines of each set intersecting among themselves at 1 point. In this case every line of each set intersects. every line of the other 2 sets, and there are 3 points besides. Therefore the numbers in the 3 sets must be multiplied together in pairs, and 3 be added to the product. Suppose the number in the 3 sets to be 4, 4, 4 6, 6, 6 2, 3, 4 66 3, 3, 3 we have (3 × 3) + (3 × 3) + (3 × 3) +3 = 30 (4 × 4) + (4 × 4) + (4 × 4) + 3= 51 (6 × 6)+(6 × 6) + (6 × 6)+3=111 (2 × 3) + (2 × 4)+(3 × 4)+3= 29 (3 × 4) + (3 × 5) + (4 × 5) +3= 50 4. The lines of the 1st set parallel to one another; those of the 2d set intersecting one another at 1 point; those of the 3d set intersecting one another at the greatest number of points. The calculation must be made accordingly. If the 3 sets are 5,5,5 we have (5×5)+1+[(5+5)×5]+(5X4)=86 ps. 6,6,6 (6×6)+1+[(6+6)×6]+(6x5)=1 =124" 8,8,8 (8x8)+1+[(8+8)×8]+(8x7) =221" (10×6)+1+[(6+10)×4]+(4x3)=131 « II. LINES. 26. Let us now calculate how many straight lines may be drawn between a given number of points, of which only two lie in the same direction. 1. Between 2 points, only 1 straight line can be drawn, and these determine the length and position of the line; a straight line, drawn from the 1st point to the 2d, will coincide with a straight line drawn from the 2d to the 1st. To coincide is to fall on, and exactly fill the same space. 2. Between 3 points. In this case we may have 3 distinct straight lines, viz., from the 1st point to the 2d, from the 2d to the 3d, from the 3d to the 1st. We can state it in this manner; from each of the 3 points straight lines may be drawn to the other 2 points; thus, from each point 2 lines; consequently, from the 3 points 3 × 2=6 lines. But each line is thus counted twice; for the straight line from the 1st point to the 2d is the same as that drawn from the 2d to the 1st; therefore the product 3 × 2 must be divided by 2. The number of straight lines that can be drawn between 3 points is 3 X 2 3. 3. Between 4 Points. Straight lines may be drawn from each of these 4 points to each of the other points; thus from each point 3 straight lines, and therefore from 4 points 4 x 312 straight lines. For the reason before given, this is twice the real number; the true num4 X 3 ber is 2 6. 4. Between five or more points. We may take any number of points, the process will be the same. Straight lines can be drawn from each point to every other point. Therefore from each point as many lines can be drawn as there are points less 1. We can now form the general rule. Multiply the whole number of points by the same number less 1, and divide the product by 2. The converse problem should be solved, namely, the number of straight lines which may be drawn between an unknown number of points being given, required the number of points. 27. Every part of a straight line which is bounded by two points in that line, is called an extent. Two or more contiguous extents form together a compound extent. We will now calculate the number of single extents between a given number of points in a straight line. Thus the number of single extents between any number of points in a straight line is one less than the number of points. We will now calculate the number of single and compound extents between a given number of points in a straight line. 2 points. Between 2 points lies 1. 3 points. From each point to every other point may lie one. Thus from 3 points there would be 3 × 2=6; but as each line goes out from 2 points, and has therefore been reckoned twice, the true number of extents is (3 × 2)=3. 4 or more points. From every point may proceed as many extents as there are points beside this one. There fore to find the number of single and compound extents in a straight line, multiply the given number of points by the same number less 1, and divide this product by 2. 28. What is the greatest number of diagonals which can be drawn in any polygon? Answer. 1. A triangle can have no diagonal, for the vertex of each angle is already connected with the vertices of each of the others. 2. A quadrilateral. One diagonal can be drawn from the vertex of each angle to the vertex of one of the others. But each diagonal connects 2 vertices, therefore there can be only two diagonals in a quadrilateral. 3. In a pentagon are 5 vertices; each is connected with 2 others by the bounding lines; therefore from each vertex only 2 diagonals can be drawn; from the 5 vertices 2 × 5=10; but this is double the real num2X5 ber, which is =5. 2 4. If we extend the examination to hexagons, we shall find that so many diagonals can be drawn from the vertex of each angle of a figure, as the figure has sides less 3; each vertex being already connected with 2 others by the bounding lines. Rule. To find the number of diagonals that may be drawn in a polygon, multiply the number of its sides or angles by the same number less 3, and divide this product by 2. |