38. What is the greatest number of convex angles which a rectilineal figure may have?

A convex angle, when it occurs in a figure, is called a re-entering angle, because the vertex is directed inwards; the concave angles are called salient angles. In all the rules and remarks, figures with salient angles only are intended, except the contrary is expressly stated.

A triangle can have no convex angle.

A quadrilateral can have 1 convex angle.

In general, in any rectilineal figure there may be, at the most, as many convex (or re-entering) angles as the figure has sides less 3.

In every rectilineal figure there must be at least 3 concave angles.

IV. FIGURES.

39. What is the greatest number of triangles which can be formed with a given number of straight lines ? 1. Three straight lines can form 1 triangle, neither more nor less.

2. Four straight lines may intersect one another at 6 points, (22.) In each line there would be 3 intersectionpoints, and between 3 points there may be 3 extents, (27.) Upon each extent there may be one triangle, and thus there would be as many triangles as there are extents. The number of extents is 3 × 4 = 12. But each triangle has 3 sides, and therefore we have reckoned each triangle 3 times in the above estimate. We must therefore divide the product 12 by 3 to obtain the real number of triangles. It is

12
= 4.
3

3. 5 straight lines. If 5 straight lines intersect one another at the greatest number of points, there will be in each straight line 4 intersection-points. Between 4 points 4×3 extents; therefore, in the 5 lines 2 5×4×3 2

there may be

there are

extents. Upon each of these ex

tents may be one triangle, and thus

5×4×3
2

triangles. But in this calculation each triangle is counted 3 times, 5×4×35×4×3 therefore the true number will beor 2×3

1×2×3

From these examples we may deduce the rule. To find the greatest number of triangles which may be formed by a given number of straight lines, multiply the number of the lines by the product of the two numbers preceding it in the natural order of the numbers, and divide the product thus obtained by 1 × 2 × 3.

7 straight lines can form at most

1×2×3

7×6×5

35 triangles.

40. What number of parallelograms can be formed by two sets of parallel lines crossing each other? (figs. 41 and 42.)

1. 3 parallels in one set, and 2 in the other set. There will be as many parallelograms as there are inter

vals between the 3 parallels; for all the intervals will be enclosed by the 2 intersecting parallels. Between the 3 parallels there will be as many intervals, single and compound, as there are extents between 3 points in a straight 3×2 line, that is 3. Consequently, in the supposed 2 case, 3 parallelograms are formed.

2. 4, 5, 6, or more parallels in one set, and 2 parallels in the other set. Find as before the number of intervals, single and compound, in each set, and multiply

2×1
2

them together. But 2 parallels give =1 interval;

we have therefore only to find the number of intervals in the other set, and it will be the required number of parallelograms.

4 parallels crossed by 2 parallels give

4×3

6 parallel'ms.

2
5×4
2
6×5
2

From these examples, the rule for finding the number of parallelograms formed by the intersection of any number of parallel lines by 2 parallel lines is apparent. Multiply the number of the lines of the former set by the same number less 1, and divide this product by 2.

3. Let there be any number of parallels in each set, and we have this rule for finding the number of parallelograms made by the intersection of the two sets. Multiply the number of intervals between the parallels of one set, by the number of intervals between the parallels of the other.

10 parallels crossed by 10 parallels give

10×910×9
X

2

2

2025 parallelograms.

6 parallels crossed by 4 parallels give

6×54×3 = 90 parallelograms.

2

2

12 parallels crossed by 30 parallels give

12 × 1130 × 29
X
2
2

= 28.710 parallelograms.

The rule may be thus expressed: Multiply the number of lines in each set taken separately by the same number less 1; take half of each product; multiply these half-products together, and this last product will be the number required.

41. We will now set down in a tabular form the number of parallelograms formed by intersecting sets of parallel lines. It may be done in the familiar manner of the multiplication table. Set down the numbers for one set in a horizontal series, and the numbers for the other set in a vertical series, and the number of parallelograms formed by the intersection of the 2 sets directly under the one, and opposite to the other.

The numbers of the first vertical series are the same as those of the first horizontal series; the numbers of the second vertical series are the same as those of the second horizontal series, and so on.

The differences between each two successive numbers

of the horizontal and of the vertical series are as follows:

Again, remark the difference between each two successive numbers of the horizontal and also of the vertical series in this table.

The horizontal series give these differences, viz. :

1st horizontal series 0 0

0

0

0 0

The vertical series give these differences, viz. :