SECTION III.

OF MAXIMA AND MINIMA.

915. DEF.-A Maximum value of a magnitude conceived to vary continuously in some specified way, is a value which is greater than the preceding and succeeding values of the magnitude.

abcde

ILL'S. Thus, suppose in a given circle, a chord passing through a fixed point, P, revolves so as to take successively the positions 1a, 2b, Ac, 3d, 4e, etc. It is at a maximum when it passes through the centre, as Ac. The chord is the magnitude which is conceived to vary in the way specified, and Ac is a value greater than the preceding and the succeeding values. Again, conceive a circle to be compressed or extended, as in the direction mn, so as to take the forms indicated by the dotted lines, its area will be diminished, the perimeter remaining the same. That is, of all figures of a given perimeter, the circle has the maxi

mum area.

916. DEF.-A Minimum value of a magnitude conceived to vary continuously in some specified way, is a value which is less than the preceding and succeeding values of the magnitude.

1

2 A 3 4

FIG. 471.

X

ILL'S. Thus, conceive the varying magnitude to be a straight line from the fixed point P to the fixed line X'X; that is, suppose such a line to start from some position P1, and move through the successive positions P2, PA, P3, P4. PA is a minimum, since it is less than the preceding and succeeding values.

PROPOSITIONS CONCERNING MAXIMA AND MINIMA.

917. Axiom.-The minimum distance between two points is a straight line.

918. Theo.-The minimum distance from a point to a line is a straight line perpendicular to the given line.

Student give proof.

919. Theo.-The maximum line which can be inscribed in a given circle is a diameter.

OF MAXIMA AND MINIMA.

303

Proof based on the fact that the hypotenuse of a right angled triangle is the greatest side.

920. Theo-The sum of the distances from two points on the same side of a line, to a point in the line, all being in the same plane, is a minimum when the lines measuring the distances make equal angles with the given line.

Student prove AP + BP < AP' + BP'.

921. Theo.--If a triangle have a constant base and altitude, its vertical angle is a maximum when the triangle is isosceles.

SUG. By what is the vertical angle measured?

922. Theo.-The base and area of a triangle being constant, its perimeter is a minimum when the triangle is isosceles.

SUG's. The area and base being constant, the vertex remains in a line parallel to the base, for all values of the other sides. The figure will suggest the demonstration, which is based on the fact that any side of a triangle is less than the sum of the other two.

FIG. 174.

923. Theo-The difference between the distances from two points on opposite sides of a fixed line to a point in that line, is a maximum, when the lines measuring these distances make equal angles with the fixed line.

SUG'S. P'O AP AP'; but P'O > A'O (= A'P) - A'P'.

QUERY.-Having the points P, P', and the fixed line given, how is the point A found by geometrical construction?

924. Theo-The lengths of two sides of a triangle being constant, the area is a maximum when the included angle is right.

FIG. 475.

F.G. 476.

925. Theo.-The sum of two adjacent sides of a rectangle being constant (AB), the area is a maximum when the sides are equal.

ISOPERIMETRY.

926. Isoperimetric Figures are such as have equal perimeters, i. e, bounding lines of equal length.

Problems in isoperimetry are a species of problems in Maxima and Minima. Thus, of all figures whose perimeters are m (say 10 inches), to find that which has the greatest area, is a problem in isoperimetry. Again, what must be the form of a pentagon whose perimeter is m, in order that its area may be a maximum?

927. Theo-Of isoperimetric triangles with a constant base, the isosceles is a maximum.

SUG's. By means of the figure to Theorem (921), we can readily show that any triangle having the same base as the isosceles triangle, and its vertex either in or beyond the line through the vertex of the isosceles triangle and parallel to its base, has a greater perimeter than the isosceles triangle. Hence, the isoperimetric triangle on the given base has its vertex below this parallel, except when isosceles; and consequently the isosceles is the maximum.

928. COR. Of isoperimetric triangles, the equilateral has the maximum area (?)

929. Prob.-Given any triangle with a constant base, to construct the maximum isoperimetric triangle.

930. Prob.-Given any triangle, to construct the maximum isoperimetric triangle.

Α

D

C

FIG. 478.

931. Theo-Of isoperimetric quadrilaterals, the square has the maximum area.

DEM.-Let ABCD be any quadrilateral. If AD is not B equal to DC, ADC can be replaced by the isosceles isoperimetric triangle AD'C, and the area of the quadrilateral increased. So ABC can be replaced by AB'C. Therefore AB'C'D > ABCD. In like manner if AD' is not equal to AB', D'AB' can be replaced by the maximum isoperimetric triangle D'A'B'. So also D'CB' can be replaced by D'C'B'. Therefore A'B'C'D' > AB'CD' > ABCD. Now, A'B'C'D' is a rhombus (?), and the student can show that the square on A'B' is greater than any rhombus with the same side.

932. Prob.--Having given a quadrilateral, to construct the maximum isoperimetric quadrilateral.

933. Theo-Of isoperimetric quadrilaterals with a constant base, the maximum has its three remaining sides equal each to each, and the angles which they include equal.

DEM.-Let ABCD be the maximum isoperimetric quadrilateral on the base AD, then AB BC = = CD, and angle ABC BCD. For, if AB is not equal to BC, draw AC, and replacing the triangle ABC with its isoperimetric isosceles triangle, we shall have a quadrilateral isoperimetric with ABCD, and greater than ABCD, i. e., greater than the maximum, which is absurd.

BC.

B

E

A

FIG. 479.

Again, if angle ABC is not equal to BCD, let ABC < BCD, whence BCE < EBC, and BE < EC. Take EF = EC, and EC EB, whence the triangles FEG and BEC are equal, and FG Also, since AB + BC + CD = AE + ED (EB + EC) + BC, and AF + FG+ GD AE + ED — (FE + EG) + FG, it follows that AFGD and ABCD are isoperimetrical, and, since ABCD AED - BEC, and AFGD = AED - FEG, that AFGD and ABCD are equal. Therefore, AFGD is a maximum, and by the preceding part of the demonstration AF FG BC AB, which is absurd; and there can be no inequality between angles ABC and BCD.

934. Theo-Of isoperimetric polygons of a given number of sides, the regular polygon has the maximum area.

DEM.-First, the polygon must be equilateral; for, if any two adjacent sides, as AB, BC, are unequal, the triangle ABC can be replaced by its isoperimetric isosceles triangle, and thus the area of the polygon be increased.

A

B

FIG. 480.

Second, the polygon must be equiangular; for, if any two adjacent angles, as B and C, are unequal, the quadrilateral ABCD can be replaced by its isoperimetric quadrilateral with B = C, and thus the area of the polygon be increased.

D

935. Theo-Of isoperimetric regular polygons, the one of the greater number of sides is the greater.

DEM.-Let ABC be an equilateral (regular) triangle. Join any vertex, as A, with any point, as D, in the opposite side. Replace the triangle ACD with the isosceles isoperimetric triangle AED. Then is the quadrilateral ABDE > the triangle ABC.

But, of isoperimetric quadrilaterals, the regular (the square) is the greater. Hence, the regular quadrilateral (the square) isoperimetric with the triangle ABC, is greater than

B

FIG. 481.

C

the triangle. In the same manner the regular pentagon isoperimetric with the square can be shown greater than the square; and thus on, ad libitum.

936. COR.-Of plane isoperimetric figures, the circle has the maximum area, since it is the limiting form of the regular polygon, as the number of its sides is indefinitely increased.

T

SECTION IV.

OF TRANSVERSALS.

937. DEF.-A Transversal is a line cutting a system of lines. A transversal of a triangle is a line cutting its sides; it either cuts two sides and the third side produced, or the three sides produced. In speaking of the transversal of a triangle (or polygon), the distances on any side (or side produced) from the intersection of the transversal with that side to the angles, are Segments. Of these there are six. Adjacent segments are such as have an extremity of each at the same point. Nonadjacent segments are such as have no extremity common.

B

FIG. 482.

B

ILL'S. TR is a transversal of the triangle ABC; aA, aC, ¿C, bB, cA, cв are adjacent segments two and two; aC, bB, cA, and aА, bC, cB

are the two groups of non-adjacent segments.

938. THE TWO FUNDAMENTAL PROPOSITIONS OF THE THEORY OF TRANSVERSALS.

939. Theo-The product of three non-adjacent segments of the sides of a triangle cut by a transversal, is equal to the product of the other three.

=

DEM.-ABC being cut by the transversal TR, aA × b°C × cB aC × bB × CA. Draw BD parallel to AC, and from the similar triangles we have

* Or, sides produced-this expression being usually omitted in higher Geometry as all lines are to be considered indefinite unless limited in the problem.