Hence either polygon can be made to differ from the area of the curve by less than any assigned quantity.

Therefore the area of the curve is the common limit to which the polygon-areas approach.

Q. E. D.

MAXIMA AND MINIMA OF PLANE FIGURES

1020. Def.-Of the values which a variable quantity assumes, the largest value is called the maximum; the smallest, the minimum.

Thus, the diameter of a circle is the maximum among all straight lines joining two points of the circumference; and among all the lines drawn from a given point to a given straight line the perpendicular is the minimum.

1021. Of all triangles having the same base and equal areas, that which is isosceles has the minimum perimeter.

GIVEN the isosceles triangle ABC and any other triangle DBC having an equal area and the same base BC.

TO PROVE the perimeter of ABC is less than the perimeter of DBC.

Outline proof.-The vertices A and D are in the straight line XY parallel to BC. (Why?)

Draw CE perpendicular to BC, meeting BA produced at F. Join DF.

The angle CAE=angle FAE, and the triangle CAE=triangle FAE. Hence AE is perpendicular to CF at its middle point.

1022. Remark.—The converse of the preceding proposition is also true, viz.: Of all triangles having the same base and equal areas, that which has the minimum perimeter is isosceles. In fact, it is practically the same theorem as the proposition itself, for there is only one isosceles triangle fulfilling the given conditions, and only one triangle of minimum perimeter fulfilling the given conditions; just as to say that John Smith is the tallest man in the room is equivalent to saying that the tallest man in the room is John Smith, provided we know that there is only one John Smith in the room and only one tallest man. 1023. Cor. Of all triangles having the same area, that which is equilateral has the minimum perimeter.

1024. Def. When two figures have equal perimeters they are called isoperimetric.

PROPOSITION IX. THEOREM

1025. Of all isoperimetric triangles having the same base, that which is isosceles has the maximum area.

GIVEN the isosceles triangle ABC and any other triangle DBC having an equal perimeter and the same base BC.

TO PROVE

the area of ABC > area DBC.

Outline proof.--Draw AE perpendicular to BC and DF parallel to BC.

Therefore the area of triangle ABC > area of triangle DBC. Q. E. V.

1026. COR. Of all isoperimetric triangles, that which is equilaterai has the maximum area.

1027. Of all triangles having two sides of one equal to two sides of the other, that in which these two sides are perpendicular to each other

1028. The locus of the vertex of a right angle whose sides pass through two fixed points is the circumference of a circle whose diameter is the straight line joining those points.

Hint.-Apply SS 202, 207, 210.

PROPOSITION XII. THEOREM

1029. Of all isoperimetric plane figures, the maximum figure is a circle.

GIVEN―among any number of isoperimetric plane figures the figure of maximum area ABCDA.

First, draw any straight line AC (Fig. 1), dividing its perimeter into two parts of equal length.

Then AC will also divide its surface into two parts of equal area. For, if not, as for example if area ABC> area ADC, form the figure AB'C' symmetrical to ABC by revolving ABC on AC as an axis.

Then the figure ABCB'A would be greater than ABCDA and would have the same perimeter.

Hence ABCDA would not be the maximum.

This would be contrary to the hypothesis.

Therefore AC does divide the surface into two equivalent parts. Secondly, take any point B in the semiperimeter ABC.

We will prove that the angle ABC is a right angle.

Form the figure AB'C symmetrical to ABC.

Then

area AB'C-area ABC.

But we have just proved area ABC=area ADC.
Therefore

area ABCB'A=area ABCDA.

That is, ABCB'A is equivalent to the maximum figure.

Ax. 2

Now, if the angle ABC were not a right angle, we could increase the area of the equal triangles ABC and AB'C by moving the points A and C nearer together or farther apart, so as to make ABC a right angle, without changing the lengths of the straight lines AB, BC, AB', B'C, and without changing the areas of the segments AMB, BNC, AM'B', B'N'C.

H

In doing this the area of the whole figure ABCB'A would be increased.

But this is impossible, since ABCB'A is equivalent to the maximum figure.

Therefore ABC must be a right angle.

In like manner, if we should choose any point D in the semiperimeter ADC, we could show that ADC would be a right angle. Therefore the figure ABCDA is a circle.

$ 1028 Q. E. D.

1030. Of all plane figures containing the same area, the circle has the minimum perimeter.

GIVEN—a circle C, and any other figure A having the same area as C.
TO PROVE
the perimeter of C is less than that of A. .

$1029

Let B be a circle having the same perimeter as the figure A.
Then area A < area B, or area C< area B.
Now, of two circles, that which is the less has the less perimeter.

$$ 491, 499

Therefore the perimeter of C is less than the perimeter of Å, or less than the perimeter of A.

Q. E. D.