The common property here is, that all positions of the point are at the same distance from a fixed point. It is evident they must lie in the circumference of a circle. Hence this circumference is the required locus.

2. Find the locus of a point in a plane at a given distance from a straight line. Parallel line.

3. Locus of a point in space at a given distance from another point. Surface of a sphere.

4. Locus of a point in space at a given distance from a straight line. Surface of a cylinder.

5. Locus of a centre of a circle which passes through two given points.

From (III. 1) all circles passing through these points must have their centres in a perpendicular to the line joining the points at its middle point. Hence the required locus is this straight line.

6. Locus of the centre of a circle which is tangent to two given lines.

Line bisecting the angle formed by the given lines.

7. Locus of the centre of a circle passing through a given point and having a given radius. Circle with given point as

Let DE be a chord, and DB.BE a given rectangle. Draw AC so that it will be bisected at B; then (V. 26) DB. BE=AB2. If AC be placed in different positions in the circle, all chords through its middle point will be divided so as to make the rectangle of the segments equal to DB.BE. Hence the locus is a circumference with radius FB.

9. Locus of point which is equally distant from two given points.

A plane perpendicular to line joining the points, through its middle point.

10. Locus of point which is equally distant from two given straight lines in same plane. Plane between the lines.

11. Locus of middle point of a line joining two perpendicular but non-intersecting lines. Parallel plane between the lines. 12. Locus of point at a distance a from a point A, and a distance b from point B.

Circle which is the intersection of two spheres.

SECTION III.

MAXIMA AND MINIMA.

DEFINITIONS.

1. Of all quantities of the same kind that which is greatest is the maximum, and that which is least is the minimum. 2. Figures which have equal perimeters are said to be isoperimetric.

Proposition 1.

Theorem. Of all equal triangles on the same base, that which is isosceles has the minimum perimeter.

Let ABC be an isosceles triangle, and DBC an equal triangle on the same base; then the perimeter of ABC will be less than the perimeter of DBC.

Produce BA, making AE= AB-AC. Join CE; also DA,

and produce it to F; join DE. Because the triangles DBC, ABC are equal, DF is parallel to BC.

B

E

F

C

EAF and CAF. Hence EF = CF, and the angles at F are right angles. Then in the triangles DEF, DCF DF, FE, and DFE are equal to DF, FC and DFC, each to each; therefore DE DC. Hence BD+DE = BD+DC; but BE<BD+DE; ... BA+AC<BD+DC; and this is true no matter in what point of the line DF, D be taken. Hence BA+AC is a minimum.

=

Corollary. Of all equal triangles, that which is equilateral has the minimum perimeter.

For it must be isosceles, whichever side is taken for a base.

Proposition 2.

Theorem. Of all isoperimetric triangles on the same base,

that which is isosceles is the maximum.

Let ABC be an isosceles triangle, and let DBC, on the same base BC, have an equal perimeter; then ▲ ABC > A DBC.

The point D will fall between the parallels BC and EF

F

A

E

For, if it fell on EF, the triangles could not be isoperimetric (Prop. I.);* still less could they be so if it fell above EF; hence it falls below EF, and the area BDC is less than the area BAC

Corollary. Of all isoperimetric triangles, that which is equilateral is a maximum.

*References such as this are to propositions in Modern Geometry Others are referred to as heretofore.

Proposition 3.

Theorem. Of all triangles having two sides equal, each to each, that which has these sides at right angles is the maximum.

Let AB, BC, at right angles to each other, be equal, each to each, to DB, BC, not at right angles to each other; ABC is greater than DBC.

Draw the altitude DE; AB is greater than DE; hence ABC is greater than DBC.

Proposition 4.

D

B

E

Theorem. Of all isoperimetric figures, the circle is the

maximum.

1. The given maximum figure must be convex; that is, a line joining any two points of the perimeter must fall inside the perimeter. For, if EA CB be a figure not convex, for the part ACB could be substituted the convex line ADB of E equal length, which would increase the area of the figure without increasing the perimeter. Hence, any figure not convex cannot be a maximum.

2. Let ABCD be the maximum figure. Draw AC bisecting the perimeter; it will also bisect the area. For, if not, let ABC be greater than ADC; then a line equal to ABC could be substituted for ADC, which would, with an equal perimeter, cut off a greater area. But ABCD was, by hypothesis, the

E

H

D

B

F

H

maximum. Hence, the area is bisected by AC.

E

D

G

Also, take any point B in the perimeter; the angle ABC is a right angle. For, if not, draw BE at right angles to BC, making it equal to BA, and on it describe a segment equal to the segment on BA. Then (Prop. 3) the triangle EBC is greater than the triangle ABC; hence, the whole figure EFBGC is greater than the whole figure AHBGC; and if symmetrical figures be described on the opposite sides of EC and AC, the figure on EC would be greater than the figure on AC. But they have equal perimeters. Hence, ABCD is not a maximum, which is contrary to the hypothesis. Therefore, ABC is a right angle.

Now, B is any point of the perimeter; hence, ABCD is a circle, and AC its diameter.

Proposition 5.

Theorem. Of all equal figures, the circle has the minimum perimeter.

Let A be a circle, and B

any other equal figure; then the perimeter of A will be less than the perimeter of B. Let C be a circle having the same perimeter as B; then (Prop. 4) C>B; ... C>A; hence, the circumference of C is greater than the circumference of A; therefore, also,

B

C

Α

the perimeter of B is greater than the circumference of A.