For, let B" be a polygon similar to B' and A", P" and N" its area, perimeter, and the number of its sides. If P"-P and N" N, then A" A (9); and therefore if the parts of B" be proportionally increased till A" A, then its perimeter= P' (VI. 20). But P'P", and therefore P'P.

=

PROPOSITION XIII. THEOREM.

Of regular polygons having the same area, that which has the greatest number of sides has the least perimeter.

Let B be a regular polygon, A, P, N, its area, perimeter, and the number of its sides; A', P', N', those of another regular polygon B'; and A", P", N", those of another B" similar to B'. If NN' then P is P'. For since N". N, if P" P, A"▲A (10); and if the parts of B" be increased proportionally till A"= A, then (VI. 20) its perimeter will=P'; but P'>P” and P"P, therefore P'⇒P.

Schol. In accordance with this and other propositions, bees instinctively construct their cells in the form of regular hexagons, which form requires the least quantity of wax.

PROPOSITION XIV. THEOREM.

The perimeter of a circle is less than that of any polygon having the same area.

Let A and P be the area and perimeter of a circle B, A' and P' those of a polygon B'; and A", P", those of a polygon B" similar to B'. Then, if P" P, A" will be ZA (11); and if the parts of B" be proportionally increased till A"= A, then its perimeter will = P'; but P'P" and P" P, therefore P'P.

PROPOSITION XV. THEOREM.

Of all isoperimetrical plane figures, the circle contains the greatest area.

E

Let ACE be the maximum figure for a given perimeter. If it is not a circle, let an equilateral polygon be described in it. It is evidently possible for such a polygon to exist in it, so that each of its angular points shall not be in the circumference of the same circle with all the other

B

C

angular points, and therefore the polygon will be irregular

Let a regular polygon P of the same number of sides and the same perimeter be described, then each of its sides will be equal to those of the inscribed polygon. On the sides of P describe segments equal to AFB, BGC, &c., then a curvilinear figure Q will thus be formed, having the same perimeter as the given figure ACE; and since the regular polygon exceeds the other polygon (9), the whole figure Q will exceed the whole ACE. When ACE therefore is not a circle, another figure as Q having the same perimeter, can always be found greater than it; hence the circle is the maximum figure (Ax.)

Schol.-No part of the figure ACE is supposed to be convex internally. Were any portion of it convex, an equal concave arc being substituted for it, the area would thus be increased, while the perimeter is unaltered.

PROPOSITION XVI. THEOREM.

Of all plane figures of the same area, the circle has the least perimeter.

Let the circle A the figure B, the perimeter of A is less than that of B. For, if not, let another circle C be assumed whose perimeter ist equal to that of B; then (15)

B

its area exceeds that of B or of A; and hence its perimeter exceeds that of A (Qu. VIII.)

EXERCISES.

1. A straight line and two points without it being given, to find a point in it such that the sum of the lines drawn from it to the two given points, shall be a minimum, that is, that they shall be less than the sum of any two lines similarly drawn from any other point in it.

2. If any point, except the centre, be taken in the diameter of a circle, of all the chords passing through this point, that is the least which is perpendicular to the diameter.

3. Of all triangles that have the same vertical angle, and whose bases pass through a given point, that whose base is bisected by the point is a minimum.

4. The sum of the four lines drawn to the angular points

of any quadrilateral from the intersection of the diagonals, is less than that of any other four lines similarly drawn from any other point.

5. To find a point in a given line such that the difference of the lines drawn to it from two given points may be a maximum.

6. A straight line and two points on the same side of it being given, to find a point in it such, that the angle contained by lines drawn from it to the given points shall be a maximum.

7. Given three points, to find a fourth such, that the sum of its distances from the given points shall be a minimum.

ON GEOMETRICAL ANALYSIS.

In the method of Geometrical Analysis, the process of demonstration follows an order the reverse of that observed in the ordinary or Synthetic Method. The latter method proceeds from established principles, and, by a chain of reasoning, deduces new principles from these; the former proceeds from the principles that are to be established considered as known, and from these, taken as premises, arrives, by reversing the chain of reasoning, at known principles. The latter method is the didactic method used in communicating instruction; the former is rather employed in the discovery of truth. This method is not exactly that pursued by the mind in analysing any proposition, but it be considered to be that method reduced to a systematic

may

form.

Magnitudes are said to be given when they are actually given or may easily be found; they are said to be given in position when their position may be determined; and rectilineal figures are said to be given in species when figures similar to them may be found. A circle is said to be given in position when its centre is given, and in magnitude when its radius is given. A ratio is said to be given when two quantities having that ratio are given.

The two following propositions are given as examples of this method.

PROPOSITION I. PROBLEM.

P

Given two points P and Q and a straight line AB, to find a point C in AB such, that lines PC, CQ, drawn to it from P and Q, may make equal angles PCA, QCD, with AB. By analysis. For either of the two given points as Q, draw QD perpendicular to AB, and produce QD to meet PC produced in E; then the angle ECD=ACP (I. 15) =QCD (Hyp.), and the angles CDE, CDQ, are equal, being right angles, and CD is common to the two triangles CDE, CDQ; therefore (I. 26) they are equal in every respect, and hence

מן

E

DE=DQ. But the perpendicular DQ is given, and therefore DE and the point E are given; hence the line PE, and consequently the point C are given.

Schol. By this analysis, the construction is discovered by which the point C is determined. That C is the required point may now be demonstrated by synthesis or composition, thus:

By composition.-Draw QD perpendicular to AB, and produce DQ till DE DQ, and join PE, then C will be the point required.

For, join CQ. Then because QD - DE and DC is common to the two triangles CDE, CDQ, and the angles CDE and CDQ are equal, being right angles, therefore (I. 4) these triangles are every way equal; and hence angle ECD = QCD. But PCA ECD (I. 15), therefore also PCA =QCD; and C is the point required.

=

PROPOSITION II. PROBLEM.

A straight line AB, and two points P and Q without it being given, to find a point C in it such, that the two lines CP, CQ, drawn to them from C shall be equal.

Analysis. Join PQ, and bisect it in

D, and join CD. Because CP = CQ ☎

B

and PD DQ and CD common to the two triangles CDP,

P

CDQ, they are every way equal (I. 8), therefore the angle PDC=QDC. Therefore CD is perpendicular to PQ. But PQ is given, therefore the point D and the perpendicular DC are given, and consequently the point C is given. It may be easily proved by composition that C is the point required.

Any of the exercises given in the preceding books will serve as exercises in geometrical analysis.

ON PLANE LOCI.

If every point in a straight or curve line, and no other point, fulfil certain conditions, the line is said to be a locus of the point.

As a simple illustration of a locus, consider that of a point which is always equally distant from a given point. This is obviously a circle, whose radius is equal to that distance. So the locus of a point, which is always equally distant from a given straight line, is a line parallel to it, and at a distance from it equal to the given distance.

EXERCISES.

1. Find the locus of the vertices of all the triangles that have the same base and one of their sides of a given length. 2. Find the locus of a point that is at equal distances from two given points.

3. Find the locus of a point that is equally distant from two given lines, either parallel or inclined to one another. 4. To find the locus of the vertices of all triangles that have the same base and equal altitudes.

5. To find the locus of the vertices of all triangles that have the same base and one of the angles at the base equal. 6. To find the locus of the angular point opposite to the hypotenuse of all the right-angled triangles that have the same hypotenuse.

7. To find the locus of the vertices of all the triangles that have the same base and equal vertical angles.