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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 NZ N', then Pis P'. For since N'<N, if P'= P, A"LA (10); and if the parts of B" be increased proportionally till A"= A, then (VI. 20) its perimeter will = P';

* but P' 7P" and P'= P, therefore P' z P. a 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 <A (11); and if the parts of B" be proportionally increased till A'=A, " then its perimeter will = ''; but P' 7 P" and P'=P, therefore P' ¿P.

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

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 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

B

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 is equal to that of B; then (15) 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 an eccentric point 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. The sum of two lines drawn from two given points to a point in the circumference of a given circle, is least when these lines are equally inclined to the radius or the tangent drawn to that point.

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

9. Given the base and altitude of a triangle, to construct it so that the sum of the squares on its sides may be a minimum. ? 10. Given the perimeter of a parallelogram, to construct it so that its area may be a maximum.

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.

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. Given two points P and Q and a straight line AB, to find a point C in AB such, that lines PC and CQ, drawn to it from P and Q, may make equal angles PCA and QCD, with AB.

By analysis.-From 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 and CDQ are equal, being right angles, and CD is common to the two triangles CDE and CDQ; therefore (I. 26) they are equal in every respect, and hence 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.-(Const.) Draw QD perpendicular to AB, and produce DQ till DE ='DQ, and join PE, then Č will be the point required. Join also CQ.

(Denn.) Then because QD = DE, and DC is common to the two triangles CDE and 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 (1. 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 and CQ drawn to them from C shall be equal.

Analysis.-Join PQ, and bisect it 'in D, and join CD. Because CP = CQ, and PD = DQ, and CD is common to the two triangles CDP, CDQ, they are every way equal (I. 8), therefore the angle PDC = QDĆ. Therefore CD is perpendicular A

с в 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.

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