A SECOND BOOK IN GEOMETRY. [Continued from Page 143.] CHAPTER VIII. THE MAXIMUM AREA. 114. I will prove only one more proposition; but I will select a difficult one, in order that it may require a number of preliminary proofs. I will select the proposition given in the "First Lessons in Geometry," chap. xxiii. § 14: "Of all isoperimetrical figures the circle is the very largest." 115. When we attempt to analyze this, we shall see that it implies that any regular polygon is less than a circle isoperimetrical with it, and that any other polygon is less than a regular one, isoperimetrical with it. 116. Let us begin, however, by defining a few of the words we shall need to use. 117. A polygon is a plane figure bounded by straight lines. 118. The perimeter of a polygon is the sum of the length of its sides. 119. Isoperimetrical polygons are those of equal perimeter. 120. Among quantities of the same kind, the largest is called a maximum. 121. A circle is a plane figure, bounded by one line that curves equally in every part. This line is called the circumference of the circle, and frequently the line itself is called the circle. Portions of the circumference are called arcs. 122. Theorem. Fig. 60. с of the circle. There is a point within the circle equally distant from every point of the circumference. This point is called the centre of the circle.-Proof. Let A D and B D be equal adjacent arcs in a circumference. Through the points A, B, and D draw lines at right angles to the curve at those points. Now, since the circle curves uniformly at every point, the point of intersection with B C must be the same point in both lines, A C and D C, and the points A, B, and D are equally distant from C. But A and B may be taken anywhere in the circle, only provided they are equally distant from D; and hence every point in the circle is equally distant from C, the centre 123. A straight line joining the centre to the circumference is called a radius. The straight line formed of two opposite radii is called a diameter. 124. All radii are of course equal to each other. 125. A straight line joining the two ends of an arc is called a chord. 126. A straight line which, however much prolonged, touches the circle in one point only, is called a tangent. 127. It is manifest that the tangent coincides in direction with the arc at the point of contact. 128. A polygon formed wholly of chords in a circle is said to be inscribed in that circle. A а 129. A polygon formed wholly by tangents to a circle is said to be circumscribed about the circle. 130. The circle is said to be inscribed in the circumscribed polygon, and to be circumscribed about the inscribed polygon. 131. If a polygon, about which a circle can be circumscribed, or in which a circle can be inscribed, has its sides equal, one to the other, the polygon is called a regular polygon, and the centre of these circles is called also the centre of the polygon. 132. Let us now attempt to analyze the proposition that the circle is the maximum among isoperimetrical polygons. This is equivalent to saying that if an isoperimetrical circle and regular polygon are laid one over the other, the polygon will be the smaller. But we see that by laying them one on the other, a circle inscribed in the polygon would be smaller than the isoperimetrical circle. The question of course suggests itself, whether the area of a regular polygon is not proportional to the radius of the inscribed circle. Now it is plain that it is. For by dividing the polygon into triangles, by lines from its centre to its vertices, we find the area of the polygon will be the sum of the areas of the triangles, and these areas will be measured by half the product of the perimeter multiplied by the radius of the inscribed circle. The area of the isoperimetrical circle will be measured by half the product of the perimeter multiplied by the radius. But as the perimeters of the polygon and the isoperimetrical circle are the same, and the radius of the inscribed circle is smaller than that of the isoperimetrical circle, it is evident that the area of the polygon is smaller than that of the isoperimetrical circle. It will now remain to show that the area of a regular polygon is greater than that of an isoperimetrical irregular polygon. It is evident that this can be done, since a polygon of given sides is manifestly largest when most nearly circular, and a polygon of given perimeter inscribed in a circle is manifestly largest when the sides are equal. We can surely have no difficulty in proving these two points, and then our proof will be complete. 133. Let us return, then, to the synthetic mode, and establish these propositions: - First, that the maximum of polygons formed of given sides may be inscribed in a circle; secondly, that the maximum of isoperimetrical polygons having a given number of sides has its sides equal; and thirdly, that such a regular polygon is of smaller area than a circle isoperimetrical with it. 134. Theorem. The area of a triangle is found by multiplying the base by half the altitude. This theorem has been already proved (Art. 111). 135. We shall need the Pythagorean proposition, which implies all the propositions into which we have already analyzed it (Arts. 64 – 113.) 136. Theorem. Of two unequal lines, from a point to a third straight line, the shorter is more nearly perpendicular to the third line. Proof. Let C be the given point, and A D the third straight line. Let B on BD is the difference of the squares on B C (which is smaller than A C) and the same CD. 137. Corollary. A perpendicular is the shortest line from a point to a given straight line. 138. Theorem. The maximum of triangles having two sides given is formed when these two sides are at right angles. Proof. Let A' B, A B, and A′′ B be equal to each other. The area of A' BC, AB C, or A" B C, being found by multiplying B C into the perpendicular height of A, A', or A", above B C, will be in proportion to that height. (Fig. 17.) be perpendicular to B C, and the height of A above the base will equal B A. of A" above the base must, by 137, be less than B A" which is equal to B A. Let then A B 139. An angle is said to be measured by an arc of a circle such as would be intercepted by radii making that angle with each other. And, since the circumference curves equally in all parts, and the radii are at right angles to it, it is evident that this measure is just, and that the angle will bear the same ratio to four right angles that the arc bears to a whole circumference, whatever be the size of the circle. Fig. 16. B b 140. Theorem. If two sides in a triangle are equal, the angles opposite those sides are equal. - Proof. Let AB and BC be equal sides in a triangle. Imagine A C divided in the centre, at the point b. The triangles ABb and C Bb will now be composed of equal sides, and we have already proved (Arts. 91–95) that they must have equal angles, 141 Theorem. If one side of of the opposite internal angles. 142 Two chords starting from one point in a circumference intercept double the arc that would be intercepted by radii making the same angle; or, the angle of the chords is measured by half the arc included between them.-Proof. If one chord, as AB, passes through the centre D of the circle, it is plain that by drawing D C the angle C D B will be equal to the sum of the angles CAD and DC A. But since D A and D C are equal, these angles are equal, and CDB is equal to twice C AD. that is, the angle at A is equal to that at C. a triangle is prolonged, the external angle is equal to the sum This has been proved in Art. 57. If neither chord passes through the centre of the circle, we can draw a third chord, starting from A, passing through the centre of the circle, and apply this reasoning to the two angles formed with this third chord by the other two. The angle of the other two chords will simply be the sum or the difference of these two angles. 143. Corollary. If the vertex of a right angle be placed in the circumference, the sides will intercept a semicircle. 144. Corollary. If a circle be circumscribed about a triangle, and one side of the triangle passes through the centre of the circle, the opposite angle is a right angle. B 145 Theorem. The maximum of polygons, having all the sides given but one, may have a circle circumscribed about it, having the unknown side for a diameter - Proof. Let A B C D E be the maximum polygon, formed of given sides, A B, B C, &c., and the unknown side, A E. Join B E by a straight line. Now, since the polygon is a maximum, we cannot, leaving BE unaltered, enlarge the triangle ABE, because that would enlarge the polygon. The angle ABE is therefore a right angle, by Art. 138, and a circumference, having AE for its diameter, would pass through the point B. In like manner it can be shown that a circumference having the same diameter would pass through each of the other points. E 146. Theorem. The maximum of polygons formed with given sides can be inscribed in a circle. Proof. Let A B C DE be a polygon formed of given sides, with a circle circumscribed about it. Draw the diameter A F, and join F C and FD. The polygons ABCF and ADEF are now maximum polygons, and therefore ABCDE must also be a maximum, since its enlargement would enlarge the sum of the other two. We have thus proved the converse of the proposition, and the proposition is true, unless there is more than one maximum form of the polygon. E Fig 9. A B The converse is more easily proved than the proposition, and I therefore proved it, on the assumption that there is but one maximum form. That is, I have proved that a polygon of given sides, when inscribed in a circle, is a maximum; but that does not strictly prove that the maximum can always be inscribed in a circle; except on the assumption, which is, however, a safe one, that a polygon formed of given sides, arranged in a given order of succession, can have but one maximum form. B 147. Theorem. Of isoperimetrical triangles with one side given, the maximum has the two undetermined sides equal, -Proof. In order to prove this we have only to show that the point A is at its greatest distance from the base, B C, when opposite the middle of it. This might seem scarcely to need proof. For when we use a string and stick to illustrate the problem, we can see that by sliding the finger from the middle of the string, it can be brought down into a line with the stick; and the greatest height from the stick is near the middle of the string. Further consideration shows it must be exactly at the centre of the string, because the finger and string have precisely the same relation to one end of the stick as to the other; and a motion towards either end must affect the height of the finger in a similar manner. This reasoning is doubtless satisfactory to every fair mind. Yet it is not a good mathematical demonstration, and I have given it to you for the purpose of illustrating the peculiar nature of mathematical reasoning. The reasoning just given leaves no real doubt on the mind, but it is rather because we see with the eye that the finger is highest in the middle, than because we see with the mind that it must be. There is another step still lacking, to prove to us that the highest points are not on each side of the exact middle, as that would satisfy the conditions of symmetry and of declination towards each end. Let us then seek a proof which shall not force us to consider the whole motion of the finger, but which shall simply compare two forms of the triangle, one with the finger in the middle of the string, and one with the finger on one side. 148. Theorem. If a straight line be drawn from the vertex of two equal sides in a triangle, at right angles to the third side, it divides the third side into equal parts. - Proof. Let c and a be equal sides in a triangle ABC. Since (Fig. 16, Art. 140) the angles at A and C are equal, the angles b B C and b B A are also equal. If, therefore, the triangle Bbc be folded over on the line Bb, the line a will take the same direction as the line c, and, being of the same length, will coincide with it. Hence, b C will also coincide with b A, and the two lines must be of equal length. 149. New proof of Art. 147. Let A B C and AB'C be isoperimetrical, and let AB and = Then, by Art. Join A E. AE BC be equal. Continue A B to D, making BD-BA BC, and join DC. 144, the angle D C A is a right angle. Draw B'E making it equal to B' C. will be less than the sum of AB' and B' E, that is less than AB' and B D E H I B' C, that is less than A B and B C, that is less than A D. But if A E is less than A D, then CE must be less than CD, because the square on CE is equivalent to the difference of the squares on AE and AC, while the square on CD is equivalent to the difference of the squares on AD and A C. Draw BH and B'I at right angles to CD; we have CI which is half CE less than CH which is half CD. But CI and CH are the altitudes of the triangles ABC and AB'C above their The triangle with the undetermined sides equal has the greatest altitude and must be the largest triangle. A common base A C. C 150. Theorem. The maximum of isoperimetrical polygons of a given number of sides is equilateral, that is, has equal sides. - Proof. Let AB CED (Fig. 9) be the maximum of isoperimetrical polygons of a given number of sides. Then AB must equal B C. For if it did not, then after joining A and C we could enlarge the triangle ABC by equalizing A B and A C, and thus enlarge the polygon without altering the number of sides or the perimeter, and the present form would not be the maximum. In like manner we may prove that B C = CE, &c. 151. Corollary. The maximum of isoperimetrical polygons of a given number of sides is regular by Arts. 150 and 146. 152. Axiom. A circle may be considered as a regular polygon having an unlimited number of sides. And this regular polygon may be considered as either inscribed in or circumscribed about the real curve. 153. Theorem. The area of a regular polygon is measured by half the product of the perimeter into the radius of the inscribed circle. - Proof. For if lines be drawn from each vertex of the polygon to the centre, the polygon will be divided into triangles having a common altitude equal to the radius of the inscribed circle, the sum of the bases of these triangles being equal to the perimeter of the polygon. 154. Corollary. The area of a circle is measured by half the product of the radius into the circumference. D B 155. Theorem. The perimeter of a circumscribed polygon is greater than the circumference of the circle.-Proof. Let A B be half a side of a circumscribed polygon, and DB the portion of arc intercepted by lines drawn from A and B to the centre of the circle. Divide DB into arcs so small that each may be considered as a short A straight line. Through the points of division draw lines extending from the line AB to the point C. At the end B, the little arcs are equal to the corresponding pieces of the line AB; but as you approach A the divisions of the line grow longer than the corresponding divisions of the arc, for two reasons; first, the little arcs are at right angles to the radii, while the portions of the line are not (Art. 136); secondly, the little arcs are nearer to the point C, towards which the radii converge. The whole of A B must therefore be longer than the whole of D B. But it is manifest that the circumference consists of as many times D B as the perimeter does of the line A B. 156. Corollary. The circle inscribed in a regular polygon is smaller than a circle isoperimetrical with the polygon and has a shorter radius. 157. Corollary. The circle is the maximum among isoperimetrical regular polygons. 158. Corollary. The circle is the maximum among isoperimetrical figures; a proposition |