APPLICATION OF ALGEBRA TO GEOMETRY. THE subject of algebra is the general science of reasoning by means of arbitrary symbols, whilst that of geometry is the science of reasoning by means of magnitudes. In applying algebra, then, to the resolution of a problem of geometry, the symbols employed must be adequate to represent all the affections and relations of which the magnitudes represented by them are susceptible. Such symbols, therefore, must be regarded as the representatives of quantities admitting of arithmetical interpretation only in the case in which the magnitudes of geometry admit of such interpretation. When, however, the geometrical magnitudes in any problem are all related to some other magnitude of the same kind, in the same manner that certain abstract numbers are related to unity, these numbers which may be taken to represent the magnitudes must no longer be considered as abstract but concrete quantities. Thus if A, B, C, D be straight lines related to a straight line U in the same manner that the abstract numbers a, b, c, d are related to unity; so that Then because A contains U, a times, B contains the same, b times, etc. if A, B, C be denoted by a, b, c, these symbols will now be concrete numbers; and hence in the solution of a problem of this kind particular care must be taken to reduce all the terms to the same denomination. The quantity U, which is called the linear unit, may be of any length: if it be a yard, a mile will be represented by 1760; if it be an inch, a foot will be represented by 12, and so on with other quantities. If upon the linear unit a square be described, that figure is called the square unit. It will moreover be obvious, that every equation of the problem under discussion will be homogeneous, that is, the sum of the exponents of each term will be the same unless one of the quantities under consideration be taken for the unit. What has been stated refers principally to the magnitude of lines; their position is to be determined by the principles laid down in the Geometry of Coordinates. In the solution of a particular problem a figure must be constructed to represent the conditions of the problem, and such other lines drawn as may seem to facilitate the solution. Then denoting as usual known quantities by a, b, c, etc., and unknown ones by z, y, x, etc., as many independent equations must be formed as there are unknown quantities to be determined. The entire number of conditions is never given, except implicitly as geometrical properties of the figure to which the problem relates. These implicit ones are geometrical theorems which must be sought for to suit the case, or investigated independently by geometrical methods. The following solutions are given for illustration :— PROBLEM I. Through a given point P in the hypothenuse A C of a right-angled triangle A B C to draw a straight line D P E, meeting A B in E, and B C produced in D, so that the triangles APE and CPD may be equal in area. = From P draw P M perpendicular to A B and put A M =b, MP = c and ME x. Then by similar triangles, = a, M B AM: MP: AB: BC, or a c :: a + b: BC and EM: MP::EB: BD, or x : c :: x + b : B D c (x + b) х Now by Euc. i. 41, and the conditions of the problem, c b2 (a − x) ах = c(ax); whence which gives the position of the required line D P E. The formula (1) points out a simple geometrical construction. For а x = b2, or AM. ME = B M2, hence ME is a third proportional to AM and MB. PROBLEM II. the inscribed square. Given the base and perpendicular of a triangle, to find the side of Let A B C be the triangle and EFGH the inscribed square. = АВ Put Equating the rectangle of the extremes to that of the means (Euc. vi. 16), we readily get— the side of the required square. Cor. 1. When cp, or when the base and perpendicular are equal, porc, that is, the side of the inscribed square is equal to half the base or half the perpendicular. x = Let us next inquire whether the angular points E, F may not be in A C and B C produced. In this case, A C and B C, unless the base and perpendicular are equal, and then one only can be described. Cor. 2. Since by (1), e+p: p::c: x ::AB: EF:: AC: CE, that is, : Produce the base A B till the produced part B L is equal to the perpendicular C D, and join the points Land C; then draw B E parallel to LC to meet AC in the point E, and this will be an angular point of the inscribed square. The formula (2) is constructed in a similar way.* PROBLEM III. The front wall of a house is of such height, that if a ladder of certain length be placed at the distance of 12 feet from it, the top of the ladder will just reach to the top of the wall; but if it be placed at a distance of 20 feet from it, its top will be 4 feet below the top of the wall. Find the height of the wall and length of the ladder. Let B E be the height of the wall, and A B, DC the first and second positions, respectively, of the ladder. Let the length of the ladder A B = CD = x, the height of the wall *All the results that are obtained algebraically by the solution of a quadratic, may be constructed in like manner, but this part of the subject must necessarily be very brief in a work like the present. PROBLEM IV. Given the base of a plane triangle, the perpendicular, and the rectangle of the two sides, to determine the sides. Let ABC be the plane triangle. Put A DD B* = AB = 2 х, АС. СВ Ꭺ Ꭰ - Ꭰ Ᏼ = m + x, a2, and CD BD m (m + x)2 + p2, B C2 +p. Hence the condition = = α. {(m + x)2 + p2} {(m − x)2 + p2} = a'. This equation readily reduces to the following A x + 2 (p2 - m2) x2 = a* — (m2 + p3)3, which gives for x the four values = ± {(m + p) (m − p) ± √√ (a2 + 2p m) (a2 − 2 pm)}*; from which the sides are easily determined. - The four resulting values of a show that the problem admits in general of four solutions, as indeed it might be expected from Euc. vi. C; for as the perpendicular and rectangle of the sides are given. the diameter of the circumscribed circle is also given by that proposition. Hence the problem is the same as that of inscribing in a given A circle a triangle of given base and altitude, as in the annexed diagram. PROBLEM V. To divide a given straight line in extreme and mean ratio. Let the given line A B other part B C = a − x. (Euc. vi. 17), x2 = a (a− x) The solution of this equation gives the two values the former being positive and the latter negative. The negative value of x gives a point to the left of A (Geo. of coordinates), and the property to which it belongs may be determined by writing - for x in the equation (1). Hence we have x2 a (a + x), which, when translated into words, gives the following problem: To produce a given straight line so that the square of the produced This method will be found, in similar cases, very effective in the solution of a problem. Should the difference of two quantities be given, instead of the sum, denote the difference by twice a given quantity and the sum by twice an unknown quantity. part shall be equal to the rectangle contained by the given line, and the line made up of the whole and the part produced. This is merely the given problem enunciated more generally, as might be expected. PROBLEM VI. Given the perimeter (774) of a right-angled triangle, to find the sides, so that a perpendicular from the right angle may cut the hypothenuse in extreme and mean ratio. Let A B C be the right-angled triangle, the hypothenuse A B of which is divided in extreme and mean ratio But (Euc. vi. 8), A C2 BA. A D = (3 − √ 5) = n x, = m x2, and B C2 BA.BD = the condition of the problem we have or x = Consequently, == x + x √ m + x √ n = 77, 77 A C = x √ m D nx2; hence by = 32.36, the side A B. PROBLEM VII. Given the sides of a plane triangle, to find— (1). The area of the triangle; (2). The radius of the inscribed circle; (3). The radius of the circumscribed circle. In this and some other discussions relative to a plane triangle, the following notation will be used: A will denote the area of the triangle; r, R, the radii of the inscribed and circumscribed circles; a, b, c, the sides opposite, respectively, to the angles A, B, C; 2s, the perimeter (a+b+c); and, consequently, b + c = a = 2 (s − a), a + c − b 2 (s—b), a + b -C= 2 (s-c). Area of the Triangle.-Put the perpendicular C D = x, and BD = y (fig. to Problem VI.). Now (Euc. i. 41), A = ex, or A2 = † c2x2 = 1 c* (a2 — y2) = ‡ c2 (a + y) (a − y) .... (1). But (Euc. ii. 13), |