APPLICATION OF ALGEBRA TO THE SOLUTION OF GEOMETRICAL PROBLEMS. PROBLEM I. In an equilateral triangle, having given the lengths of the three perpendiculars drawn from a certain point within it to the three sides, to determine its side. Let ABC be the equilateral triangle, and DE, DF, DG the perpendiculars from the point D upon the sides respectively. Denote these perpendiculars by a, b, c, in order, and the side of the triangle ABC by 2 x. Then, if the perpendicular CH be drawn, CH=√AC2 — AH2 = √4x2 — x2—x √3. A F C E D HG B The area of the triangle ADB =}AB.GD=cx. Similarly the triangle BDC= ax, the triangle CDA = bx, and the triangle ACB= AB.CH=x2 √3. Also, BDC+CDA+ADB=ABC; that is, in symbols, which is half the side of the triangle sought. Cor. From the resulting equation, we have x√3 = a+b+c: we also had CH=√3. Hence CH=a+b+c; or the whole perpendicular CH is equal to the sum of the three smaller perpendiculars from D upon the sides, whenever the point D is taken within the triangle. Had the point D been taken without the triangle, the perpendicular upon the side which subtends the angle within which the point lies would become negative. Thus, had the point been without the triangle, but between the sides AB, AC produced, then CH=DF+DG-DE. PROBLEM II. A Maypole was broken off by the wind, and its top struck the ground twenty feet from the base; and, being repaired, was broken a second time five feet lower, and its top struck the ground ten feet farther from the base. What was the height of the May-pole? Let AB be the unbroken May-pole, C and H the points in which it was successively broken, and D and F the corresponding points at which the top B struck the ground. Then will CAD and HAF be right-angled triangles. F D B C H Put, BC=CD=x, CA=y, AD=a, AF =b, and CH=c. Then AB=x+y, BH=HF=x+c, and HA=y-c; therefore (B. III., T. XIV.), we have Expanding (2) and subtracting (1) from it, we have, after a slight reduction, x+y= b2 — a2 =50 feet, the required height. PROBLEM III. A statue eighty feet high stands on a pedestal fifty feet high, and to a spectator on the horizontal plane, they subtend equal angles; required the distance of the observer from the base, the height of the eye being five feet. Let AB = a, the height of the pedestal; BC=b, the height of the statue; DE=c, the height of the eye ground, from and_DA=EF=x, the distance sought. C B F But, since the angle CEB = BEA, we have (B. III., T. XII.) ECEA: CB: BA, or, or, in symbols, x2+(a+b−c)2 : x2 + c2 : : b2 ; a2. From this proportion, we readily deduce the double sign merely indicating that this value of x may be measured either way, from A towards D, or from D towards A. PROBLEM IV. To determine the area of a triangle when the three sides are given. Let ABC be the triangle. From the angle C draw the perpendicular CD, which in the second figure falls without the triangle, and meets the base AB produced. C C A D B A B D Denote the sides respectively opposite the angles A, B, C, by a, b, c, and the perpendicular CD by p. The right-angled triangles CDA, CDB give b2=DA2+p2; a2= DB2+p2; consequently, Hence, in the first figure, b2 - a2 DA-DB= Now, since half the difference of two quantities added to half their sum gives the greater, we have, in both cases, Since the difference of two squares is equal to the product of the sum of the two roots into their difference, we have √(b2+2bc+c2 — a2) (a2-b2+2bc- c2 p= 2c √[(b+c)2 —a2] × [a2 − (b − e)2] 2c √(a+b+c) (− a + b + c) (a − b + c) (a + b − 2c (1) Multiplying this perpendicular by half the base, we have for the area, {√(a + b + c) (− a + b + c) (a − b + c) (a + b −c), or [ ( a + b + c ) ( = a + b + c) ( a − b + c ) ( a + b = 0)]3. (2) 2 2 2 2 Hence we may find the area of a triangle, when the three sides are known, by this RULE-Take half the sum of the three sides, and from this half sum subtract each side separately; then take the square root of the continued product of the half sum and the three remainders, and it will be the area. PROBLEM V. Given the three sides a, b, c of a triangle, to find: 1. The three perpendiculars from the angles upon the opposite sides ; 2. The area of the triangle; 3. The radius of the circumscribed circle; 4. The radius of the inscribed circle; 5. The radii of the escribed circles. Let ABC be the triangle; and let a, b, c denote the sides opposite the angles A, B, C respectively, and P1, P2, P3 the perpendiculars drawn from the angles A, B, C; ▲ the area of the triangle; R the radius of the circumscribing circle; that of the inscribed circle; and 71, 72, 73 the radii of the three escribed r1, circles, which touch the sides a, b, c externally. An escribed circle has already been defined as a circle which touches one of the sides of a triangle exteriorly, and the other two sides produced. (See B. II., T. XII., S. II.) We have already found (Prob. IV.) the perpendiculars to be √(a+b+c)(−a+b+c) (a−b+c) (a+b−c) (1) 2 a √(a+b+c)(−a+b+c) (a−b+c) (a+b − c) (a+b−c), (2) 26 √(a+b+c) (− a+b+c) (a−b+c) (a+b−c) (3) 2c We have, also, under the same Problem, found the area to be A: 2 2 2 We have (B. III., T. XXV.) A= we also have = {(a + b + c ) ( = a + b + c ) ( a − b + c) (a + b = 0) { }. (4) 2 abc Substituting for P, its value already found, we have √(a+b+c) (− a+b+c) (a − b + c) (a + b − c) The sum of the areas of the three triangles ADB, BDC, CDA equals the area ABC. But ADB=rc, K B (6) } } (7) − a+b+c) (a − b + c) (a + b − e) 4 (a+b+c) We will now seek the radius r, of the escribed circle, whose centre is at E. |