843. In a triangle, given the base and altitude, and the ratio of the other sides, to determine the triangle. 844. Given the base, the medial line, and the sum of the other sides of a triangle, to determine the triangle. 845. To determine a right angled triangle, knowing the perimeter and area. SUG'S. x2 + y2 = z2, x + y + z = 2p, x2 + 2xy + y2 = 4p3 4pz + 22, 22 Now use y + x = 2p 2= 4pz + 22; whence and xy = 282, give y + x = 2p − 2, + 482 = 4p2 Р 846. To determine a right angled triangle, knowing the perimeter, and the sum of the hypotenuse, and the perpendicular upon the hypotenuse from the right angle. zu. Then 4pz, and hence SUG'S. x2 + y2 = 22, x + y + z = 2p, z + u = a, 4pz + 22; whence 2xy = 4p2 xy = x2 + 2xy + y2 = 4p2 2z (a 847. The volume, the altitude, and a side of one of the bases of the frustum of a square pyramid being known, to determine a side of the other base. 848. To determine a right angled triangle, knowing the perimeter, and the perpendicular let fall from the right angle upon the hypotenuse. 849. To determine a triangle, knowing the base, the altitude, and the difference of the other sides. 850. To determine a triangle, knowing the base, the altitude, and the rectangle of the other sides. 851. To determine a right angled triangle, knowing the hypotenuse and the difference between the lines drawn from the acute angles to the centre of the inscribed circle. SUG's. Let fall CD a perpendicular upon AO produced. Now, since the the angles BAC and ACB are bisected, and COD = OAC + OCA, and ICD = IAB, they being complements of the equal angles CID, IAB, we have, COD = OCD, and CD = OD = √ Co. Hence, putting AC = h, CO = x, and AO = x + d, we have FIG. 442. (x + d + √ ‡ x)2 + (√‡ x)2 = h2. From this x is readily found. The student should then be able to complete the solution. 852. Given two sides of a triangle and the bisector of their included angle, to determine the triangle. 853. Given the three medial lines, to determine a triangle. 854. Given the three sides of a triangle, to determine the radius of the circumscribed circle. 855. Four equal balls whose radius is r are placed on a plane so that each is tangent to the other three, thus forming a pyramid; what is its altitude? 856. Given the base of a triangle, the bisector of the opposite angle, and the radius of the circumscribing circle, to determine the triangle. 2 SUG's. First to find ED = x. Since EM = r — √r2 - b2, it may be considered known and put equal to c. We then have DM = √c2 + x2; and also, DM x α = AD × DB = b2 x2, or DM = a Calling ED = 8, the student will have no difficulty in proceeding with the solution. CHAPTER II. INTRODUCTION TO MODERN GEOMETRY.* SECTION I. OF LOCI. 857. The term Locust, as used in geometry, is nearly synonymous with geometrical figure, yet having a latitude in its use which the other does not possess. The locus of a point is the line (geo With strict propriety only the latter sections of this chapter belong to the Modern Geometry, technically so called. But, as the entire chapter is composed of matter which has not hitherto found place in our common text-books, and the relative importance of which is becoming more fully appreciated in modern times, the author has ventured to embrace the whole under this title. The word Locus is the Latin for place. metrical figure) generated by the motion of the point according to some given law. In the same manner, a surface is conceived as the locus of a line moving in some determinate manner. ILL'S. The locus of a point in a plane, which point is always equidistant from the extremities of a given right line, is a straight line perpendicular to the given line at its middle point. Thus, suppose AB a fixed line, and the locus of a point equidistant from its extremities is required; that point may be anywhere in a perpendicular to AB at its middle point, and cannot be anywhere else in this plane. This perpendicular is the locus (place) of a point subject to the given law. FIG. 444. Again, a boy on the green is required to keep at just 20 feet from a certain stake; where may he be found? i. e., what is his locus (place)? Evidently, the circumference of a circle whose radius is 20 feet. Thus, the locus of a point in a plane, equidistant from a given point, is the circumference of a circle. This is the place of such a point. What is the locus in space of a point equidistant from a given point? What is the locus of a point in space equidistant from the extremities of a given line? A plane. What is the locus of a line moving so that each point in it traces a right line? In general, a plane; if it move in the direction of its length, a straight line. What is the locus of a right line parallel to and equidistant from a given line? What is the locus of a right line intersecting a given line at a constant angle? * A conical surface of revolution. What is the locus of a semicircle revolving on its diameter? PROPOSITIONS AND PROBLEMS IN DETERMINING LOCI. [NOTE. The student should be required to give every demonstration in form, and in detail. Frequent exercise in writing out demonstrations, is almost the only method of securing a good, independent style in demonstration. Plane loci are always understood if the contrary is not stated.] 858. Theo. The locus of a point in a plane, equidistant from the extremities of a given line, is a perpendicular to that line at its middle point. SUG. To prove this we have simply to show that every point in such a perpendicular is equidistant from the extremities of the given line, and that no other point has this property (PART II., 129). * That is, an angle which remains of the same size, 859. Prob.-Find the locus of a point at any constant distance m from a given straight line. Of what proposition in PART II. is this the converse? SUG's. To prove the proposition which the answer to this question asserts, it will be necessary to show that every point in the affirmed locus is at the same C a O E m E' B D' distance from the given line and that no other point is at that distance. We affirm that the locus is two right lines parallel to the given line and at a distance m therefrom. The formal demonstration is as follows: Let AB be the given line, and OE, OE', perpendiculars thereto, each equal to m. Through E and E' draw CD and C'D' parallel to AB; then is CD, C'D', the locus required.* For, by Part II. (156), every point in CD, C'D', is at the distance m from AB; and we may readily show that any other point, as P or P', is at a distance greater or less than m from AB. Hence CD, C'D', is the locus required. FIG. 445. 860. Theo.—In a circle, the locus of the centre of a chord parallel to a given line is a diameter perpendicular to the given line. DEM.-Let mn be any circle, and AB a given line. Then is the locus of the centre of a chord parallel to AB, a diameter of the circle. For, let DH be any chord parallel to AB. Through the centre of the circle C, and P, the middle point of DH, draw EL. Now EL is perpendicular to DH (?), and consequently to AB (?). Then will EL be perpendicular to any and every chord parallel to DH (?), and hence will Therefore the locus of the centre of a chord parallel to Again, any point in the circle and out of the line EL is not the middle point of chord parallel to AB. Thus, letting P' be such a point, draw a chord through P' parallel to AB. As there can be but one such chord (?), and as EL bisects it (?), P' is without the diameter (?). 861. Theo.-The locus of the centre of a circumference passing through two given points is a straight line perpendicular to the line joining the two points at its middle point. SUG.-Consult PART II. (159, 163, 197). 862. Theo.-The locus of the centre of a circle which is tan * It is important that the student think of these two lines as one locus, or as parts of one and the same locus, if this will aid the conception. A locus may consist of any number of detached parts; all that is necessary being that the given conditions be fulfilled. In this respect the word locus has a more enlarged meaning than the term geometrical figure. gent to a given circle at a given point, is a straight line passing through the centre of the given circle. PP T A DEM.-Let C be the centre of the given circle, and B the point in the circumference to which the circle shall be tangent, the locus of whose centre is required. Through B draw TL tangent x to the given circle. Now, a circle passing through B, and tangent to the given circle, will have TL for its tangent (?), and as a radius is perpendicular to a tangent at its extremity, and only one perpendicular can be drawn to TL through B, the centre of a circle tangent to the given circle at B must be in this straight line. Moreover, as the given circle is tangent to the right line TL at B, its centre is in the perpendicular AX. Hence AX is the locus required. FIG. 447. 863. Theo.—The locus of the centre of a circle of given radius R, and tangent to a given straight line, is two parallels to this line at a distance R therefrom, on each side. Give proof in form. 864. Prob.—Find the locus of the centre of a circle of given radius R, whose circumference passes through a given point. Give proof in form. 865. Theo.-The locus of the centre of a line of constant length, having its extremities in two fixed lines which cut each other at right angles, is the circumference of a circle. SUG'S.-Let MN be the length of the given line, and CD, and AB, the two lines intersecting at right angles, in which the extremities of MN are to remain. Now, in whatever position MN may be placed, its middle point, P, is at the same distance (MN) from O (?). To show that any point not in this circumference, as q, is not the middle point of a line equal to MN passing through it, and limited by the fixed lines, from as a centre, with a radius MN cut CD, as in C; and from C as a centre with ΑΝ M P N O M M M FIG. 448. the same radius strike the arc P. If p is without the circle, CB > MN, if within, less. Hence, the required locus is a circumference whose centre is O, and whose radius is MN. • Observe the form of expression. We say "the circle," and not "the circles," using the term in a generic sense, as including all which have the required property, i. e., all which are tangent to the given circle at B. |