But Cor. This is identical with the problem, "To find a geometric mean between two given segments," and it furnishes the means of constructing the segment a, when a=√bc, b and c being given. (165°) .. Ex. I. To construct an equilateral triangle equal to a given rectangle. B Let AC be the given rectangle, and suppose PQR to be the required triangle. Then AB. BC=PR.QT =PT.QT. QT=PT/3, PT.QT PT2√3 BF2=AB. BE, = (176°, Cor. 1) .. (181°, Cor.) whence PT2 AB/3.BC. And PT is the side of a square equal to the rectangle whose sides are AB√3 and BC, and is found by means of 181°, 127°, and 185°. Thence the triangle is readily constructed. Ex. 2. To bisect the area of a triangle by a line parallel to its base. E F P T R H Let ABC be the triangle, and assume PQ as the required line, and complete the parallelograms AEBC, KFBC, and let BD be the altitude to AC. Because PQ is || to AC, BD is 1 to PQ. Now EP C PC, 7FC= EQ, or PQ. BD=AC. BG. 2FQ=EC, or 2PQ. BG=AC. BD; But .. dividing one equation by the other, and reducing to one BD2=2BG2; A B K G D C (145°) (153°, 1) line, and therefore BG is one-half the diagonal of the square of which BD is the side, and the position of PQ is determined. 186°. Problem.—To find the circle which shall pass through two given points and touch a given line. Let A, B be the given points and L the given line. A B Constr.-Let the line AB cut L in O. Take OP=OP', a geometric mean between OA and OB (185°). The circles through the two sets of three points A, B, P and A, B, P' are the two solutions. The proof is left to the reader. (See 176°, Cor. 2.) S' 187°. Problem.-To find a O to pass through two given points and touch a given O. Let A, B be the points and S the given O. B P P Constr.-Through A and B draw any so as to cut S in two points C and D. Let the line CD meet the line AB in O. From O draw tangents OP and OQ to the OS (114°). P and Qare the points of contact for S the Os which pass through A and B and touch S. Therefore the Os through the two sets of three points A, B, P and A, B, Q are the Os required. D Proof. OB.OA=OC.OD=OQ2=OP2; therefore the Os through A, B, P and A, B, Q have OP and OQ as tangents (176°, Cor. 3). But these are also tangents to OS; therefore P and Q are the points of contact of the required Os. EXERCISES. 1. Describe a square that shall have twice the area of a given square. 2. Describe an equilateral triangle equal to a given square. 3. Describe an equilateral triangle having five times the area of a given equilateral triangle. 4. Construct AB/7, where AB is a given segment. 5. Construct √a2+b2 and √a2-b2, where a and b denote given line segments. Show that AC is the diagonal of the square on CB. Does this hold for external division also? 6. Divide the segment AB in C so that AC2=2CB2. 7. ABCD is a rectangle and DE, a part of DA, is equal to DC. EF, perpendicular to AD, meets the circle having A as centre and AD as radius in F. Then DF is the diagonal of a square equal to the rectangle. 8. In the Fig. of 183°, CE2=3AB. CB, CD2=CE2+3ED2=3AB(AB+CB). 9. Show that the construction of 183° solves the problem, "To divide a segment so that the rectangle on the parts is equal to the difference of the squares on the parts." 10. Show that the construction of 183° solves the problem, "To divide a given segment so that the rectangle on the whole and one of the parts is equal to the rectangle on the other part and the segment which is the sum of the whole and the first part." II. Construct an equilateral triangle when the sum of its side and altitude are given. What does the double solution mean? (See 183°, Ex. 2.) 12. Describe a square in a given acute-angled triangle, so that one side of the square may coincide with a side of the triangle. 13. Within a given square to inscribe a square having threefourths the area of the first. 14. Within an equilateral triangle to inscribe a second equilateral triangle whose area shall be one-half that of the first. 15. Produce a segment AB to C so that the rectangle on the sum and difference of AC and AB shall be equal to a given square. K 16. Draw a tangent to a given circle so that the triangle formed by it and two fixed tangents may be (1) a maximum, (2) a minimum. 17. Draw a circle to touch two sides of a given square, and pass through one vertex. Generalize this problem and show that there are two solutions. 18. Given any two lines at right angles and a point, to find a circle to touch the lines and pass through the point. 19. Describe a circle to pass through a given point and to touch a given line at a given point in the line. 20. Draw the oblique lines required to change a given square into an octagon. If the side of a square is 24, the side of the resulting octagon is approximately 10; how near is the approximation? 21. The area of a regular dodecagon is three times that of the square on its circumradius. 22. By squeezing in opposite vertices of a square it is transformed into a rhombus of one-half the area of the square. What are the lengths of the diagonals of the rhombus? 23. P, Q, R, S are the middle points of the sides AB, BC, CD, and DA of a square. Compare the area of the square with that of the square formed by the joins AQ, BR, CS, and DP. 24. ABCDEFGH is a regular octagon, and AD and GE are produced to meet in K. Compare the area of the triangle DKE with that of the octagon. 25. The rectangle on the chord of an arc and the chord of its supplement is equal to the rectangle on the radius and the chord of twice the supplement. 26. At one vertex of a triangle a tangent is drawn to its circumcircle. Then the square on the altitude from that vertex is equal to the rectangle on the perpendiculars from the other vertices to the tangent. 27. SOT is a centre-line and AT a tangent to a circle at the point A. Determine the angle AOT so that AS=AT. PART III. PRELIMINARY. 188°. By superposition we ascertain the equality or inequality of two given line-segments. But in order to express the relation between the lengths of two unequal segments we endeavour to find two numerical quantities which hold to one another the same relations in magnitude that the given segments do. Let AB and CD be two given segments. If they are commensurable (150°, 5) some u.l. can be found with respect to which the measures of AB and CD (150°, 2) are both whole numbers. Let m denote the measure of AB and n the measure of CD with respect to this unit-length. The numbers m and n hold to one another the same relations as to magnitude that the segments AB and CD do. m n The fraction is called in Arithmetic or Algebra the ratio of m to n, and in Geometry it is called the ratio of AB to CD. Now n has to m the same ratio as unity has to the fraction m But if CD be taken as u.l. its measure becomes unity, n while that of AB becomes m Therefore the ratio of AB to CD is the measure of AB with respect to CD as unit-length. When AB and CD are commensurable this ratio is expressible arithmetically either as a whole number or as a fraction; |