ON THE ALGEBRAICAL SOLUTION OF GEOMETRICAL PROBLEMS. In the case of Prop. 11 it was shown that the problem could be stated algebraically thus, a (a x) = x2; and in solving this quadratic equation we arrived at the following equation : The right-hand member of this equation might have suggested that in order to solve the problem we must find a line whose square is five times the square on half the given line. And this was done, for the square on EF is equal to five times the square on CE, which is equal to half AB. (See p. 145, Ex. 1. (vi).) Similarly in other cases a geometrical problem may often be stated algebraically, and the solution of an equation may suggest the geometrical solution of the problem. Take for example the following problem : To divide a given straight line into two parts, so that the square on one part may be equal to twice the square on the other part. Let a denote the length of the given straight line; and let x denote the length of one part; then a x will be the length of the other part. It is required to find the value of x which satisfies the equation. (2) Now equation (1) suggests that we must find a line, the square on which is equal to twice the square on the given line; and equation (2) suggests that the required part x will be found by taking a line equal in length to twice the given line, and then adding or subtracting the length of the line suggested by equation (1). We know that the square on the diagonal of a square is equal to twice the given square; therefore we get the following solution : A P B E Let AB be the given straight line; on AB describe the square ADEB, and join DB. Then add to or subtract from AC a part equal to DB, 2BQ 2. The algebraical solution will not do more than suggest that some line must be found whose length bears a certain ratio to the length of the given line. In the case given, the length of line required is given by the diagonal of the square described on the given line. In other cases other well-known figures will often give the required lengths; but it must depend on the ingenuity of the student to make the right selection. EXERCISES ON GEOMETRICAL PROBLEMS. (To be solved algebraically and geometrically.) 1. Construct a rectangle equal to a given rectangle in area, but having one of its sides three times the length of one of the sides of the given rectangle. 2. Construct a rectangle having given its area and perimeter. 3. Divide a given straight line into two parts so that the sum of the squares on them may be equal to a given square. 4. Divide a given straight line AB at C, so that the rectangle contained by BC and a given line may be equal to the square on AC. 5. Divide a given straight line into two parts so that the rectangle contained by the two parts may be equal to the rectangle contained by the sum and difference of the parts. 6. Divide a given straight line both externally and internally into two parts so that the square on one part may be equal to three times the square on the other part. 7. Divide a given straight line both externally and internally so that twice the square on one part may be equal to three times the square on the other part. 8. Divide a straight line into two parts so that the rectangle contained by the two parts may be a maximum. 9. Divide a given straight line both externally and internally into two parts so that the square on one part may be equal to twice the rectangle contained by the whole line and the other part. MISCELLANEOUS EXERCISES AND RIDERS ON BOOK II. On Propositions 1-11. 1. If there be two straight lines, each of which is divided into any number of parts, the rectangle contained by the straight lines is equal to the sum of the rectangles contained by each of the parts of the first line and each of the parts of the second line. 2. If D be a point in the hypotenuse of a right-angled triangle, such that the rectangle BD, BC is equal to the square on AC, prove that the rectangle BC, DC is equal to the square on AB. 3. In any right-angled triangle twice the sum of the squares on the three medians is equal to three times the square on the hypotenuse. 4. If the straight line AB be divided into two parts at C so that the square on AB is double the square on AC, and if D be taken in AC so that AD is equal to CB, prove that the square on CD will be double the square on AD. 5. If the straight line AB be divided into three parts at C and D so that AD is equal to CB, and if the square on CB be equal to three times the square on AC, prove that the rectangle AB, CD is equal to twice the square on AC. 6. If a straight line be divided into two pairs of unequal parts, prove that the sum of the squares on the greatest and least of the four parts is greater than the sum of the squares on the other two parts. Express this algebraically. 7. ABC is an equilateral triangle and D any point in BC; prove that the square on BC is equal to the rectangle BD, DC, together with the square on AD. 8. State and prove the theorem corresponding to No. 7 when D is in BC produced. 9. If the straight line AB be divided into any two parts at C, and if AC and CB be bisected at D and E respectively, prove that the square on AE, together with three times the square on EB, is equal to the square on BD, together with three times the square on DA. 10. If the straight line AB be bisected in P and produced to Q so that the rectangle AQ, BQ is equal to twice the rectangle AB, PQ, prove that the square on BQ is equal to the rectangle AB, AQ. 11. Any rectangle is equal to half the rectangle contained by the diagonals of the squares described on two of its adjacent sides. 12. A and B are two fixed points, and P is a point which moves so that the difference of the squares on PA and PB is constant; show that the locus of P is two straight lines at right angles to the line AB, 13. If AB be divided in medial section internally at H and externally at K, prove that: (i) The rectangle BK, HA is equal to the rectangle BA, AK. (ii) The rectangle BH, AK is equal to the rectangle BA, HA. (iii) The square on HK is equal to five times the square on AB. (v) The sum of the squares on BH and BK is equal to seven 14. If AB be divided internally at C and externally at D so that the square on AC is double the square on BC, and the square on AD is double the square on BD, prove that : (i) the rectangle AC, AD is equal to twice the square on AB. (ii) the rectangle CB, BD is equal to the square on AB. (iii) the rectangle AD, CB is equal to the rectangle AC, BD. (iv) the square on CD is equal to eight times the square on AB. (v) if P be the middle point of CD, B is the middle point of AP. (vi) the difference between the rectangle AD, BD and the rectangle AC, CB is equal to eight times the square on AB. 15. If AB be divided internally at P and externally at Q so that the square on AP is equal to twice the rectangle AB, PB, and the square on AQ is equal to twice the rectangle AB, QB, prove that :— (i) the rectangle PA, AQ is equal to twice the square on AB. (ii) the rectangle BP, BQ is equal to the square on AB. (iii) the rectangle PA, BQ is equal to the rectangle BA, AQ. (iv) the rectangle BP, AQ is equal to the rectangle BA, PA. (v) the sum of the squares on AP, AQ is equal to eight times the square on .AB. (vi) if M be the middle point of PQ, A will be the middle point of BQ. On Propositions 12 and 13. 16. In any triangle three times the sum of the squares on the sides of the triangle is equal to four times the sum of the squares on the medians. 17. ABC is a triangle having the angle BAC a right angle; prove that four times the sum of the squares on the medians BE and CF is equal to five times the square on the hypotenuse BC. 18. ABC is any triangle and its medians intersect in G; if P be any point, show that the sum of the squares on AP, BP, CP is greater than the sum of the squares on AG, BG and CG by three times the square on GP. 19. ABC is an isosceles triangle having the side AB equal to the side AC; show that if AB is produced to D so that AB equals BD, the square on CD is equal to the square on AC, together with twice the square on BC, 20. If DE be drawn parallel to the base BC of an isosceles triangle ABC, the square on BE is equal to the square on CE, together with the rectangle contained by BC, DE. 21. The rectangle contained by the sum and difference of any two sides of a triangle is equal to twice the rectangle contained by the base and the line intercepted between the middle point of the base and the foot of the perpendicular from the vertex to the base. 22. ABC is an acute-angled triangle, and BE and CF are drawn perpendicular to the opposite sides; prove that the sum of the rectangles AB, FB and AC, EC is equal to the square on BC. 23. ABC is an acute-angled triangle, and perpendiculars APD, BPE are drawn on BC, CA from the opposite angles; prove that the rectangle AD, PD is equal to the rectangle BD, DC. 24. In the figure of Prop. 13 if BE be drawn perpendicular to AC, prove that the rectangle AC, EC is equal to the rectangle BC, DC. 25. ABC is an acute-angled triangle and perpendiculars APD, BPE, CPF are drawn to the three sides from the opposite angles; prove that the rectangles AP, PD and BP, PE and CP, PF are all equal to one another. : 26. ABC is an isosceles triangle having each of the angles at the base double the vertical angle BAC; if CD be drawn perpendicular to AB, prove that :(i) the square on AC is equal to twice the rectangle AD, BC; (ii) the sum of the squares on AD, DB is equal to the square on CD ; (iii) the square on AC is equal to the square on BC, together with the rectangle AC, BC; (iv) the rectangle BA, BD is equal to half the square on BC; (v) the square on CD is equal to the square on BD, together with twice the rectangle BD, DA. 27. If AD be drawn perpendicular to BC, a side of the triangle ABC, prove that the angle BAC will be obtuse or acute according as the rectangle BD, DC is greater or less than the square on AD. 28. ABC is any triangle, having B and C acute angles, and squares are described externally on its three sides, using the same lettering as in I. 47 ; then BN and CM are drawn perpendicular to CA and BA respectively, produced if necessary; complete the rectangles HANR and GAMQ and prove them equal to each other. Hence deduce proofs of Props. 12 and 13, according as the angle BAC is obtuse or acute. 29. In the figure of I. 47, if BAC be any angle, prove that the difference of the squares on AB and AC is equal to the difference of the squares on AD and AE. |