18. In the figure ABC is an equilateral triangle. A, B, C are the centers of the arcs BC, CA, and AB. Semicircles are constructed on the diameters AB, BC, CA. Let AB = a. (See Ex. 11, p. 93.) Circles are constructed tangent to the various arcs as in the figure. (a) Find the radius r'. SUGGESTION. Find in order BM, MN, BN, HO", BO". (b) Find the radius r'. SUGGESTION. BN is known from the solution of (a). Substituting for HO', KB and KO' in (1) we may solve for r'. (c) Find the radius r. (1) SUGGESTION. Use A BLO and HLO and proceed as under (b). (d) Using the radii thus found, show how to construct the figure. (e) What fraction of the whole area is contained in the circles? 19. ABCD is a parallelogram with fixed base and altitude. Find the locus of the intersection points of the bisectors of its interior base angles. 20. Find the locus of a point P such that the sum of the squares of its distances from two fixed points is constant. SUGGESTION. By § 440, AP2 + PB2 = 2 AC2 + 2 CP2. Solve for CP. B 21. Find the locus of a point P such that the ratio of its distances from two fixed points is equal to the constant ratio m: n. SUGGESTIONS. Let A and B be the fixed points. On the line AB there are two points P' and P" on the locus, i.e. AP":P"B= AP' : P'B=m: n. Let P be any other point on the locus. Then AP: BP: AP': P'B = AP": P"B. = K B P Show by the converses of §§ 250, 253 that PP' bisects ZAPB and PP" bisects / BPK, and hence P'P and PP form a right angle. 23. The points ABCD are collinear. Find the locus of a point P from which the segments AB and CD subtend the same angle. SUGGESTIONS. The ABP, ACP, BDP, and P CDP have a common altitude, and hence their areas are to each other as their bases. Also ABP and CDP have equal vertex angles, whence by § 427 their areas are to each other as the products of the sides forming these angles. Similarly for ▲ ACP and BDP. Hence AP PB AB and AP.PC AC B C From these equations show that AP: PD is a constant ratio. 24. ABC is a fixed isosceles triangle. With center C and radius less than AC, construct a circle, and from A and B draw tangents to it meeting in P. Find the locus of P. SUGGESTIONS. (a) Show that part of the locus is the straight line PP'. (b) Show that 21=22 and hence that ZAP"B is constant. P B MAXIMA AND MINIMA. 445. Definitions. Of all geometric figures fulfilling certain conditions it often happens that some one is greater than any other, in which case it is called a maximum. Or it may happen that some one is less than any other, in which case it is called a minimum. E.g. of all chords of a circle the diameter is the maximum, and of all segments drawn to a line from a point outside it the perpendicular is the minimum. In the following theorems and exercises the terms maximum and minimum are used as above defined. However, a geometric figure is often thought of as continuously varying in size, in which case it is said to have a maximum at any position where it may cease to increase and begin to decrease, whether or not this is the greatest of all its possible values. Likewise it is said to have a minimum at any position where it may cease to decrease and begin to increase. I A B C D E F E.g. if in the figure a perpendicular from a point in the curve to the straight line be moved continuously parallel to itself, the length of this perpendicular will have maxima at A, C, and E, and minima at B, D, and F. Certain simple cases of maxima and minima problems. have already been given. Some of these will be recalled in the following exercises. 1. If from a point within a circle, not the center, a line-segment be drawn to meet the circle, show that this segment is a maximum when it passes through the center and a minimum when, if produced in the opposite direction, it would pass through the center. 2. Show that of all chords through a given point within a circle, not the center, the diameter is a maximum and the chord perpendicular to the diameter is a minimum. 3. Of all line-segments which may be drawn from a point outside a circle to meet the circle, that is a maximum which meets it after passing through the center, and that is a minimum which, if produced, would pass through the center. 4. Show that if a square and a rectangle have equal perimeters, the square has the greater area. SUGGESTION. If s is the side of the square and a and b are the altitude and base of the rectangle respectively, then 2 b + 2 a = 4s or 5. Use the preceding exercise to find that point in a given linesegment which divides it into two such parts that their product is a maximum. 6. Find the point in a given line-segment such that the sum of the squares on the two parts into which it divides the segment is a minimum. SUGGESTION. If a and b are the two parts and k the length of the segment, then or, a+b= k and a2 + 2 ab + b2 = k2, a2 + b2 k2 2 ab. = Hence, a2+b2 is least when 2 ab is greatest. Now apply Ex. 5. 7. In the preceding exercise show that a2 + b2 increases as b grows smaller, that is, as the point of division approaches one end. When is this sum a maximum? 8. Two points A and B, on opposite sides of a straight stream of uniform width, are to be connected by a road and a bridge crossing the stream at right angles. Find by construction the location of the bridge so as to make the total path from A to B a minimum. B 447. THEOREM. Of all triangles having equal perimeters and the same base, the isosceles triangle has the maximum area. G Given ABC isosceles and having the same perimeter as ▲ ABD. To prove that area ▲ ABC> area ▲ ABD. Outline of proof: Draw CE ↓ AB. Construct A AFB Draw GB and GD. The ob having its altitude FE the same as that of ▲ ABD. Prolong AF making FG = AF. ject is to prove that EF, the altitude of ▲ ABD, is less than EC, the altitude of ▲ ABC. (1) Show that ▲ AFB, FBG, and GBD are all isosceles, for which purpose it must be shown that GB AB and FD || AB. AD+DBAD + DG>AG. Then Or But Hence, AF+FB <AD+ DB. AD + DB = AC+ CB. AFFB <AC+CB. (3) Use the last step to show that area ABC > area ▲ ABD. Give all the steps and reasons in full. 448. COROLLARY. Of all triangles having the same area and standing on the same base, that which is isosceles has the least perimeter. |