Ex. 570. Find the locus of a point the sum of the squares of whose distances from two given points is constant. Let A and B be the given points, and let P be any point of the required locus. To find the locus of P such that PA2+ PB2 is constant. Draw AB and bisect AB at M. Draw PA, PB, PM. Let PA2+ PB2 = k2, a constant quantity. M Hence, the required locus is the circumference of a circle whose centre is the middle point of the line that joins the two given points. Q. E. F. Discussion. If k2 is known, and k2 <2 AM2; that is, if k<AM√2, PM is an imaginary number and there is no locus. If k2 = 2 AM2, that is, if k = AM√2, PM = 0, and the circumference of the circle is reduced to the point M. Ex. 571. Find the locus of a point the difference of the squares of whose distances from two given points is constant. Let A and B be the given points, and let P be any point of the required locus. To find the locus of P such that PA2-PB2 Hence, the locus is two straight lines perpendicular to the line that joins the two given points at equal distances from the middle point of that line. Q. E. F. Discussion. If k2 is known, the distance MH is the third proportional to the lengths 2 AB and k. The other part of the locus is the line 1 to AB erected at the distance MH on the other side of M. Ex. 572. Find the locus of the vertex of a triangle that has a given base and the other two sides in the given ratio m: n. Let AB be the given base, and let P be taken so that PA: PB: =m: n. K The circumferences of circles on CD and C'D' as diameters are the locus required. D Ex. 567 Q. E. F. Ex. 573. To divide a given trapezoid into two equivalent parts by a line parallel to the bases. To divide the trapezoid ABCD into two equivalent parts by a line || to AB. Construction. Draw a rt. A with the leg a equal to AB, and the leg b = DC. Find d so that d2: c2 = 1:2. § 427 On AB take AH equal to d, and draw HG || to AD, and FG || to AB. FG is the line required. Proof. Produce AD and BC until they meet at E. FG AH = d. § 180 Since DC, FG, and AB are ||, A ABE, FGE, and DCE are similar. From (1), by division, AFGE A DCE: A DCE = FG2 - DC2: DC2. Substituting for FG2 its equal † (AB2 + DC2), ▲ FGE — ^ DCE ; ^ DCE = † (AB2 – DC2) : DC2. · (a2 + b2) § 333 Const. (3) From (2), by division, ▲ ABE – ▲ DCE : ▲ DCE = AB2 – DC2 : DC2. (4) .. the trapezoid FGCD≈ the trapezoid ABCD. Q. E. F. Ex. 574. To divide a given trapezoid into two equivalent parts by a line through a given point in one of the bases. Let ABCD be the given trapezoid and P the given point. To divide the trapezoid ABCD into two equivalent parts by a line through P. Construction. Draw the median EF and draw the altitude NM through O, the middle point of EF. D N HC Through P and O draw PH, meeting DC at H. Proof. The trapezoids APHD and PBCH have the equal medians EO and OF, and the same altitude NM. .. the trapezoid APHD≈ the trapezoid PBCH. 2. Suppose that Construction. Draw the altitude h. Find BF, the fourth proportional to BP, † (AB + DC), and h. Draw BFL to AB at B, FE || to BA, and EG || to FB, and draw PE. § 408 .. BP: (AB + DC) = h: EG. The area of the trapezoid ABCD. Ex. 575. To construct a regular pentagon, given one of the diagonals. On AC, homologous to A'C', construct ▲ ABC similar to ▲ A'B'C'. On AC, homologous to A'C', construct the quadrilateral ACDE similar to the quadrilateral A'C'D'E'. ABCDE is the regular pentagon required. § 391 Q. E. F. Ex. 576. To divide a straight line into two segments such that their product shall be the maximum. Since the maximum of isoperimetric polygons of the same number of sides is a regular polygon (§ 489), the maximum quadrilateral whose perimeter is equal to twice the given line is a regular quadrilateral, that is, a square; and each side of the square is equal to half the given line. Therefore, bisect the given line, and the product of the two segments is the maximum. Q. E. F. Ex. 577. To find a point in a semicircumference such that the sum of its distances from the extremities of the diameter shall be the maximum. Let ACB be a semicircle, C the middle point of the arc ACB, and D any other point. To prove that AC + BC > AD + BD. Proof. Produce AC to F, making CF equal to AC. Draw BF, and produce AD to E, making DE equal to DB. Draw EB, and produce it to meet AK drawn through A 1 to EB. = Since CA CB CF, B is on the semicircumference on AF as a diameter. Ex. 578. To draw a common secant to two given circles exterior to each other such that the intercepted chords shall have the given lengths a, b. Let O and O' be the centres of two exterior to each other. To draw a common secant so that the intercepted chords shall have the given lengths a and b. Construction. In the whose centre is O, draw a chord AB equal to the given length a, and in the O whose centre is O', draw a chord CD equal to the given length b. Draw OE to AB and O'FL to CD. With centre O and radius OE, describe a circumference, and with centre O and radius O'F, describe a circumference. Now any chord in the given O with centre O tangent to the concentric O is equal to a, and any chord in the given O with centre O' tangent to the concentric O is equal to b. Draw MN a common tangent to the auxiliary ©. § 249 Ex. 245 MN is the common secant required, as is obvious from the construction. Q. E. F. Discussion. Since the auxiliary are exterior to each other, two common external and two common internal tangents may be drawn, each of which satisfies the given condition. Ex. 579. To draw through one of the points of intersection of two intersecting circles a common secant which shall have a given length. Let O and O be the centres of the two given and P one of the points of intersection of the two . M To draw through P a common secant which shall have a given length m. B N Construction. Draw the line of centres 00', and on 00' as diameter describe a semicircle. With O' as centre and radius equal to m, describe an arc, cutting the semicircle at C. Draw CO. Through P draw the common secant MN || to CO. Then MN is the common secant required. P Proof. Produce OC to meet MN at A, and draw O'B 1 to MN. |