PROBLEMS INVOLVING PROPERTIES OF THE CIRCLE. PROBLEM I.—Through a given point within a given circle to draw a chord of a given length. P H D F the diameter AB of the circle, the B Given point P, that is, AP and PB, and the length of the chord GPF. draw Analysis.-Through the given point P, and the centre C, the diameter APB. Apply the chord BE the given chord GF; and from the centre C let fall the perpendiculars CD and CH; then, since BE GF, we have (III. 8*) CD=CH. Whence this Construction. -- Having drawn the diameter APB, from its extremity B apply the chord BE― the given chord, and from the centre C let fall on it the perpendicular CD. On CP describe a semicircle, in which apply CH=CD, and through P and H draw the chord GPHF, which will be equal to BE, and be the chord required. Demonstration.—The angle PHC, being in a semicircle, is a right angle. Hence CH is perpendicular to the chord GPF, and, it being equal to CD by construction, we have (IH. 8†) GPF=BE. Q. E. D. Calculation.—1. The perpendiculars CD and CH (III. 6†) bisect the chords BE and GF respectively; hence BD-BE- FG = FHGH, and CH2 CD2BC2 — BD2, PH=√PC2—CH2. 2. Now, GP GH-PH, and PF GH+ PH=FH+ PH. PF=GH+ * I I. 14. (96) † III. 14. † III. 3.. Limits.-1. The given chord cannot be greater than the diameter ΑΒ. 2. The distance of the given chord from the centre cannot be greater than CP. When it is equal to CP, the required chord will be a perpendicular to the diameter through P. Hence the shortest chord that can be drawn through a given point is the one which is perpendicular to the diameter at that point. PROBLEM II.-Through a given point without a given circle to draw a right line so that the chord intercepted by the circumference shall be of a given length. Analysis.-Through the given point P and the centre C draw the diameter PAB. Apply the chord BE the given chord GF, and from the centre C let fall the perpendiculars CD and CH; then, since BE = GF, we have (III. 8*) CD: CH. Whence this Construction.—Having drawn PACB, from B, apply the chord BE- the length of the given chord, on which let fall from the centre C the perpendicular CD. On CP describe a semicircle, in which apply CH⇒ CD, and through the points P and H draw the line PGF; then GF will be equal to BE, and PGF be the line required. Demonstration.-The angle PHC, being in a semicircle, is a right angle. Hence CH is perpendicular to the chord GF, and, being equal to CD by construction, GF= BE (III. 8†). Q. E. D. Calculation. The perpendiculars CD and CH (III. 6) bisect the chords BE and GF respectively; hence BDBE FG Now, CH2 CD2 = BC2 Then GP= Then GP FH GH. ✔ CP2 — CH2. PH + FH. BD2, and PH PH—GH, and PF-PH+GH= Limit.—The given chord GF must not be greater than the diameter AB. * III. 14. E † III. 14. † III. 3. Scholium.-To draw a tangent from the given point P without the circle, join P with the point T where the semicircle on PC intersects the semicircle AFB, and PT will be a tangent to the circle AFB, for (III. 18, cor. 2*) angle PTC is a right angle. PROBLEM III.-With a given radius it is required to describe three equal circles which shall touch one another, and then to describe another circle which shall touch them all three. Analysis. Suppose the problem constructed as in the above figure. Join the centres A, B, and C, and an equilateral triangle ABC will evidently be formed, having each of its sides equal to twice the given radius. Bisect two of the angles A and B by the lines AO and BO, and join OC, which three lines will evidently be equal. Produce these three lines; then they will intersect the equal circles in the points G, H, and I; also, before being produced, in the points G', H', and I'. Now, since the lines OA, OB, and OC are all equal, we have OH, OI, and OG all equal, as also OH', OI', and OG' all equal. Whence this Construction.-Describe the equilateral triangle ABC, having each of its sides double the given radius. Then, with A, B, and C as centres, and the given radius, describe the three circles, which will evidently touch each other at D, E, and F, the middle points of the sides. Bisect the two angles A and B by the lines 40 and BO, intersecting in O, and join OC. These three lines, and these lines produced, intersect the given circles in the points G', H', and I' and G, H, and I. Then, with centre O, and radii OH' and OH, describe the circles H', I', G' and H, I, G respectively, and the problem is constructed as required. 3 III. 31. The demonstration is all evident from the analysis. Calculation. In triangle AOB, Case 1, find OA; then OH= OA+AH, and OH' OA-AH', the radii required. NOTE. From the above is immediately deduced the following PROBLEM IV. Having given the radius of a circle, to inscribe therein three equal circles which shall touch one another and also touch the given circle. (See figure on opposite page.) Analysis. Suppose the problem constructed with OH or OI the given radius. Join GH, HI, and IG, and HIG will evidently be an equilateral triangle. Join OG, OH, OI, and CI, and draw GL parallel to CI, meeting OI, produced, in L. Then, in similar triangles CBI and GIL, since IB = BC, we have IL IG. Whence this Construction.-Having described the circle with the given radius, draw the diameter GOK, and in the circle apply the chords KI and KH, each equal to the given radius KO. Join GH, HI, and IG; then (V. 4*) HIG will be an inscribed equilateral triangle, having the side GI-=✔ GK2 KI2 √ 4 OK2 — OK2 2 √30K2 OK√3. Also, join OG, OH, and OI, and produce OI, making IL IG. Join LG, draw IC parallel to LG, CB parallel to GI, and CA parallel to GH, and join AB. With the centres A, B, and C, and the radii AH, BI, and CG respectively, describe the circles. HED, IFE, and GDF, which will be the equal circles required, touching each other at the points D, E, and F. Demonstration.--Because of the parallel lines, it is evident that AH, BI, and CG are all equal, that ABC is an equilateral triangle, and that the triangles GIL and CBI are similar. Hence it only remains to be proved that the three circles touch each other. In the similar triangles GIL, CBI, since GI=2 IL by construction, we have CB=2 BI⇒ CF + FB = CD + DA AE + EB. Hence, the distances between the centres of the circles being equal to the sum of the radii (III. 13†), the circles touch each other externally in the points D, E, and F. Q. E. D. Calculation. By parallel lines, OL: IL:: OG: CGBI= AH, and 2 CG AC = AB = BC; also, OG – CG OC - OA OB. * IV. 15. † III. 12. PROBLEM V.-To describe a circle which shall pass through two given points and have its centre in a straight line given in position. Analysis. Suppose the figure constructed as above. Join AB, and from C, the centre of the circle, let fall on AB the perpendicular CD; then (III. 6*) AD=DB. Whence this Construction.-Join the given points A and B, bisect the line AB in D, and draw DCG perpendicular to AB, and the point in which this perpendicular cuts the given line FE will be the centre of the circle required. In the figure this point is C. Join CB, and with it as a radius, and centre C, describe the circle ABH, which will be the one required. The demonstration is evident from the preceding. Calculation. Let the line joining A and B be produced, if necessary, to meet the given line FE in P; then the angle P, and the distance PA or PB, are known from the given position. of FE. Thence PA, PD, and PB are known, by adding or subtracting, as the point P is in AB produced, or between A and B. And DB or DA = 1⁄2 AB. In triangle PDC, Case 1, find PC and CD; then radius CB= ✔ CD2 + DB2 CA. Limits.-1. When FE is parallel to AB, CD will be the perpendicular distance between the two parallel lines, and, as FE is given in position, the distance CD is known, and thence the radius CA or CB √ CD2+DB2, as before. 2. When the given angle at P is a right angle, FE and DG will be parallel, and the problem is impossible; that is, in the language of mathematicians, "the radius would be infinite, and the arc of a circle with an infinite radius is a straight line," so that the straight line passing through A and B would fulfil the conditions of the problem mathematically. * III. 3. |