622 



KNOWLEDGE 



[Mat 26, 1882. 



(Bur i^latbnnati'ral Column. 



SOLUTIONS TO PROBLEMS. 



[P. 526] — In a triangle given a-i-b,a + c, ond angle A, to construct 

 the triangle. 



Wo have received correct Bolntions to this problem (Knowledoe, 

 April I'l-) from R. F. K., E. N. Dalton, C. K., J. P. Morgan, J. B., 

 Thos. Lyons, and others. The following construction and proof is 

 somewhat simpler than any sent us; Mr. Ualton's is nearest to it, 

 and is in principle the same j but the construction of a species of 

 anxiliary triangle is unnecessary and inelegant . 



Fig. 1. 



Let AD = a+c; AE = a + !/; DAE=angleA and produce AD the 

 less to U, making AH = AE. With centre H, distance HA, describe 

 a circle cutting ED produced in K; draw DL parallel to KH to 

 meet HE, LC parallel to AH, and CB parallel to DL. Then ABC 

 is the recpiired triangle. 



For triangle ECL, being similar to EAH, is isosceles ; and CLDB 

 is a parallelogram. Now, 



DL : KH::EL : En::EC : EA 



But KH = AH = EA; 



therefore DL = EC = CL=BD 



or BC = EC = BD 



.So that AC+BC = AE = u + b 

 and AB + BC = AD = a + c. 



[33] — Tohisect a triamjle, ABC, hy a line draicn /mm n gtven 

 point, D, without it. 



The neatest solution we have received of this problem is the 

 following : — 



Continue out the side AC to meet GD 

 D / \ parallel to AB in point G ; and bisect the side 



AC in E. Then make, as GD is to AB so 

 AE to AH. And again, make as GX is to 

 AF so HF to AH ; then, it the line DLF be 

 dna^vn, it will bisect the triangle ABC ; i.e., 

 . L, ^,. C the triangle ALF will be half of the triangle 

 ABC. J. C. Kelly. 



Fig. 2. [Mr. Kelly does not give the proof, but it 



is tolerably obvious. Thus, by construction : — 



GA : AF ::HF : AH 



.-.GP : AF :: AF : AH 



GF': AF»:: GF : AH 



Now triangle ABC : triangle GDF;:AC.AB' : GF.GD 



::AC.AH : GF.AE 



And triangle GDF : triangle ALF::GF' : AF^ 



::GF : AH 



Wherefore triangle ABC : triangle .ALF::AC : AE 



::2:i — Ed.] 



Wo havo received a neat solution from C. E., and other correct 

 solations from R. L., Ezon, J. P., and others. 



[37] — From a given point to draw two equal straight lines to two 

 given lines, the straight lines thus drawn to include a given angle. 



Fig. 3. 



The neatest solution sent ns is the following : — Let AB, AC, be 

 the two given lines, D the given angle, and P the given point. 

 First let P be outside the angular space BAG. Draw PE perjjen- 

 dicnlar to AC, and make the angles EPF, EPG each equal half the 

 given angle D. At the point F make the angle PFK equal PGC, 

 aud cut off GL equal FK, and join KP, PL. KP, PL are the lines 

 required. iSince the triangles PEF, PEG are equal in every respect, 

 PF equal PG. Again PF, PK are resiiectively equal to PG, GL, and' 

 the angle PFK to tho angle PGL ; therefore, PK is equal to PL, and 

 the angle FPK to the angle GPL; therefore the whole angle FPG 

 is equal to KPL, and FPG is equal to the given angle D. 



Fig. i. 



Next, let P be -within the angular space BAC. Draw the per- 

 pendicular PE, and construct the triangles EPF, EPG as before. 

 Make tho angle PGK equal PFG, and cut off FL equal GK, and 

 join LP. Because triangles PFL, PGK are equal (Euclid i. 4) 

 in every respect, therefore, PL equal PK, and angle FPL equal 

 GPK. Take away the common angle GPL, therefore angle FPG 

 equal LPK, and FPG is equal to the given angle D. 



[D. M. tails to note, however, that there are two solutions to 

 each case. If we make angle PGK' equal to PFA in the first case, 

 and in CA produced if necessary, make FL' equal to GK', then PK' 

 will equal PL', and the angle K' PL' will be equal to D. So in the 

 second case -we may make the angle PFK' equal FGE, and take GL' 

 (towards A) equal to FK' ; then will PK' equal PL'. — Ed.] 



We have received various correct solutions of this problem from 

 T. R., Nemo, H. A. N., E. Whitley, and others. 



[38] — Let A be the centre of the disc at which its whole 

 weight may be supposed to be placed, AB the radius of the circum- 



Fig. 5. 



ferenoc described by the- centre A, AC the radios of the disc, 

 C being in contact with the plane on which the disc rolls. Draw 

 the vertical line AD to the level plane, and complete the parallelo- 

 gram ED. Let 11) be the weight of tho dhc, AC its radius = r. 

 AB = r', AB + CD = R the radius of the track, angle CAD = and 

 and vel of A = v. 



Then 



:EA : AD 



EA 



.*. -L = _- = tan0 : hence / = —1 — 

 gr' AD jtanS 



Now CD = )-sin9.*.R( = CD + AB) = rsiD0 + _il— 



t^tantf 



— D. M. 



[This solution can only be regarded as approximate ; the assump- 

 tion that the whole weight may be regarded as collected at the 

 centre of gravity being inadmissible in a problem of this sort. — Ed.] 



