Director of a Conic Inscribed in a Triangle Author(s): E. P. Rouse Source: The Mathematical Gazette, No. 2 (Jul., 1894), pp. 12-13 Published by: Mathematical Association Stable URL: http://www.jstor.org/stable/3602373 Accessed: 26-11-2015 04:44 UTC

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This content downloaded from 139.184.14.159 on Thu, 26 Nov 2015 04:44:41 UTC All use subject to JSTOR Terms and Conditions 12 1THE MilAT'HEIMATICAL GAZETTE then known, and constructed to indicate tlese correctly above referred to: " Of him it may truly be said that for 17,100 years. It lasted, in fact, but a very few he studied more to serve the public than himself, and years. During the troubles of the civil war it was though he was rich in fame and in the promises of cast away as old rubbish, and in 1646 was found by the great, yet he died poor, to the scandal of an Sir Jonas Moore, and deposited in its dilapidated state ungrateful age." G. HEPPEL. at his house in the Tower. From the time of the discovery of Algoristic arith- metic the pathway of practical science was northward NOTES ON "A. I. G. T. SYLLABUS OF ELE- across the Continent to England, and thus our early MENTARY DYNAMICS." PART II. mathematicians were on the watch to see constantly (Macmillan& Co.) what was going on in Italy and the Netherlands. The notes be found useful to those who Wright gave us a translation of a work of Stevin's following may are using the Syllabus. called The Haven-finding Art, and he was the first to Addendumto ? 24, tofollow line 6.-If two of the forces the which Mercator's explain principle upon depended are parallel, this condition only necessitates that the third projection, the inventor, Kauffman, having applied shall be parallel to them. Its line of action must be merely rule-of-thumb processes. In a historical dis- determined some other way, such as that of ? 26, or of sertation prefixed to Robertson's Navigation, it is ? 31. stated that nearly all that was done to advance navi- Addendumto ? 31.-If the two given forces are parallel, the above constructionfails. But the theorem of moments gation in the seventeenth and eighteenth centuries still holds. was the work of three men-Wright, Norwood, and For, let p, q (Fig. i.) be the given parallel forces. Halley. With Napier, the inventor of logarithms, Replacep by two componentsr, s, and let t be the resultant are associated the names of and as Briggs Wright, of q, s, so that r, t are equivalent to p, q. Then, generally, assisting in the promotion and improvement of this r, t will not be parallel. And therefore, by this section, means of calculation. Wright translated into Eng- the sum of the moments of p, q about any point = the = lish Napier's great work on the Description of sum of the moments of r, s, q the sum of the moments of t = the moment of the resultant. Logarithms.1 r, If the forces had been and There are some other matters of interest with p, q equal opposite parallel forces (Fig. ii.), the forces r, t would also have been equal which Wright had to do. He was the first to suggest and opposite parallel forces, so that in this case the new of a standard the division the establishment length by method also fails. For this case see ?? 33, 34. of a meridian circle. The determination of the longi- tude was an unsolved problem in Wright's time. The idea of doing this by means of a clock had been OF A CONIC INSCRIBED suggested, but no clocks were accurate enough for DIRECTOR CIRCLE the purpose. Wright elaborated a method which IN A TRIANGLE. the variation of the sound and depended on compass, 1. Let TP, TQ (Fig. i.) be to a conic, C its good in itself, but which improvements in clock- centre, S, H its foci, 2a and 2b its axes. From S draw a making rendered useless. The project of supplying perpendicularto TP and produce it to its image S'; then = = = London with water brought from springs near Ware, we know that S'T ST, angle S'TP PTS, and S'H major- axis = 2a. with a drawn from H in Hertfordshire, was conceived and all but carried Similarly to TQ and producedto its image H'. Thus the two triangles out by him. He was not, however, as able to finance S'TH and STH' are equal, and the angles S'TS, HTH', and schemes as to them into for others to make " put shape therefore the halves of these angles, are equal, that is, the use of. was and Sir Wright easily pushed aside, tangents from T are equally inclined to the focal distances Hugh Middleton had all the glory of the famous of T." work called the New River. No more fitting close Taking either of the above triangles, we have to a notice of this too little-known mathematician can 4a2 = ST2 + TH2- 2ST. TH cos PTQ - 2CS2 + 2CT2 - 2ST. TH cos be given than the last few lines of the Latin paper T, .. a2 + b2= CT2- ST. TH cos T. 1 other work on logarithms canonis Napier's (Mirifici Logarithmorum 2. Let the TP. TQ be cut in K, L, the constructio) was first translated into English in 1889 by W. R. tangents by Macdonald. at a third point R, and in k, I by the tangent

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This content downloaded from 139.184.14.159 on Thu, 26 Nov 2015 04:44:41 UTC All use subject to JSTOR Terms and Conditions THE MA THEMA TICAL GAZETTE 13 parallel to it. Then the rectangle kP. PK = CP v/2 describe a circle cutting OC producedboth ways = sq. on radius parallel to PK in G. G'. The point where a circle through GDG' cuts = SP. PH. Also angles SPic, HPK are equal. OD again is a vertex of the triangle, and the rest follows .. the triangles SPk, HPK are similar at once. - .. angle PkS = PHK = 2PHR = PHQ ~QHR y. Let a conic touch the sides of a quadrilateral. Com- = THQ - LHQ= THL; plete the quadrilateral. The above investigation applies and the angles kTS, HTL have been shown equal (1), to the orthocentre of each of the four triangles formed therefore these triangles kTS, HTL are similar and the by the tangents, and to each of the conics touched by = = rectangle kT. TL ST. TH constant. them; and as one of these conics can be a we 3. The result of Article 1 may now be written have this result: " The orthocentresof the four triangles of a2 + b2= CT2- kT. TL cos T. a complete quadrilaterallie on one straight line," viz. the Now draw LF (Fig. ii.) at right angles to TK, and TD directrix of the parabola. at right angles to KL, passing through the orthocentreO, 8. Except in the case of right-angled triangles, which and cutting kl in d. Also draw CV parallel to the tan- we may neglect as being easy, we can see from a figure gents, and thereforebisecting Dd at right angles: and let us that two at least of these four triangles are obtuse-angled. suppose the triangle TKL acute-angled, so that O falls Let us confine our attention to two obtuse-angledtriangles within it. and the correspondingpolar . As the directorcircles The above result now becomes of all inscribedconics cut these polar circles orthogonally,it a2 + b2= CT2 - kT. TF follows that "the centres of the director circles, and, there- = CT2- TO. Td fore, of the conics, lie on a straight line." This line is the = TO. TD + CT2- TO(TD + Td) radical axis of the polar circles, and is therefore perpen- = TO. OD + TO2+ CT2- 2TO. TV dicular to the line of orthocentres. Also, it is easily seen = TO. OD + CO2. that "each of the director circles cuts the line of ortho- [If, instead of TKL, we had taken the circumscribed centres at two fixed points." triangle Tkl and o its orthocentre,we should have found e. Each diagonal of a complete quadrilateral is the in the same way a2 + b2= To. od + Co2.] ultimate form of an inscribed . Hence " The middle Now make O the centre of a circle with radius points of the three diagonals lie on a straight line," and = /TO. OD, and the conclusionruns thus: " The director "The centre of an inscribed conic lies on the straight line the middle of circle of any conic touching the sides of an acute-angled joining points the diagonals." triangle cuts this circle at the extremities of a diameter." ~. Lastly, the director circle correspondingto each of these is If the triangle be obtuse-angled,so that O falls outside it diagonals the circle which has that diagonal for its for the (in which case the above circle is called the polar circle of diameter, expression a2 + b2becomes a2 + 0. Hence "The the triangle), a similar investigation leads to the result circles on the three diagonals of a complete quadri- lateral the a2 + b2= - TO. OD + CO2, and the interpretation is, " The pass through same two points," viz. the two fixed on the line director circle of any conic touching the sides of an obtuse- points of orthocentres. I reserve for angled triangle cuts the polar circle of the triangle another paper the propositions which can orthogonally." be deduced from a special case of Article 2. E. P. RousE. 4. Hence follow easy proofs of some well-known pro- positions. If the conic be a the directorcircle a. parabola, becomes SOLUTIONS OF EXAMINATION QUESTIONS. the directrix, and we have, "The directrix passes through the orthocentre." The Editorwill be glad to avail himself of the help of all classes readerstowards /. If the conic be an in-circle or ex-circle and r its of makingthis sectionof the Gazette as as MATHEMATICALTUTORS radius, 2 + b2 becomes 2r2, and we have the property useful possible. are invited to send neat STUDENTSto call attentionto which has been used to prove Feuerbach's theorem that solutions; classesof exceptional and the nine-point circle touches each of the in- and ex-circles problemspresenting difficulties, EXAMINERSwho sympathisewith us toforward their The (see for example Richardson and Ramsey's ModernPlane copiesof papers. help readersis requestedin ,pp. 32, 34). of foreign especially obtainingcopies of set in the examinations othercountries. From this property we can deduce a solution of a papers public of problem set in a scholarshippaper of King's College, Cam- 7. A string hangs verticallyfrom one side of a horizontal bridge, for December 1892. "Given C the centre of the circularcylinder of givenradius, and carriesa heavyparticle at inscribed circle of a triangle, and P its point of contact its lowerend. Find the least velocitywith whichthe particle with a side, and O the orthocentre,construct the triangle." mustbe projected horizontally in orderthat the stringmay wrap Through P draw a perpendicularto PC, and from O draw itself roundthe cylinder,never becoming slack. OD at right angles to it. With centre C and radius [Inter. Arts Hons. 92.)

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