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Longueyhigginsintell2.Pdf . .IB A ,1, " , , ,, " , , , , ," , ,, , , ,, , " , , , , , , , , , L, , " , , , , , , , , , , " , , , ," , , , , , ,, F A," , , , , , , , , , I , , ,, B D A' C *GB ' , , *' 'C E, and F denote the feet of the altitudes AH, BH, and CH; ",', GA,' see Figure 6. Then, HDCE, for example,is a cyclic quadri- I " " ',' , lateral;hence, LBHD = C. Therefore, " , I "':', " HD = BD cot C = ABcosBcot C. " , ',,~ " BecauseAB = 2Rsin C, this givesus .. IA H = 2R(cos B cas C, cosC cosA, cosA cosB). I' The coordinates of the point L, which is the reflection of i erties of two pencils of rays having the same cross-ratio, H in 0 (i.e., L = 2 X 0 - H), can therefore be written [ b~t with one ray in common. The proof is too long to be L = 2R(a - {3'Y,(3 - 'Ya,'Y - a{3), (2) given here. The fourth and shortest proof goes as follows. The trilinear coordinates of a point P in the plane of a where a, {3, and 'Y stand for cos A, cos B, and cos C, re- triangle ABC may be defmed [4, 10] as the lengths (x, y, z) spectively (see also [4] and [5]). I of the perpendicularsfrom P to the three sidesBC, CA, and What are the coordinatesof the Gergonnepoint G?Now, I AB. By convention, x, y, and z are positive when Plies in- from Figure 3, we see that XC = r cot(iC), and therefore side the triangle. for X, we have Thus, the incenter I, being equidistant from the three y = XC sin C = 2r COS2(lC). sides, has coordinates 2 Similarly, I = r(1, 1, 1), (1) z = XB sin B = 2r cos2(lB). where r is the radius of the incircle. For the circumcenter 2 0, since LBOC = 2 X LBAC, it will be seen that Therefore,the coordinatesof G, which lies on the line AX, must also satisfy 0 = R(cosA, cosB, cos C), where R is the circumradius and A, B, and C are the an- JL = cos2(lC)2 = 1 + cos C gles at the vertices of ABC. For the ortlwcenter H, let D, z cos2(lB) 1 + cosB' 2 Hence, G = [(1 + (3)(1+ 'Y),(1 + 'Y)(1+ a), (1 + a)(1 + {3)]g, (3) where g is a normalizing constant. Therefore, in order to prove the collinearitY of I, G, andL, we have only to show that the determinant 1 1 1 D = (1 + (3)(1+ 'Y) (1 + 'Y)(1+ a) (1 + a)(1 + fJ) a - {3'Y {3 - 'Ya a - {3'Y vanishes.However, upon adding the elementsof the third row to those of the second,we seethat eachterm becomes equalto 1 + a + {3+ 'Y,bringing the first two rows of D into H proportions,so D = O.This provesthat IG passesthrough L. 56 THEMATHEMATlCAL INTEUJGENCER Likewise,the coordinatesof the excenterIA are .O'CiII:la- IA = rA (-1, 1, 1). A In detennining the coordinates of GA, we have only to re- place B and C by 7T- B and 7T- C, respectively,also x by -x. Hence, GA= [-(1 - fJ)(l - y), (1 - y)(l + a), (1 + a)(l - fJ)]gA, where gA is another constant. The correspondingdetemti- nant DA will be found to vanish in a similar way. Hence, we have proved the following theorems: 1. Thefour lines IG, IAGA,InGB' and IcGc all meet in a point L. 2. L lies on the Euler line and is the reflection of the or- B X A thocenter H in the circumcenter O. Thus, the separa- tions of L, 0, M, 0', and H along the Euler line are in the t 6 2 1 3 Figur 5 Zc be defined similarly. Then,the three lines AXA,BY B, and raws.. ,see. e. CZc are concurrent in the pomt N; see Figure 7. Note a corollary. From Figure 5 we see that ML = 2 X To prove this, note that the length of the tangent CXA MH. Now, consider the homotheticity in which points of from C to the excircle center IA is -!ca + b - c), where a, the plane are first reflected in the medianpoint M and then b, and c are the sides of ABC. Ther:fore, enlargedby a factor of 2. All transformed points being de- 1 noted by the suffix 1, the orthocenter HI of the triangle CXA= CYB= 2(a + b - c), AlBlCl is coincident with L. But A, B, C are the midpoints which is equal to the length of the tangent BX from B to of the sides of A lBl Cl, so ABC is the median triangle of the incircle, center I. In fact, XA, YB,and Zc are the re- AlBlCl. Substituting ABC for AlBlCl, we have the follow- flections of X, Y, and Z in the midpoints A', B', and C' of ing theorems: the three sides of ABC. Hence, as before, 3. The orthocenter of a triangle ABC is coUinear with the AYB . BZc . CXA incenter I' of the median triangle (i.e., the Spieker AZc . BXA . CYB = 1, point of ABC) and the Gergonnepoint G' of the me- d le ( Fi 6) and by Ceva'stheorem, the three lines are concurrent. tan. trwng. see g. 4. If I'A, I'B, and I'G are the excentersof the median tri- angle and GA, GR, and Gb are the corresponding .O'CiIJ:I~:- Gergonnepoints, then lAG;', lEGE,and I/;G6 also pass \ IB throughthe orthocenterH. NA "'" "'"'"', , , WhenI wrote to my long-timefriend and colleague "'"'"'"' , H.S.M.Coxeter about these results, he pointed out that the "',"', I, French geometer G.A.G.de Longchamps(1842-1906) had shown [6] that the orthocenterHI of AlBIC 1has certain in- terestingproperties related naturally to the triangle AlBl Cl but having no obvious connection with I or G; see [1], [8]. The point HI has been called the de Longchampspoint of ABC; see [1] and [5]. Thus, theorems 1 and 2 can be stated alternatively as 5. Thelines IG, IAGA,InGB' and IoGc all pass through the de Longchampspoint of ABC. Or more succinctly, , B ".., 'C -. 6. L coincides with HI. "".. " " Nc ..' I The question now arises: Are there any other fourfold "" ", " .. , , concurrenciesanalogous to the one through L? .. .." , There is a somewhat similar situation involving the I ,~~, Nagelpoint N mentionedearlier. The Nagelpoint of the / IA triangle ABC may be defined in the following way [2], [11]. NB As before, let XA denote the point of contact of the excir- / cle, center lA, with the side BC opposite A, and let YB and I VOLUME22,NUMBER 1,2000 57 By similar triangles, it may be shown (see [2, pp. 161- To establish relation (A) we need, by symmetry, to con- 162]) that N is collinear with I and M and that MN = 2MI. sider only the x components of this equation. Thus, we Hence,N is the incenter of the triangle AlBIC!, or N = II. need to show only that Now, corresponding to., each excenter of ABC, say lA, &.fl,'1D(1 - cos B)(1 - cos C) + Oll-OD sm. ( -I A) sm, ( -1 B) sm-. ( 1C) we also have a Nagel pomt, Thus, if t:4, 7lA,~ are reflec- 2 2 2 tions ofXA, YA,ZA, respectively,~ in the midpoints A', B', = 2R sinB sin C. C', respectively(so t:4 = X), thenAt:4, B7lA, C~ meet in the However, from sin(lA) = cos[l(B + C)], this last result be- Nagel point NA, say; and similarly for NB and Nc. comes evident. The~efore,M ~deed divides the line IN in We can now prove two theorems somewhat analogous the ratio 1:2. to Theorems 1 and 2, namely In a similar way, we fmd 7, The!our li~s IN, lANA, InNB' and IcNc all meet in the NA = 2R[(1 + (3)(1 + y), -(1 + y)(1 - a), -(1 - a)(1 + (3)] med'/,anpomt M, 8, M divides each of IN, lANA, IBNB, and IcNc in the ra- and tio 1 : 2. I = r A A ( -1 " 1 1), Hence, where 9. NA, NB, and Nc are the three excentersof the triangle r = 4R sin (lA ) cos (l B) cog ( 1r"\ A B C A 2 2 ZVh 1 1 1. An analytic proof is as follows, Our method will be to show that N + 2/ = 3M, (A) . As earlier, the ratio of the perpendicularsfrom XA to the sidesAC and AB is given by 1L - XAC sin C - XB sin C - r cot(iB) sin C. z - XABsinB - XCsinB - rcot(lB) sinB' 2 that is to say, . 2 sm (k2'-'J 1 1L==~ z sin2(1B) 1 - {3' 2 H th trilin. dinate fN MICHAEL LONGUET-HIGGINS ence, e ear coor s 0 are Instituteof Nonlinear, Science , N = [(1 - (3)(1- y), (1 - y)(1 - a), (1 - a)(1 - {3)]n, UniversityofCalifomia LaJolla . , .. LaJolla, California 92093-0402 USA where n IS a normalizmg constant,which we need to eval- uate. Now,b' theI trilinear. coordinates of any point (x, y, z) The auth or grad uat ed In' math emat ICS' from Camb rl' d ge Unl-' must 0 VlOUSY satIsfy versity in 1946 and did 3 years' NationalService at the ax + by + cz = 211., AdmiraltyResearch Laboratory in Teddington,where he be- A d th f ABC B came interested in various geophysical phenomena. He re- where ~ enotes e area 0 . ecause .' .. tumed to Cambridge to take a Ph.D, In geophysIcs In 1952. 11. = 2R2 sin A sin B sin C He haspublished extensively on topicsin fluiddynamics, par- '1D . A d Ii rth find aft f ticularly on surface waves and ripples, wave breaking and thand a I= .&.fl, smA B' an Cso 0 th" we er some use 0 soundgeneration , In.
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