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Notes from the Mathematical Seminary JOHNS HOPKINS UNIVERSITY CIRCULARS Publis/zed wit/z t/ze approbation of t/ze Board of Trustees VOL. XXII.—No. 160.] BALTIMORE, DECEMBER, 1902. [PRICE, 10 CENTS. NOTES FROM THE MATHEMATICAL SEMINARY. EDITED BY PROFESSOR FRANK MORLEY. ON THE HYPOCYCLOIDS OF’ CLASS THREE INSCRIBED IN A 3-LINE. (4) (K—1)X+o-,X H. A. By CONVERSE. and identifying (3) and (4) we have ~ 1. Steiner has proved’ that if from any point P of the circum- cY3K. 3-y, the perpendiculars be drawn to the circle of a given triangle a/ K = — 1 gives Steiner’s case. The centre is at a,/2, and size l. three sides of the triangle, their feet lie on a line L; and as the Hence in general (4) gives a deltoid turned through an angle point P moves around the circle the envelope of the line L is a K, whose centre is ~‘ and whose size is 1 (73K hypocycloid of the fourth order, and third class, which for short- —K+1 K—i, ness we shall call a deltoid, which touches the sides of the given For the cusps 11 = t, = iXT and triangle. This curve may be obtained by the rolling of one circle within another of radius three times as great. And, if we use Eliminate K and we have 3Xo-, or 1 (X’—o-,) x+o-,o-,— — o- — circular coiirdinates, its equation is x 2t + ~, as I runs on the (5) x 3X,the locus of the cusps as X varies. unit circle. If we generalize Steiner’s problem so that the lines 1 —~ from P, instead of being perpendicular to the three sides of the Put X= t, i. e. let X run around the unit circle, and theequation triangle, make a constant angle with them, we obtain the follow- __ a + /3+y — 3t ing:— 1— Theorem I. If from a point P on the circunwircie of a given a/3y triangle cL/3 maps the unit circle into a rational cubic curve. Hence the 7, the lines be drawn making a constant angle with the Theorem IL The locus of the three cusps of the system of del- three sides of the triangle, their cut-points with the three sides will lie on a line L, and the envelope of the line L as P moves around toids touching the given triangle is a rational cubic curve given by the circle will be a deltoid. And as the angle varies we will obtain the equation. (1) a+/3+y—3t a system of all deltoids touching the given triangle. For Xr The equation of the chord /3y is 1—— a/3-y Let us examine this equation. We may unite it with its Any line KX + f3yy — constant, makes a constant angle with (1). Let this line cut the unit circle at t, and we have its equation conjugate as (2) (2) KX+/3yy~Kt (7,0~ — 3tcr, y — — o,t’ — 3cr,t t u,(cr,—-t’) Combining (1) and (2), we have (for since ~ ÷1±1 a/S +13y + ya the conjugateof o~ is (K—i) X~Kt+—~—ff1 + a, where a/3y a/3y cr,/ (3) at Let (2) be cut by any line a + /3+7, ~ c43 + /3y + y’a, (7, = a/3-y. (3) Ax + By + 1 =0 (whereA and B areconjugatequantities). We put Then from (2) and (3) we have ~2~3 — 3to _______________ ‘ A o,t’ — 3o Uber eine besondere Curve dritter kiasse (und vierten Grade), Crelle, cr 1= 0, or Vol. 58, 1857. 8—t’ ’ 2 JOHNS HOPKINS [No. 160. 2± 3A0- 2 — A~ (4) ~‘ B3B~ 3 ~ 3 1~3 ~ o. Draw the Euler line through ~1, the orthocentre, and 0, the 0-3 0-3 Ba3 0-3 B0-3 0-3 circumcentre, of the triangle as the axis of reals. Take the cir- The roots of (4), t1, t2, t~ give by (2) the points of intersection of cumcentre as the origin, and determine the positive direction of the line with the cubic. To find the involution of these points, put 3 r t1+~2±3~S1~ the axis to be from 0 to ~,. At the point ~ (0-3—.--,) erect a per- 3A — ~S’ pendicular to the axis of reals. This is the line of flexes, and cuts (5) -~ t1t2 + ~2t3~ s. 3Ao3’ 3o’ 2Bo3— — 0, the three sides of the triangle in the three points of inflexion. Bcr2 —i--3 0~—~— 03~ L t,f Ao1o3 3A—S For the asymptotes, (keeping 0-~ real) 2f3:3 ~— Bo2 4- 0~3 0io3 3aoB—S3a3+ 03— 0. ~ Eliminating A and B from these we obtain (6) 3~3—0-3~3+0-3~3—30-3 0, Let 73 -I~~o-3, -r3 — vi~9~’o-3 ~ 3 —thev~/’o-,imaginarywhere cube-~>0-~ rootsis aofselectedunity. the involution of points in which a line cuts the cubic. Callcubetheseroot quantitiesof ~ and T~v and(i~1,v 2, 3). Then if in If the three points coincide we have the triple-points of the 0~3 — 3o 3 — 3j3 involution, i. e. the points of infiexion of the cubic. (6) then x Xy~ 3f — A(o,t becomes 0303— 33Y 3- 37~37~3~ The roots of (7) are a, /3, and -y, hence the cubic has three real we put X 01Tj x Xy2 T,’cT 3) —3o-lo-3(-i31—13 )—3o- r (72 t3) —I-- 9o- 3t— t3’ri) (o,7i3—3Ti’) (0 3) infiexions, 3(Ti’—1 3(i-~ 3—t (o3 — 3a) 0-s, (0-3 3/3)0-s (0-3—37)0-3 for t — r the right side of this becomes (8) x— 3 3 0-3/3 — 0-37 — 0-3~0-3— 20- 20- 30-3’r~ — 3T~’ 30-3 To determine the flex-tangents putt1~ t3 — t3 —a; then S~ 3a 3-1-i3 and from (5) whence we have — (0- 3 — 3r~3) x (0-30-3 — 30- 20- _ 3B~3 —/3y —1 3 orB~ anditsconjugateA— Hence 3r~ 3r~) y 0-3~0-3 — 30-3 — 30-3-ri —20-3~~2. 0-3 /3-’r-y /3 7 3, remembering .that ~ i-j3, we have (3) becomes (13)Dividing(0- through by i-~ 3T~3) ~ 0- 20- 3Ti — 3) x (0-3Ti — 3?i — 37~ — 20-1. Hence the This for i — 1, 2, 3, gives the three asymptotes and is perpendicu- Theorem III. The sides of the triangle are theflex-tangents of lar to the line through -r~ and ~, for the equation of this line is the cubic curve, and thepoint of irifiexion corresponding to any vertex 3 x y 1 z~0 lies on the opposite side. 1 3) x—(~ 3) y~0- To determine the line of flexes, turn again to the involution. (14) ~ 1 or (0-31-~— 3r~—3z-~ 3ri’ —0-i. S3~a+/3 7~0-i Hencefrom(S) ~ i Hence the (10) 0-3~/3 3B~3 B— _O0-3~0-30-30-10-3 A — 3o-3—0-30-3______ Theorem V. The asymptotes of the cubic are perpendicular to Hence (3) becomes the lines joining the centroid of the triangle to the points On the (11) 0-3x 0-30-3y 30-3—0-30-3 0, circumeirele which correspond to the infinities of the cubic. the line of flexes. This is perpendicular to the Euler line of the The asymptotes are then easily drawn, for (13) and (14) cut at triangle the point ~ ~‘ and x y 1 ~0 2 are perpendicular. ~ 1 or A cubic with three real inflexions has an isolated double point 0-3 given by the neutral pair of theinvolution, theHessian of the cubic. 001 0-’0-~ —30-3 The Hessian of (7)3 is(90-3 — 0-30-3) t + (0-3’ — 30-30-3) = 0; (12) o-3x — o-3o-3y 0; and cuts this at the point x— 2o~3 for o-~ real(0-~~this— 30-is3)t Hence 0-3(0-3—30-3) t 0-3(9—0-3’) t + 0-30-3 (0-30-3—3>z~ 0. Theorem IV. The lines qf flexes of the cubic is perpendicular If h and h’ are the roots of this to the Euler line of the triangle. If now we take the Euler line of the triangle to he the axis of hh’ 0-3 (0-10-3 — 3 0-—30-1 3 reals, 0-~ is real. 0-~ is then its own conjugate and we have 0-3 h 0-~ (0-i — 0-3 — or 0-~ 0-30-3 3) 0-3 Then the double point is The Euler line (12) then becomes ~ (0-j 03 30-3 — y 0 and the line of flexes is x (ar— 3h) 3h’) (h h’ — h3— h’3 , or ____ 0-3~0-3 — 30-3 3 x+y— (15) x= 30-3 0-10-3 0-3 h’+hh’ ,(2. These meet at For further aid in constructing the curve put t’ — -r~ and we 3 2x~a have 3 —— This gives a very simple method of constructing the line of flexes and the points of inflexion, as follows - the three cusps of the deltoid of Steiner’s case. ) DECEMBER, 1902.] UNIVEI?SITY CIPCULABS. 3 y 2. If we consider again a deltoid touching the three given lines o-&(x+y) —1- ~a3crax+2ao-&(2y—ax)~3+4a0-lu3+oL~, or (the sides of the triangle a/3y) and make it touch any fourth line we can determine the point of tangency P.
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