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XLVI. On a Triangle in-arid-circumscribed to a Quartic Curve. 

 By Professor Cayley, F.R.S* 



^THE quartic curve (x?—a?)' 2 +{y 2 — b'*)' 2 =c 4 presents a simple 

 example of a triangle in- an d-circum scribed to a single 

 curve, viz. such that each angle of the triangle is situate on, and 

 each side touches, the curve. Assuming that the triangle is 

 symmetrically situate in regard to the axis of y, viz. if it be the 

 isosceles triangle a c e, the sides whereof touch 

 the curve in the points B,D,F respectively, then 

 we must have a single relation between the con- 

 stants a, b, c of the curve ; or if (as may be 

 done without loss of generality) we write a=l, 

 then there must be a single relation between b, 

 and c. The relation in question is most conve- 

 niently expressed by putting b and c equal to 

 certain functions of a parameter (/>, which is in 

 fact = tan 2 6, if 6 be the angle at the base of 

 the triangle ; the equation of the curve is thus obtained in the 

 form 



., , ,„ , / , <P l +w-i y_i , (4> 2 -i) 4 . 



and the coordinates of a, c, e, B, D, F are as follows : — 



Coordinates of a are 0, */\$i 



c, e> >, ± V% - Vi4>, 



» D, „ 0, _ \/T$ } 



b f + f- 1 , _j*y±_ 



' ". ■- V2(^+l) ^2(^+1)' 

 It is easy to verify that the points a, c, e, D are points of the 

 curve, and it is obvious that the tangent at D is the horizontal 

 line c e. It only remains to be shown that B and F are points 

 of the curve, and that the tangents at these points are the lines 

 a c and e a respectively. It is sufficient to consider one of the 

 two points, say the point F ; and taking its coordinates to be 



*~ a/2(<£ 2 + 1)' V V2{^+\) 



we have to show that (f, rj) is a point of the curve, and that the 

 equation of the tangent at this point is X \/<f>-\-Y= V\<f>, 

 where (X, Y) are current coordinates. 



First, to show that (£, rj) is a point of the curve, the equation 



* Communicated by the Author. 



