Prof. Cay ley on Gauss's Pentagramma mirificum. 311 



It will be at once apparent that in one respect the analogy be- 

 tween these chlorides and the oxides of hydrogen is incomplete, 

 inasmuch as the most stable oxide of hydrogen is water, into 

 which the dioxide is easily converted by heating — whereas the 

 reverse of this happens with the corresponding sulphur chlorides, 

 the most stable compound being the S 2 CI 2 , into which body and 

 free chlorine the S CI 2 is resolved on heating. 



XXXVI. On Gauss's Pentagramma mirificum, 

 By Professor Cayley, F.R.S * 



TAKE on a sphere (in the northern hemisphere) two points, 

 A, B, whose longitudes differ by 90°, and refer them to 

 the equator by the meridians A E and B C respectively ; join 

 A, B by an arc of great circle, and take in the southern hemi- 

 sphere the pole D of this circle ; and join D with E and C re- 

 spectively by arcs of great circle. We have a spherical pentagon 

 A B C D E, which is in fact the " Pentagramma mirificum/' con- 

 sidered by Gauss, as appearing vol. iii. pp. 481-490 of the Col- 

 lected Works. Among its properties we have 



the distance of any two non-adjacent summits ~\ _qr\o. 

 the inclination of any two non-adjacent sides j ~~ 3 



so that each summit is the pole of the opposite side, or the pen- 

 tagon is its own reciprocal. 



Each angle is the supplement of the opposite side. 



If the squared tangents of the sides (or angles) taken in 

 order are a, /3, 7, 8, e, then 



l + a = y8, l+/3 = 8e, l+y=e«, l+S = a /3, l+e=#y, 



equivalent to three independent equations, so that any three of 

 the quantities may be expressed in terms of the remaining two. 

 (This agrees with the foregoing construction, where the arbitrary 

 quantities are the latitudes of A, B respectively.) 



Projecting from the centre of the sphere upon any plane, we 

 have a plane pentagon which is such that the perpendiculars let 

 fall from the summits upon the opposite sides respectively meet 

 in a point. This (as easily seen) implies that the two portions 

 into which each perpendicular is divided by the point in question 

 have the same product. 



Conversely, starting from the plane pentagon, and erecting 

 from the point of intersection a perpendicular to the plane, the 

 length of this perpendicular being equal to the square root of 

 the product in question, we have the centre of a sphere such 

 that the projection upon it of the plane polygon is the penta- 

 gramma mirificum. 



* Communicated by the Author. 



