VOL. VIII.] PHILOSOPHICAL TRANSACTIONS. Q'/ 



A Prohlem of Alhazen solved: by M. Hay gens and M. Slusius. Translated from 



the Latin, N° 97, p. 6119. 



Mr. Huygens writes : By this opportunity I send you a solution of Alhazcn's 

 problem, which I have lately discovered, and which our colleges approve of very 

 much. The problem is this : Having given a concave or a convex speculum, 

 as also the place of the eye and the visual point ; to find the point of reflexion. 



Let the point A be the centre of the spherical speculum (fig. 1, pi. 3), B the 

 eye, and C the visible point, and let a plane drawn through A, B, C, cut the 

 sphere in the circle Dd, in which are the points of reflexion to be found. 

 Through the three points A, B, C, describe the circumference of a circle, whose 

 centre is Z, and meeting EI produced perp. to BC, in R; and let NA be a 

 third proportional to RA and OA; then NM, parallel to BC, will be one of 

 the asymptotes. Again, let EA, 4 OA, lA be proportionals, and taking lY 

 equal to IN, draw YM parallel to AZ; and it will be the other asymptote. 

 Lastly, taking IX, IS, which shall each be in power equal to half the square 

 of AO, with the square of AI ; then will the points X and S be in the hyper- 

 bola, or the opposite sections Dd, to be described to those asymptotes, the in- 

 tersections of which with the circumference DO, will show the points of re- 

 flexion sought. This construction takes place in every case in which the pro- 

 blem is solid, except in that in which not an hyperbola but a parabola is to be 

 described ; viz. when the circumference described through the points A, B, C, 

 touches the right line AE. 



The editor having sent a copy of this to M. Slusius the 24th of September 

 1670, he sent the following answer Nov. 22, the same year. — When I reduced 

 the construction of the celebrated Huygens to calculation, I found that he had 

 followed the same analysis as myself. But since two solutions are derived from 

 it, both of them by the hyperbola about the asymptotes, he made choice of the 

 one, and I of the other, as being the easier. Now it is evident, that nothing 

 more is required in this problem, if we reduce it to mere geometrical terms, 

 than in a given circle, whose centre is A and radius AP (fig. 2, 3, 4), to find 

 some point P such, that drawing the lines PE, PB, to the given points EB, 

 unequally distant from the centre A, the line AP produced, may bisect the 

 angle EPB. Now this admits of various cases. For, either the perpendicular 

 from A on the right line EB, that is AO, falls between E and B, or beyond B. 

 If beyond it, either the rectangle EOB is equal to the square of AO, or is 

 greater or less. Of the case of equality we shall speak below. Now the other 

 three cases we shall comprehend nearly under the same construction. Let a 



voii. II. O 



