62 BULLETIN OP THE 



SIXTH SOLUTION. 



Let OE = OE'=a? (Fig. 2), OD=:OD'=^, AE==a, 

 and BD — /3; the angles marked with a dot are each equal to 

 45°, and therefore E E' = xs/2, and D D' =yN/2. 



From similar triangles BO : B D = E' : D D', or 

 j3 — y : ^ : : X : yy/2. Whence 



{?—y)yV2 = ?x. (12) 



(a — x)xs/2^o.ij. (13) 



From (13) 



y~.Yjix(a — x), and substituting in (12) 



Ii-~x(a-x)=^^^^^;^y 



Which reduces to 



(a — ^)'^ X = 2^ (a — 2 X). 



Expanding, rearranging, etc., this reduces to 



x'~2aX-^+ai^^ + ^)x~^-^=^0_ (H) 

 CONSTRUCTIONS. 



First Construction. — The equations obtained in the sixth solu- 

 tion point to a simple construction of the problem, as follows : — 

 Equations (12) and (13) may be written as follows: — 



x'^ — ax -{- -r^y—0. (15) 



2/' — ^y + ;^-^ = 0. (16) 



And each of these equations is the equation of a parabola. Jf 

 these two parabolas be constructed, their intersection will deter- 

 mine X and y. The position and size of the parabola will readily 

 appear by transforming co-ordinates. In equation (15) let 



a 



x=:x^ + -g and 

 ./==/+ 2^, then 



