248 
THE DUPLICATION OF THE CUBE 
Let APG' be the corresponding position of the revolving 
semicircle, and let AG' meet the circumference ABC in i¥. 
Drawing PM perpendicular to the plane of ABC, we see 
that it must meet the circumference of the circle ABC because 
P is on the cylinder which stands on ABC as base. 
Let AP meet the circumference of the semicircle BQE in Q, 
and let AG' meet its diameter in A r . Join PC', QM, QN. 
Then, since both semicircles are perpendicular to the plane 
ABC, so is their line of intersection QN [Eucl. XI. 19]. 
Therefore QN is perpendicular to BE. 
Therefore QN 2 = BN. NE = AN. NM, [Eucl, HI. 35] 
so that the angle AQM is a right angle. 
But the angle A PC' is also right; 
therefore MQ is parallel to G'P. 
It follows, by similar triangles, that 
G'A-.AP = AP:AM= AM-.AQ; 
that is, AG: AP = AP: AM = AM: AB, 
and AB, AM, AP, AC are in continued proportion, so that 
AM, AP are the two mean proportionals required. 
In the language of analytical geometry, if AG is the axis 
of x, a line through A perpendicular to AG in the plane of 
ABC the axis of y, and a line through A parallel to PM the 
axis of z, then P is determined as the intersection of the 
surfaces 
(1) 
X 2 + y 2 + Z 2 = p X 2 , 
(the cone) 
(2) 
x 2 + y 2 = ax, 
(the cylinder) 
(3) 
x 2 + y 2 + z 2 = aV (x 2 + y 2 ), 
(the tore) 
where 
AG = a, AB=h. 
From 
the first two equations we obtain 
x 2 + y 2 + z 2 — {x 2 + y 2 ) 2 / b 2 , 
and from this and (3) we have 
a V{x 2 + y 2 + ?A) \/ (x 
2 + 2/ 2 ) 
V (x 2 + y 2 + z 2 ) ~ V {x 2 + y 2 ) 
h ’ 
or 
AG:AP = AP: AM = AM: AB