A history of Greek mathematics - Wilbourhall.org

EUTOCIUS. ANTHEMIUS 541
missing solution, promised by Archimedes in On the Sphere
and Cylinder, II. 4, of the auxiliary problem amounting
to the solution by means of conies of the cubic equation
(a — x)x 2 = be 2 , (3) the solutions (a) by Diocles of the original
problem of II. 4 without bringing in the cubic, (b) by Dionysodorus
of the auxiliary cubic equation.
Anthemius of Tralles, the architect, mentioned above, was
himself an able mathematician, as is
a work of his, On Burning-mirrors.
seen from a fragment of
This is a document of
considerable importance for the history of conic sections.
Originally edited by L. Dupuy in 1777, it was reprinted in
Westermann's IIapaSo£oy pd(f>oi (Scriptores rerum mirabiliwm
Graeci), 1839, pp. 14 9-58. The first and third portions of
the fragment are those which interest us. 1 The first gives
a solution of the problem, To contrive that a ray of the sun
(admitted through a small hole or window) shall fall in a
given spot, without moving away at any hour and season.
This is contrived by constructing an elliptical mirror one focus
of which is at the point where the ray of the sun is
admitted
while the other is at the point to which the ray is required
to be reflected at all times. Let B be the hole, A the point
to which reflection must always take place, BA being in the
meridian and parallel to the horizon. Let BO be at right
angles to BA, so that OB is an equinoctial ray ; and let BD be
the ray at the summer solstice, BE a winter ray.
Take F at a convenient distance on BE and measure FQ
equal to FA. Draw HFG through F bisecting the angle
AFQ, and let BG be the straight line bisecting the angle EBO
between the winter and the equinoctial rays. Then clearly
since FG bisects the angle QFA, if we have a plane mirror in
the position HFG, the ray BFE entering at B will be reflected
to J..
To get the equinoctial ray similarly reflected to A, join GA,
and with G as centre and GA as radius draw a circle meeting
BO in K. Bisect the angle KGA by the straight line GLM
meeting BK in L and terminated at 31, a point on the bisector
of the angle CBD. Then LM bisects the angle KLA also, and
KL = LA, and KM = MA.
the ray BL will be reflected to A.
1
If then GLM is a plane mirror,
See Bibliotheca mathematica, vii 3
, 1907, pp. 225-33.
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