Alhazen's problem history I was wondering if anyone has read '' alhazen book of optics'' and has understood his solution of Alhazen's problem.
I know a modern solution of the problem by Dörrie-100 great problems of mathematics, but I'm interested really in ibn alhaytham's solution. I'm just curious about how he went about solving the problem without the modern tools. Also I couldn't find the problem in his book, maybe I didn't look carefully for it, but it would be nice if someone could reference me to it in his book also.
 A: Here is the latin translation of Alhazen's statement, in the edition by Friedrich Risner (Basel 1572):

Visu et visibili a centro speculi sphaerici convexi inaequabiliter
distantibus, punctum reflexionis invenire. [n. 39, p. 150]

That is:

To find the reflection point, for eye and object at different
distances from the centre of a convex spherical mirror.

His solution is shown in figure below. Let $a$ be the eye and $b$ the object whose light is reflected to the eye by a spherical mirror of centre $g$ (the circle is of course the intersection between the sphere and plane $abg$). To find reflection point $d$, construct a segment $mk$ and a point $f$ on it such that
$mf:fk=bg:ga$. Let then $o$ be the midpoint of $mk$ and construct on its perpendicular bisector a point $c$ such that $\angle ock={1\over2}\angle agb$ (I'm assuming $gb>ga$).
Now, construct on line $kc$ a point $p$ such that, if $s$ is the intersection between lines $pf$ and $oc$, then the ratio $sp/pk$ is equal to the ratio between $bg$ and the radius of the sphere. This construction cannot be done by ruler a compass.  However, Alhazen gives a neusis construction for it, as a lemma (n. 38 in the same edition) and also an alternate construction (n. 34) where the neusis is solved by means of a hyperbola.
He also warns the reader that two possible points $p$ can be constructed, in general, but only one of them, at most, is such that $\angle pks>\pi/2$, and this is the one to be considered. If both solutions give $\angle pks\le\pi/2$ then no reflection point exists.
Point $d$ on the circle, such that $\angle bgd=\angle spk$, can be constructed, and Alhazen then proves this is the wanted reflection point, because $\angle pks=\angle bdg=\angle adg$.

