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.
Here is the latin translation of Alhazen's statement, in the edition by Friedrich Risner (Basel 1572):
That is:
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 $\angle pks=\angle bdg=\angle adg$.