AL HAYTHAM'S PROBLEM
In the figure A: source, B: eye, P: point of reflection. First cut: region of the circle contained in the acute angle APB, opposite cut: the symmetric part of the surface through C.
Al Haytham proved that the highest number of points that can join A and B are four. His proof is based on lemmas described in Sabra's work: Ibn Al Haytham's lemmas for solving "AlHazens'" problem (Archive for the Exact Sciences Vol. 26 December 1982)
Software was developed to have an interactive way of following the construction and the proof of the lemmas.
Let circle ABG , with diameter GB, be known ; let GB be produced the side of G; let line KE be given and let point A be given on the circumference of the circle. We wish to draw from A a line, as AHD, so that the part of it that lies between the diameter and the circle--such as HD--is equal to line KE.
The second lemma is similar to the first
Again, in triangle ABG let angle B be right; let D be given on line BG; and let the ratio of E to Z be known; we wish to draw from D a line like DTK so that the ratio of TK to TG is as the ratio of E to Z.
The 3rd lemma uses the 1st lemma in constructing
The proof is given in 9 steps.
Fourth Lemma: This Lemma is used to find the points on the
circle that solve the problem.
We present two software applications:
This is a difficult proof with 20 steps. The picture shows the final step:
On the left side we see the use of the 3rd lemma.
In the 1st text box we have explanations of what is involved on each step.
On the 2nd text box we have the calculations of several parameters that are connected with each step.
On the 20th step it appears the button PARALLEL AW, which deals with a special case.
An Excel file for the use of the 4th Lemma INDEX FILES/USE OF LEMMA 4.xlsx
On the left is presentedHuygens
Here as we check for the total path we can see a) For parabola all the paths that start at a certain distance Y from the X-axis have the same path and that the reflected ray passes through the focus of the parabola