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This was a question that was for some reason removed by the OP. I'm a prospective sophomore intrigued by the Alhazen's billiard problem. Could someone post the answer to the original question here?

I was working on a project for school, and was wondering how to go about it in an unconventional manner. I thought of solving Alhazen's Billiard Problem, that is, deriving a general equation for it, for an ellipse. This problem is usually done for a circle, and so far, I've tried to use the method provided in Heinrich Dorrie's '100 Great Problems of Elementary Mathematics.' This problem concerns reflections, and is solved using tan ratios of the angles that the incident and reflected ray strike with. Here is a diagram for your reference (of Dorrie's method that I've tried to apply here). Image used for help

When I equate the tan ratios of alpha and beta using the other angle sums mentioned, the substitution gets messy due to the ellipse equation. Any outlook or thought would be appreciated. Thank you!

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    http://mathworld.wolfram.com/AlhazensBilliardProblem.html is insoluble with straightedge and compass, so which kind of problem are you trying to solve? – 2017-01-01
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    I tried to find the cached page of the original question, and here's what the original OP said (continued in next comment) – 2017-01-01
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    "Alhazen's Problem states the following: In a given circle, find an isosceles triangle whose legs pass through two given points inside the circle. Essentially, if we have a point A anywhere inside a circle, at what point on the circumference must it be sent as a ray so that it reflects (incident angle=reflected angle) and hits another point B. The difficulty with this problem is that we must form a general equation such that points A and B can be anywhere on the circle. So, the solution to the problem is an equation which can derive points on the circumference to reflect from and hit 'B'." – 2017-01-01
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    I still do not understand what is your actual question: how to derive such equation, maybe? – 2017-01-01
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    Precisely, that's the question – 2017-01-01
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    All right, I hope my answer is enlightening, then. – 2017-01-01

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Assume that $A$ and $B$ are two points inside a circular billiard $\Gamma$ with centre $O$.
We want a point $P$ on the boundary of $\Gamma$ such that the shot from $A$ to $P$ goes through $B$: enter image description here

so we want a point $P\in\partial\Gamma$ such that $OP$ bisects $\widehat{APB}$.
We have (at least) two ways for approaching the question:

  1. We may consider a variable $P(\theta)$ on $\partial\Gamma$ and compute $Q(\theta) = OP\cap AB$.
    By the bisector theorem, $OP$ bisects $\widehat{APB}$ iff $\frac{AQ}{QB}=\frac{AP}{PB}$;
  2. We may consider the ellipse through $P$ with foci at $A,B$. $P$ is a solution iff such ellipse is tangent to $\Gamma$ at $P$, hence we may consider the pencil of ellipses with foci at $A$ and $B$ and compute the parameter for which a member of such pencil is tangent to $\Gamma$ by imposing that a discriminant is vanishing.

In both cases, the solution is given by a root of a third degree polynomial, so the problem, in general, is not solvable through straightedge and compass.

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Could you apply our recent results? and we are considering the applications to Geometry.

The division by zero is uniquely and reasonably determined as 1/0=0/0=z/0=0 in the natural extensions of fractions. We have to change our basic ideas for our space and world

Division by Zero z/0 = 0 in Euclidean Spaces Hiroshi Michiwaki, Hiroshi Okumura and Saburou Saitoh International Journal of Mathematics and Computation Vol. 28(2017); Issue 1, 2017), 1 -16.　 http://www.scirp.org/journal/alamt　 　http://dx.doi.org/10.4236/alamt.2016.62007 http://www.ijapm.org/show-63-504-1.html http://www.diogenes.bg/ijam/contents/2014-27-2/9/9.pdf http://okmr.yamatoblog.net/division%20by%20zero/announcement%20326-%20the%20divi http://okmr.yamatoblog.net/

Relations of 0 and infinity Hiroshi Okumura, Saburou Saitoh　and Tsutomu Matsuura： http://www.e-jikei.org/…/Camera%20ready%20manuscript_JTSS_A… https://sites.google.com/site/sandrapinelas/icddea-2017