(Note: This is an article from 2016 that I posted on Hubpages. In this article I used Riaz’s way of illustrating Lill’s method. To become more familiar with Lill’s method I recommend going through the links provided in my Lill’s Method page. Also see my Papers page and my Lill’s method articles/posts. )

In this hub I want to show that the problem of finding the Philo line for a right angle is equivalent to finding the solution to a specific type of cubic equations using Lill’s method. These cubic equations have the following form: αz^{3}-β=0, where α and β are the coordinates of the given point P(α,β). But to make myself more clear, first I must define some important terms.

### Philo Line

Given two intersecting lines OA and OB forming an angle with vertex at O and a point P inside the angle AOB, the Philo line is the shortest line segment passing though P reaching from line OA to line OB. In image 1, point C is on line OA and point D is on line OB such that the segment CD passes through point P. The segment CD is called a Philo line only if the length of CD has the smallest possible value. Philo line has many interesting characteristics that I will not cover in this hub. Also in this hub I will deal only with the case where mAOB=90.

When mAOB=90 , the problem of finding the Philo line can be redefined as another kind of optimization problem. Finding the shortest line through P is equivalent to finding the largest ladder that can fit a corridor with an external corner at O and an internal corner at P. In image 2, the segment CD is a Philo line if it is the longest ladder that can be maneuvered around the corner of the corridor.

If the coordinates of point P are (α,β), where α and β are positive, then the length of the longest ladder that can be maneuvered around the corner of the corridor is given by the formula (α^{2/3}+β^{2/3})^{3/2}. In this hub I will not prove this formula, but you can see sources [1] and [2] provided at the bottom of this hub.

### Cubic Equations and Lill’s Method

In a previous hub I used a specific quadratic equation in order to illustrate some general rules for finding the real roots of a polynomial equation of the form P(z)=a_{n}z^{n}+a_{n-1}z^{n-1}+…+a_{1}z+a_{0 }using Lill’s method. Lill’s method is very remarkable since the roots of a polynomial can be illustrated geometrically. In Lill’s method, z_{r} is a real root if z_{r}=-tan(θ), where θ is a constant “reflection” or “refraction” angle. It is better to read my previous hub in order to understand Lill’s method.

Now let’s have point O(0,0) as the intersection of x and y coordinates. We also start with the point P(α,β), were α and β are positive. Since x and y are perpendicular and α and β are positive, we have a 90 degrees angle and a point P inside it. In the previous section I mentioned the fact that the length of the shortest segment from x to y passing through P(α,β) is (α^{2/3}+β^{2/3})^{3/2}. Now I want to show how to obtain this formula using Lill’s method.

In the introductory paragraph I mentioned the fact that the cubic equation we want to solve using Lill’s method is of the form αz^{3 }-β=0, where α and β are the coordinates of the given point P(α,β). In image 3 I use the length and direction formula a_{k}e^{i(n-k)π/2} to construct the reflective segments. In our case a_{3}=α, a_{2}=0, a_{1}=0 and a_{0}=β. The segment OA corresponds to coefficient a_{3}=α and the segment AP corresponds to coefficient a_{0}=β. As explained in my previous hub, all the segments are extended at infinity in both directions by dotted lines and these dotted lines are refractive lines. a_{2}=0 is represented only by a dotted line that overlaps segment AP and its extension of dotted lines. Similarly a_{1}=0 is represented only by a dotted line that overlaps segment OA and its extension of dotted lines. In image 4 I illustrate exactly what it means to find a solution to our equation. We have point B on dotted line belonging to a_{2}=0, point C on dotted line belonging to a_{1}=0 and point D on y axis. Image 4 shows a solution since we have a path of 3 perpendicular “rays” or segments that start from point O and arrive at point P by making 2 “refractions” at constant angle θ. Numerically the root z_{r}=-tan(θ), and θ should be negative in this case and z_{r }should be positive. Now we want to show that CD is the Philo line passing through P.

### Equivalence

Without going into details, we know that a cubic equation of the form αz^{3 }-β=0, where where α and β are positive real numbers, has one positive real solution and 2 complex solutions. The cubic is also equivalent to αz^{3 }= β , z^{3 }= β/α or z = (β/α)^{1/3}. Thus z_{r}= -tan(θ) =(β/α)^{1/3}. Looking at image 4 we see that triangles AOB, ABC and ACP are similar. tan=opposite/ adjacent, thus z_{r}= -tan(θ)=AB/OA=AC/AB=AP/AC. We know that OA=α and AP=β. Using the previous proportions we also get AB=α z_{r}, AC=α z_{r}^{2} and AP=β=α z_{r}^{3}.

To show equivalence we need to show that length of CD= (α^{2/3}+β^{2/3})^{3/2}. Looking at image 4 we can see that z_{r}= -tan(θ)=OD/OC. OC=OA+AC=α+ α z_{r}^{2}. OD= z_{r}OC=α z_{r}+ α z_{r}^{3}. Finally we see that CD^{2 }= OC^{2 }+ OD^{2}. The final calculations are shown in image 5.

### Sources

[1] https://ckrao.wordpress.com/2010/11/07/the-ladder-around-a-corner-problem/

[2] Uncommon Mathematical Excursions: Polynomia and Related Realms pages 142-143