In geometry, the Philo line is a line segment defined from an angle and a point inside the angle as the shortest line segment through the point that has its endpoints on the two sides of the angle. Also known as the Philon line, it is named after Philo of Byzantium, a Greek writer on mechanical devices, who lived probably during the 1st or 2nd century BC. Philo used the line to double the cube;^[1][2] because doubling the cube cannot be done by a straightedge and compass construction, neither can constructing the Philo line.^[1][3]

Geometric characterization

The defining point of a Philo line, and the base of a perpendicular from the apex of the angle to the line, are equidistant from the endpoints of the line. That is, suppose that segment ${\displaystyle DE}$ is the Philo line for point ${\displaystyle P}$ and angle ${\displaystyle DOE}$, and let ${\displaystyle Q}$ be the base of a perpendicular line ${\displaystyle OQ}$ to ${\displaystyle DE}$. Then ${\displaystyle DP=EQ}$ and ${\displaystyle DQ=EP}$.^[1]

Conversely, if ${\displaystyle P}$ and ${\displaystyle Q}$ are any two points equidistant from the ends of a line segment ${\displaystyle DE}$, and if ${\displaystyle O}$ is any point on the line through ${\displaystyle Q}$ that is perpendicular to ${\displaystyle DE}$, then ${\displaystyle DE}$ is the Philo line for angle ${\displaystyle DOE}$ and point ${\displaystyle P}$.^[1]

Algebraic Construction

A suitable fixation of the line given the directions from ${\displaystyle O}$ to ${\displaystyle E}$ and from ${\displaystyle O}$ to ${\displaystyle D}$ and the location of ${\displaystyle P}$ in that infinite triangle is obtained by the following algebra:

The point ${\displaystyle O}$ is put into the center of the coordinate system, the direction from ${\displaystyle O}$ to ${\displaystyle E}$ defines the horizontal ${\displaystyle x}$-coordinate, and the direction from ${\displaystyle O}$ to ${\displaystyle D}$ defines the line with the equation ${\displaystyle y{=}mx}$ in the rectilinear coordinate system. ${\displaystyle m}$ is the tangent of the angle in the triangle ${\displaystyle DOE}$. Then ${\displaystyle P}$ has the Cartesian Coordinates ${\displaystyle (P_{x},P_{y})}$ and the task is to find ${\displaystyle E=(E_{x},0)}$ on the horizontal axis and ${\displaystyle D=(D_{x},D_{y})=(D_{x},mD_{x})}$ on the other side of the triangle.

The equation of a bundle of lines with inclinations ${\displaystyle \alpha }$ that run through the point ${\displaystyle (x,y)=(P_{x},P_{y})}$ is

${\displaystyle y=\alpha (x-P_{x})+P_{y}.}$

These lines intersect the horizontal axis at

${\displaystyle \alpha (x-P_{x})+P_{y}=0}$

which has the solution

${\displaystyle (E_{x},E_{y})=\left(P_{x}-{\frac {P_{y)){\alpha )),0\right).}$

These lines intersect the opposite side ${\displaystyle y=mx}$ at

${\displaystyle \alpha (x-P_{x})+P_{y}=mx}$

which has the solution

${\displaystyle (D_{x},D_{y})=\left({\frac {\alpha P_{x}-P_{y)){\alpha -m)),m{\frac {\alpha P_{x}-P_{y)){\alpha -m))\right).}$

The squared Euclidean distance between the intersections of the horizontal line and the diagonal is

${\displaystyle ED^{2}=d^{2}=(E_{x}-D_{x})^{2}+(E_{y}-D_{y})^{2}={\frac {m^{2}(\alpha P_{x}-P_{y})^{2}(1+\alpha ^{2})}{\alpha ^{2}(\alpha -m)^{2))}.}$

The Philo Line is defined by the minimum of that distance at negative ${\displaystyle \alpha }$.

An arithmetic expression for the location of the minimum is obtained by setting the derivative ${\displaystyle \partial d^{2}/\partial \alpha =0}$, so

${\displaystyle -2m^{2}{\frac {(P_{x}\alpha -P_{y})[(mP_{x}-P_{y})\alpha ^{3}+P_{x}\alpha ^{2}-2P_{y}\alpha +P_{y}m]}{\alpha ^{3}(\alpha -m)^{3))}=0.}$

So calculating the root of the polynomial in the numerator,

${\displaystyle (mP_{x}-P_{y})\alpha ^{3}+P_{x}\alpha ^{2}-2P_{y}\alpha +P_{y}m=0}$

determines the slope of the particular line in the line bundle which has the shortest length. [The global minimum at inclination ${\displaystyle \alpha =P_{y}/P_{x))$ from the root of the other factor is not of interest; it does not define a triangle but means that the horizontal line, the diagonal and the line of the bundle all intersect at ${\displaystyle (0,0)}$.]

${\displaystyle -\alpha }$ is the tangent of the angle ${\displaystyle OED}$.

Inverting the equation above as ${\displaystyle \alpha _{1}=P_{y}/(P_{x}-E_{x})}$ and plugging this into the previous equation one finds that ${\displaystyle E_{x))$ is a root of the cubic polynomial

${\displaystyle mx^{3}+(2P_{y}-3mP_{x})x^{2}+3P_{x}(mP_{x}-P_{y})x-(mP_{x}-P_{y})(P_{x}^{2}+P_{y}^{2}).}$

So solving that cubic equation finds the intersection of the Philo line on the horizontal axis. Plugging in the same expression into the expression for the squared distance gives

${\displaystyle d^{2}={\frac {P_{y}^{2}+x^{2}-2xP_{x}+P_{x}^{2)){(P_{y}+mx-mP_{x})^{2))}x^{2}m^{2}.}$

Location of ${\displaystyle Q}$

Since the line ${\displaystyle OQ}$ is orthogonal to ${\displaystyle ED}$, its slope is ${\displaystyle -1/\alpha }$, so the points on that line are ${\displaystyle y=-x/\alpha }$. The coordinates of the point ${\displaystyle Q=(Q_{x},Q_{y})}$ are calculated by intersecting this line with the Philo line, ${\displaystyle y=\alpha (x-P_{x})+P_{y))$. ${\displaystyle \alpha (x-P_{x})+P_{y}=-x/\alpha }$ yields

${\displaystyle Q_{x}={\frac {(\alpha P_{x}-P_{y})\alpha }{1+\alpha ^{2))))$

${\displaystyle Q_{y}=-Q_{x}/\alpha ={\frac {P_{y}-\alpha P_{x)){1+\alpha ^{2))))$

With the coordinates ${\displaystyle (D_{x},D_{y})}$ shown above, the squared distance from ${\displaystyle D}$ to ${\displaystyle Q}$ is

${\displaystyle DQ^{2}=(D_{x}-Q_{x})^{2}+(D_{y}-Q_{y})^{2}={\frac {(\alpha P_{x}-P_{y})^{2}(1+\alpha m)^{2)){(1+\alpha ^{2})(\alpha -m)^{2))))$.

The squared distance from ${\displaystyle E}$ to ${\displaystyle P}$ is

${\displaystyle EP^{2}\equiv (E_{x}-P_{x})^{2}+(E_{y}-P_{y})^{2}={\frac {P_{y}^{2}(1+\alpha ^{2})}{\alpha ^{2))))$.

The difference of these two expressions is

${\displaystyle DQ^{2}-EP^{2}={\frac {[(P_{x}m+P_{y})\alpha ^{3}+(P_{x}-2P_{y}m)\alpha ^{2}-P_{y}m][(P_{x}m-P_{y})\alpha ^{3}+P_{x}\alpha ^{2}-2P_{y}\alpha +P_{y}m]}{\alpha ^{2}(1+\alpha ^{2})(a-m)^{2))))$.

Given the cubic equation for ${\displaystyle \alpha }$ above, which is one of the two cubic polynomials in the numerator, this is zero. This is the algebraic proof that the minimization of ${\displaystyle DE}$ leads to ${\displaystyle DQ=PE}$.

Special case: right angle

The equation of a bundle of lines with inclination ${\displaystyle \alpha }$ that run through the point ${\displaystyle (x,y)=(P_{x},P_{y})}$, ${\displaystyle P_{x},P_{y}>0}$, has an intersection with the ${\displaystyle x}$-axis given above. If ${\displaystyle DOE}$ form a right angle, the limit ${\displaystyle m\to \infty }$ of the previous section results in the following special case:

These lines intersect the ${\displaystyle y}$-axis at

${\displaystyle \alpha (-P_{x})+P_{y))$

which has the solution

${\displaystyle (D_{x},D_{y})=(0,P_{y}-\alpha P_{x}).}$

The squared Euclidean distance between the intersections of the horizontal line and vertical lines is

${\displaystyle d^{2}=(E_{x}-D_{x})^{2}+(E_{y}-D_{y})^{2}={\frac {(\alpha P_{x}-P_{y})^{2}(1+\alpha ^{2})}{\alpha ^{2))}.}$

The Philo Line is defined by the minimum of that curve (at negative ${\displaystyle \alpha }$). An arithmetic expression for the location of the minimum is where the derivative ${\displaystyle \partial d^{2}/\partial \alpha =0}$, so

${\displaystyle 2{\frac {(P_{x}\alpha -P_{y})(P_{x}\alpha ^{3}+P_{y})}{\alpha ^{3))}=0}$

equivalent to

${\displaystyle \alpha =-{\sqrt[{3}]{P_{y}/P_{x))))$

Therefore

${\displaystyle d={\frac {P_{y}-\alpha P_{x)){|\alpha |)){\sqrt {1+\alpha ^{2))}=P_{x}[1+(P_{y}/P_{x})^{2/3}]^{3/2}.}$

Alternatively, inverting the previous equations as ${\displaystyle \alpha _{1}=P_{y}/(P_{x}-E_{x})}$ and plugging this into another equation above one finds

${\displaystyle E_{x}=P_{x}+P_{y}{\sqrt[{3}]{P_{y}/P_{x))}.}$

Doubling the cube

The Philo line can be used to double the cube, that is, to construct a geometric representation of the cube root of two, and this was Philo's purpose in defining this line. Specifically, let ${\displaystyle PQRS}$ be a rectangle whose aspect ratio ${\displaystyle PQ:QR}$ is ${\displaystyle 1:2}$, as in the figure. Let ${\displaystyle TU}$ be the Philo line of point ${\displaystyle P}$ with respect to right angle ${\displaystyle QRS}$. Define point ${\displaystyle V}$ to be the point of intersection of line ${\displaystyle TU}$ and of the circle through points ${\displaystyle PQRS}$. Because triangle ${\displaystyle RVP}$ is inscribed in the circle with ${\displaystyle RP}$ as diameter, it is a right triangle, and ${\displaystyle V}$ is the base of a perpendicular from the apex of the angle to the Philo line.

Let ${\displaystyle W}$ be the point where line ${\displaystyle QR}$ crosses a perpendicular line through ${\displaystyle V}$. Then the equalities of segments ${\displaystyle RS=PQ}$, ${\displaystyle RW=QU}$, and ${\displaystyle WU=RQ}$ follow from the characteristic property of the Philo line. The similarity of the right triangles ${\displaystyle PQU}$, ${\displaystyle RWV}$, and ${\displaystyle VWU}$ follow by perpendicular bisection of right triangles. Combining these equalities and similarities gives the equality of proportions ${\displaystyle RS:RW=PQ:QU=RW:WV=WV:WU=WV:RQ}$ or more concisely ${\displaystyle RS:RW=RW:WV=WV:RQ}$. Since the first and last terms of these three equal proportions are in the ratio ${\displaystyle 1:2}$, the proportions themselves must all be ${\displaystyle 1:{\sqrt[{3}]{2))}$, the proportion that is required to double the cube.^[4]

Since doubling the cube is impossible with a straightedge and compass construction, it is similarly impossible to construct the Philo line with these tools.^[1][3]

Minimizing the area

Given the point ${\displaystyle P}$ and the angle ${\displaystyle DOE}$, a variant of the problem may minimize the area of the triangle ${\displaystyle OED}$. With the expressions for ${\displaystyle (E_{x},E_{y})}$ and ${\displaystyle (D_{x},D_{y})}$ given above, the area is half the product of height and base length,

${\displaystyle A=D_{y}E_{x}/2={\frac {m(\alpha P_{x}-P_{y})^{2)){2\alpha (\alpha -m)))}$.

Finding the slope ${\displaystyle \alpha }$ that minimizes the area means to set ${\displaystyle \partial A/\partial \alpha =0}$,

${\displaystyle -{\frac {m(\alpha P_{x}-P_{y})[(mP_{x}-2P_{y})\alpha +P_{y}m]}{2\alpha ^{2}(\alpha -m)^{2))}=0}$.

Again discarding the root ${\displaystyle \alpha =P_{y}/P_{x))$ which does not define a triangle, the slope is in that case

${\displaystyle \alpha =-{\frac {mP_{y)){mP_{x}-2P_{y))))$

and the minimum area

${\displaystyle A={\frac {2P_{y}(mP_{x}-P_{y})}{m))}$.