Philo line

The philo line of a point P and angle DOE, and the defining equality of distances from P and Q to the ends of DE, where Q is the base of a perpendicular from the apex of the angle

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]

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

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

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]

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

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

• Weisstein, Eric W. "Philo Line". MathWorld.