Philo line

File:Philo line equality.svg

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 $DE$ is the Philo line for point $P$ and angle $DOE$, and let $Q$ be the base of a perpendicular line $OQ$ to $DE$. Then $DP=EQ$ and $DQ=EP$.{{r|eves}}

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

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

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

The equation of a bundle of lines with inclinations $\backslash alpha$ that

run through the point $(x,y)=(P\_x,P\_y)$ is

: $$

y=\alpha(x-P_x)+P_y.

These lines intersect the horizontal axis at

: $$

\alpha(x-P_x)+P_y=0

which has the solution

: $$

(E_x,E_y)=\left(P_x-\frac{P_y}{\alpha},0\right).

These lines intersect the opposite side $y=mx$ at

: $$

\alpha(x-P_x)+P_y=mx

which has the solution

: $$

(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

: $$

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 $\backslash alpha$.

An arithmetic expression for the location of the minimum

is obtained by setting the derivative $\backslash partial\; d^2/\backslash partial\; \backslash alpha=0$,

so

: $$

-2m^2\frac{(P_x\alpha -P_y)[(mP_x-P_y)\alpha^3+P_x\alpha^2-2P_y\alpha+P_ym]}{\alpha^3 (\alpha-m)^3}=0 .

So calculating the root of the polynomial in the numerator,

: $$

(mP_x-P_y)\alpha^3+P_x\alpha^2-2P_y\alpha+P_ym=0

determines the slope of the particular line in the line bundle which has the shortest length.

[The global minimum at inclination $\backslash alpha=P\_y/P\_x$ 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 $(0,0)$.]

$-\backslash alpha$ is the tangent of the angle $OED$.

Inverting the equation above as $\backslash alpha\_1=P\_y/(P\_x-E\_x)$ and plugging this into the previous equation

one finds that $E\_x$ is a root of the cubic polynomial

: $$

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

: $$

d^2=

\frac{P_y^2+x^2-2xP_x+P_x^2}{(P_y+mx-mP_x)^2}

x^2m^2

.

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

: $Q\_x=\backslash frac\{(\backslash alpha\; P\_x-P\_y)\backslash alpha\}\{1+\backslash alpha^2\}$

: $Q\_y=-Q\_x/\backslash alpha\; =\; \backslash frac\{P\_y-\backslash alpha\; P\_x\}\{1+\backslash alpha^2\}$

With the coordinates $(D\_x,D\_y)$ shown above, the squared distance from $D$ to $Q$ is

: $DQ^2\; =\; (D\_x-Q\_x)^2+(D\_y-Q\_y)^2\; =\; \backslash frac\{(\backslash alpha\; P\_x-P\_y)^2(1+\backslash alpha\; m)^2\}\{(1+\backslash alpha^2)(\backslash alpha-m)^2\}$.

The squared distance from $E$ to $P$ is

: $EP^2\; \backslash equiv\; (E\_x-P\_x)^2+(E\_y-P\_y)^2\; =\; \backslash frac\{P\_y\; ^2(1+\backslash alpha^2)\}\{\backslash alpha^2\}$.

The difference of these two expressions is

: $DQ^2-EP^2\; =\; \backslash frac\{[(P\_xm+P\_y)\backslash alpha^3+(P\_x-2P\_ym)\backslash alpha^2-P\_ym]$

[(P_xm-P_y)\alpha^3+P_x\alpha^2-2P_y\alpha+P_ym]}{\alpha^2(1+\alpha^2)(a-m)^2}.

Given the cubic equation for $\backslash alpha$ above, which is one of the two cubic polynomials in the numerator, this is zero.

This is the algebraic proof that the minimization of $DE$ leads to $DQ=PE$.

The equation of a bundle of lines with inclination $\backslash alpha$ that

run through the point $(x,y)=(P\_x,P\_y)$, $P\_x,P\_y>0$, has an intersection with the $x$-axis given above.

If $DOE$ form a right angle, the limit $m\backslash to\backslash infty$ of the previous section results

in the following special case:

These lines intersect the $y$-axis at

: $$

\alpha(-P_x)+P_y

which has the solution

: $$

(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

: $d^2=(E\_x-D\_x)^2+(E\_y-D\_y)^2\; =\; \backslash frac\{(\backslash alpha\; P\_x-P\_y)^2(1+\backslash alpha^2)\}\{\backslash alpha^2\}.$

The Philo Line is defined by the minimum of that curve (at

negative $\backslash alpha$).

An arithmetic expression for the location of the minimum

is where the derivative $\backslash partial\; d^2/\backslash partial\; \backslash alpha=0$,

so

: $$

2\frac{(P_x\alpha -P_y)(P_x\alpha^3+P_y)}{\alpha^3}=0

equivalent to

: $$

\alpha = -\sqrt[3]{P_y/P_x}

Therefore

: $$

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 $\backslash alpha\_1=P\_y/(P\_x-E\_x)$ and plugging this into another equation above

one finds

: $$

E_x=P_x+P_y\sqrt[3]{P_y/P_x}.

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 $PQRS$ be a rectangle whose aspect ratio $PQ:QR$ is $1:2$, as in the figure. Let $TU$ be the Philo line of point $P$ with respect to right angle $QRS$. Define point $V$ to be the point of intersection of line $TU$ and of the circle through points $PQRS$. Because triangle $RVP$ is inscribed in the circle with $RP$ as diameter, it is a right triangle, and $V$ is the base of a perpendicular from the apex of the angle to the Philo line.

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

File:Philo line.svg

Since doubling the cube is impossible with a straightedge and compass construction, it is similarly impossible to construct the Philo line with these tools.{{r|eves|kimberling}}

Given the point $P$ and the angle $DOE$, a variant of the problem may minimize the area of the triangle $OED$. With the expressions for $(E\_x,E\_y)$ and $(D\_x,D\_y)$ given above, the area is half the product of height and base length,

: $A\; =\; D\_yE\_x/2\; =\backslash frac\{m(\backslash alpha\; P\_x-P\_y)^2\}\{2\backslash alpha(\backslash alpha-m)\}$.

Finding the slope $\backslash alpha$ that minimizes the area means to set $\backslash partial\; A/\backslash partial\; \backslash alpha=0$,

: $-\; \backslash frac\{m(\backslash alpha\; P\_x-P\_y)[(mP\_x-2P\_y)\backslash alpha+P\_ym]\}\{2\backslash alpha^2(\backslash alpha-m)^2\}=0$.

Again discarding the root $\backslash alpha\; =\; P\_y/P\_x$ which does not define a triangle, the slope is in that

case

:$\backslash alpha\; =\; -\backslash frac\{mP\_y\}\{mP\_x-2P\_y\}$

and the minimum area

:$A\; =\; \backslash frac\{2P\_y(mP\_x-P\_y)\}\{m\}$.

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{{refend}}

• {{mathworld | title = Philo Line | urlname = PhiloLine}}

{{DEFAULTSORT:Philo Line}}

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