Philo line explained

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; because doubling the cube cannot be done by a straightedge and compass construction, neither can constructing the Philo line.

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

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 .

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 .

Algebraic Construction

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

\alpha thatrun through the point (x,y) = (P _{x,P} _{y)} isy = \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 - (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} ,m _{y) = \left( (\alpha P _{x-P} _{y})/(\alpha - m)} (\alpha P _{x-P} _{y})/(\alpha - m) \right).

The squared Euclidean distance between the intersections of the horizontal lineand the diagonal is

ED ^{2} = d ^{2= (E} _{x - D} ^{2+ (E} ^{y - D} ^{2} ^{y)} = (m ^{2 (\alpha} P _{x-P} ^{2 ( 1+ \alpha} ^{y)} ^{2)})/(\alpha ^{2(\alpha-m)} ^{2}) .

The Philo Line is defined by the minimum of that distance atnegative

\alpha .\partial d ^{2/\partial} \alpha =0 ,so-2 m ^{2 ((P _{x\alpha} - P _{y)[(mP} _{x-P} ^{2-2 P} ^{y\alpha + P} _{ym]})/(\alpha ^{3} (\alpha - m) ^{3})} =0 . So calculating the root of the polynomial in the numerator,(mP _{x-P} ^{2-2 P} ^{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 <math>\alpha=P_y/P_x</math> 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 <math>(0,0)</math>.]- \alpha is the tangent of the angle OED .

Inverting the equation above as

\alpha _{1=P} _{y/(P} _{x-E} _{x)} and plugging this into the previous equationone finds that E _{x} is a root of the cubic polynomialmx ^{3+ ( 2 P} _{y -3 mP} ^{2+3 P} ^{x(mP} _{x-P} _{y)x-(mP} _{x-P} _{y)(P} ^{2 )} ^{y} . 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 givesd ^{2= (P ^{2+ x} _{y} ^{2-2 xP} ^{x + P} ^{2} ^{x})/((P _{y + mx - mP} ^{2} ^{x)})} x ^{2m} ^{2 .}

Location of

Q

Since the line

OQ is orthogonal to ED , its slope is -1 /\alpha , so the points on that line are y =- x/\alpha . The coordinates of the point Q = (Q _{x,Q} _{y)} are calculated by intersecting this line with the Philo line, y = \alpha(x - P _{x)+P} _{y} . \alpha(x - P _{x)+P} _{y=-x/\alpha} yieldsQ _{x = ((\alpha P _{x-P} _{y)\alpha})/(1+ \alpha ^{2})} Q _{y=-Q} _{x/\alpha} = (P _{y-\alpha} P _{x})/(1+ \alpha ^{2}) With the coordinates (D _{x,D} _{y)} shown above, the squared distance from D to Q isDQ ^{2} = (D _{x-Q} ^{2+ (D} ^{y - Q} ^{2} ^{y)} = ((\alpha P _{x-P} ^{2 ( 1+ \alpha} ^{y)} m) ^{2})/(( 1+ \alpha ^{2)(\alpha-m)} ^{2}) .The squared distance from E to P isEP ^{2} \equiv (E _{x-P} ^{2+ (E} ^{y - P} ^{2} ^{y)} = (P ^{2 ( 1+ \alpha} _{y} ^{2)})/(\alpha ^{2}) .The difference of these two expressions isDQ ^{2-EP} ^{2} = ([(P _{xm + P} ^{3+ (P} ^{x -2 P} ^{2- P} ^{ym] [(P} _{xm-P} ^{2-2 P} ^{y\alpha + P} _{ym]})/(\alpha ^{2(1+\alpha} ^{2)(a-m)} ^{2}) .Given the cubic equation for \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 .

Special case: right angle

The equation of a bundle of lines with inclination

\alpha thatrun 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\to inf ty of the previous section resultsin 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 linesis

d ^{2= (E} _{x - D} ^{2+ (E} ^{y - D} ^{2} ^{y)} = ((\alpha P _{x-P} ^{2 ( 1+ \alpha} ^{y)} ^{2)})/(\alpha ^{2}) .

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

\alpha ).An arithmetic expression for the location of the minimumis where the derivative \partial d ^{2/\partial} \alpha =0 ,so2 ((P _{x\alpha} - P _{y)(P} ^{3+ P} ^{y)})/(\alpha ^{3}) =0 equivalent to\alpha = - \sqrt[ 3 ]{P _{y/P} _{x}} Therefored = (P _{y-\alpha} P _{x})/(|\alpha|) \sqrt{ 1+ \alpha ^{2 } = P} _{x[ 1+ (P} _{y/P} ^{2 / 3} ^{x)} ] ^{3/2} . Alternatively, inverting the previous equations as \alpha _{1=P} _{y/(P} _{x-E} _{x)} and plugging this into another equation aboveone findsE _{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

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 :\sqrt[ 3 ]{ 2 } , the proportion that is required to double the cube.

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

Minimizing the area

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} = (m(\alpha P _{x-P} ^{2} ^{y)})/(2 \alpha(\alpha - m)) .Finding the slope \alpha that minimizes the area means to set \partial A/\partial \alpha =0 ,- (m(\alpha P _{x-P} _{y)[(mP} _{x-2P} _{y)\alpha+P} _{ym]})/(2 \alpha ^{2(\alpha-m)} ^{2}) =0 .Again discarding the root \alpha = P _{y/P} _{x} which does not define a triangle, the slope is in thatcase \alpha =- (mP _{y})/(mP _{x-2P} _{y}) and the minimum areaA = (2 P _{y(mP} _{x-P} _{y)})/(m) .

Further reading

• Neovius, Eduard. Ueber eine specielle geometrische Aufgabe des Minimums. Mathematische Annalen. 31. 1888. 359–362. 10.1007/BF01206220. 3. 123120289.
• Neuberg . J. . Joseph Jean Baptiste Neuberg. Sur un minimum. Mathesis. 1907. 68–69.
• Wetterling, W. W. E.. Philon's line generalized: an optimization problem from geometry. Journal of Optimization Theory and Applications. 90. 1996. 3. 517–521. 1402620. 10.1007/BF02189793. 119699906.