††thanks: This work was supported by ADA University Faculty Research and
Development Fund.

# The dual of Philo’s shortest line segment
problem

Yagub N. Aliyev  School of IT and Engineering
ADA University
Ahmadbey Aghaoglu str. 61
Baku 1008, Azerbaijan<EMAIL_ADDRESS>
(Date: January 1, 2004)

###### Abstract.

We study the dual of Philo’s shortest line segment problem which asks to find
the optimal line segments passing through two given points, with a common
endpoint, and with the other endpoints on a given line. The provided solution
uses multivariable calculus and geometry methods. Interesting connections with
the angle bisector of the triangle are explored. A generalization of the
problem using $L_{p}$ ($p\geq 1$) norm is proposed. Particular case $p=\infty$
is studied. Interesting case $p=2$ is proposed as an open problem and related
property of a symedian of a triangle is conjectured.

###### Key words and phrases:

Philo’s line, geometric inequalities, geometric maxima and minima, angle
bisector, symedian.

###### 1991 Mathematics Subject Classification:

Primary 51M16, 51M25; Secondary 51M04, 52A38, 52A40.

##  1\. Introduction

Maximum-minimum problems in geometry played crucial role in the history of
development of mathematical methods [11], Chapter VII, [8], [21] Chapters
VIII-XIX. Many attempts to solve variety of geometry problems for optimization
lead to the develoment of differential and integral calculus and later to the
calculus of variations. Maxima-minima problems also play important role in the
pedagogy of mathematics [3], [26]. In mathematical olympiads for schools or
universities, and journals with problem solving columns such as Crux
Mathematicorum and American Mathematical Monthly one can often find at least
one geometric inequality included in its list of proposed problems. Recent
breakthough in automation of proofs for classical euclidean geometry theorems
is yet to be done for the geometric inequalities [28]. The connections of the
geometric extrema problems with $L_{p}$ spaces in functional analysis were
discussed in [24], [33]. Mathematicians noticed very early that some of the
geometric optimization problems do not have solutions and even when they do
have a solution it is not always possible to construct it with euclidean
instruments (a compass and a straightedge). The problem about minimal line
segment with fixed point and inscribed into a given angle is one of them [9],
p. 405-406, [23], p. 36-37, [25] p. 33, 179-181, [18], [5], [19]. In this
problem an angle $ABC$ and a point $D$ in it is given and it is asked to find
the line $EF$ through this point which has shortest line segment $EF$ cut by
this angle. Sir I. Newton mentioned this problem as an example for his method
of fluxions (see [20], p. 45, Problem 5.III). The solution of this problem is
known in the literature as Philo’s line [13], [14] p. 39 and 234-238, [30] p.
182-183, [29], [16] p. 115–116. If $BG$ is the perpendicular to $EF$ from
point $B$, then $EF$ is shortest when $|ED|=|FG|$ (see Figure 1).

Figure 1. Philo’s line and the dual problem

Philo or Philon of Byzantium found solution of the classical problem of the
dublication of cube using this line [7], p. 107, [32], [22] p. 24, [27]. The
digits of the length of Philo’s line through the incenter of triangle with
sidelengths 3,4,5 appear in the OEIS as A195284 [17]. Generalizations for non-
euclidean geometry and higher dimensions appeared in [12], [31]. Related
problem of the longest or shortest chords of conics and curves in general have
similar solutions [4], [6], [2].

In projective geometry the principle of duality is used to explain the
similarity between theorems where the roles of lines and points are
interchanged. If we use this idea in the above extremum problem, then we
obtain the following dual problem ([2], p. 4). Given line $l$ and points $A$
and $B$ find point $C$ with ray $CA$ and $CB$ intersecting line $l$ at $D$ and
$E$, respectively, such that the sum of lengths of $CD$ and $CE$ is minimal
(see Figure 1 and Table 1). This problem is surprisingly much more challenging
than Philo’s shortest line problem. In the following text this problem is
solved using combination of methods from calculus and geometry. At the end of
the paper we mentioned other interesting formulae about the configuration and
several open problems.

| Philo’s line problem | Dual problem
---|---|---
Fixed point(s) | 1 ($D$) | 2 ($A$ and $B$)
Fixed line(s) | 2 ($BA$ and $BC$) | 1 ($l$)
Moving point(s) | 2 ($E$ and $F$) | 1 ($C$)
Moving line(s) | 1 ($EF$) | 2 ($CD$ and $CE$)
Table 1. The number of elements in Philo’s line problem and its dual.

##  2\. Main results.

As was mentioned before the optimal solution of geometric extrema problems can
sometimes be non-constructible with the help of a compass, a ruler, and the
given data. Without the proof of the existence of the optimal solution the
proof is incomplete. For example, in Hadamard’s classical geometry problem
book some of such extrema problems are provided with a comment like "assume
that a maximum exists" (see e.g. Exercise 366 in p. 313, Exercise 418b in p.
322 of [15]). Most of the time it is possible to refer to calculus results for
more rigorous proof with compact sets and Weierstrass theorem but it makes the
solution less appealing from the geometric point of view. Another possible way
out from this situation is to give or construct the optimal solution first in
the diagram and then ask to prove that the other possibilities are not
optimal. In the problem about Philo’s line we can give $\triangle BEF$ and
points $D$ and $G$ on its side $EF$ such that $|ED|=|FG|$, $BG\perp EF$, and
then prove that any other line segment $E_{1}F_{1}$ through point $D$ has a
greater length (see Figure 1). We can formulate the mentioned dual problem
above in the same way by giving the optimal solution first. It will be clear
from the proof why it is optimal and how it was found in the first place.

######  Theorem 2.1.

Let $CDE$ be a triangle with acute angles at vertices $D$ and $E$. Let $CK$ be
the angle bisector of triangle $CDE$. Drop perpendiculars $KL$ and $KM$ to
sides $CD$ and $CE$, respectively. Take points $A$ and $B$ on sides $CD$ and
$CE$, respectively, such that $|AC|=|LD|$, $|BC|=|ME|$. Then for any point
$C_{1}$ different from $C$, such that rays $C_{1}A$ and $C_{1}B$ intersect
line $DE$ at points $D_{1}$ and $E_{1}$, respectively, inequality
$|C_{1}D_{1}|+|C_{1}E_{1}|>|CD|+|CE|$ holds true (see Figure 2).

Figure 2. Inequality $|C_{1}D_{1}|+|C_{1}E_{1}|>|CD|+|CE|$

###### Proof.

Let us denote the coordinates of points as $A(a,b),\ B(c,d),\ D_{1}(x_{1},0)$
and $E_{1}(x_{2},0)$. Without loss of generality we can assume that $a<c$ and
$b\geq d>0$. If we denote coordinates of $C_{1}\ $as $(x,y)$, then
$x_{1}=\frac{ay-bx}{y-b}$ and $x_{2}=\frac{cy-dx}{y-d}$. Here
$(x,y)\in\mathbb{D}$, where $\mathbb{D}=\\{(x,y)|x\in(-\infty,+\infty),\
y>b\\}$. The last inequality is needed to guarantee the condition in the
statement of the problem which says that rays $C_{1}A$ and $C_{1}B$ intersect
line $DE$. We obtain that

$\left|C_{1}D_{1}\right|+\
\left|C_{1}E_{1}\right|=\sqrt{\left(x-x_{1}\right)^{2}+y^{2}}+\sqrt{\left(x-x_{2}\right)^{2}+y^{2}}$
$=y\left(\sqrt{\left(\frac{x-a}{y-b}\right)^{2}+1}+\sqrt{\left(\frac{x-c}{y-d}\right)^{2}+1}\right)=:f(x,y).$

Note that $f(x,y)$ is continuous in $\mathbb{D}$ and
$f(x,y)\rightarrow+\infty$ as $x\rightarrow\pm\infty$, $y\rightarrow b$ when
$x\neq a$. Similarly, $f(x,y)\rightarrow+\infty$ as $y\rightarrow+\infty$.
Also $\lim_{(x,y)\rightarrow(a,b)}f(x,y)$ does not exist and depends on the
direction of approach to point $(a,b)$. Indeed, if $b>d$, then

$\lim_{t\rightarrow
0^{+}}f(a+k_{1}t,b+k_{2}t)=b\left(\sqrt{\frac{k_{1}^{2}+k_{2}^{2}}{k_{2}^{2}}}+\frac{\sqrt{(b-d)^{2}+\left(a-c\right)^{2}}}{b-d}\right),$

where $k_{2}>0$. This limit is minimal when $k_{1}=0$ which corresponds to the
direction of line $x=a$. If $b=d$, then the last limit is also infinite.
Function $f(x,y)$ is also bounded from below because it is a positive
function. Then either there is a point $(x^{*},y^{*})\in\mathbb{D}$ where
function $f(x,y)$ is minimal or this function approaches its infimum when
$(x,y)\rightarrow(a,b)$ $(y>b)$. We will study the last case later. In the
first case $f^{\prime}_{x}(x^{*},y^{*})=0$, $f^{\prime}_{y}(x^{*},y^{*})=0$.
Obviously, $a\leq x^{*}\leq c$, because if $x<a$, then $f(x,y)>f(a,y)$, and if
$x>c$, then $f(x,y)>f(c,y)$. Note also that if $y$ is fixed, then function
$f(x,y)$ is convex with respect to $x$. Indeed,

$f_{xx}=\frac{y(y-b)}{\sqrt{{\left(\left(x-a\right)^{2}+(y-b)^{2}\right)^{3}}}}+\frac{y\left(y-d\right)}{\sqrt{{\left(\left(x-c\right)^{2}+(y-d)^{2}\right)^{3}}}}>0.$

We calculate

$f_{x}(a,y)=\frac{y\left(a-c\right)}{\left(y-d\right)\sqrt{{\left(a-c\right)^{2}+\left(y-d\right)^{2}}}}<0,$
$f_{x}(c,y)=\frac{y\left(c-a\right)}{\left(y-b\right)\sqrt{{\left(a-c\right)^{2}+\left(y-b\right)^{2}}}}>0.$

Therefore $a<x^{*}<c$. In order to locate point $x^{*}$, note that equality
$f^{\prime}_{x}(x,y)=0$ implies

$\frac{x-a}{(y-b)^{2}\sqrt{\left(\frac{x-a}{y-b}\right)^{2}+1}}=\frac{c-x}{(y-d)^{2}\sqrt{\left(\frac{x-c}{y-d}\right)^{2}+1}}\Leftrightarrow$
$\frac{x-a}{(y-b)\sqrt{(x-a)^{2}+(y-b)^{2}}}=\frac{c-x}{(y-d)\sqrt{(x-c)^{2}+(y-d)^{2}}}.$

Figure 3. $|K_{1}L_{1}|=|K_{2}M_{1}|.$

The last equality can be interpreted geometrically in the following way. Take
points $L_{1}$ and $M_{1}$ on sides $C_{1}D_{1}$ and $C_{1}E_{1}$,
respectively, such that $|AC_{1}|=|L_{1}D_{1}|$, $|BC_{1}|=|M_{1}E_{1}|$. Take
points $K_{1}$ and $K_{2}$ on line $D_{1}E_{1}$ such that $K_{1}L_{1}\perp
C_{1}D_{1}$ and $K_{2}M_{1}\perp C_{1}E_{1}$. Equality $f^{\prime}_{x}(x,y)=0$
is equivalent to say that $|K_{1}L_{1}|=|K_{2}M_{1}|$ (see Figure 3). Indeed,
if we draw the line through $C_{1}$ and parallel to $D_{1}E_{1}$ and take
points $F$ and $G$ such that $AF\perp AC_{1}$ and $BG\perp BC_{1}$, then
$|AF|=|K_{1}L_{1}|$ and $|BG|=|K_{2}M_{1}|$. On the other hand,

$|AF|=|AC_{1}|\cdot\tan{\angle
AC_{1}F}=\sqrt{(x-a)^{2}+(y-b)^{2}}\cdot\frac{y-b}{a-x}.$

Similarly,

$|BG|=\sqrt{(x-c)^{2}+(y-d)^{2}}\cdot\frac{y-d}{c-x}.$

Therefore

$|K_{1}L_{1}|=|K_{2}M_{1}|.$ $None$

The other condition ($f^{\prime}_{y}(x,y)=0$) can also be interpreted
geometrically but its little harder than the previous condition. First, drop
perpendicular $C_{1}H$ to side $D_{1}E_{1}$. Next, we calculate

$f^{\prime}_{y}(x,y)=\left(\sqrt{\left(\frac{x-a}{y-b}\right)^{2}+1}+\sqrt{\left(\frac{x-c}{y-d}\right)^{2}+1}\right)+$
$y\left(-\frac{\left(x-a\right)^{2}}{\sqrt{1+\frac{\left(x-a\right)^{2}}{\left(y-b\right)^{2}}}\,\left(y-b\right)^{3}}-\frac{\left(x-c\right)^{2}}{\sqrt{1+\frac{\left(x-c\right)^{2}}{\left(y-d\right)^{2}}}\,\left(y-d\right)^{3}}\right).$

If $a<x<c$, then equality $f^{\prime}_{y}(x,y)=0$ can be interpreted as

$|C_{1}D_{1}|+|C_{1}E_{1}|=\frac{|D_{1}H|^{2}}{|L_{1}D_{1}|}+\frac{|E_{1}H|^{2}}{|M_{1}E_{1}|}.$
$None$

Indeed, first note that

$\frac{1}{y}\left(|C_{1}D_{1}|+|C_{1}E_{1}|\right)=\sqrt{\left(\frac{x-a}{y-b}\right)^{2}+1}+\sqrt{\left(\frac{x-c}{y-d}\right)^{2}+1}.$

On the other hand note that

$\frac{\left(x-a\right)^{2}}{\sqrt{1+\frac{\left(x-a\right)^{2}}{\left(y-b\right)^{2}}}\,\left(y-b\right)^{3}}+\frac{\left(x-c\right)^{2}}{\sqrt{1+\frac{\left(x-c\right)^{2}}{\left(y-d\right)^{2}}}\,\left(y-d\right)^{3}}$
$=\frac{\frac{\left(x-a\right)^{2}}{\left(y-b\right)^{2}}}{\sqrt{{\left(x-a\right)^{2}}+{\left(y-b\right)^{2}}}}+\frac{\frac{\left(x-c\right)^{2}}{\left(y-d\right)^{2}}}{\sqrt{{\left(x-c\right)^{2}}+{\left(y-d\right)^{2}}}}$
$=\frac{\tan^{2}{\angle AC_{1}F}}{|AC_{1}|}+\frac{\tan^{2}{\angle
BC_{1}G}}{|BC_{1}|}.$

We know that $\angle AC_{1}F=\angle D_{1}$, $\angle BC_{1}G=\angle E_{1}$,
$|AC_{1}|=|L_{1}D_{1}|$ and $|BC_{1}|=|M_{1}E_{1}|$. Therefore,

$|C_{1}D_{1}|+|C_{1}E_{1}|=|C_{1}H|^{2}\left(\frac{\tan^{2}{\angle
D_{1}}}{|L_{1}D_{1}|}+\frac{\tan^{2}{\angle E_{1}}}{|M_{1}E_{1}|}\right),$

Figure 4. One possible order of points on line $D_{1}E_{1}$.

from which it follows that (2) holds true. Since
$\frac{|D_{1}H|^{2}}{|L_{1}D_{1}|}=|D_{1}H|\cdot\frac{|C_{1}D_{1}|}{|D_{1}K_{1}|}$
and
$\frac{|E_{1}H|^{2}}{|M_{1}E_{1}|}=|E_{1}H|\cdot\frac{|C_{1}E_{1}|}{|E_{1}K_{2}|}$,
equality (2) can also be written as

$|C_{1}D_{1}|+|C_{1}E_{1}|=|C_{1}D_{1}|\cdot\frac{|D_{1}H|}{|D_{1}K_{1}|}+|C_{1}E_{1}|\cdot\frac{|E_{1}H|}{|E_{1}K_{2}|},$

which we can rewrite as

$|C_{1}D_{1}|\cdot\left(1-\frac{|D_{1}H|}{|D_{1}K_{1}|}\right)=|C_{1}E_{1}|\cdot\left(\frac{|E_{1}H|}{|E_{1}K_{2}|}-1\right),$
$None$

or finally as

$|C_{1}D_{1}|\cdot\frac{|K_{1}H|}{|D_{1}K_{1}|}=|C_{1}E_{1}|\cdot\frac{|K_{2}H|}{|E_{1}K_{2}|}.$

From this we obtain

$\frac{|D_{1}C_{1}|}{|C_{1}E_{1}|}\cdot\frac{|E_{1}K_{2}|}{|K_{2}H|}\cdot\frac{|HK_{1}|}{|K_{1}D_{1}|}=1.$
$None$

If $AU$ is the angle bisector of triangle $D_{1}C_{1}E_{1}$, then by the angle
bisector theorem
$\frac{|D_{1}C_{1}|}{|C_{1}E_{1}|}=\frac{|D_{1}U|}{|UE_{1}|}$. By noting this
we can write (4) as

$\frac{|D_{1}U|}{|UE_{1}|}\cdot\frac{|E_{1}K_{2}|}{|K_{2}H|}\cdot\frac{|HK_{1}|}{|K_{1}D_{1}|}=1.$
$None$

If we denote $|D_{1}H|=v$, $|K_{2}H|=w$, $|K_{2}U|=z$, $|K_{1}U|=t$,
$|K_{1}E_{1}|=u$ (see Figure 4), equation (5) can be written simply as

$\frac{v+w+z}{t+u}\cdot\frac{u+t+z}{w}\cdot\frac{w+z+t}{v+w+z+t}=1,$

which after simplifications gives

$t^{2}v+t^{2}z+tuv+tuz+2tvz+2twz+2tz^{2}$
$+uvz+uwz+uz^{2}+vwz+vz^{2}+w^{2}z+2wz^{2}+z^{3}$
$=t(tv+tz+uv+uz+2vz+2wz+2z^{2})$ $+z(uv+uw+uz+vw+vz+w^{2}+2wz+z^{2})=0.$

It is possible only when $z=t=0$, which means that points $K_{1},K_{2},$ and
$U$ coincide. The other cases when points $K_{1}$ or $K_{2}$ are outside of
the side $D_{1}E_{1}$, or when point $K_{1}$ is closer to point $D_{1}$ than
point $K_{2}$ can be considered similarly. For example, if the order of points
is as in Figure 5, where the notations for the lengths of the line segments
has slightly changed, then

$\frac{v+w+t}{z+u}\cdot\frac{u}{z+t+w}\cdot\frac{w}{v+w}=1,$

which is not possible. Indeed, in this case since
$\frac{v+w+t}{z+u}=\frac{|D_{1}C_{1}|}{|C_{1}E_{1}|}<1,$ $v+w+t<z+u$, and
since

$u=|K_{2}M_{1}|\cdot\csc{\angle E_{1}}=|K_{1}L_{1}|\cdot\csc{\angle
E_{1}}<|K_{1}L_{1}|\cdot\csc{\angle D_{1}}=v+w,$

$u<v+w$, and therefore

$\frac{v+w+t}{z+u}\cdot\frac{u}{v+w}\cdot\frac{w}{z+t+w}<1.$

Figure 5. Another order of points on line $D_{1}E_{1}$.

Finally, if the order of points is as in Figure 6, where again the notations
for the lengths of the line segments has slightly changed, then

$\frac{w+v+z}{t}\cdot\frac{t+z}{v}\cdot\frac{v+z+t+u}{w+v+z+t+u}=1,$

which again simplifies to an impossible equality

$t^{2}w+t^{2}z+tuw+tuz+2tvz+2twz+2tz^{2}$
$+uvz+uwz+uz^{2}+v^{2}z+vwz+2vz^{2}+wz^{2}+z^{3}=0.$

Figure 6. One more order of points on line $D_{1}E_{1}$.

Note that because of equality (1), we only need to consider the cases when
point $U$ is between points $K_{1}$ and $K_{2}$. Otherwise, if $U$ is not
between points $K_{1}$ and $K_{2}$, then one can drop perpendiculars from
point $U$ to sides $C_{1}D_{1}$ and $C_{1}E_{1}$ and one of them will be
larger than the other one, which is impossible. Similarly, we do not need to
consider the cases when points $K_{1}$ and $K_{2}$ are on different sides of
point $H$, because in these cases one of the sides of equality (3) is positive
while the other side is negative and therefore (3) does not hold.

If $\angle D_{1}=\frac{\pi}{2}$ ($x=a$), then point $K_{1}$ goes to infinity
and equation (5) simplifies to
$\frac{|D_{1}U|}{|UE_{1}|}\cdot\frac{|E_{1}K_{2}|}{|K_{2}H|}=1,$ which simply
means that points $K_{2}$ and $U$ coincide. Similarly, if $\angle
E_{1}=\frac{\pi}{2}$ ($x=c$), then point $K_{2}$ goes to infinity and equation
(5) simplifies to
$\frac{|D_{1}U|}{|UE_{1}|}\cdot\frac{|HK_{1}|}{|K_{1}D_{1}|}=1,$ which implies
that points $K_{1}$ and $U$ coincide.

We proved that if $C_{1}(x^{*},y^{*})$ is a minimum point of $f(x,y)$, then
endpoint $U$ of angle bisector $C_{1}U$ coincides with points $K_{1},K_{2}$.
Note that triangle $CDE$ and its angle bisector $CK$ satisfy all these
conditions together with (1) and (2). In order to complete the proof, we need
to show that there are no minimum points other than $C$.

We will now study the locus of intersection point of line $L_{1}K_{1}$ and
$M_{1}K_{2}$ when $|AD_{1}|=|BE_{1}|$ and $D_{1}(x_{1},0)$ changes on line
$DE$, and then intersection of this locus with line $DE$ (see Figure 7). If
$b=d$, then by symmetry this locus is the perpedicular bisector of the line
segment $AB$, which always intersect line $DE$ at a unique point. So, we can
assume that $b>d$. Let us denote $|AD_{1}|=|BE_{1}|=t$. Then
$x_{1}=a-\sqrt{t^{2}-b^{2}}$ and $x_{2}=c+\sqrt{t^{2}-d^{2}}$. Then the
equations of lines $AD_{1}$ and $BE_{1}$ are
$\frac{y-b}{x-a}=\frac{-b}{x_{1}-a}$ and $\frac{y-d}{x-c}=\frac{-d}{x_{2}-c}$,
respectively. The coordinates of intersection point $C_{1}(x_{0},y_{0})$ of
these lines are

${x_{0}}=-\frac{d\sqrt{t^{2}-b^{2}}\,\sqrt{t^{2}-d^{2}}+bd^{2}-bt^{2}}{t^{2}\left(b^{2}-d^{2}\right)\left(d^{2}-t^{2}\right)}\times$
$\times\left(\left(-cd\sqrt{t^{2}-d^{2}}-\left(b-d\right)\left(d^{2}-t^{2}\right)\right)\sqrt{t^{2}-b^{2}}+ab\left(d^{2}-t^{2}\right)\right),$
${y_{0}}=\frac{db\left(\sqrt{t^{2}-d^{2}}\,b-d\sqrt{t^{2}-b^{2}}\right)\left(\sqrt{t^{2}-d^{2}}+\sqrt{t^{2}-b^{2}}-a+c\right)}{t^{2}\left(b^{2}-d^{2}\right)}.$

Figure 7. Locus of point $(x^{\prime},y^{\prime})$.

The coordinates of point $L_{1}(x_{3},y_{3})$ are $x_{3}=x_{0}+x_{1}-a$,
$y_{3}=y_{0}-b$. Similarly, the coordinates of point $M_{1}(x_{4},y_{4})$ are
$x_{4}=x_{0}+x_{2}-c$, $y_{3}=y_{0}-d$. Then the equations of lines
$L_{1}K_{1}$ and $M_{1}K_{2}$ are $\frac{y-y_{3}}{x-x_{3}}=\frac{x_{1}-a}{b}$
and $\frac{y-y_{4}}{x-x_{4}}=\frac{x_{2}-c}{d}$, respectively. By solving the
last two equations for $y$ we calculate ordinate $y^{\prime}$ of intersection
point of lines $L_{1}K_{1}$ and $M_{1}K_{2}$ as

$y^{\prime}=\phi(t):=\frac{(bd-t^{2})\left(\sqrt{1-\frac{b^{2}}{t^{2}}}+\sqrt{1-\frac{d^{2}}{t^{2}}}\right)+\frac{bd(c-a)}{t}}{b\sqrt{1-\frac{d^{2}}{t^{2}}}+d\sqrt{1-\frac{b^{2}}{t^{2}}}},$

where the denominator is always positive and the numerator is a decreasing
function for $t\geq b$. The abscissa $x^{\prime}$ of this intersection point
can also be determined but we will not need it:

$x^{\prime}=-\frac{\left(b^{2}d-dt^{2}+b\sqrt{t^{2}-d^{2}}\,\sqrt{t^{2}-b^{2}}\right)}{\left(b^{2}-t^{2}\right)t^{2}\left(b^{2}-d^{2}\right)}\times$
$\times\left(\left(ab\sqrt{t^{2}-b^{2}}+\left(b^{2}-t^{2}\right)\left(b-d\right)\right)\sqrt{t^{2}-d^{2}}\\\
+t^{2}\left(b-d\right)\sqrt{t^{2}-b^{2}}-cd\left(b^{2}-t^{2}\right)\right).$

The formula for $x^{\prime}$ and $y^{\prime}$ determine a parametric curve
which is locus of intersection point of lines $L_{1}K_{1}$ and $M_{1}K_{2}$.
This locus can intersect line $DE$ at maximum 1 point for $t\geq b$. Indeed,
note that $\phi(b)=d-b+\frac{d(c-a)}{\sqrt{b^{2}-d^{2}}}$ and
$\phi(+\infty)=-\infty$. Therefore, if
$d-b+\frac{d(c-a)}{\sqrt{b^{2}-d^{2}}}\geq 0$, then equation $\phi(t)=0$ has a
unique solution in interval $t\geq b$. Otherwise, if
$d-b+\frac{d(c-a)}{\sqrt{b^{2}-d^{2}}}<0$, then equation $\phi(t)=0$ has no
solutions in interval $t\geq b$. In the statement of the problem we are given
acute angled triangle $CDE$ satisfying all the conditions for extrema.
Therefore, $d-b+\frac{d(c-a)}{\sqrt{b^{2}-d^{2}}}>0$ and the only solution of
equation $\phi(t)=0$ in interval $t>b$ is $t_{0}=|AD|$. Note that
$t_{0}^{2}=d^{2}+\lambda^{2}$, where $\lambda$ is one of the solutions of
fourth degree equation

$\left(b^{2}-d^{2}\right)x^{4}+\left(2bda-2bc\right)x^{3}+\left(-2b^{3}d+2b^{2}d^{2}+2b\,d^{3}-2d^{4}\right)x^{2}$
$+\left(-2a\,b^{2}d^{2}+2ab\,d^{3}+2b^{2}cd-2bc\,d^{2}\right)x$
$+a^{2}b^{2}d^{2}+b^{4}d^{2}-2b^{3}d^{3}+2b\,d^{5}-d^{6}-2a\,b^{2}cd+b^{2}c^{2}=0,$

and therefore can be solved in radicals using Ferrari’s method.

It remains to show that if $b>d$ and $k_{2}>0$, then

$\lim_{t\rightarrow 0^{+}}f(a+k_{1}t,b+k_{2}t)>|CD|+|CE|.$

Since the limit is minimal when $k_{1}=0$, it is sufficient to show that

$\lim_{y\rightarrow b^{+}}f(a,y)>|CD|+|CE|.$

Note that since $2d^{3}+y^{3}\geq 3d^{2}y$,

$f_{yy}(a,y)=\frac{\left(a-c\right)^{2}\left(2d^{3}+{y^{3}}-{3d^{2}y}+2d\left(a-c\right)^{2}\right)}{\sqrt{\left(\left(y-d\right)^{2}+\left(a-c\right)^{2}\right)^{3}}\,\left(y-d\right)^{3}}>0,$

and therefore $f(a,y)$ is convex in interval $y\geq b$. Also
$\lim_{y\rightarrow+\infty}f_{y}(a,y)=2>0$ and

$\lim_{y\rightarrow
b^{+}}f_{y}(a,y)=1+\sqrt{1+\frac{\left(a-c\right)^{2}}{\left(b-d\right)^{2}}}-\frac{b\left(a-c\right)^{2}}{\sqrt{1+\frac{\left(a-c\right)^{2}}{\left(b-d\right)^{2}}}\,\left(b-d\right)^{3}}<0.$

Indeed,

$\lim_{y\rightarrow
b^{+}}f_{y}(a,y)=\left(1+\sqrt{1+\frac{\left(a-c\right)^{2}}{\left(b-d\right)^{2}}}\right)\left(1-\frac{d}{b-d}\left(\sqrt{1+\frac{\left(a-c\right)^{2}}{\left(b-d\right)^{2}}}-1\right)\right),$

where the first factor is always positive. The second factor can be simplified
as

$\frac{b-d\sqrt{1+\frac{\left(a-c\right)^{2}}{\left(b-d\right)^{2}}}}{b-d}=\frac{b(b-d)-d\sqrt{\left(a-c\right)^{2}+\left(b-d\right)^{2}}}{\left(b-d\right)^{2}}$
$=\frac{(b^{2}-d^{2})(b-d)^{2}-d^{2}(a-c)^{2}}{(b-d)^{2}\left(b(b-d)+d\sqrt{\left(a-c\right)^{2}+\left(b-d\right)^{2}}\right)},$

where the denominator is always positive. The numerator can be factorised as

$\left((b-d)\sqrt{b^{2}-d^{2}}-d(a-c)\right)\left((b-d)\sqrt{b^{2}-d^{2}}+d(a-c)\right),$

which is negative because as was mentioned before
$d-b+\frac{d(c-a)}{\sqrt{b^{2}-d^{2}}}>0$. Therefore $f(a,y)$ is decreasing
and then increasing for $y>b$ and consequently, $f(a,y)$ can not have infimum
at $(a,b)$ and the only minimum of $f(x,y)$ is point $C$. ∎

As an immediate application of Theorem 1 we obtain the following particular
result.

######  Corollary 2.2.

Let $CDE$ be a triangle with acute angles at vertices $D$ and $E$, such that
$|CE|>|CD|$. Let $CK$ be the angle bisector of triangle $CDE$. Drop
perpendiculars $KL$ and $KM$ to sides $CD$ and $CE$, respectively. Take points
$A$ and $B$ on sides $CD$ and $CE$, respectively, such that $|AC|=|LD|$,
$|BC|=|ME|$. Drop perpendicular $AA_{0}$, $BB_{0}$, and $CC_{0}$ to side $DE$.
Extend line $AB$ to intersect line $DE$ at point $P$. Then
$|AA_{0}|+|AP|>|CD|+|CE|$ (see Figure 9).

Note that we also solved the following more general problem.

######  Problem 1.

Given $A(a,b),\ B(c,d),$ with $a<c$ and $b\geq d>0$ find the minimum of
$|C_{1}D_{1}|+|C_{1}E_{1}|$, where $C_{1}(x,y)$ is a point such that $y>b$ and
rays $C_{1}A$ and $C_{1}B$ intersect line $y=0$ at points $D_{1}$ and $E_{1}$,
respectively.

Answer. If $b=d$ or if $b>d$ and $d-b+\frac{d(c-a)}{\sqrt{b^{2}-d^{2}}}>0$,
then the minimum exists and it is equal to $|CD|+|CE|$, from the previous
result. If $b>d$ and $d-b+\frac{d(c-a)}{\sqrt{b^{2}-d^{2}}}\leq 0$, then the
minimum does not exist. In this case there is infimum of
$|C_{1}D_{1}|+|C_{1}E_{1}|$ at point $(a,b)$ which is equal to
$b\left(1+\sqrt{1+\frac{\left(a-c\right)^{2}}{\left(b-d\right)^{2}}}\right)$.
See Figure 8 for the graph of $f(x,y)$ in these two cases.

Figure 8. Graph of $f$ in cases $d-b+\frac{d(c-a)}{\sqrt{b^{2}-d^{2}}}\leq 0$
(left, $a=0.1$, $b=0.3$, $c=0.3$, $d=0.1$) and
$d-b+\frac{d(c-a)}{\sqrt{b^{2}-d^{2}}}>0$ (right, $a=0.1$, $b=0.3$, $c=0.3$,
$d=0.2$). Black point $\left(a,b,\lim_{y\rightarrow b^{+}}{f(a,y)}\right)$.

Remark 1\. Note that in general $f(x,y)$ is not a convex function. So, we can
not use the well know result in analysis which says that a local minimum of a
convex function is also the absolute minimum (see e.g. [10], p. 543).
Actually, each of functions

$f_{1}\left(x,y\right)=y\cdot\sqrt{1+\frac{\left(x-a\right)^{2}}{\left(y-b\right)^{2}}},\
f_{2}\left(x,y\right)=y\cdot\sqrt{1+\frac{\left(x-c\right)^{2}}{\left(y-d\right)^{2}}},$

have negative Hessian. Indeed,

$(f_{1})_{xx}(f_{1})_{yy}-(f_{1})_{xy}(f_{1})_{yx}=-\frac{b^{2}\left(x-a\right)^{2}}{\left(y-b\right)^{4}\left(\left(x-a\right)^{2}+(y-b)^{2}\right)}<0,$

and similarly for $f_{2}$. As some numerical examples show there can be points
$(x,y)$ for which Hessian of $f=f_{1}+f_{2}$ is also negative.

Remark 2\. We proved that triangle $CDE$ has optimal $|CD|+|CE|$ within all
other triangles $C_{1}D_{1}E_{1}$, which share common pair of points $A$ and
$B$ on their sides. Triangle $CDE$ has also other interesting properties worth
mentioning. First drop perpendiculars $AA_{0}$, $BB_{0}$, and $CC_{0}$ to side
$DE$ and suppose that $|CD|<|CE|$. Extend lines $AB$ and $ML$ to intersect
line $DE$ at points $P$ and $Q$, respectively (see Figure 9).

Figure 9. Properties of $\triangle CDE$.

  * •

$|PE|=|QD|=\frac{|DE|^{2}}{|B_{0}C_{0}|-|A_{0}C_{0}|}$. In particular, this
implies that if points $D$, $E$, and $C_{0}$ are fixed, and point $C$ moves
vertically along $CC_{0}$, then lines $AB$ and $LM$ pass though fixed points
$P$ and $Q$, respectively.

  * •

$|DE|=|A_{0}B_{0}|+\frac{[CDE]}{R}$, where $[CDE]$ and $R$ are the area and
the circumradius, respectively, of $\triangle CDE$.

  * •

$\frac{|AA_{0}|}{|BB_{0}|}=\frac{|CE|}{|CD|}$.

  * •

$|CA|+|CB|=\frac{|DE|^{2}}{|CE|+|CD|}$.

  * •

$|PQ|=\frac{|DE|^{3}}{|CE|^{2}-|CD|^{2}}$.

  * •

$\frac{|CA|}{|CB|}=\frac{|DC_{0}|}{|EC_{0}|}$.

Remark 3\. It is possible to generalize Problem 1 by asking to find the
minimum of $\left(|C_{1}D_{1}|^{p}+|C_{1}E_{1}|^{p}\right)^{\frac{1}{p}}$,
which can be interpreted as $L_{p}$ ($p\geq 1$) norm. We already solved case
$p=1$. In analogy with [24], [33] we can look at case $p=\infty$. In this case

$\left(|C_{1}D_{1}|^{p}+|C_{1}E_{1}|^{p}\right)^{\frac{1}{p}}=\max{\left(|C_{1}D_{1}|,|C_{1}E_{1}|\right)}$

and the solution is essentially the same. We will only give the answer and
some implications.

######  Problem 2.

Given $A(a,b),\ B(c,d),$ with $a<c$ and $b\geq d>0$ find the minimum of
$\max{\left(|C_{1}D_{1}|,|C_{1}E_{1}|\right)}$, where $C_{1}(x,y)$ is a point
such that $y>b$ and rays $C_{1}A$ and $C_{1}B$ intersect line $y=0$ at points
$D_{1}$ and $E_{1}$, respectively.

Answer. If
$k_{0}:=\left(\frac{b+d}{c-a}\right)^{\frac{1}{3}}>\frac{b-d}{c-a}$, then the
minimum of function $f(x,y)$ is $f(x^{*},y^{*})$, where

$x^{*}=\frac{a+c}{2}-\frac{b-d}{2k_{0}},\
y^{*}=\frac{b+d}{2}-k_{0}\frac{a-c}{2}.$

If $k_{0}\leq\frac{b-d}{c-a}$, then the minimum of function $f(x,y)$ does not
exist. In this case there is infimum of function $f(x,y)$ at some points
infinitely close to $(a,b)$ and the value of the infimum is equal to

$b\sqrt{\left(\frac{c-a}{b-d}\right)^{2}+1}.$

Using this result we can prove the following theorem.

######  Theorem 2.3.

Let $A$ and $B$ be points on sides $CD$ and $CE$, respectively, of isosceles
triangle $CDE$ such that $\tan{\angle D}=\tan{\angle
E}=\left(\frac{|AA_{0}|+|BB_{0}|}{|A_{0}B_{0}|}\right)^{\frac{1}{3}}$, where
$AA_{0}$ and $BB_{0}$ are perpendiculars from points $A$ and $B$,
respectively, to side $DE$, and $|AA_{0}|>|BB_{0}|$. Then for any point
$C_{1}$ different from $C$, such that rays $C_{1}A$ and $C_{1}B$ intersect
line $DE$ at points $D_{1}$ and $E_{1}$, respectively, inequality
$\max{\left(|C_{1}D_{1}|,|C_{1}E_{1}|\right)}>|CD|$ holds true. In particular,
if lines $AB$ and $DE$ intersect at point $P$, then $|AP|>|CD|$ (see Figure
10).

Figure 10. Inequalities $\max{\left(|C_{1}D_{1}|,|C_{1}E_{1}|\right)}>|CD|$
and $|AP|>|CD|$.

I leave as an open question to study case $p=2$ and other values of $p$. In
particular, prove that if $CK$ in Theorem 2.1 is a symedian, then

$|C_{1}D_{1}|^{2}+|C_{1}E_{1}|^{2}>|CD|^{2}+|CE|^{2}.$

As a problem for further exploration it would also be interesting to study
these questions for more than two points instead of just points $A$ and $B$.
One can also replace line $DE$ with a plane and ask the same question for more
than two points in space.

##  3\. Conclusion

In the paper we solved geometric optimization problems inspired by Philo’s
line. The obtained solution reveals interesting properties of the angle
bisector and its projections on the sides of this triangle. The provided proof
uses multivariable optimization and geometrical methods. Some generalizations
and an open problem about symedians is proposed.

## Acknowledgments

##  4\. Declarations

Ethical Approval. Not applicable.
Competing interests. None.
Authors’ contributions. Not applicable.
Funding. This work was completed with the support of ADA University Faculty
Research and Development Fund.
Availability of data and materials. Not applicable

## References

  * [1]
  * [2] Aliyev Y.N., Construction of Tangents of Conchoid of Nicomedes and Limacon of Pascal, and Solution of Extremal Chord Problem in Ellipse Using Ophiuride Curve, Tr. J. Nature Sci., Vol. 2 No. 2 (2013) 1-6. https://dergipark.org.tr/tr/download/issue-full-file/21277
  * [3] T. Andreescu, O. Mushkarov, L. Stoyanov, Geometric Problems on Maxima and Minima, Birkhäuser, Berlin, 2006.
  * [4] Anghel N., Geometric loci associated to certain minimal chords in convex regions, J. Geom. 66, No.1- 2, 1-16 (1999).
  * [5] Anghel N., Minimal chords in angular regions., Forum Geom. 4, 111-115, electronic only (2004).
  * [6] Anghel N., On the constructability with ruler and compass of a minimum chord in a parabola, Libertas Math. 17, 9-12 (1997).
  * [7] B. Bollobás, The Art of Mathematics – Take Two: Tea Time in Cambridge, Cambridge University Press, 2022.
  * [8] Boltyansky V. G., Yaglom I. M., Geometrical problems on maximum and minimum, in Encyclopedia of Elementary Mathematics in Five Volumes. Volume 5. (in Russian: Entsiklopediya elementarnoy matematiki v pyati tomakh. Tom 5) Under the editorship of P.S. Aleksandrov, A.I. Markushevich, A.I. Khinchin, Moscow, Nauka, 1966. p. 270-348. https://math.ru/lib/61
  * [9] Booth J., A treatise on some new geometrical methods, Vol. 2, Taylor and Francis, London, 1873. https://hdl.handle.net/2027/coo.31924059412936?urlappend=%3Bseq=439
  * [10] P.G. Ciarlet, Linear and Nonlinear Functional Analysis with Applications, The Society for Industrial and Applied Mathematics, SIAM, Philadelphia, 2013.
  * [11] R. Courant, H.E. Robbins, Was ist Mathematik?, Springer-Verlag, Berlin, 2010.
  * [12] Coxeter, H.S.M., van de Craats, J. Philon lines in non-Euclidean planes. J Geom 48, 26–55 (1993). https://doi.org/10.1007/BF01226799
  * [13] Eves, H. Philo’s Line. Scripta Math. 24, 141-148, 1959.
  * [14] Eves, H. W. A Survey of Geometry, Vol. 2. Boston, MA: Allyn and Bacon, 1965.
  * [15] J. Hadamard, Lessons in Geometry, Vol. 1: Plane Geometry, American Mathematical Society, 2008.
  * [16] Kimberling, C. Geometry in Action: A Discovery Approach Using The Geometer’s Sketchpad. Emeryville, California: Key College Publishing. 2003.
  * [17] Kimberling C., A195284, The On-Line Encyclopedia of Integer Sequences (OEIS). https://oeis.org/A195284
  * [18] Neovius, E. (1888). "Ueber eine specielle geometrische Aufgabe des Minimums". Mathematische Annalen. 31 (3): 359–362.
  * [19] Neuberg, J. (1907). Sur un minimum. Mathesis: recueil mathématique, Volume 27, 68–69.
  * [20] Newton I., The method of fluxions and infinite series : with its application to the geometry of curve-lines, Printed by Henry Woodfall, London, 1736. https://archive.org/details/methodoffluxions00newt/page/44/mode/2up
  * [21] G. Polya Mathematics and Plausible Reasoning, Volume 1: Induction and Analogy in Mathematics, Princeton University Press, 1990.
  * [22] V.V. Prasolov, Three classical construction problems. Doubling the cube, trisection of an angle, squaring the circle. (in Russian: Tri klassicheskiye zadachi na postroyeniye. Udvoyeniye kuba, trisektsiya ugla, kvadratura kruga) Moscow, Nauka, 1992. https://math.ru/lib/plm/62
  * [23] Protasov V.Yu., Maxima and Minima in geometry (in Russian), MCCME, Moscow, 2005. https://math.ru/lib/files/pdf/mp-seria/book.31.pdf
  * [24] V. Protasov, V. Tikhomirov, $L_{p}$ spaces and remarkable points of a triangle (in Russian: Prostranstvo $L_{p}$ i zamechatel’nye tochki treugol’nika), Kvant, 2012, no. 2, 2–11. http://mi.mathnet.ru/kvant1860
  * [25] Shklyarsky D.O., Chensov N.N., Yaglom I.M., Geometric inequalities and problems on maxima and minima (in Russian), Nauka, Moscow, 1970.
  * [26] V.M. Tikhomirov, Stories about Maxima and Minima, AMS, Providence, RI, 1990\.
  * [27] G.T. Toussaint, Scanning the Issue: Computational Geometry, Proceedings of the IEEE, Vol. 80, No. 9, September 1992. https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=163405
  * [28] Trinh, T.H., Wu, Y., Le, Q.V. et al. Solving olympiad geometry without human demonstrations. Nature 625, 476–482 (2024). https://doi.org/10.1038/s41586-023-06747-5
  * [29] Weisstein, Eric W. "Philo Line." From MathWorld–A Wolfram Web Resource. https://mathworld.wolfram.com/PhiloLine.html
  * [30] Wells, D. The Penguin Dictionary of Curious and Interesting Geometry. London: Penguin, 1991.
  * [31] Wetterling,W.W.E., Philon’s line generalized: An optimization problem from geometry, J. Optim. Theory Appl. 90 (1996) 517–521.
  * [32] Wikipedia: Philo line (the free encyclopedia). https://en.wikipedia.org/wiki/Philo_line
  * [33] Zaslavskii A., One more time on $L_{p}$ spaces and remarkable points of a triangle (in Russian: Eshhe raz o prostranstvah $L_{p}$ i zamechatel’nyh tochkah treugol’nika), Kvant, 2013, no. 5-6, 45–47. http://mi.mathnet.ru/kvant2002