The following result was proved by G.
Toussaint and B.
Bhattacharya in 1990. They needed to prove this result in a paper on
the shortest transversals. This problem consist of computing the shortest
line segment that intersects a set of n given line segments or lines in
the plane.

Let QOR be the angle in consideration, P the point we are interested in and let H be the intercept of the philon line on OQ and K be the intercept of the philon line on OR. Let be the angle subtended by RKH as in Fig. 1. And let l denote the length of line segment HK. Let l() denote l as a function of where =< <=< . Then l() is a unimodal function on the interval.

**Figure 1. **Illustrating the fundamental geometric minimization
problem

**Figure 2.** Illustrating the proof of the lemma. O
represents the origin and x = u + u'

In order to simplify analysis it is convenient to
break up the problem into two cases depending on whether
is greater than
/ 2 or not, and to parameterize the problem not as a function of
as in Fig 1, but rather as a function of x =u + u' as in Fig. 2. It is
straightforward to demonstrate that unimodality under the first parameterization
implies unimodality under the second. Furthermore, we restrict our analysis
to the more difficult case of
greater than
/ 2. The analysis for
less than or equal to
/ 2 is similar and less involved. Accordingly, let p(,ß)
be the point in the interior of the cone in question. Thus ß > 0.
H'p has length a and is parallel to OK. K'p is parallel to OH. Let l(x)
denote the length of line segment HK as a function of x. We then have the
following lemma.

**Lemma:** For x > u > 0, l(x)is a unimodal function.

*Proof:*Without loss of generality we assume that
> 0. For otherwise we may construct a symmetrical diagram where ß
represents the perpendicular drop from p to OH rather than OK. First we
determine the value of x such that HK is a minimum.

Since pHH' and pK'K are similar it follows that pH / a = pK / (x - u) and thus, (pH+ pK) / pK = 1+ a / (x - u). Since l(x) = pH+ pK we can write l(x) as follows:

l(x)=(1+a/(x - u)) (1)

Differentiating (1) with respect to x we obtain,

l'(x)=[- 2u+ (- au)x - a].

Since x > u it follows that l'(x)=0 implies that

x^{3}- 2ux^{2}+ (-
au)x - a=
0 (2)

Rewriting (2) in standard form we have that

x^{3}+ ax^{2}+ bx + c = 0

where a = -2u < 0; b = u(u - a) < 0, since a > u, and c = -a
< 0.

Let , , and be roots of (2). We assume that , , and are all real. Then we can write

x^{3}+ ax^{2}+ bx + c = (x -r_{1})(x - r_{2})(x
- r_{3}) (3)

We now show that we need only examine one root in more detail. Since a, b, and c, are all less than zero we argue that precisely one of the roots must be greater than zero and the other two must be less than zero. We start by rewriting (3) as follows:

x^{3}+ ax^{2}+ bx + c = x^{3}- ( r_{1}+
r_{2}+ r_{3})x^{2}+ (r_{1}r_{2}
+ r_{2}r_{3} + r_{3}r_{1})x - r_{1}r_{2}r_{3}=
0

where a = -( r_{1}+ r_{2}+ r_{3}), b = (r_{1}r_{2}
+ r_{2}r_{3} + r_{3}r_{1}) and c= - r_{1}r_{2}r_{3}.
Now c < 0 and the only way - r_{1}r_{2}r_{3}<
0 is for either all three roots
to be positive or one to be positive and the other two negative. In addition
since b < 0 we must have (r_{1}r_{2} + r_{2}r_{3}
+ r_{3}r_{1}) < 0. Thus consider the case in which all
three roots
are positive. This implies (r_{1}r_{2} + r_{2}r_{3}
+ r_{3}r_{1}) > 0, a contradiction. It follows that one
root
must be positive and the other two negative. We can in a similar way show
that when only one root
of (2) is real that root
has to be a positive one. For assume that we have, as before,

x^{3}+ ax^{2}+ bx + c = 0 with a,b,c < 0,

and that we have two complex
roots
and one real. Let r_{1}be the real root.
Then we can rewrite this equation as

(x -r_{1})(x^{2}+ ex + f ) = 0.

For there to be two complex
roots
we must have e^{2}- f4 < 0, which implies in turn that f > 0.

Since c = -r_{1}f < 0 and f > 0 it follows that r_{1}>
0.

We conclude from the above discussion that we need only be concerned with the positive root. Let x = r be positive root of (2). We now show that l(x) attains a minimum value at x = r. We do this by demonstrating that the second derivative of l(x) evaluated at x = r must be greater than zero.

Let A = . Then l'(x) =A. It is sufficient to show that dA/dx > 0 at x = r. We see that

dA/dx = 3x^{2}- 4ux + u^{2} - au (4)

Evaluating (4) at x = r we obtain

dA/dx(r) = 3r^{2}- 4ur + u^{2}- au

= (2r^{2}- 2ur) + (r^{2}- 2ur + (u^{2}- au))

= (2r^{2}- 2ur) + a/r, since r^{3}- 2ur^{2}+ (u^{2}- au)r - a= 0.

= 2r(r - u) +a/r. (5)

Since all term in (5) are positive and r > u, dA/dx at x = r > 0,
and l(x) attains its minimum value at x = r. Since this minimum is the
only local
minimum in the region of interest, unimodality in the region of interest
follows.