5
$\begingroup$

Definition: A line segment with both end points on the boundary of a planar convex region $C$ is called a chord of $C$.

  1. Consider any point $P$ within a given planar convex region $C$. From among all chords of $C$ that pass thru $P$, find that chord for which the ratio between the 2 segments into which $P$ divides the chord (longer seg : shorter seg) is a maximum and note this maximum. Now, how does one find and characterize the position of $P$ inside $C$ such that this maximum ratio is a minimum?

Remarks: Numerical experiments indicate that when $C$ is a triangle, the optimal position of $P$ is always the centroid of the triangle. But for general convex $C$, $P$ may not be at the center of mass.

  1. Consider any $P$ within a given planar convex region $C$. For every angle of orientation measured from $P$, there is a unique chord of $C$ that passes thru $P$. Consider the average over the orientation of the length of the corresponding chord. Now, how does one find and characterize that position of $P$ such that the average length of chord thru it is maximum?

Remarks: Numerically, when $C$ is a triangle, the optimal position of $P$ does not seem to be on the centroid or incenter of $C$. Note also that the average length of chord thru $P$ can also be considered over each chord starting at each point on the boundary of $C$ and passing thru $P$ – in this case we might have a different optimal $P$.

Note 1: Both questions above appear to have natural higher dimensional analogs.

Note 2: (this is actually, a further question) We can also look for that interior point that minimizes the ratio between the max distance from it to the boundary of C and the min distance from it to the boundary of C. Numerically, when C is a triangle, this 'center' is not seen to coincide with centroid or incenter.

$\endgroup$
2
  • $\begingroup$ A reasonable start would be calculating the centers for a 6-9-13 triangle, and then finding them in or adding them to the Encyclopedia of Triangle Centers: faculty.evansville.edu/ck6/encyclopedia/ETC.html $\endgroup$ – Matt F. Feb 8 at 13:52
  • 2
    $\begingroup$ For question 1 and with C as a triangle, the desired optimal center is (numerically) coincident with centroid. For question 2, with C a 6,9,13 triangle (ie one with coordinates (0,0), (13,0), (8.2307692, 3.64066448), the optimal center is found to be approx (6.549397, 1.194187) and I could not find a triangle center coincident with this point in the 'encyclopedia'. However, am new to that resource. $\endgroup$ – Nandakumar R Feb 9 at 7:13
3
$\begingroup$

This is not really an answer to the questions but 2 observations about the first definition. I only have proof sketches for both, but if I can always try to fill in more details.

  1. for any point P define its depth as the maximum of (longer seg : shorter seg). So the depth of a point in a convex region is some value between 1 and infinity, where the points of depth infinity are those on the boundary. We are looking for a point which has minimal depth. I claim that there is always a point of depth 2, which is tight, as witnessed by the centroid in an equilateral triangle. Sketch of a proof: for any point P we define a vector as follows. If there is a chord through P on which the ratio is >2, we put in a vector along the chord towards the larger side (we call those the chord vectors). We then average over all such vectors. Intuitively, this vector „drags“ P into the direction of the solution. It can be shown that no 3 chord vectors can have P in their convex hull (using convexity and the fact that we only define a chord vector if the ratio is >2), implying that the average vector can only be 0 if the depth of P is at most 2. The existence of such a point now follows from the Brouwer fixed point theorem.

  2. let A(x) be the region of all points with depth at most x. Then these regions are nested (by definition) and convex, implying that the optimal point is unique. To see that the regions are convex, consider points a and b of depth x and assume there is a point c on the segment ab of larger depth. This means there is a chord gh through c s.t. the ratio gc/ch>x. Consider now the chords defined by g and a, and g and b. On both these chords the ratio must be at most x. But from this it follows that the chord through the two other endpoints does not cross the chord gh, implying that c is not on the boundary of the convex hull, which is a contradiction.

$\endgroup$
5
  • $\begingroup$ Thanks. I could follow this much: (1)If one plots the 'chord vector' at each interior point with depth > 2, it will drag the search to lower depth regions. (2) The regions of progressively lower depth are nested into the interior. So starting from near the boundary, one will reach a region of depth of depth<=2 points within C if such a region is guaranteed to exist. You obtain the guarantee I guess, like this: Indeed, If every point within C has depth > 2, then the chord vectors will not lead to progressively smaller nested regions - an impossibility due to your proposition 2.. $\endgroup$ – Nandakumar R Feb 10 at 9:48
  • $\begingroup$ Yes, I think you can use the second part to show that a region of depth <=2 exists, like you mentioned, I didn‘t notice that. My idea was that the vectors define a continuous function from the convex set to itsself, which maps each point to the endpoint of its vector. By the Brouwer fixed point theorem there thus has to be a fixed point. However, as the vectors are non-zero for points of depth >2, such a fixed point must have depth <=2. $\endgroup$ – Patrick Schnider Feb 10 at 10:35
  • $\begingroup$ Thanks again for that intuitive clarification. Hope you could give answers to the other questions (q 1 and the additional one posed in Note 2 above) as a separate answer, $\endgroup$ – Nandakumar R Feb 11 at 7:40
  • $\begingroup$ I think both properties above (convexity of the regions and existence of a „deep“ point) should also go through for the definition in Note 2 and for higher dimension (where for the tightness we replace the equilateral triangle by a simplex whose edges all have the same lenghts). As for your main question, the characterization of the optimal point, I have no idea so far, but I‘ll post an answer should I find something :) $\endgroup$ – Patrick Schnider Feb 11 at 8:03
  • $\begingroup$ Thanks. I can now see that the arguments you gave for question 1 essentially apply to the additional question as well - although (numerically) even when C is a triangle, the optimal point does not coincide with a well-known center of the triangle. $\endgroup$ – Nandakumar R Feb 11 at 10:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.