Ref: Convex curves with many inscribed triangles maximizing perimeter

Given a planar convex region C. Let P be a variable point on its boundary.

Observations: When C is an ellipse, the variation in the perimeter of the max perimeter inscribed triangle with one vertex constrained to be at P is found to be within around 10% as P runs around C - even when ratio a/b of C tends to infinity (thinking of an ellipse beginning as a circle and then getting stretched along major axis, we have a very 'physical' function of a/b that grows from 1 to just about 1.1 as the a/b goes all the way from 1 (circle) to infinity!). When C is an equilateral triangle, the variation in perimeter of max perimeter inscribed triangle with one node fixed is almost 22%. For a square, this variation is found to be only 5%.

Question: Among all planar convex regions of given area and perimeter, which shapes minimize and maximize the variation in the perimeter of the max perimeter inscribed triangle with one vertex constrained to be at P - as P varies around the boundary of the convex region?

Note 1: One can ask same question with minimum perimeter triangles that contain C such that one side of the triangle has to be a tangent to C at P. And also consider, say inscribed quadrilaterals with 2 vertices fixed.

Note 2: When C is an ellipse, the area of the max area inscribed triangle with one vertex at P remains constant as P moves around boundary of C - at each position of P, the max area inscribed triangle is one with centroid coincident with the center of C and it has C as its Steiner circumellipse.

  • 1
    $\begingroup$ Edited the question with more experimental data in an attempt to address your concerns. $\endgroup$ Commented Oct 10, 2022 at 10:27
  • $\begingroup$ Thanks, this is much improved! $\endgroup$
    – user44143
    Commented Oct 10, 2022 at 12:44
  • $\begingroup$ Do you know a nice one-parameter family of convex shapes which all have the same area and perimeter? Without that it’s hard to find good comparisons to use here. $\endgroup$
    – user44143
    Commented Oct 10, 2022 at 12:51
  • $\begingroup$ for a suitably chosen A and P, one can form a continuous range of triangles beginning with isosceles all of which have area A and perimeter P. but not sure if it is useful here. $\endgroup$ Commented May 19 at 6:11

1 Answer 1


The maximal variation is $4/3$, achieved by the convex region which is a $5-5-6$ triangle. In that region, inscribed triangles have perimeter of at most $16$, and inscribed triangles with a vertex at the midpoint of the base side have a perimeter of at most $12$.

Let us prove that this is the maximum. Let $P$ be any point on the boundary of a convex region, let $ABC$ be a triangle inscribed in that region with maximum perimeter, and assume that the perimeter is $16$. We show that the perimeter of one of the triangles $PAB$, $PBC$, $PCA$ is at least $12$, which yields the desired estimate.

Assume the converse. Replace $P$ with its metric projection $P_1$ to the triangle $ABC$ (i.e., $P_1$ is the point in $ABC$ closest to $P$; all the perimeters decrease!); assume that $P_1$ lies on $AB$.

Next, replace $P_1$ with the point $P’$ on $AB$ such that $P’A+AC=P’B+BC$ (that is the tangency point of the excircle with $AB$). Then the perimeters of $P'AC$ and $P'AB$ become equal, and they do not exceed the maximum of the perimeters of $P_1AC$ and $P_1BC$ (e.g. in the picture, the perimeter of $P'AC$ does not exceed that of $P_1AC$).

Assume that $P’A\geq P’B$. Let $Q$ be the midpoint of $AB$, and $D$ be the point such that $AQ+BQ=AC+BC$ and $AQ=BQ$. Then $P’D\leq P’C$ (the circle centered at $P'$ with radius $P'D$ meets the ellipse with foci $A$ and $B$ and passing through $C$ and $D$ at two points); so $QD\leq P’D\leq P’C$, and hence the point $Q$ in $ABD$ provides a ratio no greater than $P’$ in $ABC$.

Finally, if $AB\geq 6$, then the perimeter of $QAB$ is at least $12$, otherwise $AC=BC>5$ and $QC>4$, so the perimeter of $QAC$ is larger than $12$. Therefore, one of the perimeters of $P'AC$, $P_1AC$ and $PAC$ isalso at least $12$.

enter image description here

NB. Clearly, the minimum variation is $1$ achieved on a circle.

  • $\begingroup$ @MattF. Surely, an inscribed triangle here is inscribed into the given region, as, e.g., in the title of the question.By a projection I mean the metric projection, I.e., the closest point to $P$ in the triangle; sorry for that. $\endgroup$ Commented Oct 11, 2022 at 4:10
  • $\begingroup$ Could you give a figure? That would help me understand the arguments better. $\endgroup$ Commented Oct 13, 2022 at 16:14
  • $\begingroup$ I've added a figure and expanded some arguments. $\endgroup$ Commented Oct 14, 2022 at 8:25
  • $\begingroup$ Thanks very much. I think this is a very nice proof! I understand this takes care of the global extreme case of the question. With both {area,perimeter} specified, the shape of C that achieves the maximum of the ratio between inscribed triangle perimeters might still be a triangle - for those {A,P} pairs for which triangles can be drawn. I hope this question too would soon find an answer: mathoverflow.net/questions/432417/… $\endgroup$ Commented Oct 18, 2022 at 7:33
  • $\begingroup$ This makes sense to me now with the edits that I made -- I hope they align with your intentions, and that if not you'll correct them. $\endgroup$
    – user44143
    Commented Oct 22, 2022 at 14:19

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