On maximum perimeter triangles inscribed in convex regions with one vertex fixed

Question

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.

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 $12$.

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