Smallest triangles that contain 2D convex regions with reflection symmetry Given any 2D convex region $C$ with a mirror symmetry. Two pairs of questions: 


*

*We need to find the smallest area (likewise, smallest perimeter) triangle that contains $C$. Is it sufficient to only search among isosceles triangles aligned along the direction of mirror symmetry of $C$ for answers to both questions? 

*Similarly, if we seek the largest area (largest perimeter) triangle contained within $C$, is it enough to look only among isosceles triangles aligned along the direction of symmetry of $C$? 
 A: *

*No.  Consider a region $C$ that looks like this (in blue): a square with two small, slightly different "bulges" at the top and bottom (so the only mirror symmetry is across a vertical axis).  An enclosing triangle of least area is shown in red, whose area is twice the area of the square.  But an enclosing triangle with the same mirror symmetry would need more area, to accommodate the bulges.



A: The example below seems to suggest No for the inscribed question 
as well. 
The line of symmetry is horizontal (dashed).
It seems the best aligned isosceles triangle (pink) has
area $A_1=\frac{1}{2} (c+\epsilon) b$,
while the unaligned, non-isosceles triangle (green)
has area $A_2 = \frac{1}{2} c (b+5 \epsilon)$.
$A_2 > A_1$ when $5c > b$, which clearly holds (because $c > b$),

          


          

Green $\Delta$ area $>$ pink $\Delta$ area.


I say "seems" because I have not proved that the
pink isosceles triangle is the largest such.
(Nor have I proved that the green triangle is the
largest inscribed triangle.)
