# Smallest triangles that contain 2D convex regions with reflection symmetry

Given any 2D convex region $$C$$ with a mirror symmetry. Two pairs of questions:

1. 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?

2. 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 $$C$$?

1. 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.

• The example is quite simple and nice; it also appears to show that the smallest containing triangle need not even be isosceles! – Nandakumar R Nov 8 '19 at 16:51
• It is not clear immediately if a similar negative answer holds for the second question - the largest triangle contained within C. Even in the case of the first question, one could ask for that mirror symmetric C for which taking the smallest isosceles container aligned along the axis of C gives the worst answer to the question of the smallest containing triangle and also when the answers to the area and perimeter questions will differ most. – Nandakumar R Nov 8 '19 at 16:59

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.)

• Checked numerically with a sequence of such 'coffin' shapes got by moving the position of the widest point in the heptagon leftwards from the middle. There is a small but finite range of positions for the widest point for which the green triangle beats all axis-aligned isosceles triangles contained inside the 'coffin' (including the pink triangle) in BOTH area and perimeter. That should settle both guesses in the negative. Note: b was chosen as 4 units, c as 10 and epsilon as 0,1 – Nandakumar R Nov 9 '19 at 11:29
• @NandakumarR: Thanks for performing the calculations. – Joseph O'Rourke Nov 9 '19 at 13:40