Consider the following:

Conjecture. Every convex set of area $1$ is contained in a triangle of area $A = 2$.

I can prove it if $A$ is changed to $4$. The convex set being a square shows that $A = 2$ would be tight if true.

It seems like this must be known are at least considered in the literature, but googling hasn't worked for me. Can anyone supply a proof of this or give me a reference?

  • $\begingroup$ It doesn't have to be the same triangle for all convex sets? $\endgroup$
    – Brian Tung
    Mar 26, 2018 at 21:53
  • $\begingroup$ What context does this question have? How did you come across this question? Even simple, fundamental-sounding questions don't tend to have answers when they don't serve some greater purpose. $\endgroup$ Mar 26, 2018 at 21:57
  • $\begingroup$ I suspect that you can consider polytopes without loss of generality, as polytopes are dense in the Hausdorff metric amongst non-empty bounded convex sets. It's worth noting that, although I suspect the lebesgue measure is continuous with respect the Hausdorff distance over convex sets, this is known not to be true over arbitrary sets, see math.stackexchange.com/questions/2552127/… $\endgroup$ Mar 26, 2018 at 22:37
  • $\begingroup$ @BrianTung: Right, different triangles for different sets (otherwise the conjecture is false: consider a circle and a long thin rectangle -- the same triangle can't possibly work for both) $\endgroup$ Mar 26, 2018 at 22:45
  • 1
    $\begingroup$ @Theo: If "triangle" is replaced by "rectangle", then it's a problem from "The Art of Mathematics: Coffee Time in Memphis" by Bollobás. Also see this link from cut-the-knot. I thought it was a beautiful proof and was personally curious about triangles. $\endgroup$ Mar 26, 2018 at 22:49

1 Answer 1


Okay, after thinking about it last night I think I have a proof. Let $a(X)$ be the area of $X$.

Consider a counter-example to the conjecture, with convex set $M$ and minimum enclosing triangle $T$ such that $a(M) < 2 a(T)$. Without loss of generality, assume that one side of $T$ is contained in the $x$-axis of the plan with its center on $(0, 0)$. For every point $(x, y)$ in the plane, perform a shear transformation $x \mapsto x + k_1 y$ so that $T$ is isosceles, and then perform a horizontal stretch / compression $x \mapsto k_2 x$ so that $T$ is an equilateral triangle. Let $T$ and $M$ now refer to these new figures under these transformations. Since these transformations preserve relative area, take triangles to triangles, take convex sets to convex sets, and are invertable,we know that $T$ and $M$ must still be a counter-example to the conjecture.

So we now have a diagram that looks like this:

enter image description here

Notice I've drawn $T$ intersecting $M$ at the midpoint of each side ($D, E, F$ in the diagram). This must be if $T$ is a minimal enclosing triangle (see this paper - basically if there was some side that doesn't meet $M$ at the midpoint, you could make a very small rotation of that one side towards the larger part yielding a smaller enclosing triangle).

Now let's flip $T$ upside-down, create a "star of David" diagram.

enter image description here

Consider $\triangle ADE$. Notice that $T'$ cuts this triangle into two pieces exactly $1/3$ of the way from the base $DE$ to $A$. Also notice that $M$ fails to reach the $T'$ line by a distance of $x$. This is important: if $M$ did reach this line, than the part of $M$ inside $\triangle ADE$ would be at least $1/3$ of the area of $\triangle ADE$. If, furthermore, $M$ also reached the line for triangles $\triangle BDF$ and $\triangle CEF$, then $M$ would have area at least $$ a(M) \geq a(\triangle DEF) + \frac{1}{3}\bigg(a(\triangle ADE) + a(\triangle BDF) + a(\triangle CEF)\bigg) \geq \frac{1}{2} a(T) $$ This is a clear contradiction. So, in at least one of the three triangles, $M$ does not reach the $1/3$ mark. In our example, it fails to reach $T'$ in $\triangle BDF$ and $\triangle ADE$ by a distance of $x$ and $y$ respectively. However, $M$ exceeds $T'$ by a distance of $z$ in $\triangle CEF$. We may still have the contradictory statement $a(M) \geq \frac{1}{2} a(T)$ if $z$ is too big compared to $x$ and $y$. In fact, since we can bound the area of $M$ in each triangle proportional to height, we can say explicity $x + y > z$, since otherwise we could show $a(M) \geq \frac{1}{2} a(T)$.

With the statement $x + y > z$, we can now slide $T'$ without rotating so that it now strictly encloses $M$. Why? Well, imagine sliding $T'$ in the direction $\overrightarrow{DF}$. As $x$ gets smaller, $z$ gets smaller at the same rate, and $y$ stays the same. Now, if you slide $T'$ in the direction of $\overrightarrow{DE}$, then $y$ will get smaller at the same rate $z$ gets smaller, and since $x + y > z$, we can slide $T'$ so it totally encloses $M$: enter image description here

This contradicts the fact that $T$ was a minimum triangle (since we can now make $T'$ a bit smaller and still enclose $M$.

Now, one final worry you might have is that there is no absolute minimum enclosing triangle $T$ (perhaps because of an infinite sequence of triangles that get smaller and smaller). However, do not worry -- any convex set does have an absolute minimum triangle that contains it. Why? Well consider the function $F(\alpha, \beta, \gamma)$ that takes in three angles on the range $[0, 2 \pi]$. For angle $\alpha$, we take a line perpendicular to $\alpha$ and approach $M$ (from infinity) with this line until we hit $M$. If we do this with $\beta$ and $\gamma$, then we create a triangle. The output of $F(\alpha, \beta, \gamma)$ is the area of this triangle. While this function may take on infinite values, if we cut these sections out we get a continuous function on a compact domain and therefore must actually achieve it's minimum values.


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