7
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ It doesn't have to be the same triangle for all convex sets? $\endgroup$ – Brian Tung Mar 26 '18 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. $\endgroup$ – Theo Bendit Mar 26 '18 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/… $\endgroup$ – Theo Bendit Mar 26 '18 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) $\endgroup$ – Tyler Seacrest Mar 26 '18 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. $\endgroup$ – Tyler Seacrest Mar 26 '18 at 22:49
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.