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The points $A$, $B$ and $C$ lie on the sides of a square of side $1$ cm and no two points lie on the same side. Show that the length of at least one side of the triangle $ABC$ must be less than or equal to $(√6−√2)$ cm.

The given result can easily be derived by asserting that (1) the triangle is equilateral and (2) touches one of the corners.

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This is a perfectly intuitive result, but how can I prove (1) and (2) rigorously?

twentyyears
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2 Answers2

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A more "geometric" proof than the one @mechanodroid provided:

Claim: Fix $A,B$ on the side of the square. The point $C$ maximizing $\min(|AC|,|BC|)$ must either be a corner or $|AC|=|BC|$.

Proof: Suppose $C$ is not a corner, so we have a direction $\mathbf{d}$ of the edge which we can perturb $C$ both ways. Suppose $|AC|<|BC|$, then either $\angle(\mathbf{d},\overrightarrow{AC})\geq 90^\circ$ or $\angle(-\mathbf{d},\overrightarrow{AC})\geq 90^\circ$. So we can increase $|AC|$ either by moving $C$ to $C'=C+\varepsilon\mathbf{d}$ or $C'=C-\varepsilon\mathbf{d}$, while $|AC'|<|BC'|$ still holds, by the cosine rule and continuity. So we can move $C$ until either $|AC|=|BC|$ or $C$ hits a corner and increases $\min(|AC|,|BC|)$ in the process. QED.

So we have either an equilateral triangle, or we must have some corners.

  • 3 corners gives minimum side length 1
  • 2 corners gives either diagonally opposite corners, in which case the maximizing third vertex must be another corner; or adjacent corners, in which case we get minimum side length $\leq 1$.
  • Now 1 corner gives an equilateral triangle as the maximizer, and the case of equilateral triangles inscribed in a square is done in the link provided by @Invisible in the above comment.
user10354138
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Here is an entirely analytical attempt. Let $(x_1,0), (1,y),(x_2,1)$ be the vertices of the triangle inscribed in the unit square $[0,1]^2$. Since only one of them can be in a vertex of the square, we will allow only $x_1 \in [0,1\rangle$ and $x_2,y \in \langle 0,1\rangle$. Also we can assume $x_1 \le x_2$. The length of the smallest side is then given by the function $f : \{(x_1,x_2,y) \in [0,1\rangle \times \langle 0,1\rangle^2 : x_1\le x_2\}\to \Bbb{R}$ defined as $$f(x_1,x_2,y) = \min\left\{\sqrt{(x_2-x_1)^2+1},\sqrt{(1-x_1)^2+y^2},\sqrt{(1-x_2)^2+(1-y)^2}\right\}.$$ Let $(x_1,x_2,y)$ be the triple which maximizes $f$. Assume that this triangle is not equilateral. Then we can offset the vertices a bit so that the same side remains the smallest but it is a bit larger than before. For example if $$\sqrt{(x_2-x_1)^2+1} < \sqrt{(1-x_1)^2+y^2}, \sqrt{(1-x_2)^2+(1-y)^2}$$ then by continuity there exists $\varepsilon > 0$ such that $$\sqrt{((x_2+\varepsilon)-x_1)^2+1} < \sqrt{(1-x_1)^2+y^2}, \sqrt{(1-(x_2+\varepsilon))^2+(1-y)^2}$$ and hence $$f(x_1,x_2+\varepsilon,y) =\sqrt{((x_2+\varepsilon)-x_1)^2+1} > \sqrt{(x_2-x_1)^2+1} = f(x_1,x_2,y)$$ which contradicts maximality of $(x_1,x_2,y)$. Similarly in the other cases.

We conclude that the triangle is equilateral, so in particular $$f(x_1,x_2,y) = \sqrt{(x_2-x_1)^2+1}=\sqrt{(1-x_1)^2+y^2}=\sqrt{(1-x_2)^2+(1-y)^2}.$$ The term $\sqrt{(1-x_1)^2+y^2}$ is maximized when $x_1 = 0$ (since $y=1$ is not allowed). Therefore $$f(x_1,x_2,y) = \sqrt{x_2^2+1}=\sqrt{1+y^2}=\sqrt{(1-x_2)^2+(1-y)^2}.$$ and hence $y=x_2$ and $1+y^2=2(y-1)^2$. This yields $$x_1=0,\quad x_2=y=2-\sqrt{3}$$ and $f(x_1,x_2,y) = \sqrt{1+y^2} = \sqrt{6}-\sqrt{2}$.

mechanodroid
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