Finding the smallest possible perimeter of a 4-gon inscribed in a square

Question

"Take the square $ABCD$. Let $E_1$ be a point on the side $AB$ such that $AE_1 : E_1B = 2 : 1$. Find the points $E_2, E_3, E_4$ on the sides $BC, CD$, and $DB$ respectively such that the sum $|E_1E_2| + |E_2E_3| + |E_3E_4| + |E_4E_1|$ is the smallest. Find this sum."

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My Attempt:

I first took the two triangles (formed by the diagonal $DB$) individually. I reflected $E_1$ in each of the sides and considered the line joining these two reflected points. Where that line hit the triangle on $DB$, I labelled as $E_4$. I did the same for the second triangle and formed a path.

I don't know if this method is correct and, if it is, I do not know how to find the perimeter of this shape, as the question requires. Any help?

Answer 1

Whatever $E_2$, $E_3$, $E_4$ you choose,

1. Reflect square $ABCD$ wrt side $BC$ to get square $A'BCD'$.
   The quadrilateral $E_1E_2E_3E_4$ get mapped to quadrilateral $F_1E_2F_3F_4$.

2. Reflect square $A'BCD'$ wrt side $CD'$ to get square $A''B'CD'$.
   The quadrilateral $F_1E_2F_3F_4$ get mapped to quadrilateral $G_1G_2F_3G_4$.

3. Reflect square $A''B'CD'$ wrt side $A''D'$ to get square $A''B''C'D'$.
   The quadrilateral $G_1G_2F_3G_4$ get mapped to quadrilateral $H_1H_2H_3G_4$.

The key is independent of choice of $E_2,E_3,E_4$, the point $H_1$ always mapped to same spot. In fact, if one choose a coordinate system so that $A = (0,0), B = (1,0), C = (1,1)$ and $D = (0,1)$, one find $E_1 = (\frac23,0)$ and $H_1 = (\frac83,2)$. In general, we always have $|E_1H_1| = 2\sqrt{2}|AB|$.

If one stare at above diagram long enough, one will notice

• $|E_2E_3| = |E_2F_3|$,
• $|E_3E_4| = |F_3F_4| = |F_3G_4|$,
• $|E_4E_1| = |F_4F_1| = |G_4G_1| = |G_4H_1|$.

From this, we can deduce $$\begin{align} &|E_1E_2|+|E_2E_3|+|E_3E_4|+|E_4E_1|\\ = &\; |E_1E_2| + |E_2F_3| + |F_3G_4|+|G_4H_1|\\ \ge &\; |E_1H_1| = 2\sqrt{2}|AB| \end{align}$$

This means the perimeter we seek is bounded from below by $2\sqrt{2}|AB|$. To see this is the actual minimum perimeter, we can repeatedly "fold" segment $E_1H_1$ back to square $ABC$ to construct the "optimal" quadrilateral. We find when

$$BE_2 : E_2C = DE_4 : E_4A = 1 : 2\quad\text{ and }\quad CE_3 : E_3D = 2 : 1$$

the quadrilateral $E_1E_2E_3E_4$ does have $2\sqrt{2}|AB|$ as perimeter.

Answer 2

Hint: Start by drawing the $ABCD$ square, and the point $E_1$ on $AB$. Now let's choose a random point $E_3$ on $CD$. How do you get the shortest path from $E_1$ to $E_3$ so that it touches $BC$ in some $E_2$? Now repeat the procedure for the $AD$ side as well. Once you have these steps, there is just one more (similar) step to find the final answer.

Answer 3

Let $ABC'D'$, $A''BC'D''$, $A'''B''C'D''$ and $A'''B'''C'''D''$ be squares

and $E_1'\in A'''B'''$ such that $A'''E_1':E_1'B'''=2:1.$

Now, let $E_1E_1'\cap BC'=\{E_2'\},$ $E_1E_1'\cap D''C'=\{E_3'\}$ and $E_1E_1'\cap A'''D''=\{E_4'\}.$

Now, easy to see that $$E_1E_2+E_2E_3+E_3E_4+E_4E_1\geq E_1E_1'=E_1E_2'+E_2'E_3'+E_3'E_4'+E_4'E_1'$$ and $$BE_2:E_2C=BE_2':E_2C'=1:2,$$ $$CE_3:E_3D=C'E_3':E_3'D''=2:1$$ and $$DE_4:E_4A=D''E_4':E_4'A'''=1:2.$$

Answer 4

Let $A,B,C$ be three corners of the square in rotational order. Draw a linecsegment $PQ$ from point $P$ on $AB$ to point $Q$ on $BC$. Using SOH CAH TOA on right triangle $PQB$ infer that

$|PB|=|PQ|\cos\angle QPB$

$|QB|=|PQ|\sin\angle QPB$

$\dfrac{|PQ|}{|PB|+|QB|}=\dfrac{1}{\cos\angle QPB+\sin\angle QPB}\ge \dfrac{1}{\sqrt{2}}$

We know that $\cos\angle QPB+\sin\angle QPB \le\sqrt{2}$ because $(\cos\theta+\sin\theta)^2+(\cos\theta-\sin\theta)^2=2(\cos^2\theta+\sin^2\theta)=2$.

Applying this result to all four sides of the inscribed quadrilateral leads to the perimeter of this quadrilateral being $\ge 1/\sqrt{2}$ times that of the square. That this bound is sharp is proved from the simple case of the sides of the inscribed quadrilateral being parallel to the angle bisectors of the square (always possible given the symmetries involved).