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I'm supposed to find a quadrangle of the smallest perimeter possible inscribed in a rectangle. The inscribed quadrangle has each of its four vertices on another side of the rectangle.

Let's call the rectangle ABCD and let the shorter side be $a$ and the larger $b$. Let's call the quadrangle $KLMN$.

So for example $K$ lies on $AB$, $L$ on $BC$, $M$ on $CD$, $N$ on $AD$.

I think that the sides of $KLMN$ would be the shortest of its diagonals intersected at the right angle, because then by the law of cosines, we have $KL^2 + LM^2= KM^2 + 2 KL \cdot LM \cdot \cos \angle KLM$, and $\cos \angle KLM \le 0$ if $\angle KLM \ge 90 ^{\circ}$ and $KM$ is the shortest if it is parallel = equal to the proper side of the rectangle.

Could you tell me if I'm right or correct me if I'm wrong?

Thank you.

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There's no such thing, since you could make it arbitrarily small, with perimeter tending to $0$. – Lucian Mar 7 '14 at 15:18
Are you, for example, looking for a vertex on each side of the rectangle? Or every vertex on the perimeter of the rectangle? – Mark Bennet Mar 7 '14 at 15:31
The inscribed quadrangle has $4$ vertices on $4$ sides of the rectangle. I'll add it to my question. I'm sorry for the confusion. – Don Mar 7 '14 at 15:32
up vote 3 down vote accepted

Unflod your rectangle to make a grid and place K, L, M and N on it. You goal is to shorten the distance $KL+LM+MN+NK'$.

You need to align $KLMNK'$ to reach the minimal distance for $KK'$. That's $2.AC$

Here is a picture of a non-optimal situation. If it were optimal, then $KLM'N'K'$ would be a straight line.


Here is a picure of an optimal situation:

enter image description here

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Thank you for your answer. Could you explain what you mean by unfolding a rectangle? – Don Mar 7 '14 at 15:52
Does it mean "disconnecting" it at a vertex? – Don Mar 7 '14 at 15:54
I can't tell if "unfold" is the proper term, here is what I mean: An a plan: draw $ABCD$, then by symmetry, around $BC$, draw $BCD'A'$, an so one in every direction for every rectangle. It will make a grid. KL+LM is the same if you place M on $CD'$, etc. – Sylvain Biehler Mar 7 '14 at 15:55
I see. Thank you. – Don Mar 7 '14 at 16:02
@Bilou06: I hope you don't mind my editing your post to add an illustration. And I hope I correctly depicted what you had in mind, at least in principle. I believe that there might be a slight difference in where to put these primes, but I couldn't come up with a way that matches your post and seemed consistent to me. And it really doesn't matter that much, it's all the same orbit in any case. Nice answer, by the way! – MvG Mar 7 '14 at 16:03

I will prove the following: Let $ABCD$ a rectangle having $|AB|=a$ and $|BC|=c$ and let $K$, $L$, $M$, $N$ be points on $AB$, $BC$, $CD$ and $DA$, respectively. Then the perimeter of $KLMN$ is at least $2 \sqrt{a^2+b^2}$. This is achievable by letting $K$, $L$, $M$ and $N$ the midpoints of the sides of $ABCD$.

Write $k = |KB|$, $l = |LC|$, $m = |MD|$, $n = |NA|$. Then the perimeter of $KLMN$ equals $\sqrt{k^2+(b-l)^2} + \sqrt{l^2 + (a-m)^2} + \sqrt{m^2+(b-n)^2} + \sqrt{n^2+(a-k)^2}$.

We use Minkowski's inequality $$\left({\sum_{k \mathop = 1}^n \left({a_k + b_k + c_k + d_k}\right)^p}\right)^{1/p} \le \left({\sum_{k \mathop = 1}^n a_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n b_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n c_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n d_k^p}\right)^{1/p}$$ for $p=n=2$ and sequences $(k,b-l)$, $(a-m,l)$, $(m,b-n)$ and $(a-k,n)$.

We then find $$\sqrt{k^2+(b-l)^2} + \sqrt{l^2 + (a-m)^2} + \sqrt{m^2+(b-n)^2} + \sqrt{n^2+(a-k)^2} \geq \sqrt{(k+(a-m)+m+(a-k))^2+((b-l)+l+(b-n)+n)^2} = 2 \sqrt{a^2+b^2}.$$ The proof is complete.

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Very nice answer. Thank you a lot! – Don Mar 7 '14 at 16:08

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