Smallest inscribed quadrilateral

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Smallest inscribed quadrilateral

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Source: Spring Mathematics Olympiad Problems- St. Petersburg, Russia

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279 posts

#1 h Oct 11, 2004, 1:09 PM • 2 Y

Y by Adventure10, Mango247

Let S be a square, Q be the perimeter of the square, and P be the perimeter of a quadrilateral T inscribed within S such that each of its vertices lies on a different edge of S.
What is the smallest possible ratio of P to Q?

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#2 h Oct 12, 2004, 9:19 PM • 2 Y

Y by Adventure10, Mango247

Is it $\frac 1{\sqrt 2}$ ? That's what I keep getting, and I think I can prove it, but I'm not sure it's Ok. I'd like to know if what I'm trying to show is correct first

.

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#3 h Oct 13, 2004, 7:17 AM • 2 Y

Y by Adventure10, Mango247

grobber wrote:

Is it $\frac 1{\sqrt 2}$ ? That's what I keep getting, and I think I can prove it, but I'm not sure it's Ok. I'd like to know if what I'm trying to show is correct first

.

Yes! Your claim is correct, or to be more precise is the same result I found:-) : I don't know the "ufficial" solution.

I faced this nice problem few years ago: it was proposed as one of selection tests for the admission to the "Scuola Normale di Pisa".

Just this days I seen it again on the web from the source I referenced. But I don't know who is the original proposer of the problem: russian or italian or others.

I found a very short, purely geometric and extremely elementar solution of this problem.

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#4 h Oct 13, 2004, 7:41 AM • 2 Y

Y by Adventure10, Mango247

I had a rather ugly (although not too ugly

). I think I also have a geometric solution now: use three reflections in the sides of the square: reflect $ABCD$ in $CD$ to get $A'B'CD$ , then $A'B'CD$ in $B'C$ to get $A''B'CD''$ , and then this last square in $A''B'$ to get $A''B'C'''D'''$ . The perimeter we're of the quadrilateral is the length of a broken line $MN$ with $M\in AD,\ N\in A''D'''$ s.t. $MD=ND'''$ , and this line has the smallest length when it's straight, and it's easy to see that its length is, in this case, the length of a diagonal of a square with the side $2AB$ , which is $2\sqrt 2 AB$ , and since the perimeter of the square is $4AB$ , that's pretty much it.

The equality is reached when the inscribed quadrilateral is a rectangle having its sides parallel to the diagonals of the square.

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#5 h Oct 13, 2004, 3:46 PM • 2 Y

Y by Adventure10, Mango247

very beautiful proof! Fagnano style?

I have a different one.

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#6 h Oct 19, 2004, 8:22 AM • 2 Y

Y by Adventure10, Mango247

If ABCD is the square and EFGH the quadrilateral inscribed with E on AB, F on BC, G on CD and H on DA, let E',F', G', H' be the orthogonal projection of EFGH on diagonal AC.

It holds that:

HE > H'H + EE' = AH' + AE'

EF > E'F'

FG > F'F + GG' = CF' + CG'

GH > G'H'.

Given that CG' + G'H' + H'A = CA and AE' + E'F' + F'C = AC

it results that P >= 2AC.

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