Metric relation in a hexagon

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Metric relation in a hexagon

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Source: Tuymaada 2013, Day 1, Problem 2 Juniors

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#1 h Jul 20, 2013, 10:02 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

$ABCDEF$ is a convex hexagon, such that in it $AC \parallel DF$ , $BD \parallel AE$ and $CE \parallel BF$ . Prove that
$AB^2+CD^2+EF^2=BC^2+DE^2+AF^2.$

N. Sedrakyan

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#2 h Jul 20, 2013, 10:37 PM • 2 Y

Y by Adventure10, Mango247

Perpendiculars from $B,D$ and $F$ to $DF \parallel AC,$ $FB \parallel CE$ and $BD \parallel EA$ obviously concur at the orthocenter $H$ of $\triangle BDF.$ Then $AB^2-BC^2=HA^2-HC^2,$ $CD^2-DE^2=HC^2-HE^2$ and $EF^2-FA^2=HE^2-HA^2.$ Adding these expressions together gives

$AB^2-BC^2+CD^2-DE^2+EF^2-FA^2=0 \Longrightarrow$

$AB^2+CD^2+EF^2=BC^2+DE^2+AF^2.$

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#3 h Jul 21, 2013, 5:04 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

This vectorial solution captures the same phenomenon as the more synthetic one above.

Associate position vectors to the vertices of the hexagon (with $v$ being the position vector associated to a point $V$ ). Then we can take advantage of the writing $XY^2 = \|y-x\|^2 = \left < y-x, y-x\right > = \|x\|^2 + \|y\|^2 - 2\left < y, x\right >$ . If we write these relations for the sides of the hexagon, the required equality comes to
$\left < b, a\right > + \left < d, c\right > + \left < f, e\right > = \left < b, c\right > + \left < d, e\right > + \left < f, a\right >,$
that is
$\left < b, a - c\right > + \left < d, c - e\right > + \left < f, e - a\right > = 0.$
But if we take the origin on the perpendicular from $B$ on $AC$ , then $b \perp a - c$ , and so $\left < b, a - c\right > = 0$ . Similar consequences will hold for the other two terms. Therefore, if we take the origin at the orthocentre of the triangle $BDF$ , all three terms will be null, and so the required equality will be proved.

We can in fact proceed without considering the orthocentre of $\triangle BDF$ , by leaving the origin at an arbitrary point. The equivalent relation obtained may also be written (using $\left < b, a - c\right > = \left < b, (a - e) + (e - c)\right >$ as
$\left < b - f, a - e\right > + \left < b - d, e - c\right > = 0.$
But triangles $BDF$ and $EAC$ are similar, in a similarity ratio $\rho$ , and then $\dfrac {BD} {AE} = \dfrac {FB} {EC} = \rho$ , whence $b-d = \rho(a-e)$ and $b-f = -\rho(e-c)$ , therefore
$\left < b - f, a - e\right > + \left < b - d, e - c\right > = -\rho\left < e-c, a - e\right > + \rho\left < a - e, e - c\right > = 0.$
One more proof, if needed, of the strength of these dot-product manipulations.

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#4 h Jul 26, 2013, 12:05 PM • 1 Y

Y by Adventure10

it's easy from the Carnot's theorem!

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