Convex hexagon with AD=BC+EF, BE=AF+CD, CF=DE+AB

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Convex hexagon with AD=BC+EF, BE=AF+CD, CF=DE+AB

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Source: Kazakhstan international contest 2006, Problem 6

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Valentin Vornicu

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Valentin Vornicu

#1 h Jan 22, 2006, 3:43 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCDEF$ be a convex hexagon such that $AD = BC + EF$ , $BE = AF + CD$ , $CF = DE + AB$ . Prove that:
$\frac {AB}{DE} = \frac {CD}{AF} = \frac {EF}{BC}.$

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324 posts

#2 h Feb 5, 2006, 6:45 PM • 2 Y

Y by Adventure10 and 1 other user

It can be proved by using vectors. Actually the same idea was used in IMO 2003 #3.

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#3 h Feb 5, 2006, 10:55 PM • 5 Y

Y by BSJL, Adventure10, Crazy4Hitman, pokpokben, and 1 other user

I had a similar idea, but I was very DUMB not to see what happened if I went on this path.

I got the INEQUALITIES, but I thought that adding them would never give something useful:

$\left| \overrightarrow{AD} \right| = \left| \overrightarrow{BC} \right| + \left| \overrightarrow{EF} \right| \geq \left| \overrightarrow{FC} + \overrightarrow{BE} \right|$ . (I don't know why, but this inequality seemed too natural for me to yield anything; really don't know why)

Squaring and adding: $\left( \overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF} \right)^2 \leq 0$ . Thus, $BC \| EF$ etc. Also, $BC \| DA$ . The conclusion is now obvious.

Thanks, Sung-yoon Kim.

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#4 h Feb 6, 2006, 12:14 AM • 2 Y

Y by Adventure10 and 1 other user

Yeah I think you're on the right way..

Quote:

$\left| \overrightarrow{AD} \right| = \left| \overrightarrow{BC} \right| + \left| \overrightarrow{EF} \right| \geq \left| \overrightarrow{FC} + \overrightarrow{BE} \right|$

Consider a parellelogram BCFX. Then $BC+EF=XF+FE \geq XE=|\overrightarrow{XB}+\overrightarrow{BE}|=|\overrightarrow{CF}+\overrightarrow{BE}|$ . The rest is same as yours.

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Tiks

#5 h Feb 9, 2006, 12:13 PM • 1 Y

Y by Adventure10

Guys I think we shoulld find sinthetical solution for this one,I havn't do it still

,but I will try again

.

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Davron

#6 h Apr 10, 2006, 12:45 PM • 2 Y

Y by Adventure10, Mango247

Yes it is a good idea to find a synthetic proof for this problem.

Davron

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jin

#7 h Apr 11, 2006, 9:45 AM • 1 Y

Y by Adventure10

That's really hard to find a such solution.

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Tiks

#8 h Apr 13, 2006, 4:46 PM • 2 Y

Y by Adventure10, Mango247

Yes,it is realy hard job,if that can help you I know that ther are sintetic solution,I know a man who has found it,but I want find it from misalf,so I havn't ask him about that solution.

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#9 h Dec 28, 2006, 10:19 AM • 2 Y

Y by Adventure10, Mango247

perfect_radio wrote:

The conclusion is now obvious.

Well, after proving all these parallels, I still can't see why it's obvious...

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#10 h Dec 28, 2006, 11:45 PM • 3 Y

Y by Adventure10, Crazy4Hitman, Mango247

Let $AD \cap CF = \left\{ T \right\}$ . The following statements hold:
- $ABCT$ is a parallelogram;
- $DEFT$ is a parallelogram;
- $\triangle ATF \sim \triangle DTC$ ;
- the conclusion.

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maggot

#11 h Mar 8, 2007, 4:10 PM • 1 Y

Y by Adventure10

perfect_radio wrote:

Also, $BC \| DA$ .

And why is this?

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#12 h Mar 8, 2007, 5:22 PM • 2 Y

Y by Adventure10 and 1 other user

$\left( \overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF}\right)^{2}\leq 0$ implies $\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF}= \overrightarrow 0$ . Also note that $\overrightarrow{EB}+\overrightarrow{CF}= \overrightarrow{CB}+\overrightarrow{EF}$ .
Hence,
$\overrightarrow{AD}= \overrightarrow{BC}+\overrightarrow{FE}.$

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1327 posts

#13 h Mar 9, 2007, 5:44 PM • 1 Y

Y by Adventure10

A simple construction of the configuration, as the problem states, is as follows:

A triangle $\bigtriangleup KLM$ is given and let $A$ be, a fixed point on the extension of the sideline $KL$ $($ $K,$ between $A,$ $L$ $).$

The line through the $A$ and parallel to $LM,$ intersects the sideline $KM$ at a point, so be it $B.$

The line through the $B$ and parallel to $KL,$ intersects the sideline $LM$ at a point, so be it $C.$

The line through the $C$ and parallel to $KM,$ intersects the sideline $KL$ at a point, so be it $D.$

The line through the $D$ and parallel to $LM,$ intersects the sideline $KM$ at a point, so be it $E.$

Through $A,$ $E$ now, we draw to lines parallel to $KM\equiv BE$ and $KL\equiv AD$ respectively, which intersect the sideline $LM,$ at points $F,$ $F'$ and we will prove that $F'\equiv F.$

It is easy to show that $CF = CM+MF = DE+AB$ $,(1)$ and $CF' = CL+LF' = AB+DE$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $CF' = CF$ $\Longrightarrow$ $F'\equiv F.$

So, it has already been constructed the configuration as the problem states.

$\bullet$ From $AB\parallel DE$ $\Longrightarrow$ $\frac{AB}{DE}= \frac{KA}{KD}= \frac{KB}{KE}$ $,(3)$

But, $\frac{KA}{KD}= \frac{EF}{BC}$ $,(4)$ and $\frac{KB}{KE}= \frac{CD}{AF}$ $,(5)$

From $(3),)$ $(4),$ $(5)$ $\Longrightarrow$ $\frac{AB}{DE}= \frac{CD}{AF}= \frac{EF}{BC}$ and the proof is completed.

Kostas Vittas.

Attachments:

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#14 h Mar 10, 2007, 3:31 AM • 2 Y

Y by Adventure10, Mango247

Lemma: Let $ABCD$ be arbitrary quadrilateral. Denote
$a,b,c,d$ it's sides and $e,f$ diagonals, then $a^{2}+c^{2}+2bd\geq e^{2}+f^{2}$ .

Proof: Denote $M, N, P$ midpoints of $BC, AD, BD$ , from
triangle inequality $MN\leq MP+NP$ or $2MN\leq b+d$ . We will
remember Euler theorem for quadrilateral and midpoint of opposite
sides or diagonals: $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}\leq a^{2}+c^{2}+(b+d)^{2}$ . Equality is hold when $BC \parallel AD$ and our lemma is
proved.

Now we will solve the problem.

Denote $AB=a,BC=b,CD=c, DE=d, EF=e, AF=f, AC=x_{1}, BD=x_{2}, CE=x_{3}, DF=x_{4}, EA=x_{5}, FB=x_{6}, AD=l_{1}, BE=l_{2}, CF=l_{3}$ .

We have
$l_{1}=b+e, l_{2}= c+f, l_{3}=a+f$ . Apply our theorem for $ABCD$ and
$DEFA$ we have $a^{2}+c^{2}+2bl_{1}\geq x_{1}^{2}+x_{2}^{2}$ and
$d^{2}+f^{2}+2el_{1}\geq x_{4}^{2}+x_{5}^{2}$ . Adding all of them:
$a^{2}+c^{2}+d^{2}+f^{2}+2l_{1}^{2}\geq x_{1}^{2}+x_{2}^{2}+x_{4}^{2}+x_{5}^{2}$
Summing all such quadrilaterals we get
$2(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})+2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2})\geq 2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2})$
Next step is to apply our lemma for quadrilateral $BCEF$ :
$x_{3}^{2}+x_{6}^{2}+2be\geq l_{2}^{2}+l_{3}^{2}$ . Summing with inequalities for
$CDAF$ , $DEAB$ :
$(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2})+2ad+2be+2cf\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2})$
Multiplying second inequality by $2$ and adding with first one
we get
$2(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})+4(ad+be+cf)\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2})$
or just $2(a+d)^{2}+2(b+e)^{2}+2(c+f)^{2}\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2})$

As we see the equality is hold and thus from lemma's case
of equality we get $BC \parallel AD \parallel EF$ and similarly other lines are
parallel. Let a line through $C$ parallel to $AB$ intersects $AD$ at
$P$ , then $AP=b$ and thus $PD=e$ and $PD\parallel EF$ . Hence $PDEF$ is
parallelogram and $C,P,F$ are collinear as are $A,P,D$ . So triangles
$APF$ and $DPC$ are similar and from here
$\frac{AP}{DP}=\frac{PF}{PC}=\frac{AF}{CD}$
Rewriting in hexagon's sides:
$\frac{BC}{EF}=\frac{DE}{AB}=\frac{AF}{CD}$

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#15 h Mar 10, 2007, 3:42 AM • 2 Y

Y by Adventure10, Mango247

This was problem was proposed for Mathematical Reflections 2 in 2006,
by Nairi Sedrakyan (Armenia).

Mathematical Reflections: http://www.awesomemath.org

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#16 h Dec 26, 2009, 10:39 AM • 1 Y

Y by Adventure10

in a convex hexagon with AD=BC+EF, CF=DE+AB,BE=AF+CD.THERE WE CAN FIND EASILY.A simple construction of the configuration, as the problem states, is as follows:

A triangle \bigtriangleup KLM is given and let A be, a fixed point on the extension of the sideline KL ( K, between A, L ).

The line through the A and parallel to LM, intersects the sideline KM at a point, so be it B.

The line through the B and parallel to KL, intersects the sideline LM at a point, so be it C.

The line through the C and parallel to KM, intersects the sideline KL at a point, so be it D.

The line through the D and parallel to LM, intersects the sideline KM at a point, so be it E.

Through A, E now, we draw to lines parallel to KM\equiv BE and KL\equiv AD respectively, which intersect the sideline LM, at points F, F' and we will prove that F'\equiv F.

It is easy to show that CF = CM+MF = DE+AB ,(1) and CF' = CL+LF' = AB+DE ,(2)

From (1), (2) \Longrightarrow CF' = CF \Longrightarrow F'\equiv F.

So, it has already been constructed the configuration as the problem states.

\bullet From AB\parallel DE \Longrightarrow \frac{AB}{DE}= \frac{KA}{KD}= \frac{KB}{KE} ,(3)

But, \frac{KA}{KD}= \frac{EF}{BC} ,(4) and \frac{KB}{KE}= \frac{CD}{AF} ,(5)

From (3),) (4), (5) \Longrightarrow \frac{AB}{DE}= \frac{CD}{AF}= \frac{EF}{BC} and the proof is completed.

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dizzy

#17 h Dec 22, 2013, 9:08 AM • 2 Y

Y by Adventure10, Mango247

prowler wrote:

Lemma: Let $ABCD$ be arbitrary quadrilateral. Denote
$a,b,c,d$ it's sides and $e,f$ diagonals, then $a^{2}+c^{2}+2bd\geq e^{2}+f^{2}$ .

Proof: Denote $M, N, P$ midpoints of $BC, AD, BD$ , from
triangle inequality $MN\leq MP+NP$ or $2MN\leq b+d$ . We will
remember Euler theorem for quadrilateral and midpoint of opposite
sides or diagonals: $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}\leq a^{2}+c^{2}+(b+d)^{2}$ . Equality is hold when $BC \parallel AD$ and our lemma is
proved.

Sorry for reviving the old topic, but your lemma is not true, and therefore the whole solution as well. From Euler theorem we have not as you wrote $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}$ but $a^{2}+b^{2}+c^{2}+d^{2}=e^{2}+f^{2}+4MN^{2}$ . So your lemma rewrites as $e^2+f^2+2bd\geq a^2+c^2$ .

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