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Source: Polish Second Round 2004

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116 posts

#1 h May 14, 2008, 5:45 PM • 3 Y

Y by Vietjung, Adventure10, Mango247

In convex hexagon $ABCDEF$ all sides have equal length and

$\angle A+\angle C+\angle E=\angle B+\angle D+\angle F$ .

Prove that the diagonals $AD,BE,CF$ are concurrent.

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#2 h May 15, 2008, 2:47 PM • 2 Y

Y by Adventure10, Mango247

Anybody? Any idea?

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#3 h May 16, 2008, 9:44 AM • 2 Y

Y by Adventure10, Mango247

Carlez Tevos wrote:

In convex hexagon $ABCDEF$ all sides have equal length and

$\angle A + \angle C + \angle E = \angle B + \angle D + \angle F$ .

Prove that the diagonals $AD,BE,CF$ are concurrent.

It isn't difficult to show that $\angle A=\angle D, \angle B=\angle E,\angle C=\angle F$
This problem is related to the prolem which is referred in International Olympiad 2005, Mexico.

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#4 h May 16, 2008, 12:54 PM • 2 Y

Y by Adventure10, Mango247

If it really isn't difficult to show, please show it.

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#5 h May 16, 2008, 1:53 PM • 2 Y

Y by Adventure10, Mango247

See the topic: http://www.mathlinks.ro/viewtopic.php?p=1129352#1129352
I hope you will find the solution soon.

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#6 h May 16, 2008, 2:30 PM • 2 Y

Y by Adventure10, Mango247

mr.danh wrote:

See the topic: http://www.mathlinks.ro/viewtopic.php?p=1129352#1129352
I hope you will find the solution soon.

It's a different problem! Please show my problem.

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341 posts

#7 h Mar 10, 2012, 9:08 PM • 1 Y

Y by Adventure10

Prove that ABCDEF is a quadrilateral with an inscribed circle. By definition, a tangential quadrilateral has its diagonals $AB, BE, CF$ concurrent.

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#8 h Mar 11, 2012, 1:54 AM • 2 Y

Y by Adventure10, Mango247

Umm.. I guess it's not necessarily a tangential quadrilateral.. It was just a quick thought.

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#9 h Jun 9, 2017, 2:31 AM • 2 Y

Y by Adventure10, Mango247

...........

This post has been edited 1 time. Last edited by Math1331Math, Jan 13, 2019, 3:03 AM
Reason: ,

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InCtrl

#10 h Jan 11, 2019, 12:39 AM • 1 Y

Y by Adventure10

Consider the following isometry: translate the entire hexagon until $C$ is mapped to $C'=A$ , then rotate about $C'=A$ until the image of $D$ coincides with $F$ . Notice now that, since $\angle B+\angle D+\angle F=2\pi$ , we have $\angle EFE'=2\pi -\angle AFE - \angle AFE'=2\pi-\angle AFE-\angle CDE=\angle ABC$ . Thus, $EE'=AC$ . Also, by construction, $AE'=CE$ . Thus, $CAE'E$ is a parallelogram, so $\angle AE'E=\angle ACE.$

We now seek to prove that $\angle BCD=\angle AFE$ , and analogously will follow $\angle B=\angle E, \angle D=\angle A$ .

In the following angle chase we use the fact that $\triangle ABC$ and $\triangle CDE$ are isosceles to establish $\angle BAC=\angle BCA$ and $\angle CED=\angle DCE$ . We also know that the isometry preserves angles between corresponding sides, so $\angle AE'F=\angle C'E'D'=\angle CED$ . Since $\triangle EFE'\cong \triangle CBA$ , $\angle EFE'=\angle BAC$ .

Now notice that $F$ is the circumcenter of $AEE'$ since $FA=FE=DE=FE'$ .

Thus, $\angle AE'E=\frac{1}{2}\angle AFE=\angle AE'F+\angle FE'E=\angle CED+\angle BAC=\angle ECD+\angle BCA=\angle BCD-\angle ACE=\angle BCD-\angle AE'E\iff$ $\angle AFE=2\angle AE'E=\angle BCD.$
Thus, opposite pairs of angles are equal. From here, it is easy to show that opposite sides are parallel.
Now, notice that since $AF=CD$ , and from $\triangle BAC\cong \triangle EDC$ we must have $AC=FD$ , $ACDF$ must be a parallelogram. It follows that the diagonals $AD,CF$ and $BE$ all bisect each other and are thus concurrent.

This post has been edited 4 times. Last edited by InCtrl, Jan 11, 2019, 12:50 AM

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