Concurrent lines in hexagon - 2015 PAMO Problem 2

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Concurrent lines in hexagon - 2015 PAMO Problem 2

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Source: 2015 Pan-African Mathematics Olympiad Problem 2

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#1 h Aug 26, 2015, 1:56 PM • 2 Y

Y by Adventure10, Mango247

A convex hexagon $ABCDEF$ is such that
$AB=BC \quad CD=DE \quad EF=FA$
and
$\angle ABC=2\angle AEC \quad \angle CDE=2\angle CAE \quad \angle EFA=2\angle ACE$
Show that $AD$ , $CF$ and $EB$ are concurrent.

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375 posts

#2 h Aug 26, 2015, 5:21 PM • 2 Y

Y by Adventure10, Mango247

My solution:
Let $O$ be the circumcenter of $\triangle ACE$ . Easy to get $B, D, E$ be the symmetry of $O$ in $AC, CE, EA$ $\Rightarrow ABCO$ and $OCDE$ are the parallelograms $\Rightarrow AB\overset{\parallel }{=}OC\overset{\parallel }{=}ED$ $\Rightarrow ABED$ is the parallelogram $\Rightarrow BE$ passing through midpoint of $AD$ $(*)$
Similarly $CE$ passing through midpoint of $AD$ $(**)$ .
Since $(*), (**)$ deduced $AD$ , $CF$ and $EB$ are concurrent(at midpoint of $AD$ ). DONE

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#3 h Jun 5, 2017, 5:02 PM • 2 Y

Y by Adventure10, Mango247

Hi

How did you prove that $B, D, E$ are the symmetry of $O$ in $AC, CE, EA.$

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#4 h Aug 8, 2018, 3:47 PM • 1 Y

Y by Adventure10

This can also be solved easily by Ceva's Theorem in trigonometric form.

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#5 h Nov 29, 2019, 5:22 AM • 4 Y

Y by GeoMetrix, Pakistan, Adventure10, Mango247

Here's a different Solution:-

2015 PAMO P2 wrote:

A convex hexagon $ABCDEF$ is such that
$AB=BC \quad CD=DE \quad EF=FA$ and
$\angle ABC=2\angle AEC \quad \angle CDE=2\angle CAE \quad \angle EFA=2\angle ACE$ Show that $AD$ , $CF$ and $EB$ are concurrent.

Solution:-

Brianchon's Theorem reflects in our mind when we see that we have to prove $AD,BE,CF$ are concurrent, so before applying it we must prove that $ABCDEF$ has an inscribed circle. So let's prove it!

Introduce the Circumcenter of $\triangle ACE$ , let it be $O$ . Let the perpendiculars from $O$ to $AB,BC,CD,DE,EF,FA$ be $l_1,l_2,l_3,l_4,l_5,l_6$ respectively. So it suffices to prove that $l_1=l_2=l_3=l_4=l_5=l_6$ .

Note that by the angle and length conditions we get that $\angle ABC=\angle AOC$ also $AB=BC$ and $AO=OC$ . So, $AO=OC=AB=BC$ .Similarly you get that $AO=OE=AF=FE$ and $CO=OE=ED=DC$ . Hence, $AB=BC=CD=DE=EF=FA$ . Now it's trivial to see that $l_1=l_2=l_3=l_4=l_5=l_6$ . So, $ABCDEF$ must have an inscribed circle. So, by Brianchon we conclude.

This post has been edited 1 time. Last edited by amar_04, Nov 29, 2019, 5:23 AM

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