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Given acute $\Delta{ABC} (AB<AC)$ with the incenter $I$. $AI$ intersects $BC$ at $D$. Put two extra points $E$, $F$ that satisfy: $IC$ is the perpendicular bisector of $DE$ and $IB$ is the perpendicular bisector of $DF$. The circumcircles of $\Delta{AFN}$ and $\Delta{AEM}$ intersects at $P$. Let $M,N,J$ be the midpoints of $DE,DF,EF$ respectively. Prove that $JMPN$ is a cyclic quadrilateral.

My attempt:

enter image description here

  • Sorry, I forgot to connect the lines of the quadrilateral $JMPN$ on Geometer Sketchpad (mine is the trial version, so I can't save it).

  • Because $AD$ is the angle bisector of $\widehat{BAC}$, we have $\frac{AB}{AC}=\frac{BD}{DC}=\frac{FB}{EC}\Rightarrow \frac{AF}{FB}=\frac{AE}{EC}$, following the converse of the induction theorem, we now know that $EF$ is parallel to $BC$.

  • Because $MN$ is the midline of $\Delta{DEF}$, we can conclude that $MN,EF,BC$ are all parallel.

  • Let $MN$ intersects the circumcircle of $\Delta{AEM}$ at $K$. Because $AEMK$ is cyclic and $MN||BC$, we will have $\widehat{NMD}=\widehat{MDC}=\widehat{CED}=\widehat{AKM}$, so $AK$ is parallel to $DE$.

  • $NJ$ is another midline of $\Delta{DEF}$, so $NJ||DE \Rightarrow NJ||AK.$ Then we can prove $JMPN$ is cyclic if $\widehat{JNM}=\widehat{AKM}=\widehat{APM}$, this is only true if $A,J,P$ are collinear, which I'm stuck and can't prove this property.

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  • $\begingroup$ By chance, is this a training exercise for EGMO 2018? $\endgroup$ Apr 2 '18 at 21:48
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    $\begingroup$ @JackD'Aurizio No, this is part of my homework (which will be graded by the teacher). No one in my class has solved this problem. $\endgroup$
    – user061703
    Apr 3 '18 at 10:42
  • $\begingroup$ @JackD'Aurizio Also, EGMO is for females only and I'm a boy, not a girl. $\endgroup$
    – user061703
    Apr 3 '18 at 14:44
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After a few days thinking about this problem, one of my classmates has finally helped me to solve it:

  • Because $AFNP$ is a cyclic quadrilateral and $IB$ is the perpendicular bisector of $DF$ (so $BD=BF$), we will have $\widehat{BDF}=\widehat{BFD}=\widehat{APN}$ because $\widehat{BFD}+\widehat{AFN}=\widehat{APN}+\widehat{AFN}=180^\circ$.

  • Similarly, we will also have $\widehat{CDE}=\widehat{CED}=\widehat{APM}.$

  • We should be able to prove that $JNDM$ is a parallelogram because of the midlinesof $\Delta{DEF}$, so $\widehat{NJM}=\widehat{NDM}$.

Finally, we will calculate some angles:

$\widehat{NPM}=\widehat{APN}+\widehat{APM}=\widehat{BDF}+\widehat{CDE}=180^\circ-\widehat{NDM}$

$\Rightarrow \widehat{NPM}+\widehat{NJM}=\widehat{NPM}+\widehat{NDM}=180^\circ$

$\Rightarrow JMPN$ is a cyclic quadrilateral.

The final part of our homework asked us to prove that $A,J,P$ are collinear, which can be proved using my attempts above.

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$\angle AME=\angle AKE=\angle KEM$ by circle and parallel condition.

Then $\angle APM=\angle AKM=\angle FEM=\angle JNM.$//

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  • $\begingroup$ $\widehat{APM}=\widehat{JNM}$ does not imply $\widehat{JPM}=\widehat{JNM}$, because I said in the problem that I need to prove $A,J,P$ are collinear. $\endgroup$
    – user061703
    Apr 4 '18 at 11:08
  • $\begingroup$ @TrầnThúcMinhTrí Sorry, if J is midpoint of AP, $AF'J≡PE'J$ and $FJE$ is collinear. Then since $F'E'∦FE$, $AJP$ is on one line. $\endgroup$ Apr 4 '18 at 21:09

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