Nine-point Center is awesome

by XmL, Aug 15, 2014, 9:43 PM

As we all know, a lot of properties concerning the nine-point center are hard to prove synthetically. An example of this can be found here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=550000&p=3549742#p3549742. Anyway I've stumble upon a few more while working on a problem that Arab gave me(Correlary 3) and here they are:

We start with a pretty awesome Generalization:

Generalization 1: $A'B'C',DEF$ are the medial, altitude triangle of $ABC$, it's well known that the nine-point circle is their circumcircle. Let $P\in l$ with $l$ the Euler line of $ABC$, $XYZ$ is the circumcevian triangle of $P$ wrt $A'B'C'$. Prove that $XYZ$ is the circumcevian triangle of some point on $l$ wrt $DEF$.

Proof: Let $XYZ$ be the circumcevian triangle. It's well known that $FB'\cap EC'=W\in $the Euler line of $ABC$, hence by Pascal's theorem $FZ\cap EY\in PW=l\Rightarrow DX,EY,FZ$ concur on the Euler line. $\Box$.

With this Generalization in mind, we examine the special case where $P$ becomes the nine-point center of $A'B'C'$. If we let $N$ denote the nine-point center of $ABC$, then $AN\parallel A'X$. It's easy to prove by Ceva's theorem that $DX,EY,FZ$ concur implies $AX,BY,CZ$ concur.

Correlary 1: The reflection of the $X$-altitude over $XN$ concur, where $X=A,B,C$.

Proof: Let $H_A$ be the foot of $A$-altitude on $BC$, we will show that $\angle NAX=\angle NAH_A$ and our assertion follows. Let $O_A$ denote the reflection of the circumcenter of $ABC$ over $BC$, hence $N$ is the midpoint of $AO_A$ and that $AX,O_AA'$ are reflesive over the perpendicular bisector of $AO_A$, therefore $\angle XAN=\angle A'O_AA=\angle NAH_A$ and we are done. $\Box$.

What a beautiful property regarding the nine-point center!

Correlary 2: Let $N_A,N_B,N_C$ denote the circumcenter of $NBC,NAC,NAB$, let $k_A$ denote the reflection of $ON_A$ over $N_BN_C$, similarly define $k_B,k_C$. Prove that $k_A,k_B,k_C$ concur.

Proof: Since $O_A\in ON_A$, we reflect $O_A,A'$ over $N_BN_C$ to get $O_A',A''$, hence $O_A'A''=k_A$. Since $N_CN_B$ perpendicularly bisects $AN$, hence $AN=NO_A=AO_A'$. Moreover it's easy to prove that $A''X=AO_A'$, hence $AXA''O_A'$ is parallelogram$\Rightarrow AX\parallel O_A'A''$ If we similarly define $O_B',O_C',B'',C''$, then $ABC,O_A'O_B'O_C'$ are homothetic and $A''O_A',B''O_B',C''O_C'$ concur $\iff AX,BY,CZ$ concur and we are done. $\Box$.

Note that $N_BN_AN_C$ is homothetic to the pedal triangle of the Konita's point of $ABC$ and $O$ corresponds to the Konita's points, therefore we now have a cool property regarding Konita's point!

Also if we let $U'$ denote the point at the concurrence, then the circumcenter of $N_AN_BN_C$ lies on $NU'$. To prove this, just note that $U'$ lies on the Neuberg Cubic and the assertion is simply a property of the cubic.

Correlary 3: Reflect $N_A$ over $BC$ to obatin $N_A'$, similarly define $N_B',N_C'$, then $AN_A',BN_B',CN_C'$ concur.

Proof: Construct $A^*B^*C^*$ such that its medial triangle is $ABC$, hence now we just have to prove $A^*N_A, B^*N_B,C^*N_C$ concur. By Desargue's theorem this is equivalent to $N_BN_C\cap B^*C^*$ and the cyclic contructions are collinear. Apply an inversion at $N$ with an arbitrary radius, Let $A,B,C\rightarrow A_1,B_1,C_1$. Hence the pole of $N_BN_C$ wrt the inversion circle is the reflection of $N$ over $A_1$ and the pole of $B^*C^*$ is the interestion of the polar of $A$ with $NA_2$ where $A_2,B_2,C_2$ are the poles of $BC,AC,AB$. Together the pole of $N_BN_C\cap B^*C^*$ is the reflection of $NA_2$ over $B_2C_2$. Since it's easy to prove that $A_2B_2C_2$ is homothetic to $N_AN_BN_C$ and $O,N$ are obviously two corresponding points, therefore we have the poles of $N_BN_C\cap B^*C^*$ and the cyclic contructions concur$\Rightarrow $ the intersection points are collinear and we are done. $\Box$.
This post has been edited 2 times. Last edited by XmL, Aug 16, 2014, 2:22 AM

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Still an immature proof writer :)

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  • Good blog! I hope to see more posts here!

    by WizardMath, Feb 3, 2018, 6:57 PM

  • Thanks Harry Potter :D

    by XmL, Jul 30, 2014, 10:47 PM

  • Hey dude. Nice blog. I like your work.

    by gamjawon, Jul 30, 2014, 7:43 AM

  • Thanks...although most of the posts here won't be about geometry...rightnow I'm trying to improve my skills in other areas such as number theory and combo.

    by XmL, Jun 5, 2014, 1:11 AM

  • Hi Geometer,,nice blog..:D

    by vanu1996, Jun 4, 2014, 1:27 PM

  • Check your private massage :)

    by zabihpourmahdi, Jan 11, 2013, 5:34 PM

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