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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Good blog! I hope to see more posts here!
by WizardMath, Feb 3, 2018, 6:57 PM
Thanks Harry Potter
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..
by vanu1996, Jun 4, 2014, 1:27 PM
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by zabihpourmahdi, Jan 11, 2013, 5:34 PM

6 shouts

Geometry Treasures

Nine-point Center is awesome

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

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