A New Proof to the IO Nine-point center Problem

by XmL, Aug 17, 2014, 12:17 AM

In this Blog I will present a new synthetic proof to a problem that IDMasterz and I had worked on a while ago. The old proof can be found here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=550000

Problem: The nine-point center of the incentral triangle of a triangle lies on the line connecting the incenter and the circumcenter of the triangle.

Before we begin, I will get the preliminary notations out of the way: $\triangle ABC$ will be the main triangle. $H,O,N,K$ are respectively the orthocenter, cirumcenter, nine-point center, kosnita's point(Isogonal conjugate of $N$ ) of $\triangle ABC$ .

Lemma 1: Let $\triangle A_KB_KC_K$ be the pedal triangle of $K$ , the Kosnita's point $K$ of $ABC$ . Prove that the Euler line of $\triangle AC_KB_K$ is perpendicular to $BC$ .

Proof: Since $AK$ is the diameter of circumcircle of the cyclic quadrilateral $AC_KKB_K$ , therefore the midpoint of $AK$ ,denoted by $O_{Ka}$ , is the circumcenter of $\triangle AC_KB_K$ . It's well known that the orthocenter of the triangle, denoted by $H_{Ka}$ is the reflection of $K$ over the midpoint of $C_KB_K$ , hence if we reflection $O_{Ka}$ over $C_KB_K$ to obtain $O_{Ka}'$ , then $KO_{Ka}'H_{Ka}O_{Ka}$ is a parallelogram $\Rightarrow KO_{Ka}'\parallel O_{Ka}H_{Ka}$ $\Rightarrow$ now it suffices to prove $O_{Ka}'\in KA_K$ because $KA_K\perp BC$ .

Let $AN\cap (O)=A_N$ , $O'$ is the reflection of $O$ over $BC$ . It's a property of the nine-point center that $O'\in AN$ . Since $AN,AK$ are isogonals wrt $\angle BAC$ , and $\angle C_KO_{Ka}'B_K=\angle BO'C$ we can see that $BCA_NO'\sim B_KC_KKO_{Ka}'$ $\Rightarrow \angle C_KKO_{Ka}'=\angle CA_NO'=180-\angle B$ $=\angle C_KKA_K \Rightarrow O_{Ka}'\in KA_K$ which is what we need to prove. $\Box$

To discover more properties regarding the current diagram, we similarly define $O_{Kb},O_{Kc},H_{Kb},H_{Kc}$ . Note that since $\triangle O_{Ka}O_{Kb}O_{Kc}$ is the result of the homothety $(K,\frac {1}{2})$ applied to $\triangle ABC$ . Hence the midpoint of $HK$ , denoted by $H_K$ is the orthocenter of $\triangle O_{Ka}O_{Kb}O_{Kc}$ . Moreover, recall that $H_{Ka}$ is simply the reflection of $K$ over the midpoint of $C_KB_K$ and likewise for $H_{Kb},H_{Kc}$ , hence by congruence $\triangle A_KB_KC_K$ , $\triangle H_{Ka}H_{Kb}H_{Kc}$ are homothetically congruent. By Lemma 1 we know that $H_K\in O_{Ki}H_{Ki}$ with $i=a,b,c$ , therefore $H_K,K$ are two corresponding points wrt the two congruent triangles, hence $KA_KH_KH_{Ka}$ is a parallelogram.

With these properties in mind, it's time to prove this big result:

Lemma 2: Let $N_K$ denote the nine-point center of $A_KB_KC_K$ , the pedal triangle of the $K$ , prove that $KN_K$ is parallel to the Euler line of $\triangle ABC$

Proof: Let $M$ denote the midpoint of $C_KB_K$ . It's a well known property of isogonal conjugates that the circumcenter of $A_KB_KC_K$ ,denoted by $O_K$ , is the midpoint of $KN$ .

We apply homothety $(A_K,2)$ and it maps $N_K,K\rightarrow O_K',K'$ . Hence $O_K'$ is the reflection of $O_K$ over $C_KB_K$ . Recall our newly discovered property that $KA_KH_KH_{Ka}$ is a parallelogram, which implies $MK\parallel H_KA_K, 2MK=H_KA_K$ , since $K'K=KA_K$ , therefore $M$ is the midpoint of $N_KK'$ $\Rightarrow O_K'K'O_KH_K$ is a parallelogram. Recall that $H_K$ is the midpoint of $KH$ , hence $H_KO_K$ is the $K-$ midline of $KHN$ and altogether we have $KN_K\parallel K'O_K'\parallel H_KO_K\parallel NK$ , the Euler line of $\triangle ABC$ and we are done. $\Box$

Remark: In addition to the parallelism, we've also proven the metric property that $4KN_K=HN$ or $8KN_K=HO$ .

Main Proof: Now let us consider the orthic triangle $DEF$ of $\triangle ABC$ , let $A_1B_1C_1$ be the cevian triangle of $H$ wrt $DEF$ (So $A_1=AD\cap EF$ and etc.) It's well known that $B_1C_1\perp AN$ and since $B_KC_K\perp AN$ , therefore $B_1C_1\parallel B_KC_K$ and consequently $\triangle A_1B_1C_1, \triangle A_KB_KC_K$ are homothetic with respective corresponding points $H,K$ . Let $N^*$ denote the nine-point center of $\triangle A_1B_1C_1$ . By Lemma 2, $HN^*\parallel KN_K\parallel HN$ $\Rightarrow N^*$ lies on the Euler line.

Note that $H,N$ are the incenter and circumcenter of $\triangle DEF$ , therefore we've proven the original problem. $\Box$ .

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xelich $\quad$

by qwerty123456asdfgzxcvb, Sep 24, 2024, 9:06 PM

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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

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Geometry Treasures

A New Proof to the IO Nine-point center Problem

by XmL, Aug 17, 2014, 12:17 AM

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