More on The Kosnita Point

by XmL, Apr 25, 2016, 10:09 PM

In this post, we dive into the realm of Kosnita point to examine its properties and relations with other special points. We will look at many interpretations of the point and how it plays a role with respect to different triangles. In particular, we will disect the significance of the intersections of line $KN$ with the circumcircle of $ABC$ .

We will make $ABC$ our main reference triangle, $O,H,N,K$ are respectively the circumcenter, orthocenter, nine-point center, and Kosnita point of $ABC$ . I will also label the propositions with trivial or non-trivial. This is done for personal note so do not be alarmed if it does not appear that way.

Section 1 Key Configuration

Proposition 1(non-trivial): Let $K_AK_BK_C$ denote the reflection triangle of the Kosnita point $K$ of $ABC$ . Reflect $K$ about the midpoints of $K_BK_C$ to get $K_1$ ; define $K_2, K_3$ similarly. Surprisingly, it is true that $AK_1\perp BC$ .

Proof: Let $O_A$ be the reflection of $O$ over $BC$ . Define $AO_A\cap (O)=D\ne A$ . The thing that makes Kosnita special is that we have $BCDO_A\sim K_CK_BK_1A$ . The perpendicular now follows because $\angle K_1AO_A=180-\angle OO_AA\implies (AK_1||OO_A)\perp BC$ . $\Box$

A immediate consequence of this is that triangles $ABC, K_1K_2K_3$ are perspective at $H$ . In addition, we have $K_1K_2K_3H\cong K_AK_BK_CK$ and particularly $HK_A|| KK_1$ . These facts will become important later in discussions about the perpendiculars to $HK_A$ .

Of course, if you've read my previous blogs, then these results are nothing new and were used in my proof for the infamous $IO$ problem. In my collaborated solution with Ivan Zelich, one of our Lemmas was proving $HK_A\perp AG$ , where $G$ is the intersection of $BC$ with the side of the Ceva triangle of $N$ that faces $A$ . This fact, however, is immediate from our observations above by considering the harmonic pencil $K(K_B,K_C;P_{\infty}, K_1)$ (point at infinity is on $K_BK_C$ ) and its perpendicular counterpart through $A$ .

Section 2 Perspective at $H$

For a moment we will take Kosnita out of the picture and look at somethings more intrinsic to $H$ :

Proposition 2(non-trivial): Let $X$ be on $BC$ such t hat $HX||AO$ ; define $Y,Z$ similarly. Prove that $XYZ$ and the tangential triangle of $ABC$ are orthologic to each other at $H$ , the orthocenter of $ABC$ .

Proof: It suffices to show that $HA'\perp YZ$ , where $A'$ is the intersection of the tangents to $(O)$ through $B,C$ . Perhaps unexpectedly, I will call upon poles and polar to show this.

Let $DEF$ be the orthic triangle of $ABC$ , so $H$ is the center of the incircle $\omega$ of $DEF$ . Let $E',F'$ denote the poles of $AC,AB$ wrt $\omega$ . The pole of $YZ$ is a point $P$ that satisfies $PE'||DF||A'B, PF'||DE||A'C$ . It is not hard to see that $E'F'||BC$ ; thus $PE'F', A'BC$ are homothetic through H $\implies \overline {A'HP}\perp YZ$ and we are done. $\Box$

Remark: $XYZ$ is also the Ceva triangle of the isotomic conjugate of $O$ wrt $ABC$ . Also, this might not be the easiest way to prove something about $HA'$ since there are more natural constructions that are related to that line.

Let $O_AO_BO_C$ be the reflection triangle of $O$ wrt $ABC$ . Suppose $O_AO_C\cap AB=L, O_AO_B\cap AC=J$ . Since $AJO_AL$ is clearly a parallelogram( $ABC, O_AO_BO_C$ are congruent and homothetic through $N$ ), thus $JL\cap AO_A=N$ . We now show a result that can be generalized:

Proposition 3(Trivial): $LJ||YZ$ .

Proof: Note that $AO_B||HO_C, AC||O_CO_A$ , thus $\angle O_BAJ=\angle ZO_CL, \angle AJO_B=\angle O_CLZ \implies AJO_B\sim O_CLZ$ $\implies AJ\cdot LZ=JO_B\cdot LO_C$ . Similarly, we can show that $AL\cdot JY= JO_B\cdot LO_C$ . Thus $AJ\cdot LZ=AL\cdot JY$ , which means $ZY||LJ$ . $\Box$

We now bring back the Kosnita point $K$ . Recall in our first discussion that $HK_A\perp AG$ , where $G$ is defined as follows: Let $BN\cap AC=B_N, CN\cap AB=C_N$ , $B_NC_N\cap BC=G$ . Since $A(C,B;G,N)=-1$ and $LN=NJ$ , by properties of harmonic division we have $LJ||AG$ . Because $HA'\perp (YZ||LJ)$ , we have $HA'\parallel HK_A\implies H,A',K_A$ are collinear. In other words, we have just proven the following:

Main Result 1(Non-trivial): $K_AK_BK_C$ , the reflection triangle of Kosnita point $K$ , is perspective with the tangential triangle of $ABC$ at $H$ .

Section 3 $KO$ and $P_K$

We now go deeper into how Kosnita relates to various triangles. With just Proposition 1, we can actually prove:

Main Result 2(Non-trivial):The orthocenter of $K_AK_BK_C$ , denoted by $H_K$ , lies on $OK$ . In fact, $K$ is the midpoint of $OH_K$ .

Proof: By properties of isogonal conjugates, $N$ is the circumcenter of $K_AK_BK_C$ . Let $S$ denote the midpoint of $K_BK_C$ . Reflect $H$ over $S$ to get $H'$ . Thus $OH'=2SN=K_AH_K$ , $H'K=HK_1=KK_A$ , and the corresponding parallels give us $KH'O\cong KK_AH_K\implies O,K,H_K$ are collinear and $KO=KH_K$ . $\Box$

A corollary of this is, if $H'$ is the reflection of $K_A$ over $K$ , then $OH'||AN$ . The construction of $H'$ is analogous to the perspective approach in proving the Liang-Zelich Theorem, so it is safe to say that $H'$ is a very special point and we can show the following:

Proposition 4(Trivial): $A,K,O,H'$ are concyclic and the isogonal of $AH'$ wrt $\angle BAC$ is parallel to $KO$ .

Proof: We showed that $OH'||AN$ , so $\angle (H'K,OH')=\angle (HA, AN)=\angle (AK,AO)$ because $AK,AN$ are isogonal wrt $\angle HAO$ . Thus the cyclic follows. Suppose $\ell, AH'$ are isogonal wrt $\angle BAC$ . Since $AH,AO$ are isogonals of the same angle, we have $\angle (KH',KO)=\angle (AH',AO)=\angle (AH, \ell)$ $\implies \ell || KO$ because $KH'||AH$ . $\Box$

If we construct the analogs of $AH'$ for $B,C$ , then they all concur on $(O)$ at $P_K$ , specifically at the isogonal conjugate of the point at infinity on $KO$ (this then implies that the Simson's line of $P_K$ wrt $ABC$ is perpendicular to $KO$ .) Furthermore, $P_K, K,N$ are collinear by Sondat's Theorem. To see this, relabel $H'$ as $K_A'$ and define $K_B', K_C'$ similarly. We showed that $K_A',K_B'K_C'$ and $ABC$ are perspective at $P_K$ (which is actually immediate by Liang-Zelich). Since $N,K$ are the clearly the orthologic centers for the two triangles, Sondat's gives $P_K\in KN$ . What a found! The following result summarizes this discussion:

Main result 3(non-trivial): $P_K, N,K$ are collinear, where $P_K$ is the isogonal conjugate of the point at infinity on $PK$ wrt $ABC$ .

To prepare the construction for investigating the point $P_K$ , we need the following:

Lemma(Trivial): $O$ is the circumcenter of $ABC$ . Let $P$ be an arbitrary point. Define $D,E,F$ on the circle $\odot (O,OP)$ such that $PD\perp BC, PE\perp AC, PF\perp AB$ . Prove that the orthologic center of $ABC$ wrt $DEF$ lies on $(ABC)$ and is the isogonal conjugate of the point at infinity on $PO$ wrt $ABC$ .

Proof: Let $\ell$ denote the line through $A$ and parallel to $PO$ , it suffices to show that the isogonal of $\ell$ wrt $\angle BAC$ is perpendicular to $EF$ . Select a point $Q\ne A\in \ell$ and let $Y,Z$ be its feet of projection on $AC,AB$ . By property of isogonals, it amounts to showing $ZY||EF$ , which is equivalent to $QYZ, PEF$ are homothetic. If we let $PO\cap \odot (O,OP)=R\ne P$ , then with $PO||\ell$ we can easily show that $QZAY\sim PFRE$ and the homothety follows. $\Box$

We now introduce the crux construction:

Proposition 5(non-trivial): Reflect $A$ over $K_BK_C$ to get $D$ ; similarly construct $E,F$ . From our first discussion, it is clear that $D\in KK_A$ . We will now show that $DEFK$ is a cyclic quadrilateral with center $H_K$ .

Proof: It suffices to prove that $DH_K=KH_K$ . Reflect $H_K$ over $S$ , the midpoint of $K_BK_C$ to get $T$ . Since $N$ is the circumcenter of $K_AK_BK_C$ (isogonal conjugates), we know $K_AT$ is a diameter of $(K_AK_BK_C)$ $\implies N$ is the midpoint of $K_AT$ . Note that $N$ is also the midpoint of $AO_A$ , so $DH_K=AT=K_AO_A=KO=KH_K$ and we are done. $\Box$

Now consider the Lemma with $P=K$ . Observe that since $O,H_K$ are reflexive about $K$ , the corresponding triangle $DEF$ in the Lemma is congruent and homothetic to $DEF$ in the current diagram. Because $P_K$ is the isogonal conjugate of the point ot infinity on $KO$ wrt $ABC$ , therefore $P_K$ is the orthologic center of $ABC$ wrt $DEF$ . In particular, $EF\perp AK_A'$ $\implies EF\perp DH\implies H$ is the orthocenter of $DEF$ .

It is not hard to see that $DEFK\sim ABCP_K$ . Hence we have

Main result 4(Very Non-trivial): $KHH_K\sim P_KHO$

From this we can obtain some pretty extraordinary facts:

1. $\frac {OH}{R}=\frac {HH_K}{KH_K}=\frac {2KN}{KO}\implies \frac {ON\cdot OK}{KN}=R$ , where $R$ is the circumradius of $ABC$ .
2. $\angle HKO=180-\angle OP_KH$
3. We can avoid Sondats and use this to deduce $P_K,K,N$ are collinear
4. $OHP_KK$ is similar to well known configuration, in which $(HKP_K), (OKP_K)$ are tangent to $HO$ . It is surprising for a bunch of special points to exhibit this beautiful property!

To end this section, we reveal the identity of the second intersection between $NK$ and $(ABC)$ :

Main Result 5(Non-trivial):Let $KN\cap (ABC)=E_r\ne P_K$ . We claim that $E_r$ is the Euler reflection point of $ABC$ , i.e. $\overline {E_rH_A}, \overline {OH}$ are reflexive over $BC$ .

Proof: Since $AO$ is parallel to the reflection of $\overline {OH_A}$ over $BC$ , thus it suffices to show that $\angle AOH=\angle OH_AE_r=\angle OE_rH_A$ .

We have shown before that $\angle P_KAH_A=\angle AOK$ , thus $\angle P_KE_rH_A=\angle H_AAP_K=\angle AOK$ . By the tangent result (fact 4) above, we have $\angle OE_rP_K=\angle OP_KN=\angle KON$ ; hence our claim follows. $\Box$

This post has been edited 6 times. Last edited by XmL, May 19, 2016, 4:10 AM
Reason: Thank You Telv!

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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
Check your private massage
by zabihpourmahdi, Jan 11, 2013, 5:34 PM

6 shouts

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More on The Kosnita Point

by XmL, Apr 25, 2016, 10:09 PM

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