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Difference between revisions of "Kimberling’s point X(24)"

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Kimberling defined point X(24) as perspector of <math>\triangle ABC</math> and Orthic Triangle of the Orthic Triangle of <math>\triangle ABC</math>.
 
Kimberling defined point X(24) as perspector of <math>\triangle ABC</math> and Orthic Triangle of the Orthic Triangle of <math>\triangle ABC</math>.
  
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<i><b>Theorem 1</b></i>
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==Theorem 1==
  
 
Denote <math>T_0</math> obtuse or acute <math>\triangle ABC.</math> Let <math>T_0</math> be the base triangle, <math>T_1 = \triangle DEF</math> be Orthic triangle of <math>T_0, T_2 = \triangle UVW</math> be Orthic Triangle of <math>T_1</math>. Let <math>O</math> and <math>H</math> be the circumcenter and orthocenter of <math>T_0.</math>  
 
Denote <math>T_0</math> obtuse or acute <math>\triangle ABC.</math> Let <math>T_0</math> be the base triangle, <math>T_1 = \triangle DEF</math> be Orthic triangle of <math>T_0, T_2 = \triangle UVW</math> be Orthic Triangle of <math>T_1</math>. Let <math>O</math> and <math>H</math> be the circumcenter and orthocenter of <math>T_0.</math>  
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<cmath>\angle BSH = 90 ^\circ  – \angle QSB = 90 ^\circ  – \angle CSS' =\angle CSH.</cmath>
 
<cmath>\angle BSH = 90 ^\circ  – \angle QSB = 90 ^\circ  – \angle CSS' =\angle CSH.</cmath>
  
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<i><b>Theorem 2</b></i>
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==Theorem 2==
  
 
Let <math>T_0 = \triangle ABC</math> be the base triangle, <math>T_1 = \triangle DEF</math> be orthic triangle of <math>T_0, T_2 = \triangle KLM</math> be Kosnita triangle. Then <math>\triangle T_1</math> and <math>\triangle T_2</math> are homothetic, the point <math>P,</math> center of this homothety lies on Euler line of <math>T_0,</math> the ratio of the homothety is <math>k =  \frac {\vec PH}{\vec OP} = 4 \cos A \cos B \cos C.</math>
 
Let <math>T_0 = \triangle ABC</math> be the base triangle, <math>T_1 = \triangle DEF</math> be orthic triangle of <math>T_0, T_2 = \triangle KLM</math> be Kosnita triangle. Then <math>\triangle T_1</math> and <math>\triangle T_2</math> are homothetic, the point <math>P,</math> center of this homothety lies on Euler line of <math>T_0,</math> the ratio of the homothety is <math>k =  \frac {\vec PH}{\vec OP} = 4 \cos A \cos B \cos C.</math>

Revision as of 13:02, 12 October 2022

Kimberling's point X(24)

2016 USAMO 3g.png

Kimberling defined point X(24) as perspector of $\triangle ABC$ and Orthic Triangle of the Orthic Triangle of $\triangle ABC$.

Theorem 1

Denote $T_0$ obtuse or acute $\triangle ABC.$ Let $T_0$ be the base triangle, $T_1 = \triangle DEF$ be Orthic triangle of $T_0, T_2 = \triangle UVW$ be Orthic Triangle of $T_1$. Let $O$ and $H$ be the circumcenter and orthocenter of $T_0.$

Then $\triangle T_0$ and $\triangle T_2$ are homothetic, the point $P,$ center of this homothety lies on Euler line $OH$ of $T_0.$

The ratio of the homothety is $k = \frac {\vec {PH}}{\vec {OP}}= 4 \cos A \cos B \cos C.$

Proof

WLOG, we use case $\angle A = \alpha > 90^\circ.$

Let $B'$ be reflection $H$ in $DE.$ In accordance with Claim, $\angle BVD = \angle HVE \implies B', V,$ and $B$ are collinear.

Similarly, $C, W,$ and $C',$ were $C'$ is reflection $H$ in $DF,$ are collinear.

Denote $\angle ABC = \beta = \angle CHD, \angle ACB = \gamma = \angle BHD \implies$

$\angle HDF = \angle HDE = \angle DHB' = \angle DHC' = 180^\circ - \alpha.$

$B'C' \perp HD, BC \perp HD \implies BC|| B'C'.$ $OB = OC, HB' = HC', \angle BOC = \angle B'HC' = 360^\circ - 2 \alpha \implies OB ||HB', OC || HC' \implies$

$\triangle HB'C' \sim \triangle OBC, BB', CC'$ and $HO$ are concurrent at point $P.$

In accordance with Claim, $\angle HUF = \angle AUF \implies$ points $H$ and $P$ are isogonal conjugate with respect $\triangle UVW.$

\[\angle HDE = \alpha - 90^\circ, \angle HCD = 90^\circ - \beta \implies\]

\[HB' = 2 HD \sin (\alpha - 90^\circ) = - 2 CD \tan(90^\circ- \beta) \cos \alpha = - 2 AC \cos \gamma \frac {\cos \beta}{\sin \beta} \cos \alpha = - 4 OB \cos A \cos B \cos C.\] \[k = \frac {HB'}{OB} = \frac {HP}{OP}= - 4 \cos A \cos B \cos C \implies \frac {\vec {PH}}{\vec {OP}}= 4 \cos A \cos B \cos C.\]

Claim

2016 3 Lemma.png

Let $\triangle ABC$ be an acute triangle, and let $AH, BD',$ and $CD$ denote its altitudes. Lines $DD'$ and $BC$ meet at $Q, HS \perp DD'.$ Prove that $\angle BSH = \angle CSH.$

Proof

Let $\omega$ be the circle $BCD'D$ centered at $O (O$ is midpoint $BC).$

Let $\omega$ meet $AH$ at $P.$ Let $\Omega$ be the circle centered at $Q$ with radius $QP.$

Let $\Theta$ be the circle with diameter $OQ.$

Well known that $AH$ is the polar of point $Q,$ so $QO \cdot HO = QP^2 \implies QB \cdot QC = (QO – R) \cdot (QO + R) = QP^2$ \[\implies P \in \Theta, \Omega \perp \omega.\]

Let $I_{\Omega}$ be inversion with respect $\Omega, I_{\Omega}(B) = C, I_{\Omega}(H) = O,I_{\Omega}(D) = D'.$

Denote $I_{\Omega}(S) = S'.$

\[HS \perp DD' \implies S'O \perp BC \implies BS' = CS' \implies \angle OCS' = \angle OBS'.\] \[\angle QSB = \angle QCS' = \angle OCS' = \angle OBS' = \angle CSS'.\] \[\angle BSH = 90 ^\circ – \angle QSB = 90 ^\circ – \angle CSS' =\angle CSH.\]

Theorem 2

Let $T_0 = \triangle ABC$ be the base triangle, $T_1 = \triangle DEF$ be orthic triangle of $T_0, T_2 = \triangle KLM$ be Kosnita triangle. Then $\triangle T_1$ and $\triangle T_2$ are homothetic, the point $P,$ center of this homothety lies on Euler line of $T_0,$ the ratio of the homothety is $k = \frac {\vec PH}{\vec OP} = 4 \cos A \cos B \cos C.$ We recall that vertex of Kosnita triangle are: $K$ is the circumcenter of $\triangle OBC, L$ is the circumcenter of $\triangle OAB, M$ is the circumcenter of $\triangle OAC,$ where $O$ is circumcenter of $T_0.$

vladimir.shelomovskii@gmail.com, vvsss

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