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

(Kimberling’s point X(24))
 
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<i><b> Kimberling point X(24) </b></i>
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[[File:2016 USAMO 3g.png|450px|right]]
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Perspector of Triangle <math>ABC</math> and Orthic Triangle of the Orthic Triangle.
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<i><b>Theorem 1</b></i>
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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 the Orthic Triangle of <math>T_0</math>. Let <math>O</math> and <math>H</math> be the circumcenter and orthocenter of <math>T_0.</math>
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Then <math>\triangle T_0</math> and <math>\triangle T_2</math> are homothetic, the point <math>P,</math> center of this homothety lies on Euler line <math>OH</math> of <math>T_0.</math>
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The ratio of the homothety is <math>k = \frac {\vec {PH}}{\vec {OP}}= 4 \cos A \cos B \cos C.</math>
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<i><b>Proof</b></i>
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WLOG, we use case <math>\angle A = \alpha > 90^\circ.</math>
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Let <math>B'</math> be reflection <math>H</math> in <math>DE.</math>
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In accordance with Claim, <math>\angle BVD = \angle HVE \implies B', V,</math> and <math>B</math> are collinear.
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Similarly, <math>C, W,</math> and <math>C',</math> were <math>C'</math> is reflection <math>H</math> in <math>DF,</math> are collinear.
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Denote  <math>\angle ABC = \beta = \angle CHD, \angle ACB = \gamma = \angle BHD \implies</math>
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<math>\angle HDF = \angle HDE = \angle DHB' = \angle DHC' = 180^\circ - \alpha.</math>
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<math>B'C' \perp HD, BC \perp HD \implies BC|| B'C'.</math>
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<math>OB = OC, HB' = HC', \angle BOC = \angle B'HC' = 360^\circ - 2 \alpha \implies OB ||HB', OC || HC' \implies</math>
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<math>\triangle HB'C' \sim \triangle OBC, BB', CC'</math> and <math>HO</math> are concurrent at point <math>P.</math>
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In accordance with Claim, <math>\angle HUF = \angle AUF \implies</math> points <math>H</math> and <math>P</math> are isogonal conjugate with respect <math>\triangle UVW.</math>
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<math>\angle HDE = \alpha - 90^\circ, \angle HCD = 90^\circ - \beta \implies</math>
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<math>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.</math> 
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<math>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.</math>
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<i><b>Claim</b></i>
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[[File:2016 3 Lemma.png|400px|right]]
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Let <math>\triangle ABC</math> be an acute triangle, and let <math>AH, BD',</math> and <math>CD</math> denote its altitudes. Lines <math>DD'</math> and <math>BC</math> meet at <math>Q, HS \perp DD'.</math> Prove that <math>\angle BSH = \angle CSH.</math>
  
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[[File:2016 USAMO 3g.png|450px|right]]
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<i><b>Proof</b></i>
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1. Perspector of Triangle ABC and Orthic Triangle of the Orthic Triangle.
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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 the Orthic Triangle of <math>T_0</math>. Let <math>O</math> and <math>H</math> be the circumcenter and orthocenter of <math>T_0.</math> 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 <math>OH</math> of <math>T_0,</math> the ratio of the homothety is <math>k = \frac {\vec PH}{\vec OP}= 4 cosA cosB cosC.</math>
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Let <math>\omega</math> be the circle <math>BCD'D</math> centered at <math>O (O</math> is midpoint <math>BC).</math>
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Let <math>\omega</math> meet <math>AH</math> at <math>P.</math>  
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Let <math>\Omega</math> be the circle centered at <math>Q</math> with radius <math>QP.</math>
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Let <math>\Theta</math> be the circle with diameter <math>OQ.</math>
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We know that <math>OB = OP = OC = R, PH^2 = R^2 – OH^2 \implies</math>
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<math>QP^2 + R^2 = (QH+ HO)^2 \implies P \in \Theta,  \Omega \perp \omega.</math>
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Let <math>I_{\Omega}</math> be inversion with respect <math>\Omega, I_{\Omega}(B) = C.</math>
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Denote <math>I_{\Omega}(D) = D',  I_{\Omega}(S) = S',</math>
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<math>QH \cdot QO = QP^2 \implies I_{\Omega}(H) = O.</math>
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<math>HS \perp DD' \implies S'O \perp BC \implies BS' = CS' \implies \angle OCS' = \angle OBS'.</math>
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<math>\angle QSB = \angle QCS' = \angle OCS' = \angle OBS' = \angle CSS'.</math>
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<math>\angle BSH = 90 ^\circ  – \angle QSB = 90 ^\circ  – \angle CSS' =\angle CSH.</math>
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<i><b>Theorem 2</b></i>
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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>
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We recall that vertex of Kosnita triangle are:  <math>K</math> is the circumcenter of <math>\triangle OBC, L</math> is the circumcenter of <math>\triangle OAB, M</math> is the circumcenter of <math>\triangle OAC,</math> where <math>O</math> is circumcenter of <math>T_0.</math>
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'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 08:50, 12 October 2022

Kimberling point X(24)

2016 USAMO 3g.png

Perspector of Triangle $ABC$ and Orthic Triangle of the Orthic Triangle. 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 the Orthic Triangle of $T_0$. 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 $OQ.$

We know that $OB = OP = OC = R, PH^2 = R^2 – OH^2 \implies$ $QP^2 + R^2 = (QH+ HO)^2 \implies P \in \Theta, \Omega \perp \omega.$

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

Denote $I_{\Omega}(D) = D', I_{\Omega}(S) = S',$ $QH \cdot QO = QP^2 \implies I_{\Omega}(H) = O.$ $HS \perp DD' \implies S'O \perp BC \implies BS' = CS' \implies \angle OCS' = \angle OBS'.$

$\angle QSB = \angle QCS' = \angle OCS' = \angle OBS' = \angle CSS'.$

$\angle BSH = 90 ^\circ – \angle QSB = 90 ^\circ – \angle CSS' =\angle CSH.$

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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