View source for Kimberling’s point X(24)

== Kimberling's point X(24) ==
[[File:2016 USAMO 3g.png|450px|right]]

Kimberling defined point X(24) as perspector of <math>\triangle ABC</math> and Orthic Triangle of the Orthic Triangle of <math>\triangle ABC</math>.

==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> 

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> 

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

<i><b>Proof</b></i> 

WLOG, we use case <math>\angle A = \alpha > 90^\circ.</math>

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

Similarly, <math>C, W,</math> and <math>C',</math> were <math>C'</math> is reflection <math>H</math> in <math>DF,</math> are collinear.

Denote  <math>\angle ABC = \beta = \angle CHD, \angle ACB = \gamma = \angle BHD \implies</math>

<math>\angle HDF = \angle HDE = \angle DHB' = \angle DHC' = 180^\circ - \alpha.</math>

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

<math>\triangle HB'C' \sim \triangle OBC, BB', CC'</math> and <math>HO</math> are concurrent at point <math>P.</math>

In accordance with <i><b>Claim,</b></i> <math>\angle HUF = \angle AUF \implies</math> points <math>H</math> and <math>P</math> are isogonal conjugate with respect <math>\triangle UVW.</math> 

<cmath>\angle HDE = \alpha - 90^\circ, \angle HCD = 90^\circ - \beta \implies</cmath>

<cmath>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.</cmath>   
<cmath>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.</cmath>

<i><b>Claim</b></i>
[[File:2016 3 Lemma.png|400px|right]]
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>

<i><b>Proof</b></i>
 
Let <math>\omega</math> be the circle <math>BCD'D</math> centered at <math>O (O</math> is midpoint <math>BC).</math>

Let <math>\omega</math> meet <math>AH</math> at <math>P.</math> 
Let <math>\Omega</math> be the circle centered at <math>Q</math> with radius <math>QP.</math>

Let <math>\Theta</math> be the circle with diameter <math>OQ.</math>

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

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

Denote <math>I_{\Omega}(S) = S'.</math> 

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

==Theorem 2==
[[File:2016 USAMO 3e.png|400px|right]]
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 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 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>
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>

<i><b>Proof</b></i>

Let <math>H</math> be orthocenter of <math>T_0, Q</math> be the center of Nine-point circle of <math>T_0, HQO</math> is the Euler line of <math>T_0.</math>
Well known that <math>EF</math> is antiparallel <math>BC</math> with respect <math>\angle A.</math>

<math>LM</math> is the bisector of <math>AO,</math> therefore <math>LM</math> is antiparallel <math>BC</math> with respect <math>\angle A</math>
<cmath>\implies LM||EF.</cmath>
Similarly, <math>DE||KL, DF||KM \implies \triangle DEF</math> and <math>\triangle KLM</math> are homothetic.

Let <math>P</math> be the center of homothety.

<math>H</math> is <math>D</math>-excenter of <math>\triangle DEF, O</math> is <math>K</math>-excenter of <math>\triangle KLM \implies</math>
<math>P \in HO.</math>

Denote <math>a = BC, \alpha = \angle A, \beta = \angle B, \gamma = \angle C, R</math> circumradius <math>\triangle ABC.</math> 
<math>\angle EHF = 180^\circ - \alpha, EF = BC |\cos \alpha| = 2R \sin \alpha |\cos \alpha|.</math> 
<math>LM = \frac {R}{2} (\tan \beta + \tan \gamma) = \frac {R \sin (\beta + \gamma)}{2 \cos \beta \cdot \cos \gamma} \implies k = \frac {DE}{KL} = 4\cos \alpha \cdot \cos \beta \cdot \cos \gamma.</math>
<math>\frac {\vec {PH}}{\vec {OP}}= 4 \cos A \cos B \cos C \implies P</math> is point  <math>X(24).</math>

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