Difference between revisions of "Kimberling’s point X(24)"
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+ | <i><b> Kimberling point X(24) </b></i> | ||
+ | [[File:2016 USAMO 3g.png|450px|right]] | ||
+ | |||
+ | Perspector of Triangle <math>ABC</math> and Orthic Triangle of the Orthic Triangle. | ||
+ | <i><b>Theorem 1</b></i> | ||
+ | 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_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 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> | ||
+ | |||
+ | <math>\angle HDE = \alpha - 90^\circ, \angle HCD = 90^\circ - \beta \implies</math> | ||
+ | |||
+ | <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> | ||
+ | <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> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | We know that <math>OB = OP = OC = R, PH^2 = R^2 – OH^2 \implies</math> | ||
+ | <math>QP^2 + R^2 = (QH+ HO)^2 \implies P \in \Theta, \Omega \perp \omega.</math> | ||
+ | |||
+ | Let <math>I_{\Omega}</math> be inversion with respect <math>\Omega, I_{\Omega}(B) = C.</math> | ||
+ | |||
+ | Denote <math>I_{\Omega}(D) = D', I_{\Omega}(S) = S',</math> | ||
+ | <math>QH \cdot QO = QP^2 \implies I_{\Omega}(H) = O.</math> | ||
+ | <math>HS \perp DD' \implies S'O \perp BC \implies BS' = CS' \implies \angle OCS' = \angle OBS'.</math> | ||
+ | |||
+ | <math>\angle QSB = \angle QCS' = \angle OCS' = \angle OBS' = \angle CSS'.</math> | ||
+ | |||
+ | <math>\angle BSH = 90 ^\circ – \angle QSB = 90 ^\circ – \angle CSS' =\angle CSH.</math> | ||
+ | |||
+ | <i><b>Theorem 2</b></i> | ||
+ | |||
+ | 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> | ||
+ | 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> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 08:50, 12 October 2022
Kimberling point X(24)
Perspector of Triangle and Orthic Triangle of the Orthic Triangle.
Theorem 1
Denote
obtuse or acute
Let
be the base triangle,
be Orthic triangle of
be Orthic Triangle of the Orthic Triangle of
. Let
and
be the circumcenter and orthocenter of
Then and
are homothetic, the point
center of this homothety lies on Euler line
of
The ratio of the homothety is
Proof
WLOG, we use case
Let
be reflection
in
In accordance with Claim, and
are collinear.
Similarly, and
were
is reflection
in
are collinear.
Denote
and
are concurrent at point
In accordance with Claim, points
and
are isogonal conjugate with respect
Claim
Let be an acute triangle, and let
and
denote its altitudes. Lines
and
meet at
Prove that
Proof
Let be the circle
centered at
is midpoint
Let meet
at
Let
be the circle centered at
with radius
Let be the circle with diameter
We know that
Let be inversion with respect
Denote
Theorem 2
Let be the base triangle,
be orthic triangle of
be Kosnita triangle. Then
and
are homothetic, the point
center of this homothety lies on Euler line of
the ratio of the homothety is
We recall that vertex of Kosnita triangle are:
is the circumcenter of
is the circumcenter of
is the circumcenter of
where
is circumcenter of
vladimir.shelomovskii@gmail.com, vvsss