Difference between revisions of "Kimberling’s point X(24)"
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− | + | == Kimberling's point X(24) == | |
[[File:2016 USAMO 3g.png|450px|right]] | [[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>. | |
− | < | + | |
− | 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 | + | ==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> | 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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WLOG, we use case <math>\angle A = \alpha > 90^\circ.</math> | WLOG, we use case <math>\angle A = \alpha > 90^\circ.</math> | ||
− | |||
− | In accordance with Claim, <math>\angle BVD = \angle HVE \implies B', V,</math> and <math>B</math> are collinear. | + | 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. | 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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<math>\triangle HB'C' \sim \triangle OBC, BB', CC'</math> and <math>HO</math> are concurrent at point <math>P.</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> | + | 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> | <i><b>Claim</b></i> | ||
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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> | 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.</math> | + | 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>\ | + | <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> | ||
− | <math>\angle | + | 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> | ||
+ | <cmath>LM = \frac {R}{2} (\tan \beta + \tan \gamma) = \frac {R \sin (\beta + \gamma)}{2 \cos \beta \cdot \cos \gamma} \implies</cmath> | ||
+ | <cmath>k = \frac {DE}{KL} = 4\cos \alpha \cdot \cos \beta \cdot \cos \gamma \implies</cmath> | ||
+ | <math>\frac {\vec {PH}}{\vec {OP}}= 4 \cos A \cos B \cos C \implies P</math> is the point <math>X(24).</math> | ||
− | < | + | '''vladimir.shelomovskii@gmail.com, vvsss''' |
+ | ==Theorem 3== | ||
+ | [[File:X24 as Exeter.png|450px|right]] | ||
+ | Let <math>\triangle ABC</math> be the reference triangle (other than a right triangle). Let the altitudes through the vertices <math>A, B, C</math> meet the circumcircle <math>\Omega</math> of triangle <math>ABC</math> at <math>A_0, B_0,</math> and <math>C_0,</math> respectively. Let <math>A'B'C'</math> be the triangle formed by the tangents at <math>A, B,</math> and <math>C</math> to <math>\Omega.</math> (Let <math>A'</math> be the vertex opposite to the side formed by the tangent at the vertex A). Prove that the lines through <math>A_0A', B_0B',</math> and <math>C_0C'</math> are concurrent, the point of concurrence <math>X_{24}</math> lies on Euler line of triangle <math>ABC, X_{24} = O + \frac {2}{J^2 + 1} (H – O), J = \frac {|OH|}{R}.</math> | ||
− | Let <math> | + | <i><b>Proof</b></i> |
− | + | ||
+ | At first one can prove that lines <math>A_0A', B_0B',</math> and <math>C_0C'</math> are concurrent. This follows from the fact that lines <math>AA_0, BB_0,</math> and <math>CC_0</math> are concurrent at point <math>H</math> and <i><b>Mapping theorem</b></i> (see Exeter point <math>X_{22}).</math> | ||
+ | |||
+ | Let <math>A_1, B_1,</math> and <math>C_1</math> be the midpoints of <math>BC, AC,</math> and <math>AB,</math> respectively. | ||
+ | |||
+ | Let <math>A_2, B_2,</math> and <math>C_2</math> be the midpoints of <math>AH, BH,</math> and <math>CH,</math> respectively. | ||
+ | |||
+ | Let <math>A_3, B_3,</math> and <math>C_3</math> be the foots of altitudes from <math>A, B,</math> and <math>C,</math> respectively. | ||
+ | |||
+ | The points <math>A, A_2, H,</math> and <math>A_3</math> are collinear. Similarly the points <math>B, B_2, H, B_3</math> and <math>C, C_2, H, C_3</math> are collinear. | ||
+ | |||
+ | Denote <math>I_{\Omega}</math> the inversion with respect <math>\Omega.</math> It is evident that <math>I_{\Omega}(A') = A_1, I_{\Omega}(B') = B_1, I_{\Omega}(C') = C_1, I_{\Omega}(A_0) = A_0, I_{\Omega}(B_0) = B_0, I_{\Omega}(C_0) = C_0.</math> | ||
+ | |||
+ | Denote <math>\omega_A = I_{\Omega}(A'A_0), \omega_B = I_{\Omega}(B'B_0) \omega_C = I_{\Omega}(C'C_0) \implies</math> | ||
+ | <cmath>A_0 \in \omega_A, A_1 \in \omega_A, O \in \omega_A, B_0 \in \omega_B, B_1 \in \omega_B, O \in \omega_B \implies\ O = \omega_A \cap \omega_B \cap \omega_C.</cmath> | ||
+ | |||
+ | It is known that <math>A_2O = HA_1 = A_0A_1, A_2O || HA_1 \implies \angle OA_2A_0 = \angle A_1A_0A_2, OA_1 ||A_0A_2 \implies A_2 \in \omega_A.</math> | ||
+ | |||
+ | Similarly, <math>B_2 \in \omega_B, C_2 \in \omega_C.</math> | ||
+ | |||
+ | We use <i><b>Claim</b></i> and get that the power of point <math>H</math> with respect each circle <math>\omega_X</math> is | ||
+ | <cmath>HA_2 \cdot HA_0 = HB_2 \cdot HB_0 = HC_2 \cdot HC_0 = \frac {R^2 \cdot (1-J^2)} {2}.</cmath> | ||
+ | <math>H = AA_0 \cap BB_0 \cap CC_0 \implies H</math> lies on common radical axis of <math>\omega_A, \omega_B,</math> and <math>\omega_C.</math> | ||
+ | |||
+ | Therefore second crosspoint of these circles point <math>E</math> lies on line <math>OH</math> which is the Euler line of <math>\triangle ABC \implies</math> <math>X_{24} = I_{\Omega}(E)</math> lies on the same Euler line as desired. | ||
+ | |||
+ | Last we will find the length of <math>OX_{24}.</math> | ||
+ | <cmath>OH \cdot HE = \frac {R^2 \cdot (1–J^2)} {2}.</cmath> | ||
+ | <cmath>OE \cdot OX_{24} = (OH + HE)\cdot OX_{24} = R^2.</cmath> | ||
+ | <math>\frac {OX_{24}}{OH} = \frac {R^2}{OH^2 + OH \cdot HE} = \frac {1}{J^2 + \frac {1– J^2} {2}} = | ||
+ | \frac {2}{1+J^2},</math> as desired. | ||
+ | |||
+ | <i><b>Claim</b></i> | ||
+ | [[File:AH HF.png|400px|right]] | ||
+ | Let <math>AD, BE,</math> and <math>CF</math> be the heights of <math>\triangle ABC, H = AD \cap BE \cap CF.</math> | ||
+ | |||
+ | Prove that <math>AH \cdot HD = \frac {R^2 (1 – J^2)}{2},</math> | ||
+ | |||
+ | where <math>R</math> and <math>O</math> are circumradius and circumcenter of <math>\triangle ABC, J = \frac {OH}{R}.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | It is known that <math>\triangle ABC \sim \triangle AEF.</math> | ||
+ | <math>k = \frac {AF} {AC} = \cos A \implies AH = 2 R k = 2 R \cos A.</math> | ||
+ | |||
+ | Similarly <math>BH = 2 R \cos B, DH = BH \sin \angle CBE = BH \cos C = 2R \cos B \cos C.</math> | ||
+ | |||
+ | Therefore <math>AH \cdot HD = 4R^2 \cos A \cos B \cos C.</math> | ||
+ | |||
+ | <cmath>J^2 = \frac {HO^2}{R^2} = 1 – 8 \cos A \cos B \cos C \ \implies AH \cdot HD = \frac {R^2 (1 – J^2)}{2}.</cmath> | ||
+ | |||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Latest revision as of 15:34, 25 November 2022
Kimberling's point X(24)
Kimberling defined point X(24) as perspector of and Orthic Triangle of the Orthic Triangle of
.
Theorem 1
Denote obtuse or acute
Let
be the base triangle,
be Orthic triangle of
be 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
Well known that is the polar of point
so
Let be inversion with respect
Denote
Theorem 2
Let be the base triangle,
be orthic triangle of
be Kosnita triangle of
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
Proof
Let be orthocenter of
be the center of Nine-point circle of
is the Euler line of
Well known that
is antiparallel
with respect
is the bisector of
therefore
is antiparallel
with respect
Similarly,
and
are homothetic.
Let be the center of homothety.
is
-excenter of
is
-excenter of
Denote circumradius
is the point
vladimir.shelomovskii@gmail.com, vvsss
Theorem 3
Let be the reference triangle (other than a right triangle). Let the altitudes through the vertices
meet the circumcircle
of triangle
at
and
respectively. Let
be the triangle formed by the tangents at
and
to
(Let
be the vertex opposite to the side formed by the tangent at the vertex A). Prove that the lines through
and
are concurrent, the point of concurrence
lies on Euler line of triangle
Proof
At first one can prove that lines and
are concurrent. This follows from the fact that lines
and
are concurrent at point
and Mapping theorem (see Exeter point
Let and
be the midpoints of
and
respectively.
Let and
be the midpoints of
and
respectively.
Let and
be the foots of altitudes from
and
respectively.
The points and
are collinear. Similarly the points
and
are collinear.
Denote the inversion with respect
It is evident that
Denote
It is known that
Similarly,
We use Claim and get that the power of point with respect each circle
is
lies on common radical axis of
and
Therefore second crosspoint of these circles point lies on line
which is the Euler line of
lies on the same Euler line as desired.
Last we will find the length of
as desired.
Claim
Let and
be the heights of
Prove that
where and
are circumradius and circumcenter of
Proof
It is known that
Similarly
Therefore
vladimir.shelomovskii@gmail.com, vvsss