Difference between revisions of "Gossard perspector"
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==Gossard perspector X(402) and Gossard triangle== | ==Gossard perspector X(402) and Gossard triangle== | ||
− | Euler proved that the Euler line of a given triangle together with two of its sides forms a triangle whose Euler line is parallel with the third side of the given triangle. | + | In <math>1765</math> Leonhard Euler proved that in any triangle, the orthocenter, circumcenter and centroid are collinear. We name this line the Euler line. Soon he proved that the Euler line of a given triangle together with two of its sides forms a triangle whose Euler line is in parallel with the third side of the given triangle. |
− | Gossard proved that | + | Professor Harry Clinton Gossard in <math>1916</math> proved that three Euler lines of the triangles formed by the Euler line and the sides, taken by two, of a given triangle, form a triangle which is perspective with the given triangle and has the same Euler line. The center of the perspective now is known as the Gossard perspector or the Kimberling point <math>X(402).</math> It is the crosspoint of Gauss line and Euler line. |
+ | |||
+ | Let triangle <math>\triangle ABC</math> be given. The Euler line crosses lines <math>AB, BC,</math> and <math>AC</math> at points <math>D, E,</math> and <math>F.</math> | ||
+ | |||
+ | On <math>13/01/2023</math> it was found that the Gossard perspector is the centroid of the points <math>A, B, C, D, E, F.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Gossard perspector of right triangle== | ==Gossard perspector of right triangle== | ||
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The Gossard triangle of the right <math>\triangle ABC</math> is the reflection of <math>\triangle ABC</math> in the Gossard perspector. | The Gossard triangle of the right <math>\triangle ABC</math> is the reflection of <math>\triangle ABC</math> in the Gossard perspector. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Gossard perspector and Gossard triangle for isosceles triangle== | ||
+ | [[File:Gossard equilateral.png|500px|right]] | ||
+ | It is clear that the Euler line of isosceles <math>\triangle ABC (AB = AC)</math> meet the sidelines <math>BC, CA</math> and <math>AB</math> of <math>\triangle ABC</math> at <math>A'</math> and <math>A,</math> where <math>A'</math> is the midpoint of <math>BC.</math> | ||
+ | |||
+ | Let <math>\triangle A'B'C'</math> be the triangle formed by the Euler lines of the <math>\triangle AA'B, \triangle AA'C,</math> and the line <math>l</math> contains <math>A</math> and parallel to <math>BC,</math> the vertex <math>B'</math> being the intersection of the Euler line of the <math>\triangle AA'C</math> and <math>l,</math> the vertex <math>C'</math> being the intersection of the Euler line of the <math>\triangle AA'B</math> and <math>l.</math> | ||
+ | |||
+ | We call the triangle <math>\triangle A'B'C'</math> as the Gossard triangle of <math>\triangle ABC.</math> | ||
+ | |||
+ | Let <math>\triangle ABC</math> be any isosceles triangle and let <math>\triangle A'B'C'</math> be its Gossard triangle. Then the lines <math>AA', BB',</math> and <math>CC'</math> are concurrent. We call the point of concurrence <math>Go</math> as the Gossard perspector of <math>\triangle ABC.</math> Let <math>H</math> be the orthocenter of <math>\triangle ABC, O</math> be the circumcenter of <math>\triangle ABC.</math> | ||
+ | |||
+ | It is clear that <math>Go</math> is the midpoint of <math>AA'.</math> | ||
+ | <math>M = A'C' \cap AB</math> is the midpoint <math>AB, N = A'B' \cap AC</math> is the midpoint <math>AC.</math> | ||
+ | |||
+ | <math>\triangle A'BM = \triangle CNA' \sim \triangle CBA</math> with coefficient <math>k = \frac {1}{2}.</math> | ||
+ | |||
+ | Any isosceles triangle and its Gossard triangle are congruent. | ||
+ | |||
+ | Any isosceles triangle and its Gossard triangle have the same Euler line. | ||
+ | |||
+ | The Gossard triangle of the isosceles <math>\triangle ABC</math> is the reflection of <math>\triangle ABC</math> in the Gossard perspector. | ||
+ | Denote <math>\angle BAC = \alpha, \angle ABC = \angle ACB = \beta, AO = BO = R \implies</math> | ||
+ | <cmath>\sin \beta = \cos \frac {\alpha}{2}, GoO = AO – AGo = R \cdot \frac{1 – \cos {\alpha}}{2},</cmath> | ||
+ | <cmath>OH = AH – AO = R(2 \cos \alpha – 1) \implies \vec {GoO} = \vec {OH} \cdot \frac {1 – \cos \alpha}{2(2 \cos \alpha – 1)}.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Euler line of the triangle formed by the Euler line and the sides of a given triangle== | ||
+ | [[File:Euler Euler line.png|500px|right]] | ||
+ | Let the Euler line of <math>\triangle ABC</math> meet the lines <math>AB, AC,</math> and <math>BC</math> at <math>D, E,</math> and <math>F,</math> respectively. | ||
+ | |||
+ | Euler line of the <math>\triangle ADE</math> is parallel to <math>BC.</math> Similarly, Euler line of the <math>\triangle BDF</math> is parallel to <math>AC,</math> Euler line of the <math>\triangle CEF</math> is parallel to <math>AB.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>\angle A = \alpha, \angle B = \beta, \angle C = \gamma,</math> smaller angles between the Euler line and lines <math>BC, AC,</math> and <math>AB</math> as <math>\theta_A, \theta_B,</math> and <math>\theta_C,</math> respectively. WLOG, <math>AC > BC > AB.</math> | ||
+ | It is known that <math>\tan \theta_A = \frac{3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma}, \tan \theta_B = \frac{3 – \tan \alpha \cdot \tan \gamma}{\tan \alpha – \tan \gamma}, \tan \theta_C = \frac{3 – \tan \beta \cdot \tan \alpha}{\tan \beta – \tan \alpha}.</math> | ||
+ | *[[Euler line]] | ||
+ | Let <math>O'</math> be circumcenter of <math>\triangle ADE, KO'</math> be Euler line of <math>\triangle ADE, K \in DE</math> (line). | ||
+ | |||
+ | Similarly, <math>\tan \angle O'KF = \frac{3 – \tan \theta_B \cdot \tan \theta_C}{\tan \theta_C – \tan \theta_B}.</math> | ||
+ | <cmath>3(\tan\alpha – \tan \gamma) (\tan\alpha – \tan \beta) – (3 – \tan \alpha \cdot \tan \gamma) (3 – \tan \alpha \cdot \tan \beta) = (\tan^2 \alpha – 3) \cdot (3 –\tan \beta \cdot \tan \gamma),</cmath> | ||
+ | <cmath>(3 – \tan \alpha \cdot \tan \gamma) \cdot (\tan\alpha – \tan \beta) – (3 – \tan \alpha \cdot \tan \beta) \cdot (\tan\alpha – \tan \gamma) = (\tan^2 \alpha – 3) \cdot (\tan \beta – \tan \gamma).</cmath> | ||
+ | Suppose, <math>\tan^2 \alpha \ne 3</math> which means <math>\alpha \ne 60^\circ</math> and <math>\alpha \ne 120^\circ.</math> In this case | ||
+ | <cmath>\tan \angle O'KF = \frac{3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma} = \tan \theta_A \implies \angle O'KF = \theta_A \implies O'K||BC.</cmath> | ||
+ | |||
+ | Similarly one can prove the claim in the other cases. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Gossard triangle for triangle with angle 60== | ||
+ | [[File:Gossard 60.png|500px|right]] | ||
+ | Let <math>\angle A</math> of the triangle <math>ABC</math> be <math>60^\circ, \angle B \ne 60^\circ.</math> | ||
+ | Let the Euler line of <math>\triangle ABC</math> meet the lines <math>AB, AC</math> and <math>BC</math> at points <math>D, E,</math> and <math>F,</math> respectively. | ||
+ | Prove that <math>\triangle ADE</math> is an equilateral triangle. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>\angle ABC = \beta, \angle ACB = \gamma.</math> | ||
+ | It is known that <cmath>\tan \angle AED = \tan \theta_B = \frac{3 – \tan \alpha \cdot \tan \gamma}{\tan \alpha – \tan \gamma}.</cmath> | ||
+ | *[[Euler line]] | ||
+ | <cmath>\tan \angle AED = \frac{3 – \sqrt{3} \tan \gamma}{\sqrt{3} – \tan \gamma} = \sqrt{3} \implies \angle AED = 60^\circ.</cmath> | ||
+ | Therefore <math>\triangle ADE</math> is equilateral triangle. | ||
+ | |||
+ | Let <math>\triangle A'B'C'</math> be the triangle formed by the Euler lines of the <math>\triangle BDF, \triangle CEF,</math> and the line <math>l</math> contains centroid <math>G</math> of the <math>\triangle ADE</math> and parallel to <math>BC,</math> the vertex <math>B'</math> being the intersection of the Euler line of the <math>\triangle CEF</math> and <math>l,</math> the vertex <math>C'</math> being the intersection of the Euler line of the <math>\triangle BDF</math> and <math>l,</math> the vertex <math>A'</math> being the intersection of the Euler lines of the <math>\triangle BDF</math> and <math>\triangle CEF.</math> | ||
+ | |||
+ | We call the triangle <math>\triangle A'B'C'</math> as the Gossard triangle of the <math>\triangle ABC.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Gossard triangle for triangle with angle 120== | ||
+ | [[File:Gossard 120.png|500px|right]] | ||
+ | Let <math>\angle A</math> of the triangle <math>ABC</math> be <math>120^\circ, \angle B \ne 30^\circ.</math> | ||
+ | Let the Euler line of <math>\triangle ABC</math> meet the lines <math>AB, AC</math> and <math>BC</math> at points <math>D, E,</math> and <math>F,</math> respectively. Then <math>\triangle ADE</math> is an equilateral triangle. | ||
+ | |||
+ | One can prove this claim using the same formulae as in the case <math>\angle A = 60^\circ.</math> | ||
+ | |||
+ | Let <math>\triangle A'B'C'</math> be the triangle formed by the Euler lines of the <math>\triangle BDF, \triangle CEF,</math> and the line <math>l</math> contains centroid <math>G</math> of the <math>\triangle ADE</math> and parallel to <math>BC,</math> the vertex <math>B'</math> being the intersection of the Euler line of the <math>\triangle CEF</math> and <math>l,</math> the vertex <math>C'</math> being the intersection of the Euler line of the <math>\triangle BDF</math> and <math>l,</math> the vertex <math>A'</math> being the intersection of the Euler lines of the <math>\triangle BDF</math> and <math>\triangle CEF.</math> | ||
+ | |||
+ | We call the triangle <math>\triangle A'B'C'</math> as the Gossard triangle of the <math>\triangle ABC.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Gossard perspector== | ||
+ | [[File:Gossard complete.png|450px|right]] | ||
+ | Let non equilateral triangle <math>ABC</math> be given. The Euler line of <math>\triangle ABC</math> crosses lines <math>AB, BC,</math> and <math>AC</math> at points <math>D, E,</math> and <math>F,</math> respectively. | ||
+ | |||
+ | Let the point <math>X</math> be the centroid of the set of points <math>A, B, C, D, E, F.</math> | ||
+ | |||
+ | Let Gossard triangle <math>A'B'C'</math> be defined as described above. | ||
+ | |||
+ | Prove that <math>\triangle A'B'C'</math> and <math>\triangle ABC</math> are homothetic and congruent, and the homothetic center is the point <math>X,</math> the Euler line of <math>\triangle A'B'C' </math> coincide with the Euler line of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>G_A, G_B,</math> and <math>G_C</math> centroids of the triangles <math>ADE, BDF,</math> and <math>CEF,</math> respectively. | ||
+ | It is clear that <math>G_A \in B'C', G_B \in A'C', G_C \in A'B', X \in</math> Euler line. | ||
+ | |||
+ | Let point <math>G'_A</math> be symmetric to the point <math>G_A</math> with respect to the point <math>X.</math> | ||
+ | |||
+ | Similarly we define points <math>G'_B</math> and <math>G'_C.</math> | ||
+ | <cmath>\vec {G'_A} = 2 \vec X – \vec G_A = \frac {\vec A+\vec B+\vec C+ \vec D+\vec E + \vec F}{3} – \frac {\vec A+ \vec D+\vec E }{3} = \frac {\vec B+\vec C+ \vec F}{3} \in BC.</cmath> | ||
+ | Similarly <math>G'_B \in AC</math> and <math>G'_C \in AB.</math> | ||
+ | |||
+ | <math>AB||A'B', AC || A'C', BC|| B'C' \implies</math> the crosspoints of lines <math>A'B', A'C',</math> and <math>B'C'</math> are symmetric to the crosspoints of lines <math>AB, AC,</math> and <math>BC,</math> therefore points <math>A', B',</math> and <math>C',</math> are symmetric to points <math>A, B,</math> and <math>C</math> with respect to the point <math>X \implies X</math> is the Gossard perspector of the <math>\triangle ABC.</math> | ||
+ | |||
+ | It is clear that the Gossard perspector lyes on Euler line of the <math>\triangle ABC</math> and <math>\triangle A'B'C'</math> is congruent to <math>\triangle ABC</math>. | ||
+ | |||
+ | The Euler line of <math>\triangle A'B'C'</math> is symmetric to the Euler line of <math>\triangle ABC</math> with respect to <math>X.</math> Therefore these lines coincide. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Zeeman’s Generalisation== | ||
+ | [[File:Generalization 1.png|500px|right]] | ||
+ | Let <math>l</math> be any line parallel to the Euler line of non equilateral triangle <math>ABC.</math> Let <math>l</math> intersect the sidelines <math>AB, CA, BC</math> of <math>\triangle ABC</math> at points <math>D, E, F,</math> respectively. Let <math>\triangle A'B'C'</math> be the triangle formed by the Euler lines (as in previous sections) of the triangles <math>\triangle ADE, \triangle BDF,</math> and <math>\triangle CEF.</math> Let the point <math>X</math> be the centroid of the set of points <math>A, B, C, D, E, F.</math> | ||
+ | |||
+ | Then <math>\triangle A'B'C'</math> and <math>\triangle ABC</math> are homothetic and congruent, and the homothetic center is the point <math>X,</math> the Euler line of <math>\triangle A'B'C'</math> coincide with the line <math>DE</math> and the point <math>X</math> is equidistant from the Euler lines. | ||
+ | |||
+ | In this case <math>X</math> usually called the Zeeman–Gossard perspector. | ||
+ | |||
+ | One can prove this claim using the method of previous section. | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Paul Yiu's Generalization== | ||
+ | [[File:Yu generalization.png|500px|right]] | ||
+ | Let <math>P</math> be any point in the plane of non equilateral triangle <math>ABC</math> different from its centroid <math>G.</math> | ||
+ | |||
+ | Let the line <math>PG</math> meet the sidelines <math>AB, CA,</math> and <math>BC</math> at <math>D, E,</math> and <math>D,</math> respectively. | ||
+ | |||
+ | Let the centroids of the triangles <math>AED, BDF,</math> and <math>CFE</math> be <math>Ga, Gb,</math> and <math>Gc,</math> respectively. | ||
+ | |||
+ | Let <math>Pa</math> be a point such that <math>EPa</math> is parallel to <math>CP</math> and <math>DPa</math> is parallel to <math>BP.</math> | ||
+ | Symilarly, <math>Pb: DPb||AP, FPb||CP, Pc: FPc||BP, EPc||AP.</math> | ||
+ | |||
+ | Let <math>\triangle A'B'C'</math> be the triangle formed by the lines <math>GaPa, GbPb,</math> and <math>GcPc.</math> | ||
+ | Let the point <math>X</math> be the centroid of the set of points <math>A, B, C, D, E, F.</math> | ||
+ | Then <math>\triangle A'B'C'</math> and <math>\triangle ABC</math> are homothetic and congruent, and the homothetic center is the point <math>X,</math> the points <math>P, G,</math> and <math>G'</math> are collinear. | ||
+ | |||
+ | One can prove this claim using the method of previous section. | ||
+ | |||
+ | The points <math>P, Pa, Pb,</math> and <math>Pc</math> are collinear. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | == Dao Thanh Oai's Generalization== | ||
+ | [[File:Generalization.png|400px|right]] | ||
+ | Let triangle <math>ABC</math> and line <math>\ell,</math> non parallel to sidelines be given. Let line <math>\ell</math> meets sidelines <math>AB, AC, BC</math> of <math>\triangle ABC</math> at points <math>D, E, F,</math> respectively. | ||
+ | |||
+ | Let <math>Q \in \ell.</math> Let <math>K, L, M</math> be the points such that <math>QA||EM||DL, QB||KD||FM, QC||KE||FL.</math> | ||
+ | |||
+ | Similarly define points <math>P \in \ell, K_0, L_0, M_0.</math> | ||
+ | |||
+ | Let triangle <math>\triangle A'B'C'</math> be the triangle formed by the lines <math>KK_0, LL_0, MM_0.</math> | ||
+ | |||
+ | Prove that <math>\triangle A'B'C'</math> and <math>\triangle ABC</math> are homothetic and congruent, and the homothetic center lies on <math>\ell.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | [[File:Dao Generalization.png|400px|right]] | ||
+ | Let <math>G</math> and <math>G_0</math> be the midpoints <math>BE</math> and <math>CD,</math> respectively. | ||
+ | |||
+ | Let <math>X</math> be the crosspoint of <math>\ell</math> and Gauss line <math>GG_0.</math> | ||
+ | |||
+ | Let <math>A', B', C', D', E',</math> and <math>K'</math> be the points simmetric to <math>A, B, C, D, E,</math> and <math>K</math> with respect to <math>X,</math> respectively. | ||
+ | |||
+ | We will prove that <math>K' \in BC</math> which means that <math>K \in B'C' \implies K_0 \in BC, L \in AC</math> and so on. | ||
+ | |||
+ | <math>G</math> midpoint <math>BE, X</math> midpoint <math>EE' \implies GX||BE'.</math> | ||
+ | |||
+ | <math>G_0</math> midpoint <math>DC, X</math> midpoint <math>DD' \implies G_0X||CD' \implies BE'||CD'.</math> | ||
+ | <math>KD||K'D'||BQ, KE||K'E'||CQ, CD'||BE', Q \in D'E' \implies K' \in BC</math> according the Claim. | ||
+ | |||
+ | <i><b>Claim (Parallel lines in trapezium)</b></i> | ||
+ | [[File:Pappus.png|300px|right]] | ||
+ | [[File:Pappus non convex.png|300px|right]] | ||
+ | Let <math>ABCD</math> be the quadrungle such that <math>AD||BC.</math> | ||
+ | Let <math>M \in AB, MD||BN, AN||CM.</math> | ||
+ | Prove that point <math>N</math> lyes on <math>CD.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | We prove Claim in the case <math>AB</math> non parallel to <math>CD.</math> | ||
+ | Denote <math>Q = AB \cap CD.</math> | ||
+ | |||
+ | <math>AD||BC \implies \frac {QB}{QA}= \frac{QC}{QD}.</math> | ||
+ | |||
+ | Let <math>BN||MD</math> cross <math>CD</math> at <math>N.</math> Then <math>\frac {QN}{QD}=\frac {QB}{QM}.</math> | ||
+ | |||
+ | <cmath>\frac {QC}{QN} = \frac {QC}{QD} \cdot \frac {QD}{QN} = \frac {QB}{QA} \cdot \frac {QM}{QB} = \frac {QM}{QA} \implies AN||CM.</cmath> | ||
+ | The Claim is correct in the case of non convex <math>ABCD.</math> One can simplify the proof of Dao Generalization using this variant of the Claim. | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Latest revision as of 06:44, 19 January 2023
Contents
- 1 Gossard perspector X(402) and Gossard triangle
- 2 Gossard perspector of right triangle
- 3 Gossard perspector and Gossard triangle for isosceles triangle
- 4 Euler line of the triangle formed by the Euler line and the sides of a given triangle
- 5 Gossard triangle for triangle with angle 60
- 6 Gossard triangle for triangle with angle 120
- 7 Gossard perspector
- 8 Zeeman’s Generalisation
- 9 Paul Yiu's Generalization
- 10 Dao Thanh Oai's Generalization
Gossard perspector X(402) and Gossard triangle
In Leonhard Euler proved that in any triangle, the orthocenter, circumcenter and centroid are collinear. We name this line the Euler line. Soon he proved that the Euler line of a given triangle together with two of its sides forms a triangle whose Euler line is in parallel with the third side of the given triangle.
Professor Harry Clinton Gossard in proved that three Euler lines of the triangles formed by the Euler line and the sides, taken by two, of a given triangle, form a triangle which is perspective with the given triangle and has the same Euler line. The center of the perspective now is known as the Gossard perspector or the Kimberling point
It is the crosspoint of Gauss line and Euler line.
Let triangle be given. The Euler line crosses lines
and
at points
and
On it was found that the Gossard perspector is the centroid of the points
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Gossard perspector of right triangle
It is clear that the Euler line of right triangle meet the sidelines
and
of
at
and
where
is the midpoint of
Let be the triangle formed by the Euler lines of the
and the line
contains
and parallel to
the vertex
being the intersection of the Euler line of the
and
the vertex
being the intersection of the Euler line of the
and
We call the triangle as the Gossard triangle of
Let be any right triangle and let
be its Gossard triangle. Then the lines
and
are concurrent. We call the point of concurrence
as the Gossard perspector of
is the midpoint of
is orthocenter of
is circumcenter of
so
is midpoint of
is the midpoint
is the midpoint
with coefficient
Any right triangle and its Gossard triangle are congruent.
Any right triangle and its Gossard triangle have the same Euler line.
The Gossard triangle of the right is the reflection of
in the Gossard perspector.
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Gossard perspector and Gossard triangle for isosceles triangle
It is clear that the Euler line of isosceles meet the sidelines
and
of
at
and
where
is the midpoint of
Let be the triangle formed by the Euler lines of the
and the line
contains
and parallel to
the vertex
being the intersection of the Euler line of the
and
the vertex
being the intersection of the Euler line of the
and
We call the triangle as the Gossard triangle of
Let be any isosceles triangle and let
be its Gossard triangle. Then the lines
and
are concurrent. We call the point of concurrence
as the Gossard perspector of
Let
be the orthocenter of
be the circumcenter of
It is clear that is the midpoint of
is the midpoint
is the midpoint
with coefficient
Any isosceles triangle and its Gossard triangle are congruent.
Any isosceles triangle and its Gossard triangle have the same Euler line.
The Gossard triangle of the isosceles is the reflection of
in the Gossard perspector.
Denote
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Euler line of the triangle formed by the Euler line and the sides of a given triangle
Let the Euler line of meet the lines
and
at
and
respectively.
Euler line of the is parallel to
Similarly, Euler line of the
is parallel to
Euler line of the
is parallel to
Proof
Denote smaller angles between the Euler line and lines
and
as
and
respectively. WLOG,
It is known that
Let be circumcenter of
be Euler line of
(line).
Similarly,
Suppose,
which means
and
In this case
Similarly one can prove the claim in the other cases.
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Gossard triangle for triangle with angle 60
Let of the triangle
be
Let the Euler line of
meet the lines
and
at points
and
respectively.
Prove that
is an equilateral triangle.
Proof
Denote
It is known that
Therefore
is equilateral triangle.
Let be the triangle formed by the Euler lines of the
and the line
contains centroid
of the
and parallel to
the vertex
being the intersection of the Euler line of the
and
the vertex
being the intersection of the Euler line of the
and
the vertex
being the intersection of the Euler lines of the
and
We call the triangle as the Gossard triangle of the
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Gossard triangle for triangle with angle 120
Let of the triangle
be
Let the Euler line of
meet the lines
and
at points
and
respectively. Then
is an equilateral triangle.
One can prove this claim using the same formulae as in the case
Let be the triangle formed by the Euler lines of the
and the line
contains centroid
of the
and parallel to
the vertex
being the intersection of the Euler line of the
and
the vertex
being the intersection of the Euler line of the
and
the vertex
being the intersection of the Euler lines of the
and
We call the triangle as the Gossard triangle of the
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Gossard perspector
Let non equilateral triangle be given. The Euler line of
crosses lines
and
at points
and
respectively.
Let the point be the centroid of the set of points
Let Gossard triangle be defined as described above.
Prove that and
are homothetic and congruent, and the homothetic center is the point
the Euler line of
coincide with the Euler line of
Proof
Denote and
centroids of the triangles
and
respectively.
It is clear that
Euler line.
Let point be symmetric to the point
with respect to the point
Similarly we define points and
Similarly
and
the crosspoints of lines
and
are symmetric to the crosspoints of lines
and
therefore points
and
are symmetric to points
and
with respect to the point
is the Gossard perspector of the
It is clear that the Gossard perspector lyes on Euler line of the and
is congruent to
.
The Euler line of is symmetric to the Euler line of
with respect to
Therefore these lines coincide.
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Zeeman’s Generalisation
Let be any line parallel to the Euler line of non equilateral triangle
Let
intersect the sidelines
of
at points
respectively. Let
be the triangle formed by the Euler lines (as in previous sections) of the triangles
and
Let the point
be the centroid of the set of points
Then and
are homothetic and congruent, and the homothetic center is the point
the Euler line of
coincide with the line
and the point
is equidistant from the Euler lines.
In this case usually called the Zeeman–Gossard perspector.
One can prove this claim using the method of previous section. vladimir.shelomovskii@gmail.com, vvsss
Paul Yiu's Generalization
Let be any point in the plane of non equilateral triangle
different from its centroid
Let the line meet the sidelines
and
at
and
respectively.
Let the centroids of the triangles and
be
and
respectively.
Let be a point such that
is parallel to
and
is parallel to
Symilarly,
Let be the triangle formed by the lines
and
Let the point
be the centroid of the set of points
Then
and
are homothetic and congruent, and the homothetic center is the point
the points
and
are collinear.
One can prove this claim using the method of previous section.
The points and
are collinear.
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Dao Thanh Oai's Generalization
Let triangle and line
non parallel to sidelines be given. Let line
meets sidelines
of
at points
respectively.
Let Let
be the points such that
Similarly define points
Let triangle be the triangle formed by the lines
Prove that and
are homothetic and congruent, and the homothetic center lies on
Proof
Let and
be the midpoints
and
respectively.
Let be the crosspoint of
and Gauss line
Let and
be the points simmetric to
and
with respect to
respectively.
We will prove that which means that
and so on.
midpoint
midpoint
midpoint
midpoint
according the Claim.
Claim (Parallel lines in trapezium)
Let be the quadrungle such that
Let
Prove that point
lyes on
Proof
We prove Claim in the case non parallel to
Denote
Let cross
at
Then
The Claim is correct in the case of non convex
One can simplify the proof of Dao Generalization using this variant of the Claim.
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