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Gossard perspector

Gossard perspector X(402) and Gossard triangle

In $1765$ 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 $1916$ 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 $X(402).$ It is the crosspoint of Gauss line and Euler line.

Let triangle $\triangle ABC$ be given. The Euler line crosses lines $AB, BC,$ and $AC$ at points $D, E,$ and $F.$

On $13/01/2023$ it was found that the Gossard perspector is the centroid of the points $A, B, C, D, E, F.$

vladimir.shelomovskii@gmail.com, vvsss

Gossard perspector of right triangle

Gossard 90.png

It is clear that the Euler line of right triangle $ABC (\angle A = 90 ^\circ)$ meet the sidelines $BC, CA$ and $AB$ of $\triangle ABC$ at $A'$ and $A,$ where $A'$ is the midpoint of $BC.$

Let $\triangle A'B'C'$ be the triangle formed by the Euler lines of the $\triangle AA'B, \triangle AA'C,$ and the line $l$ contains $A$ and parallel to $BC,$ the vertex $B'$ being the intersection of the Euler line of the $\triangle AA'C$ and $l,$ the vertex $C'$ being the intersection of the Euler line of the $\triangle AA'B$ and $l.$

We call the triangle $\triangle A'B'C'$ as the Gossard triangle of $\triangle ABC.$

Let $\triangle ABC$ be any right triangle and let $\triangle A'B'C'$ be its Gossard triangle. Then the lines $AA', BB',$ and $CC'$ are concurrent. We call the point of concurrence $Go$ as the Gossard perspector of $\triangle ABC.$

$Go$ is the midpoint of $AA'.$ $A$ is orthocenter of $\triangle ABC, A'$ is circumcenter of $\triangle ABC,$ so $Go$ is midpoint of $OH.$

$M = A'C' \cap AB$ is the midpoint $AB, N = A'B' \cap AC$ is the midpoint $AC, \triangle A'BM = \triangle CNA' \sim \triangle CBA$ with coefficient $k = \frac {1}{2}.$

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 $\triangle ABC$ is the reflection of $\triangle ABC$ in the Gossard perspector.

vladimir.shelomovskii@gmail.com, vvsss

Gossard perspector and Gossard triangle for isosceles triangle

Gossard equilateral.png

It is clear that the Euler line of isosceles $\triangle ABC (AB = AC)$ meet the sidelines $BC, CA$ and $AB$ of $\triangle ABC$ at $A'$ and $A,$ where $A'$ is the midpoint of $BC.$

Let $\triangle A'B'C'$ be the triangle formed by the Euler lines of the $\triangle AA'B, \triangle AA'C,$ and the line $l$ contains $A$ and parallel to $BC,$ the vertex $B'$ being the intersection of the Euler line of the $\triangle AA'C$ and $l,$ the vertex $C'$ being the intersection of the Euler line of the $\triangle AA'B$ and $l.$

We call the triangle $\triangle A'B'C'$ as the Gossard triangle of $\triangle ABC.$

Let $\triangle ABC$ be any isosceles triangle and let $\triangle A'B'C'$ be its Gossard triangle. Then the lines $AA', BB',$ and $CC'$ are concurrent. We call the point of concurrence $Go$ as the Gossard perspector of $\triangle ABC.$ Let $H$ be the orthocenter of $\triangle ABC, O$ be the circumcenter of $\triangle ABC.$

It is clear that $Go$ is the midpoint of $AA'.$ $M = A'C' \cap AB$ is the midpoint $AB, N = A'B' \cap AC$ is the midpoint $AC.$

$\triangle A'BM = \triangle CNA' \sim \triangle CBA$ with coefficient $k = \frac {1}{2}.$

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 $\triangle ABC$ is the reflection of $\triangle ABC$ in the Gossard perspector. Denote $\angle BAC = \alpha, \angle ABC = \angle ACB = \beta, AO = BO = R \implies$ \[\sin \beta = \cos \frac {\alpha}{2}, GoO = AO – AGo = R \cdot \frac{1 – \cos {\alpha}}{2},\] \[OH = AH – AO = R(2 \cos \alpha – 1) \implies \vec {GoO} = \vec {OH} \cdot \frac {1 – \cos \alpha}{2(2 \cos \alpha – 1)}.\]

vladimir.shelomovskii@gmail.com, vvsss

Euler line of the triangle formed by the Euler line and the sides of a given triangle

Euler Euler line.png

Let the Euler line of $\triangle ABC$ meet the lines $AB, AC,$ and $BC$ at $D, E,$ and $F,$ respectively.

Euler line of the $\triangle ADE$ is parallel to $BC.$ Similarly, Euler line of the $\triangle BDF$ is parallel to $AC,$ Euler line of the $\triangle CEF$ is parallel to $AB.$

Proof

Denote $\angle A = \alpha, \angle B = \beta, \angle C = \gamma,$ smaller angles between the Euler line and lines $BC, AC,$ and $AB$ as $\theta_A, \theta_B,$ and $\theta_C,$ respectively. WLOG, $AC > BC > AB.$ It is known that $\tan \theta_A = \frac{3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma}, \tan \theta_B = \frac{3 – \tan \alpha \cdot \tan \gamma}{\tan \alpha – \tan \gamma}, \tan \theta_C = \frac{3 – \tan \beta \cdot \tan \alpha}{\tan \beta – \tan \alpha}.$

Let $O'$ be circumcenter of $\triangle ADE, KO'$ be Euler line of $\triangle ADE, K \in DE$ (line).

Similarly, $\tan \angle O'KF = \frac{3 – \tan \theta_B \cdot \tan \theta_C}{\tan \theta_C – \tan \theta_B}.$ \[3(\tan\alpha – \tan \gamma) (\tan\alpha – \tan \beta) – (3 – \tan \alpha \cdot \tan \gamma) (3 – \tan \alpha \cdot \tan \beta) = (\tan^2 \alpha – 3) \cdot (3 –\tan \beta \cdot \tan \gamma),\] \[(3 – \tan \alpha \cdot \tan \gamma) \cdot (\tan\alpha – \tan \beta) – (3 – \tan \alpha \cdot \tan \beta) \cdot (\tan\alpha – \tan \gamma) = (\tan^2 \alpha – 3) \cdot (\tan \beta – \tan \gamma).\] Suppose, $\tan^2 \alpha \ne 3$ which means $\alpha \ne 60^\circ$ and $\alpha \ne 120^\circ.$ In this case \[\tan \angle O'KF = \frac{3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma} = \tan \theta_A \implies \angle O'KF = \theta_A \implies O'K||BC.\]

Similarly one can prove the claim in the other cases.

vladimir.shelomovskii@gmail.com, vvsss

Gossard triangle for triangle with angle 60

Gossard 60.png

Let $\angle A$ of the triangle $ABC$ be $60^\circ, \angle B \ne 60^\circ.$ Let the Euler line of $\triangle ABC$ meet the lines $AB, AC$ and $BC$ at points $D, E,$ and $F,$ respectively. Prove that $\triangle ADE$ is an equilateral triangle.

Proof

Denote $\angle ABC = \beta, \angle ACB = \gamma.$ It is known that \[\tan \angle AED = \tan \theta_B = \frac{3 – \tan \alpha \cdot \tan \gamma}{\tan \alpha – \tan \gamma}.\]

\[\tan \angle AED = \frac{3 – \sqrt{3} \tan \gamma}{\sqrt{3} – \tan \gamma} = \sqrt{3} \implies \angle AED = 60^\circ.\] Therefore $\triangle ADE$ is equilateral triangle.

Let $\triangle A'B'C'$ be the triangle formed by the Euler lines of the $\triangle BDF, \triangle CEF,$ and the line $l$ contains centroid $G$ of the $\triangle ADE$ and parallel to $BC,$ the vertex $B'$ being the intersection of the Euler line of the $\triangle CEF$ and $l,$ the vertex $C'$ being the intersection of the Euler line of the $\triangle BDF$ and $l,$ the vertex $A'$ being the intersection of the Euler lines of the $\triangle BDF$ and $\triangle CEF.$

We call the triangle $\triangle A'B'C'$ as the Gossard triangle of the $\triangle ABC.$

vladimir.shelomovskii@gmail.com, vvsss

Gossard triangle for triangle with angle 120

Gossard 120.png

Let $\angle A$ of the triangle $ABC$ be $120^\circ, \angle B \ne 30^\circ.$ Let the Euler line of $\triangle ABC$ meet the lines $AB, AC$ and $BC$ at points $D, E,$ and $F,$ respectively. Then $\triangle ADE$ is an equilateral triangle.

One can prove this claim using the same formulae as in the case $\angle A = 60^\circ.$

Let $\triangle A'B'C'$ be the triangle formed by the Euler lines of the $\triangle BDF, \triangle CEF,$ and the line $l$ contains centroid $G$ of the $\triangle ADE$ and parallel to $BC,$ the vertex $B'$ being the intersection of the Euler line of the $\triangle CEF$ and $l,$ the vertex $C'$ being the intersection of the Euler line of the $\triangle BDF$ and $l,$ the vertex $A'$ being the intersection of the Euler lines of the $\triangle BDF$ and $\triangle CEF.$

We call the triangle $\triangle A'B'C'$ as the Gossard triangle of the $\triangle ABC.$

vladimir.shelomovskii@gmail.com, vvsss

Gossard perspector

Gossard complete.png

Let non equilateral triangle $ABC$ be given. The Euler line of $\triangle ABC$ crosses lines $AB, BC,$ and $AC$ at points $D, E,$ and $F,$ respectively.

Let the point $X$ be the centroid of the set of points $A, B, C, D, E, F.$

Let Gossard triangle $A'B'C'$ be defined as described above.

Prove that $\triangle A'B'C'$ and $\triangle ABC$ are homothetic and congruent, and the homothetic center is the point $X,$ the Euler line of $\triangle A'B'C'$ coincide with the Euler line of $\triangle ABC.$

Proof

Denote $G_A, G_B,$ and $G_C$ centroids of the triangles $ADE, BDF,$ and $CEF,$ respectively. It is clear that $G_A \in B'C', G_B \in A'C', G_C \in A'B', X \in$ Euler line.

Let point $G'_A$ be symmetric to the point $G_A$ with respect to the point $X.$

Similarly we define points $G'_B$ and $G'_C.$ \[\vec {G'_A} = 2 \vec X – \vec G_A = \frac {\vec A+\vec B+\vec C+ \vec D+\vec E + \vec F}{3} – \frac {\vec A+ \vec D+\vec E }{3} = \frac {\vec B+\vec C+ \vec F}{3} \in BC.\] Similarly $G'_B \in AC$ and $G'_C \in AB.$

$AB||A'B', AC || A'C', BC|| B'C' \implies$ the crosspoints of lines $A'B', A'C',$ and $B'C'$ are symmetric to the crosspoints of lines $AB, AC,$ and $BC,$ therefore points $A', B',$ and $C',$ are symmetric to points $A, B,$ and $C$ with respect to the point $X \implies X$ is the Gossard perspector of the $\triangle ABC.$

It is clear that the Gossard perspector lyes on Euler line of the $\triangle ABC$ and $\triangle A'B'C'$ is congruent to $\triangle ABC$.

The Euler line of $\triangle A'B'C'$ is symmetric to the Euler line of $\triangle ABC$ with respect to $X.$ Therefore these lines coincide.

vladimir.shelomovskii@gmail.com, vvsss

Zeeman’s Generalisation

Generalization 1.png

Let $l$ be any line parallel to the Euler line of non equilateral triangle $ABC.$ Let $l$ intersect the sidelines $AB, CA, BC$ of $\triangle ABC$ at points $D, E, F,$ respectively. Let $\triangle A'B'C'$ be the triangle formed by the Euler lines (as in previous sections) of the triangles $\triangle ADE, \triangle BDF,$ and $\triangle CEF.$ Let the point $X$ be the centroid of the set of points $A, B, C, D, E, F.$

Then $\triangle A'B'C'$ and $\triangle ABC$ are homothetic and congruent, and the homothetic center is the point $X,$ the Euler line of $\triangle A'B'C'$ coincide with the line $DE$ and the point $X$ is equidistant from the Euler lines.

In this case $X$ 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

Yu generalization.png

Let $P$ be any point in the plane of non equilateral triangle $ABC$ different from its centroid $G.$

Let the line $PG$ meet the sidelines $AB, CA,$ and $BC$ at $D, E,$ and $D,$ respectively.

Let the centroids of the triangles $AED, BDF,$ and $CFE$ be $Ga, Gb,$ and $Gc,$ respectively.

Let $Pa$ be a point such that $EPa$ is parallel to $CP$ and $DPa$ is parallel to $BP.$ Symilarly, $Pb: DPb||AP, FPb||CP, Pc: FPc||BP, EPc||AP.$

Let $\triangle A'B'C'$ be the triangle formed by the lines $GaPa, GbPb,$ and $GcPc.$ Let the point $X$ be the centroid of the set of points $A, B, C, D, E, F.$ Then $\triangle A'B'C'$ and $\triangle ABC$ are homothetic and congruent, and the homothetic center is the point $X,$ the points $P, G,$ and $G'$ are collinear.

One can prove this claim using the method of previous section.

The points $P, Pa, Pb,$ and $Pc$ are collinear.

vladimir.shelomovskii@gmail.com, vvsss

Dao Thanh Oai's Generalization

Generalization.png

Let triangle $ABC$ and line $\ell,$ non parallel to sidelines be given. Let line $\ell$ meets sidelines $AB, AC, BC$ of $\triangle ABC$ at points $D, E, F,$ respectively.

Let $Q \in \ell.$ Let $K, L, M$ be the points such that $QA||EM||DL, QB||KD||FM, QC||KE||FL.$

Similarly define points $P \in \ell, K_0, L_0, M_0.$

Let triangle $\triangle A'B'C'$ be the triangle formed by the lines $KK_0, LL_0, MM_0.$

Prove that $\triangle A'B'C'$ and $\triangle ABC$ are homothetic and congruent, and the homothetic center lies on $\ell.$

Proof

Dao Generalization.png

Let $G$ and $G_0$ be the midpoints $BE$ and $CD,$ respectively.

Let $X$ be the crosspoint of $\ell$ and Gauss line $GG_0.$

Let $A', B', C', D', E',$ and $K'$ be the points simmetric to $A, B, C, D, E,$ and $K$ with respect to $X,$ respectively.

We will prove that $K' \in BC$ which means that $K \in B'C' \implies K_0 \in BC, L \in AC$ and so on.

$G$ midpoint $BE, X$ midpoint $EE' \implies GX||BE'.$

$G_0$ midpoint $DC, X$ midpoint $DD' \implies G_0X||CD' \implies BE'||CD'.$ $KD||K'D'||BQ, KE||K'E'||CQ, CD'||BE', Q \in D'E' \implies K' \in BC$ according the Claim.

Claim (Parallel lines in trapezium)

Pappus.png
Pappus non convex.png

Let $ABCD$ be the quadrungle such that $AD||BC.$ Let $M \in AB, MD||BN, AN||CM.$ Prove that point $N$ lyes on $CD.$

Proof

We prove Claim in the case $AB$ non parallel to $CD.$ Denote $Q = AB \cap CD.$

$AD||BC \implies \frac {QB}{QA}= \frac{QC}{QD}.$

Let $BN||MD$ cross $CD$ at $N.$ Then $\frac {QN}{QD}=\frac {QB}{QM}.$

\[\frac {QC}{QN} = \frac {QC}{QD} \cdot \frac {QD}{QN} = \frac {QB}{QA} \cdot \frac {QM}{QB} = \frac {QM}{QA} \implies AN||CM.\] The Claim is correct in the case of non convex $ABCD.$ One can simplify the proof of Dao Generalization using this variant of the Claim.

vladimir.shelomovskii@gmail.com, vvsss

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