# Gossard perspector

## 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.
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## 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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