# Spieker center

The Spieker center is defined as the center of mass of the perimeter of the triangle. The Spieker center of a is the center of gravity of a homogeneous wire frame in the shape of The Spieker center is a triangle center and it is listed as the point

## Contents

## Incenter of medial triangle

Prove that the Spieker center of triangle is the incenter of the medial triangle of a

**Proof**

Let's hang up the in the middle of side Side is balanced.

Let's replace side with point (the center of mass of the midpoint Denote the linear density of a homogeneous wire frame.

The mass of point is equal to the shoulder of the gravity force is

The moment of this force is

Similarly the moment gravity force acting on AB is

Therefore, equilibrium condition is and the center of gravity of a homogeneous wire frame lies on each bisector of

This point is the incenter of the medial triangle

**vladimir.shelomovskii@gmail.com, vvsss**

## Intersection of three cleavers

Prove that the Spieker center is located at the intersection of the three cleavers of triangle. A cleaver of a triangle is a line segment that bisects the perimeter of the triangle and has one endpoint at the midpoint of one of the three sides.

**Proof**

We use notation of previous proof. is the segment contains the Spieker center, WLOG, Similarly,

So is cleaver.

Therefore, the three cleavers meet at the Spieker center.

**vladimir.shelomovskii@gmail.com, vvsss**

## Radical center of excircles

Prove that the Spieker center of triangle is the radical center of the three excircles.

**Proof**

Let be given, be the midpoints of respectively.

Let be A-excircle, B-excircle, C-excircle centered at respectively.

Let be the incenter of Let be the radical axis of and be the radical axis of and be the radical axis of and respectively.

It is known that the distances from to the tangent points of is equal to the distances from to the tangent points of therefore lies on the radical axis of and Similarly,

is cleaver. Similarly, and are cleavers.

Therefore the radical center of the three excircles coinside with the intersection of the three cleavers of triangle.

**vladimir.shelomovskii@gmail.com, vvsss**

## Nagel line

Let points be the incenter, the centroid and the Spieker center of triangle respectively. Prove that points are collinear, and the barycentric coordinates of S are

The Nagel line is the line on which points and Nagel point lie.

**Proof**

Let be the midpoints of respectively. Bisector is parallel to cleaver Centroid divide the median such that

and points are collinear.

The barycentric coordinates of are The barycentric coordinates of are

**vladimir.shelomovskii@gmail.com, vvsss**

## Shatunov triangle

Let be given. Let be incircle, A-excircle, B-excircle, C-excircle centered at points respectively.

Let be the radical axes of the inscribed circle and one of the excircles of

The triangle whose sides are we name the Shatunov triangle. Accordingly, the vertices of the Shatunov triangle are the radical centers of a pair of excircles and an inscribed circle.

Prove:

a) the heights of the Shatunov triangle lie on the bisectors of the medial triangle. The orthocenter of the Shatunov triangle is the Steiner point of

b) The Shatunov triangle is homothetic to the anticomplementary triangle of with respect to the centroid with coefficient

**Proof**

a) Let be the midpoints of respectively.

The distances from to the tangent points of and are the same, so Similarly

Let and be the points of tangency of and and respectively.

It is clear that lies on the radical axis

Similarly, lies on the radical axis lies on the radical axis

D is the radical center of

Similarly

Therefore are the heights of is the orthocenter of

is the medial triangle of is the bisector of is the Steiner point of

b)

is the orthocenter of is the orthocenter of

Points and where is the centroid are collinear, sides of is the are parallel to the respective sides of is homothetic to the with respect to

the coefficient of homothety is

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