# Kimberling’s point X(24)

## 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**