This is a paragraph.

Kosnita point in a right triangle II

Let be a right triangle in which , is the foot of the altitude from , is the Kosnita point, and is the symmedian point. Then the lengths of the segments form a geometric progression:

Place the vertices of the right triangle at convenient points: , , and . Then:

• is the point ;
• is the point ;
• is the point .
Find the distance from the Kosnita point to the foot of the altitude from .

Using the given coordinates, we find:

Find the distance from the Kosnita point to the symmedian point.

Using the given coordinates, we find:

Find the distance from vertex to the Kosnita point.

Using the given coordinates, we find:

PROVE that .

Follows from

PROVE that .

Follows from

Takeaway

In any right triangle, the following statements are equivalent:

1. the right triangle is isosceles
2. the Kosnita point coincides with the centroid.

Task

• (Foot of the symmedian) Let be a triangle having vertices at , , . VERIFY that:
1. the foot of the symmedian from is
2. the foot of the symmedian from is equidistant from the feet of the altitudes from and .

Kosnita point in a right triangle I

In a right triangle, the Kosnita point:

• is on the altitude through the -vertex
• internally divides this altitude in the ratio .

Construction

Let be the circumcenter of triangle . The Kosnita point of is constructed as follows:

• find the circumcenter of triangle , then join to vertex
• find the circumcenter of triangle , then join to vertex
• find the circumcenter of triangle , then join to vertex

The lines , , concur at the Kosnita point.

If is a right triangle in which , then its circumcenter is the midpoint of , and so one of the three triangles above is degenerate, namely . To somewhat compensate for this, the Kosnita point in a right triangle behaves nicely.

Calculations

Find the Kosnita point of whose vertices are located at , , and .

Observe that the given triangle is right-angled at :

• the circumcenter of is , the midpoint of
• the circumcenter of is , call this
• the circumcenter of is , call this
• the equation of line is
• the equation of line is
• the two lines and meet at .

Thus, the Kosnita point of the given triangle is located at .

has vertices located at , , and . Find the coordinates of the foot of the altitude from .

The same triangle in the previous example:

• the equation of side is
• the equation of the altitude through is
• thus, the foot of the altitude from is .
Given with vertices located at , , and , verify that the Kosnita point lies on the altitude from .

Same triangle as in the previous examples where we obtained as the Kosnita point.

The equation of the altitude through was obtained in the previous example as . The coordinates of the Kosnita point satisfy this equation, since

and so the Kosnita point lies on the altitude through .

Given with vertices located at , , and , verify that the Kosnita point divides the altitude from in the ratio .

We have and .

Explicit calculation gives and . Thus .

Coordinates

Suppose that has vertices located at , , . Then the Kosnita point is . Use this to reobtain the Kosnita point of the triangle in example 1.

In example 1 we were given , , and . So: and . The Kosnita point is then:

as before.

Takeaway

In any right triangle, the following statements are equivalent:

1. the right triangle is isosceles
2. the Kosnita point coincides with the centroid.

Task

• (Harmonic division) Let be a right triangle having vertices at , , . PROVE that:
1. the foot of the altitude from is
2. the symmedian point is
3. the Kosnita point is
4. and divide harmonically in the ratio .