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Kosnita point in a right triangle II

Let ABC be a right triangle in which \angle C=90^{\circ}, F_c is the foot of the altitude from C, K is the Kosnita point, and S is the symmedian point. Then the lengths of the segments SK,KF_c, CK form a geometric progression:

    \[KF_c^2=(CK)(SK)\]

Place the vertices of the right triangle at convenient points: A(0,2u), B(2v,0), and C(0,0). Then:

  • F_c is the point F_c\left(\frac{2u^2v}{u^2+v^2},\frac{2uv^2}{u^2+v^2}\right);
  • K is the point \left(\frac{4u^2v}{3(u^2+v^2)},\frac{4uv^2}{3(u^2+v^2)}\right);
  • S is the point \left(\frac{u^2v}{u^2+v^2},\frac{uv^2}{u^2+v^2}\right).

Find the distance from the Kosnita point to the foot of the altitude from C.

Using the given coordinates, we find:

    \[KF_c=\frac{2uv}{3\sqrt{u^2+v^2}}\]

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

Using the given coordinates, we find:

    \[KS=\frac{uv}{3\sqrt{u^2+v^2}}\]

Find the distance from vertex C to the Kosnita point.

Using the given coordinates, we find:

    \[CK=\frac{4uv}{3\sqrt{u^2+v^2}}\]

PROVE that KF_c^2=(CK)(SK).

Follows from

    \[KF_c=\frac{2uv}{3\sqrt{u^2+v^2}},~SK=\frac{uv}{3\sqrt{u^2+v^2}},~CK=\frac{4uv}{3\sqrt{u^2+v^2}}\]

PROVE that CK=2SK.

Follows from

    \[SK=\frac{uv}{3\sqrt{u^2+v^2}},~CK=\frac{4uv}{3\sqrt{u^2+v^2}}.\]

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 ABC be a triangle having vertices at A(-6,0), B(0,0), C(2,4). VERIFY that:
    1. the foot of the symmedian from C is S_c\left(-\frac{6}{5},0\right)
    2. the foot of the symmedian from C is equidistant from the feet of the altitudes from A and B.