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Category: General

Introductory; impromptu.

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.

Kosnita point in a right triangle I

In a right triangle, the Kosnita point:

  • is on the altitude through the 90^{\circ}-vertex
  • internally divides this altitude in the ratio 2:1.

Construction

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

  • find the circumcenter O_a of triangle BCO, then join O_a to vertex A
  • find the circumcenter O_b of triangle CAO, then join O_b to vertex B
  • find the circumcenter O_c of triangle ABO, then join O_c to vertex C

The lines AO_a, BO_b, CO_c concur at the Kosnita point.

If \triangle ABC is a right triangle in which \angle C=90^{\circ}, then its circumcenter O is the midpoint of AB, and so one of the three triangles above is degenerate, namely \triangle ABO. To somewhat compensate for this, the Kosnita point in a right triangle behaves nicely.

Calculations

Find the Kosnita point of \triangle ABC whose vertices are located at A(0,6), B(8,0), and C(0,0).

Observe that the given triangle is right-angled at C:

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  • the circumcenter of \triangle ABC is O(4,3), the midpoint of AB
  • the circumcenter of \triangle BCO is O\left(4,-\frac{7}{6}\right), call this O_a
  • the circumcenter of \triangle CAO is O\left(\frac{7}{8},3\right), call this O_b
  • the equation of line AO_a is y=-\frac{43}{24}x+6
  • the equation of line BO_b is y=-\frac{8}{19}x+\frac{64}{19}
  • the two lines AO_a and BO_b meet at \left(\frac{48}{25},\frac{64}{25}\right).

Thus, the Kosnita point of the given triangle is located at K\left(\frac{48}{25},\frac{64}{25}\right).

\triangle ABC has vertices located at A(0,6), B(8,0), and C(0,0). Find the coordinates of the foot of the altitude from C.

The same triangle in the previous example:

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  • the equation of side AB is y=-\frac{3}{4}x+6
  • the equation of the altitude through C is y=\frac{4}{3}x
  • thus, the foot of the altitude from C is F_c\left(\frac{72}{25},\frac{96}{25}\right).

Given \triangle ABC with vertices located at A(0,6), B(8,0), and C(0,0), verify that the Kosnita point lies on the altitude from C.

Same triangle as in the previous examples where we obtained K\left(\frac{48}{25},\frac{64}{25}\right) as the Kosnita point.

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The equation of the altitude through C was obtained in the previous example as y=\frac{4}{3}x. The coordinates of the Kosnita point K\left(\frac{48}{25},\frac{64}{25}\right) satisfy this equation, since

    \[\frac{64}{25}=\frac{4}{3}\times\frac{48}{25}\]

and so the Kosnita point lies on the altitude through C.

Given \triangle ABC with vertices located at A(0,6), B(8,0), and C(0,0), verify that the Kosnita point divides the altitude from C in the ratio 2:1.

We have K\left(\frac{48}{25},\frac{64}{25}\right) and F_c\left(\frac{72}{25},\frac{96}{25}\right).

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Explicit calculation gives CK=\frac{16}{5} and KF_c=\frac{8}{5}. Thus CK:KF_c=2:1.

Coordinates

Suppose that \triangle ABC has vertices located at A(0,2u), B(2v,0), C(0,0). Then the Kosnita point is \left(\frac{4u^2v}{3(u^2+v^2)},\frac{4uv^2}{3(u^2+v^2)}\right). Use this to reobtain the Kosnita point of the triangle in example 1.

In example 1 we were given A(0,6), B(8,0), and C(0,0). So: u=3 and v=4. The Kosnita point is then:

    \[\left(\frac{4u^2v}{3(u^2+v^2)},\frac{4uv^2}{3(u^2+v^2)}\right)=\left(\frac{4\times 3^2\times 4}{3(3^2+4^2)},\frac{4\times 3\times 4^2}{3(3^2+4^2)}\right)=\left(\frac{48}{25},\frac{64}{25}\right)\]

as before.

Caption this:

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 ABC be a right triangle having vertices at A(0,2u), B(2v,0), C(0,0). PROVE that:
    1. the foot of the altitude from C is F_c\left(\frac{2u^2v}{(u^2+v^2)},\frac{2uv^2}{(u^2+v^2)}\right)
    2. the symmedian point is S\left(\frac{u^2v}{(u^2+v^2)},\frac{uv^2}{(u^2+v^2)}\right)
    3. the Kosnita point is \left(\frac{4u^2v}{3(u^2+v^2)},\frac{4uv^2}{3(u^2+v^2)}\right)
    4. S and F_c divide CK harmonically in the ratio 3:1.