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 $C$ :

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:

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.

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 $F_c\left(\frac{72}{25},\frac{96}{25}\right)$ .

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:

the right triangle is isosceles
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 $C$ 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 $3:1$ .