- 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
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(0,6)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-438f8ec0b492067f5f8d4afd1dd5a706_l3.png)
![Rendered by QuickLaTeX.com B(8,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-85aceaf0cf186bb7126f29637e7caa5d_l3.png)
![Rendered by QuickLaTeX.com C(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-36581d8d833c92827ae6e6709908f73d_l3.png)
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 .
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(0,6)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-438f8ec0b492067f5f8d4afd1dd5a706_l3.png)
![Rendered by QuickLaTeX.com B(8,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-85aceaf0cf186bb7126f29637e7caa5d_l3.png)
![Rendered by QuickLaTeX.com C(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-36581d8d833c92827ae6e6709908f73d_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
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
.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(0,6)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-438f8ec0b492067f5f8d4afd1dd5a706_l3.png)
![Rendered by QuickLaTeX.com B(8,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-85aceaf0cf186bb7126f29637e7caa5d_l3.png)
![Rendered by QuickLaTeX.com C(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-36581d8d833c92827ae6e6709908f73d_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
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 .
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(0,6)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-438f8ec0b492067f5f8d4afd1dd5a706_l3.png)
![Rendered by QuickLaTeX.com B(8,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-85aceaf0cf186bb7126f29637e7caa5d_l3.png)
![Rendered by QuickLaTeX.com C(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-36581d8d833c92827ae6e6709908f73d_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com 2:1](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-1f180cff8bc4b45e70adc4f8612f18d8_l3.png)
We have and
.
Explicit calculation gives and
. Thus
.
Coordinates
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(0,2u)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-a8281446e017b75853ec306f765cb52b_l3.png)
![Rendered by QuickLaTeX.com B(2v,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-9d92c46e1060a60b84d2213718038f12_l3.png)
![Rendered by QuickLaTeX.com C(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-36581d8d833c92827ae6e6709908f73d_l3.png)
![Rendered by QuickLaTeX.com \left(\frac{4u^2v}{3(u^2+v^2)},\frac{4uv^2}{3(u^2+v^2)}\right)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-a4f0562b4948b37e522958d9c378cd75_l3.png)
In example 1 we were given ,
, and
. So:
and
. The Kosnita point is then:
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:
- the foot of the altitude from
is
- the symmedian point is
- the Kosnita point is
and
divide
harmonically in the ratio
.
- the foot of the altitude from