![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com H](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d2121613062e686e294d3867f5da2955_l3.png)
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
Last time we saw that the circle with diameter passes through vertex
if and only if the equation below holds:
(1)
Let this circle also pass through vertex . Then the nine-point center has to be
.
Updated equivalence
We now add the following four equivalent statements to what we had at the end of our post on August 14:
- a circle with diameter
passes through
and
- the nine-point center coincides with vertex
- the reflection of
over
is
- the reflection of
over
is
Only triangle having
satisfies these.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com OH](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-66e39f97fc7017c68984ed04a1f092fa_l3.png)
![Rendered by QuickLaTeX.com A](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3404ae68a19d10407c93ea3824613e4e_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3f8c40a7f801883b444525b6eef00398_l3.png)
Since this circle passes through vertex , our previous post shows that equation (1) is satisfied:
Further, with as diameter, the two triangles
and
are right triangles with
and
.
By a result in this post, we conclude that the nine-point center is .
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3f8c40a7f801883b444525b6eef00398_l3.png)
![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com AC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-2509ee19f0ee692f93e2b93071b678f5_l3.png)
![Rendered by QuickLaTeX.com B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3f8c40a7f801883b444525b6eef00398_l3.png)
Easy.
![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com AC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-2509ee19f0ee692f93e2b93071b678f5_l3.png)
![Rendered by QuickLaTeX.com B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3f8c40a7f801883b444525b6eef00398_l3.png)
![Rendered by QuickLaTeX.com H](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d2121613062e686e294d3867f5da2955_l3.png)
![Rendered by QuickLaTeX.com BC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c4f3ce61859dcd00e52fbf771b86cb55_l3.png)
![Rendered by QuickLaTeX.com A](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3404ae68a19d10407c93ea3824613e4e_l3.png)
Easy.
![Rendered by QuickLaTeX.com H](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d2121613062e686e294d3867f5da2955_l3.png)
![Rendered by QuickLaTeX.com BC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c4f3ce61859dcd00e52fbf771b86cb55_l3.png)
![Rendered by QuickLaTeX.com A](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3404ae68a19d10407c93ea3824613e4e_l3.png)
![Rendered by QuickLaTeX.com OH](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-66e39f97fc7017c68984ed04a1f092fa_l3.png)
![Rendered by QuickLaTeX.com A](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3404ae68a19d10407c93ea3824613e4e_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
Easy.
Usual example
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(-4,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-bc3f15725e0a4923e8ba6e2df2ad7df3_l3.png)
![Rendered by QuickLaTeX.com B(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-7045877de833cb7e53ba1ef4ddfaeb24_l3.png)
![Rendered by QuickLaTeX.com C(2,2\sqrt{3})](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-51f4dd5600aa397f923da5141eeb676f_l3.png)
![Rendered by QuickLaTeX.com B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3f8c40a7f801883b444525b6eef00398_l3.png)
The diagram below shows the circumcircle, the nine-point circle, and the circle with diameter .
Or this one:
Takeaway
Consider with side-lengths
, circumradius
, circumcenter
, orthocenter
, and nine-point center
. If equation (1) is satisfied, then the following statements are equivalent:
- a circle with diameter
passes through
and
and
are both equilateral
is tangent to the circumcircle at
- the orthic triangle is equilateral
- the reflection of
over
is
- the reflection of
over
is
is equilateral
coincides with
These are some of the many equivalent descriptions of the isosceles triangle in which
.
Task
- (Early fifties) In a non-right triangle
, let
be the side-lengths,
the altitudes,
the feet of the altitudes from the respective vertices,
the circumradius,
the circumcenter,
the nine-point center,
the orthocenter,
the midpoint of side
, and
the reflection of
over side
. PROVE that the following fifty-two statements are equivalent:
is congruent to
is isosceles with
is isosceles with
is right angled at
is the circumcenter of
is right-angled at
is right-angled at
- quadrilateral
is a rectangle
- the points
are concyclic with
as diameter
- the reflection of
over
lies internally on
- the reflection of
over
lies externally on
- radius
is parallel to side
is the reflection of
over side
- the nine-point center lies on
- the orthic triangle is isosceles with
- the geometric mean theorem holds
- the bisector of
has length
, where
- the orthocenter is a reflection of vertex
over side
- segment
is tangent to the circumcircle at point
- median
has the same length as the segment
- the bisector
of
is tangent to the nine-point circle at
is a convex kite with diagonals
and
- altitude
is tangent to the nine-point circle at
- segment
is tangent to the nine-point circle at
.
(short of the target.)
- (Extra feature) If
satisfies equation (??), PROVE that its nine-point center
divides
in the ratio
.