If we now reflect the circumcenter over the sides of the triangle, the reflections do not lie on the circumcircle, unless a certain condition is met.
Sixty degrees
In the presence of an interior angle of or
, the reflection of the circumcenter lies on the circumcircle.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com O^a](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-67e5e14447698c4cf3cf6e13bc91c292_l3.png)
![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com BC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c4f3ce61859dcd00e52fbf771b86cb55_l3.png)
![Rendered by QuickLaTeX.com O^a](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-67e5e14447698c4cf3cf6e13bc91c292_l3.png)
![Rendered by QuickLaTeX.com \angle A=60^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-f31c68b74aa95f2a13158c53688dc4cc_l3.png)
![Rendered by QuickLaTeX.com \angle A=120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b0c144f290b1040367479152aa05c09a_l3.png)
Let be the circumradius of the triangle, and let
be its orthocenter. In the diagram below
suppose that actually lies on the circumcircle. Then
must equal
, right? We showed previously that
in any triangle with orthocenter
. But then
always equals
. And so:
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com \angle A=60^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-f31c68b74aa95f2a13158c53688dc4cc_l3.png)
![Rendered by QuickLaTeX.com \angle A=120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b0c144f290b1040367479152aa05c09a_l3.png)
![Rendered by QuickLaTeX.com BC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c4f3ce61859dcd00e52fbf771b86cb55_l3.png)
As before, let’s denote by the reflection of the circumcenter
over side
. We’ll establish the result for the
case.
The angle which the minor arc through and
subtends at the center is equal to
; that is, obtuse
, and so reflex
. Note that
is a rhombus, and so obtuse
.
Since reflex , we conclude that
lies on the circumcircle.
Simple deduction
As a consequence of the preceding examples, we can characterize equilateral triangles via reflections of the circumcenter.
Easy.
Sample diagrams
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-a4c057a4f58741a4e3fc4d5121ae3540_l3.png)
![Rendered by QuickLaTeX.com B(4,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c59102c57471d38c158f44677f558f39_l3.png)
![Rendered by QuickLaTeX.com C(1,\sqrt{3})](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d2c6ce980b5b985e11dd35d893c7fc20_l3.png)
Note that this is a right triangle in which ,
, and
. The circumcenter is the point
, and its reflection over side
is the point
.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-a4c057a4f58741a4e3fc4d5121ae3540_l3.png)
![Rendered by QuickLaTeX.com B(-4,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d8f0e38b842ca113d3ad7b5c43da44da_l3.png)
![Rendered by QuickLaTeX.com C(2,2\sqrt{3})](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-51f4dd5600aa397f923da5141eeb676f_l3.png)
Here, , and
. An extremely pleasant case.
Takeaway
In , let
be the circumcenter, and let
be the reflection of
over side
. Then the following statements are equivalent:
.
And a few more, if you view this post.
Task
- (Late 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
,
the reflection of
over side
,
the reflection of
over side
, and
the reflection of
over side
. PROVE that the following fifty-seven statements are equivalent:
is congruent to
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
.