![Rendered by QuickLaTeX.com ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b5002f05fd90c80f81a9e0e9c845e02d_l3.png)
![Rendered by QuickLaTeX.com \angle A=\angle C=30^{\circ},\angle B=120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d5db8b07deae978718d8dd3e4a643a5c_l3.png)
![Rendered by QuickLaTeX.com A](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3404ae68a19d10407c93ea3824613e4e_l3.png)
![Rendered by QuickLaTeX.com BC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c4f3ce61859dcd00e52fbf771b86cb55_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com AB](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0b7694a0a3a6ff6caddf255cb508f628_l3.png)
If we now reflect vertex over side
we obtain the circumcenter.
It follows, in the case of the -triangle, that the reflection triangle is degenerate; in addition, it degenerates to a special line connecting the centroid, circumcenter and orthocenter, known as the Euler line.
Is the -triangle the only triangle whose reflection triangle degenerates to the Euler line?
![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=\angle C=30^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-408e55b18738970715ba119fa33d44b9_l3.png)
![Rendered by QuickLaTeX.com \angle B=120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-09aa04cb2daf52ea1c2a9cb2b19f19f8_l3.png)
![Rendered by QuickLaTeX.com A](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3404ae68a19d10407c93ea3824613e4e_l3.png)
![Rendered by QuickLaTeX.com BC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c4f3ce61859dcd00e52fbf771b86cb55_l3.png)
To see how this holds, we’ll show that the length of the altitude through
is half the distance from
to the orthocenter
. Indeed:
and
This shows that the reflection of vertex over side
is the orthocenter
.
![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=\angle C=30^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-408e55b18738970715ba119fa33d44b9_l3.png)
![Rendered by QuickLaTeX.com \angle B=120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-09aa04cb2daf52ea1c2a9cb2b19f19f8_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com AB](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0b7694a0a3a6ff6caddf255cb508f628_l3.png)
Since , we have that
and
. Thus, the reflection of vertex
over side
is also the orthocenter.
![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=\angle C=30^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-408e55b18738970715ba119fa33d44b9_l3.png)
![Rendered by QuickLaTeX.com \angle B=120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-09aa04cb2daf52ea1c2a9cb2b19f19f8_l3.png)
![Rendered by QuickLaTeX.com B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3f8c40a7f801883b444525b6eef00398_l3.png)
![Rendered by QuickLaTeX.com CA](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-250879d8959d81e68a48d0fcd4f28fdf_l3.png)
To see this, note that since triangle is isosceles with
as the apex, the circumcenter lies on the axis of symmetry through
. We just have to show that the radius through
is twice the altitude through
. The altitude through
is:
![Rendered by QuickLaTeX.com 30^{\circ}-30^{\circ}-120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0a10497c2335a5cd89bea71b5766548a_l3.png)
We’ve already seen this by explicitly determining the reflection triangle from the preceding three examples. Alternatively, the necessary and sufficient condition for a degenerate reflection triangle is . This condition holds in the case of the
-triangle:
![Rendered by QuickLaTeX.com 30^{\circ}-30^{\circ}-120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0a10497c2335a5cd89bea71b5766548a_l3.png)
Let be the circumradius of the triangle. If the reflection of vertex
over side
is the orthocenter, then we must have
, as per the equivalent statements here. Similarly, if the reflection of vertex
over side
is the orthocenter, then we must have
. These two relations yield
. Since
is also equivalent to
, we obtain
. Thus, the triangle is the
-triangle.
Takeaway
In any triangle , let
be the side-lengths,
the orthocenter,
the circumcenter, and
the circumradius. Then the following statements are equivalent:
- the reflection of
over
is
- the reflection of
over
is
and the reflection of
over
is
Task
- (Aufbau) In triangle
, let
be the side-lengths,
the circumradius,
the circumcenter,
the nine-point center, and
the orthocenter. PROVE that the following statements are equivalent:
- the reflection of
over
is
- radius
is parallel to side
and radius
is parallel to side
- the reflection of
over
is
and the reflection of
over
is