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# From reflection triangle to Euler line

If we begin with triangle in which and reflect vertex over side , we obtain the orthocenter. If we reflect vertex over side we obtain the orthocenter again.

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?

In , let and . PROVE that the reflection of vertex over side coincides with the orthocenter.

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 .

In , let and . PROVE that the reflection of vertex over side coincides with the orthocenter.

Since , we have that and . Thus, the reflection of vertex over side is also the orthocenter.

In , let and . PROVE that the reflection of vertex over side coincides with the circumcenter.

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:

PROVE that the -triangle has a degenerate reflection triangle.

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:

If a triangle has the property that the reflections of two of its vertices over the corresponding opposite sides coincide with the orthocenter, PROVE that such a triangle must be the -triangle.

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:

1. the reflection of over is
2. the reflection of over is and the reflection of over is