# The Euler line as a diameter

There is this orthocentroidal circle, whose definition is a circle having as diameter the line segment joining the orthocenter with the centroid.

Then there’s this modification we want to make. Consider instead a circle having as diameter the line segment connecting the orthocenter with the circumcenter (the Euler line). This circle passes through vertex of a parent triangle if, and only if, the side-lengths satisfy

(1)

The reflection of the circumcenter over side also passes through this circle, under the above equivalence.

In , let and be the circumcenter and orthocenter, respectively. Let be the reflection of over . PROVE that . (Similarly, if were the reflection of over , then ; and if were the reflection of over , then .)

Let be the circumradius of the circle. In the diagram below

is the midpoint of and is the reflection of over , and so . Since and , we have:

In , let and be the circumcenter and orthocenter, respectively. Let be the reflection of over . PROVE that .

In the diagram below, is the midpoint of , and so is a median in triangle .

So:

Similarly, is the midpoint of , and again is a median in triangle :

In , let and be the circumcenter and orthocenter, respectively. Let be the reflection of over . PROVE that the quadrilateral is a parallelogram.

This follows from the two preceding examples, since we now have and .

#### (Main goal)

In , let and be the circumcenter and orthocenter, respectively. Let be the reflection of over . PROVE that the quadrilateral is a rectangle, if the side-lengths of satisfy equation (1).

If equation (1) is satisfied, then from one of the equivalent statements here, the segment is perpendicuclar to the radius .

Since the quadrilateral is a parallelogram, it follows that it is a rectangle. Consequently, the points are concyclic. And the circle through these four points shares the same center with the nine-point center of the parent triangle.

In , let and be the circumcenter and orthocenter, respectively. Let be the reflection of over . If the points are concyclic with as diameter, PROVE that equation (1) holds.

Since is a diameter, the triangle is right-angled at . By one of the equivalent statements here, we conclude that equation (1) holds.

## Takeaway

In a non-right , let be the side-lengths, the orthocenter, the circumcenter, and the reflection of the circumcenter over side . Then the following statements are equivalent:

1. is right-angled at
2. is right-angled at
4. points are concyclic with as diameter.

• (Half century) 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 statements are equivalent:
1. is congruent to
2. is isosceles with
3. is isosceles with
4. is right angled at
5. is the circumcenter of
6. is right-angled at
7. is right-angled at
9. the points are concyclic with as diameter
10. radius is parallel to side
11. is the reflection of over side
12. the nine-point center lies on
13. the orthic triangle is isosceles with
14. the geometric mean theorem holds
15. the bisector of has length , where
16. the orthocenter is a reflection of vertex over side
17. segment is tangent to the circumcircle at point
18. median has the same length as the segment
19. the bisector of is tangent to the nine-point circle at
20. is a convex kite with diagonals and
21. altitude is tangent to the nine-point circle at
22. segment is tangent to the nine-point circle at .
(Quite plenty, but can we reach ?)
• (Extra feature) If satisfies equation (??), PROVE that its nine-point center divides in the ratio .