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The Euler line as a diameter II

Let and be the circumcenter and orthocenter of .

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:

1. a circle with diameter passes through and
2. the nine-point center coincides with vertex
3. the reflection of over is
4. the reflection of over is

Only triangle having satisfies these.

In , if the circle with diameter passes through and , PROVE that the nine-point center is .

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 .

If the nine-point center of coincides with vertex , PROVE that the reflection of the circumcenter over side is .

Easy.

If the reflection of the circumcenter over is , PROVE that the reflection of the orthocenter over is .

Easy.

If the reflection of over is , PROVE that the circle with diameter passes through and .

Easy.

Usual example

Consider with vertices at , , and . Its nine-point center is , and so it satisfies all the equivalent statements listed above.

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:

1. a circle with diameter passes through and
2. and are both equilateral
3. is tangent to the circumcircle at
4. the orthic triangle is equilateral
5. the reflection of over is
6. the reflection of over is
7. is equilateral
8. coincides with

These are some of the many equivalent descriptions of the isosceles triangle in which .

• (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:
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. the reflection of over lies internally on
11. the reflection of over lies externally on
12. radius is parallel to side
13. is the reflection of over side
14. the nine-point center lies on
15. the orthic triangle is isosceles with
16. the geometric mean theorem holds
17. the bisector of has length , where
18. the orthocenter is a reflection of vertex over side
19. segment is tangent to the circumcircle at point
20. median has the same length as the segment
21. the bisector of is tangent to the nine-point circle at
22. is a convex kite with diagonals and
23. altitude is tangent to the nine-point circle at
24. 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 .

As always, this poster utilizes the national thanksgiving event to record his personal thanksgiving to God.

Among other things, the underlying reason dates back to a certain Thursday, June 14, 2018: beautiful, beautiful day. The poster’ll never forget that day. Also, the poster wishes that you’ll encounter a definite day you’ll always remember for good — in case you haven’t already.

All things being equal, the next iteration of the poster’s appreciation comes up on Tuesday, June 14, 2022. Until then: stay safe, do math, and give thanks.