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# A special chord of the nine-point circle

In the above diagram we have the nine-point circle of triangle going through the midpoints (of sides respectively), and the feet of the altitudes (from in that order).

It will be shown that

(1)

if (and only if) the side-lengths satisfy

(2)

It will also be shown, in our first example, that we have the relations

(3)

in any triangle (append absolute values if need be).

In the meantime, note that the right side of equation (1) evaluates to for a right triangle, where is the circumradius, whereas both sides of equation (1) evaluate to under (2). And so in the latter case, the segment is not just a chord, but a diameter of the nine-point circle.

Derive the three equations in (3).

Let’s do this for an acute triangle. Slight modification for an obtuse triangle will be needed. By drawing an appropriate diagram we have:

We can re-write the expression for entirely in terms of the side-lengths using the cosine formula:

Similarly we obtain the expressions for and .

Let be an isosceles triangle in which . PROVE that the midpoint of side coincides with the foot of the altitude from vertex .

Since , we have by the usual notation. Using example 1 above we have

and so .

In a right triangle with hypotenuse of length , PROVE that the right side of equation (1) is the square of the circumradius.

We first have . Moreover, if is the circumradius, then for a right triangle. Now:

If the side-lengths of a triangle satisfy equation (2), PROVE that equation (1) holds.

Such a triangle is necessarily non-right. Let be its circumradius. By one of the equivalent statements here, we know that equation (2) then becomes equivalent to . Re-arrange equation (2) in the form and consider both sides of equation (1):

PROVE that the segment is a diameter of the nine-point circle, if equation (2) holds.

By example 4 above, we had . Since the nine-point circle goes through and and has radius equal to , the fact that means that the chord is a diameter.

No other triangle has this property.

## Takeaway

In triangle , let denote the Euler point of vertex , the foot of the altitude from vertex , and the midpoint of side . Then the following statements are equivalent:

1. the chord is a diameter of the nine-point circle
2. .

The equivalent conditions range from extremely simple to moderately involved.

• (Late sixties) In a non-right triangle , let be the side-lengths, the altitudes, the feet of the altitudes from the respective vertices, the midpoints of sides , the Euler points, the circumradius, the circumcenter, the nine-point center, the orthocenter, the reflection of over side , the reflection of over side , and the reflection of over side . PROVE that the following sixty-eight statements are equivalent:
1. or
2. is congruent to
3. is congruent to
4. is isosceles with
5. is isosceles with
6. is right angled at
7. is the circumcenter of
8. is right-angled at
9. is right-angled at
11. the points are concyclic with as diameter
12. the reflection of over lies internally on
13. the reflection of over lies externally on
14. radius is parallel to side
15. is the reflection of over side
16. the nine-point center lies on
17. the orthic triangle is isosceles with
18. the geometric mean theorem holds
19. the bisector of has length , where
20. the orthocenter is a reflection of vertex over side
21. segment is tangent to the circumcircle at point
22. median has the same length as the segment
23. the bisector of is tangent to the nine-point circle at
24. is a convex kite with diagonals and
25. altitude is tangent to the nine-point circle at
26. chord is a diameter of the nine-point circle
27. segment is tangent to the nine-point circle at .
( short of the target. Next target now is to extend the initial target.)
• (Extra feature) If satisfies equation (??), PROVE that its nine-point center divides in the ratio .