Next, a polar circle is a circle centred at the orthocenter of a triangle, with radius related to the circumradius and the side-lengths via

(1)

Now suppose that satisfy the identity:

(2)

Notice that any triangle in which equation (2) holds is necessarily obtuse, and so it’s polar circle is defined, with equation (1) simplifying to:

(3)

Not only this, it happens that precisely when is the silver ratio.

## Reduced attacks

### The barrage was quite something prior to two weeks ago:

Grateful to find a barricade. It’s a different story now.

## Refocused attention

Sorry we had to bother you with the above. Once in a while we’ll let you see what happens behind our scenes.

We’re to show that if , then the radius of the polar circle is . Use the fact that in a non-right triangle, we have .

We had from the preceding example. And also . If we set we get after some simplification. In addition, . Thus: . Further simplification: . Take the positive value: , the silver ratio.

Similar to the preceding example.

We had . Since , we then obtain . Cosine formula: .

## Radical axis

Both the circumcircle and the polar circle have the same radius, under the given conditions. Thus the radical axis is the right bisector of the line joining their centers. But then the center of one is the circumcenter, while the center of the other is the orthocenter. The midpoint of these two centers is the nine-point center.

## Takeaway

Suppose that triangle has side-lengths , radius of polar circle , and circumradius . If , then the following statements are *equivalent*:

- is the silver ratio.

## Tasks

- (Geometric mean) Suppose that the side-lengths of satisfy . PROVE that the following statements are equivalent:
- is the silver ratio
- the nine-point center coincides with the foot of the symmedian from vertex
- the radius of the nine-point circle is the geometric mean of the two segments into which the nine-point center divides side .

- (Growing membership) 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 in that order, the Euler points, the circumradius, the circumcenter, the nine-point center, the orthocenter, the reflection of over side , the reflection of over side , the reflection of over side , the symmedian point, the foot of the symmedian from vertex , and the radius of the polar circle. PROVE that the following*eighty-three*statements are*equivalent*:- or
- is the reflection of over side
- is the reflection of over side
- is congruent to
- is congruent to
- is isosceles with
- is isosceles with
- is right angled at
- is the circumcenter of
- is right-angled at
- is right-angled at
- quadrilateral is a rectangle
- the points are concyclic with as diameter
- the reflection of over lies internally on
- the reflection of over lies externally on
- radius is parallel to side
- is the reflection of over side
- segment is perpendicular to side
- the nine-point center lies on
- the orthic triangle is isosceles with
- the geometric mean theorem holds
- the bisector of has length , where
- the orthocenter is a reflection of vertex over side
- segment is tangent to the circumcircle at point
- median has the same length as the segment
- the bisector of is tangent to the nine-point circle at
- is a convex
*kite*with diagonals and - altitude is tangent to the nine-point circle at
- chord is a diameter of the nine-point circle
- segment is tangent to the nine-point circle at .

(Five more members were recently added.)