This is a paragraph.

# The silver ratio and a polar circle

Numerically, the silver ratio is equal to , and may be denoted by .

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

A little bit of our private life before we proceed to something less irrelevant: We’re so glad that, as it stands, we have the upper hand in our fight against unwanted 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.

If the side-lengths of satisfy equation (2), PROVE that the radius of its polar circle satisfies equation (3).

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 .

Suppose that , and that the radius of the polar circle coincides with the circumradius. PROVE that is the silver ratio.

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.

If is the silver ratio and , PROVE that the radius of the polar circle coincides with the circumradius.

Similar to the preceding example.

Determine the value of under the conditions and .

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

Under the conditions and , show that the radical axis of the circumcircle and the polar circle goes through the nine-point center.

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:

1. is the silver ratio.

• (Geometric mean) Suppose that the side-lengths of satisfy . PROVE that the following statements are equivalent:
1. is the silver ratio
2. the nine-point center coincides with the foot of the symmedian from vertex
3. 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:
1. or
2. is the reflection of over side
3. is the reflection of over side
4. is congruent to
5. is congruent to
6. is isosceles with
7. is isosceles with
8. is right angled at
9. is the circumcenter of
10. is right-angled at
11. is right-angled at
13. the points are concyclic with as diameter
14. the reflection of over lies internally on
15. the reflection of over lies externally on
16. radius is parallel to side
17. is the reflection of over side
18. segment is perpendicular to side
19. the nine-point center lies on
20. the orthic triangle is isosceles with
21. the geometric mean theorem holds
22. the bisector of has length , where
23. the orthocenter is a reflection of vertex over side
24. segment is tangent to the circumcircle at point
25. median has the same length as the segment
26. the bisector of is tangent to the nine-point circle at
27. is a convex kite with diagonals and
28. altitude is tangent to the nine-point circle at
29. chord is a diameter of the nine-point circle
30. segment is tangent to the nine-point circle at .
(Five more members were recently added.)

There are milestones that one cannot afford to miss; otherwise, life may become boring, dry, insipid, lifeless. Surprisingly, the milestone may be as simple as learning a new thing, acquiring a new skill, and so on. In a nutshell: don’t miss your milestones; we didn’t miss ours.

That came on Thursday, June 14, 2018: beautiful, beautiful day. This person will never forget that day. By means of God’s merciful and meticulous redirection, there came a simple realization: learn HTML. Personally, such a simple thing is a big deal because an ingrained inclination meant such a step was outside regular imagination.

This then implies constant gratitude to God for that remarkable day. All things being equal, the next iteration of the public gratitude comes up on Friday, October 14, 2022. Until then: stay safe, do math, and give thanks.