This is a paragraph.

## An extension of the extended sine law

In any triangle with circumcenter and circumradius , the extended sine law is:

(1)

If we now denote by the radii of the circumcircles of triangles in that order, then:

(2)

## Extension confirmed

In one line of equations, the extension that arises from a combination of (1) and (2) is:

(3)

Please don’t take equation (3) too seriously: there are many relations in Euclidean geometry that can be expressed in terms of the circumradius of a triangle, and equation (3) is just one of them. Meanwhile, note that:

• one of the cosine terms in (2) (or (3)) requires an absolute value if the parent triangle is obtuse
• only right triangles fail some of the equalities in (2) (or (3)).
Let be an acute triangle with circumcenter . Denote by the circumradii of the triangles , in that order. Verify equation (2), namely: , where is the circumradius of the parent triangle .

Whether the parent triangle is acute or obtuse, at least one of the associated triangles has to be obtuse. However, this won’t affect the calculations since all we have to do is use the extended sine law.

Here is acute, so it’s circumcenter is located inside the triangle as shown above. Consider . By the inscribed angle theorem, we have:

Now apply the extended sine law to triangle , noting that :

and so . In the case where one of the triangles is obtuse, we also obtain the same result.

Let be an obtuse triangle with circumcenter . Denote by the circumradii of the triangles , in that order. Verify equation (2), namely: , where is the circumradius of the parent triangle .

In this case, the circumcenter is located outside the triangle. A sample diagram is shown below:

In , angle is equal to , while angles and are both equal to . Apply the extended sine law again to get:

This is where an absolute value is needed.

## Exception circumvented

If the parent triangle is right-angled, let’s say at , then and so we need modify equation (2) slightly.

Let be a right triangle in which . PROVE that .

We would have and . Using equation (2) we have:

This gives .

Let be a right triangle in which . PROVE that and .

Note that since triangle is right-angled at , we have . From isolate :

Similarly, . Together we get . And

## Exploration continues

We’ll have need to use equation (2) in future. For now, consider what happens in the isosceles triangle.

In the isosceles triangle, PROVE that .

Suppose that . Then . Using equation (2):

## Equation changes

An update about our ongoing hide-and-seek game with some hackers. As you can see below, we were initially hoping to use an exponential equation to model their menace, but the data suddenly changed after October 28, and so did our equation.

How wonderful would it be if the hackers spent all that time focusing on analytic geometry?

## Takeaway

In an obtuse triangle , let be as mentioned in the post. Then the following statements are equivalent:

1. .

If instead triangle is acute, then the following statements are equivalent:

1. .

• (Similar triangles) Let be a non-right triangle with circumcenter . Denote by the circumcenters of the triangles , in that order.
1. PROVE that is similar to the orthic triangle
2. Find the similarity ratio.
• (Similar triangles) Let be a right triangle with circumcenter and . Denote by the circumcenters of the triangles , in that order.
1. PROVE that is similar to the parent triangle
2. Find the similarity ratio.

## An inverted Pythagorean identity II

Today’s focus is on the equation

(1)

that can be used to characterize right triangles.

To this end, let be a right triangle in which , and

• the circumcenter
• and the diameters of triangles and .

Then equation (1) holds.

## Preliminary calculations

Consider a right triangle in which . Let be the circumcenter of , and let and denote the circumcenters of triangles and , respectively. PROVE that triangle is similar to the parent triangle .

A sample diagram is shown below:

Since is the circumcenter of , it lies somewhere on the right bisector of side and so when is joined to , the segment is then parallel to side . Similarly, is parallel to side . Thus . Now join to and use angle chasing to complete the similarity argument.

Consider a right triangle in which . Let be the circumcenter of , and let and denote the circumradii of triangles and , respectively. PROVE that .

From the previous example, we saw the similarity between triangle and the parent triangle :

So:

We obtain as required. Notice also that we have

(2)

## Partial characterization

Partial in the sense that there’s another triangle for which the equivalence in the next two examples works. We’ll see that later.

Consider a right triangle in which . Let be the circumcenter of , and let and denote the circumdiameters of triangles and , respectively. PROVE that equation (1) holds, namely: , where is the circumradius of the parent triangle .

Apply the extended sine law to triangle below, noting that :

This would give:

Similarly, applying the extended sine law to triangle gives

Since and , we now get:

Consider an acute triangle . Let be the circumcenter of , and let and denote the circumdiameters of triangles and , respectively. If equation (1) holds, namely: , where is the circumradius of the parent triangle , PROVE that is right-angled at .

We’ll show in a later post that the following extension of the extended sine law holds:

(3)

An absolute value may be needed in one of the cosine terms if the parent triangle is obtuse. However, since we’re working with the squares for now, there’s no need for an absolute value. We have from (3) that

(Or .) Assuming , we then have:

Thus, either the triangle is right-angled at , or it’s an obtuse triangle that shares several properties in common with a right-triangle.

## Pertinent consideration

We emphasize that the equation also holds in a certain non-right triangle setting. More on this later.

Consider with vertices at , , . Verify that equation (1) holds for this triangle.

We’re to show that , where is the circumradius, is the diameter of the circumcircle of , and is the diameter of the circumcircle of .

By calculation, the circumcenter is . Thus the circumcenter of is , while the circumcenter of is . Notice that the nine-point center shown in the diagram is co-linear with and , just like in a right triangle. Furthermore:

So:

## Takeaway

Suppose that triangle has side-lengths , circumcenter , circumradius , and the radii of the circumcircles of and . Then the following statements are equivalent:

1. is right-angled at
2. .

• (Geometric means) Suppose that the side-lengths of satisfy . Using the notation in this post, PROVE that:
• (Geometric means) Suppose that the side-lengths of satisfy . Using the notation in this post, PROVE that:
• (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-five statements are equivalent:
1. or
2. and divide harmonically
3. is the reflection of over side
4. is the reflection of over side
5. is congruent to
6. is congruent to
7. is isosceles with
8. is isosceles with
9. is right angled at
10. is the circumcenter of
11. is right-angled at
12. is right-angled at
14. the points are concyclic with as diameter
15. the reflection of over lies internally on
16. the reflection of over lies externally on
17. radius is parallel to side
18. is the reflection of over side
19. segment is perpendicular to side
20. the nine-point center lies on
21. the orthic triangle is isosceles with
22. the geometric mean theorem holds
23. the bisector of has length , where
24. the orthocenter is a reflection of vertex over side
25. segment is tangent to the circumcircle at point
26. median has the same length as the segment
27. the bisector of is tangent to the nine-point circle at
28. is a convex kite with diagonals and
29. altitude is tangent to the nine-point circle at
30. chord is a diameter of the nine-point circle
31. segment is tangent to the nine-point circle at .
(More statements reserved for later. We aim to stop at statements, if possible.)

Twenty twenty two’s edition of the annual Thanksgiving was observed on Monday this week, but our own thanksgiving takes place every day.

This is largely because of a certain Thursday, June 14, 2018: beautiful, beautiful day. This person is always grateful to God for Thursday, June 14, 2018, among other days and other dealings.

The next iteration of the public gratitude comes up on Wednesday, June 14, 2023 — all things being equal. Until then: stay safe, do math, and give thanks.