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# 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. 2. .

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

1. 2. .

• (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
• (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 