# On the geometric mean theorem II

Earlier, we considered the “altitude version” of the the geometric mean theorem, which states that

(1)

as per the diagram above. There is also the “leg version” of the geometric mean theorem:

(2)

where is the hypotenuse.

If denotes the circumradius, then in a right triangle and so the relations in (2) can be re-written:

(3)

Since and are merely the distances from the foot of the altitude, nothing makes equation (3) exclusive to right triangles. As such, we’ll now derive similar relations for non-right triangles that satisfy our modified Pythagorean identity

(4)

(5)

(6)

You may jump straight to the exercises after the sixth example.

Let be the circumradius of a non-right , and let be the foot of the altitude from vertex . If the side-lengths satisfy equation (4), PROVE that .

Apply Pythagorean theorem to and remember that equation (4) is equivalent to . This gives:

## Co-linearity

Let be the circumradius of a non-right , and let be the foot of the altitude from vertex . If the side-lengths satisfy equation (4), PROVE that .

As before apply Pythagorean theorem to , bearing in mind that equation (4) is equivalent to . This gives:

Let be the circumradius of a non-right , and let be the foot of the altitude from vertex . If , PROVE that the side-lengths satisfy equation (4).

As per the diagram below

we normally have

which we know to be equivalent to equation (4).

## Coincidence

Let be the circumradius of a non-right , and let be the foot of the altitude from vertex . If , PROVE that the side-lengths satisfy equation (4).

As per the diagram below

we normally have

which we know to be equivalent to equation (4).

From now on, we may be reducing the “volume” of each post (the number of examples per post) for two main reasons:

• we feel that the usual ten examples appear too much to digest at once, especially because we really want the reader to get the best out of every post;
• we will be returning to in-class teaching, after having taken full advantage of teaching from home — and saving some time — for the past few months.

Assuming the reader actually puts mind to the things we write, probably reducing the volume will aid better understanding of the material. We may end up reducing the “volume” and increasing the “pressure”, to obey Boyle’s law.

Consider with vertices at , , and . If its circumradius is , and is the foot of the altitude from vertex , PROVE that .

From the diagram below

we get . Notice that equation (4) holds:

and so we can compute the circumradius using , an equation that’s equivalent to (4).

Also, . Then:

Consider with vertices at , , and . If its circumradius is , and is the foot of the altitude from vertex , PROVE that .

Here, the only modification to the previous example is that this time around. Then:

Consider with vertices at , , and . If its circumradius is , and is the foot of the altitude from vertex , PROVE that .

From the diagram below

we get . Notice that equation (4) holds:

and so we can compute the circumradius using , an equation that’s equivalent to (4).

Also, and . Then:

Consider with vertices at , , and . If its circumradius is , and is the foot of the altitude from vertex , PROVE that .

Recall from the previous example that the foot of the altitude from is . The only modification is the calculation of , and it is

Then

Consider with vertices at , , and . If its circumradius is , and is the foot of the altitude from vertex , PROVE that .

From the diagram below

we get . Notice that equation (4) holds:

and so we can compute the circumradius using , an equation that’s equivalent to (4).

Also, , being the distance from to . Then:

Consider with vertices at , , and . If its circumradius is , and is the foot of the altitude from vertex , PROVE that .

As in the preceding example, the foot of the altitude from vertex is . The distance is then . The circumradius is as before: . Then:

## Takeaway

In a non-right , let be the side-lengths, the circumradius, and the foot of the altitude from vertex . Then the following statements are equivalent:

1. satisfy equation (4)
2. .

And many more.

• (Late thirties) In a non-right triangle , let be the side-lengths, the altitudes, the feet of the altitudes from the respective vertices, the circumradius, the circumcenter, the nine-point center, and the orthocenter. PROVE that the following thirty seven statements are equivalent:
1. radius is parallel to side
2. the nine-point center lies on
3. the orthic triangle is obtuse isosceles
4. the geometric mean theorem holds
5. the bisector of has length , where
6. the orthocenter is a reflection of vertex over side
7. segment is tangent to the circumcircle at point
8. is a convex kite with diagonals and
9. segment is tangent to the nine-point circle at .
(In terms of sheer numbers, the conditions above have now surpassed the number of statements in the Invertible Matrix Theorem. Next stop is to top the number of statements characterizing tangential quadrilaterals. Next stop after that? To stop pestering you with the same exercises over and over.)