# On the geometric mean theorem

Given a right triangle with an altitude as shown below:

the geometric mean theorem states that

(1)

As shown here, equation (1) is equivalent to the Pythagorean identity:

(2)

However, the equivalence holds because the altitude is internal. In the case of an external altitude, we present an analogous geometric mean theorem:

(3)

Here, equation (3) is equivalent to:

(4)

Consider with vertices at , , . Let be the foot of the altitude from vertex , and let be the length of this altitude. Show that .

We have

## External altitudes

In examples 2 and 3 below, we prove the equivalence between equations (3) and (4).

Establish an analogous geometric mean theorem for obtuse triangles.

Apply the Pythagorean theorem to triangles and :

Note that and are similar, so one can actually use this fact to obtain an easier proof.
PROVE that if an obtuse triangle enjoys the geometric mean theorem, then a relation of the form (4) must hold.

Assume that in the diagram above one has . As before, , giving

Thus, we have a characterization of obtuse triangles that satisfy the geometric mean theorem.

## Extra attributes

When a triangle satisfies equation (4), it is peculiar in its own right, but also shares some things in common with the right triangle.

If satisfies equation (4), PROVE that and , where are the feet of the altitudes from .

Note that (by definition) and , where and are the lengths of the altitudes from and . From the above diagram:

Similarly, .

Suppose that satisfies equation (4). Using the diagram below, PROVE that .

From the previous example we had and , so:

From the diagram

Suppose that satisfies equation (4). Using the diagram below, PROVE that .

Note that (example 4) and , where is the circumradius.

And

(Along the line we used equation (4) and its equivalent representation .)

Conclusion: or .

If satisfies equation (4), PROVE that .

The relation also holds for right triangles. Using equation (4) and the cosine formula, we obtain

If satisfies equation (4), PROVE that or .

This is probably the easiest way to see the relationship between right triangles and triangles satisfying equation (4).

From example 7 above we had . But then also, giving .

Our triangle is obtuse so we choose .

If , PROVE that is either a right triangle or else satisfies equation (4).

Using , the cosine formula, and skipping a couple of steps, we have:

The second possibility is just equation (4) re-arranged.

## Essential awareness

The preceding examples show that our newest arrival — any satisfying equation (4) –“rivals” the right triangle in many aspects. More on that later. For now, any triangle whose side-slopes form a geometric progression is also comparable to a right triangle. There, a modified geometric mean theorem holds (see example 10).

In , let the slopes of the sides be . Let be the length of the altitude from vertex . Then

(5)

where and .

Earlier aim

Our original plan was to focus on equation (5) throughout today, but we were redirected along the way. Kindly bear with us for not completely fulfilling the promise made in example 8 of our post on December 28, 2020.

Consider triangle with vertices at , , and . Verify that equation (5) is satisfied.

The foot of the altitude from is , as shown above. By the distance formula:

The slopes of the sides of form a geometric progression with and , so we can now alulate :

Finally we compare both and :

Equal. Equation (5) is satisfied.

## Takeaway

For any obtuse triangle , the following statements are equivalent:

• the orthic triangle is isosceles
• the geometric mean theorem holds.

We have a longer chain of equivalence in the exercises below.

1. (Long chain) For any obtuse triangle , PROVE that each of the following statements implies the others:
• or
• the orthic triangle is isosceles
• the geometric mean theorem holds.
(Some of the statements above are satisfied in a right triangle, but the entire chain of statements are no longer equivalent in that case.) For something different, some of the above statements are clearly redundant; notwithstanding, just admire the sheer quantity. The aim is to increase the numbers until as many statements as there are in the Invertible Matrix Theorem are reached (and “breached”).
2. (Linear combination) Suppose that satisfies equation (4).
• PROVE that .
• Write in the form .
3. Consider with vertices at , , . Denote its orthocenter and circumcenter by and as usual.
• Verify that . Hence, deduce that (in other words, equation (4) is satisfied)
• PROVE that . Hence, deduce that radius is parallel to side
• PROVE that . Deduce that the orthocenter is the reflection of vertex over side .
4. PROVE that the two statements below are equivalent for a triangle :
• it is a right triangle
• its area is , the product of the altitude and median from the same vertex.
5. Consider with side-lengths and altitudes .
• If , PROVE that and (notice the difference in the right members of both equations).
• Deduce that , under the condition above.
• Suppose that the triangle satisfies equation (4) instead. PROVE that .
• Deduce that if, and only if, is the golden ratio, for any triangle satisfying equation (4).
Bonus Question: is a right triangle if . What name would you give to a triangle in which ?

If you’ve been conversant with this poster’s style, you must have realized that he doesn’t hide from expressing gratitude to God publicly. The underlying reason dates back to a certain Thursday, June 14, 2018: beautiful, beautiful day. The poster’ll never forget that day.

By the way, if you’ve been observant, the poster often tries to post on the 14th and 28th (), and the reason for that should now be easy to understand.

All things being equal, the next iteration of the poster’s appreciation comes up on Thursday, October 14, 2021. Until then: stay safe, do math, and give thanks.