Writing for the third time on the geometric mean theorem, after an external eye scrutinized and stamped the previous two writings on the same theme, seems a joy thing.
With that said, today’s post focuses on numerical problems that concretize the geometric mean theorem in obtuse triangles.
Wow! Thanks for clicking.
Note that the triangle in this first example is a 30-30-120 triangle; it satisfies the geometric mean theorem twice.
From the above diagram, . Also and . Thus:
Similarly, if denotes the foot of the altitude from vertex , then:
Another unique property of the triangle.
From the above diagram, . And so:
In words: the altitude is the geometric mean of the segments and .
From the above diagram, . And so:
In words: the altitude is the geometric mean of the segments and .
From the above diagram, . And so:
In words: the altitude is the geometric mean of the segments and .
From the above diagram, . And so:
In words: the altitude is the geometric mean of the segments and .
Takeaway
The two statements below, the geometric mean theorem, and (more than) 75 other statements, are equivalent in any non-right triangle with circumradius and side-lenths :
- .
Task
- (Aufbau) In triangle , let be the side-lengths, the circumradius, the circumcenter, the nine-point center, and the orthocenter. PROVE that the following statements are equivalent:
- the reflection of over is
- the circle with diameter passes through vertices and
- radius is parallel to side and radius is parallel to side
- the reflection of over is and the reflection of over is
- is the geometric mean of and , and is the geometric mean of and . ( and are the feet of the altitudes from and , respectively.)