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On the geometric mean theorem III

What’s new? Exciting news (from our point of view).

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.

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Consider \triangle ABC with vertices at A(-4,0), B(0,0), and C(2,2\sqrt{3}). Let F_c be the foot of the altitude from vertex C. Verify that the altitude CF_c is the geometric mean of AF_c and BF_c.

Note that the triangle in this first example is a 30-30-120 triangle; it satisfies the geometric mean theorem twice.

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From the above diagram, CF_c=2\sqrt{3}. Also AF_c=6 and BF_c=2. Thus:

    \[AF_c\times BF_c=12=CF_c^2\]

Similarly, if F_a denotes the foot of the altitude from vertex A, then:

    \[AF_a^2=BF_a\times CF_a\]

Another unique property of the 30^{\circ}-30^{\circ}-120^{\circ} triangle.

Consider \triangle ABC with vertices at A(0,0), B(3,6), and C(0,10). If F_c is the foot of the altitude from vertex C, verify that CF_c=\sqrt{AF_c\times BF_c}.

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From the above diagram, AF_c=4\sqrt{5},~BF_C=\sqrt{5}, CF_c=2\sqrt{5}. And so:

    \[AF_c\times BF_c=20,~CF_c^2=20\]

In words: the altitude CF_c is the geometric mean of the segments AF_c and BF_c.

Consider \triangle ABC with vertices at A(0,0), B(3,-9), and C(5,5). If F_c is the foot of the altitude from vertex C, verify that CF_c=\sqrt{AF_c\times BF_c}.

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From the above diagram, CF_c^2=40, ~AF_c=\sqrt{10},~BF_c=4\sqrt{10}. And so:

    \[AF_c\times BF_c=40=CF_c^2\]

In words: the altitude CF_c is the geometric mean of the segments AF_c and BF_c.

Consider \triangle ABC with vertices at A(-6,0), B(0,0), and C(2,4). If F_c is the foot of the altitude from vertex C, verify that CF_c=\sqrt{AF_c\times BF_c}.

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From the above diagram, CF_c^2=16, ~AF_c=8,~BF_c=2. And so:

    \[AF_c\times BF_c=8\times 2=16=CF_c^2\]

In words: the altitude CF_c is the geometric mean of the segments AF_c and BF_c.

Consider \triangle ABC with vertices at A(4,4), B(0,0), and C(1,-2). If F_c is the foot of the altitude from vertex C, verify that CF_c=\sqrt{AF_c\times BF_c}.

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From the above diagram, CF_c^2=\frac{9}{2}, ~AF_c=\frac{9}{2}\sqrt{2}, ~BF_C=\frac{1}{2}\sqrt{2}. And so:

    \[AF_c\times BF_c=\frac{9}{2}=CF_c^2\]

In words: the altitude CF_c is the geometric mean of the segments AF_c and BF_c.

Takeaway

The two statements below, the geometric mean theorem, and (more than) 75 other statements, are equivalent in any non-right triangle with circumradius R and side-lenths a,b,c:

  1. a^2+b^2=4R^2
  2. (b^2-a^2)=(ac)^2+(cb)^2.

Task

  • (Aufbau) In triangle ABC, let a,b,c be the side-lengths, R the circumradius, O the circumcenter, N the nine-point center, and H the orthocenter. PROVE that the following statements are equivalent:
    1. B=N
    2. a=c=R
    3. \angle A=\angle C=30^{\circ}
    4. the reflection of B over AC is O
    5. the circle with diameter OH passes through vertices A and C
    6. radius OA is parallel to side CB and radius OC is parallel to side AB
    7. the reflection of A over BC is H and the reflection of C over AB is H
    8. AF_a is the geometric mean of BF_a and CF_a, and CF_c is the geometric mean of AF_c and BF_c. (F_a and F_c are the feet of the altitudes from A and C, respectively.)