On the geometric mean theorem III – High School Geometry

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.

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.

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 $CF_c=\sqrt{AF_c\times BF_c}$

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 $CF_c=\sqrt{AF_c\times BF_c}$

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 $CF_c=\sqrt{AF_c\times BF_c}$

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 $CF_c=\sqrt{AF_c\times BF_c}$

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 $a,b,c$ :

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

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:
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.)