Slopefessor Fr1day – Page 9 – High School Geometry

Earlier we saw that the nine-point center of a triangle coincides with one of the vertices of the triangle precisely when the parent triangle is isosceles with an apex angle of $120^{\circ}$

Equivalently, two radii are parallel to two sides of the parent triangle, as will be shown in examples 1 and 2 below.

Everything we’ve been considering so far seems to work smoothly for the $30^{\circ}-30^{\circ}-120^{\circ}$ -triangle.

In $\triangle ABC$

, let $\angle A=\angle C=30^{\circ}$

and $\angle B=120^{\circ}$

. PROVE that the radius through $C$

is parallel to side $AB$

and the radius through $A$

is parallel to side $BC$

Since the parent triangle is obtuse, its circumcenter $O$ lies outside, opposite the obtuse angle. Thus $OABC$ is a (convex) quadrilateral.

If $R$ is the radius of the circumcircle of triangle $ABC$ , then we have that $R=a=BC$ and $R=c=AB$ (see here). Since $O$ is the circumcenter, we also have that $OA=OC=R$ . Hence, $OABC$ is a rhombus; in fact, a special rhombus with the property that the diagonal $OB$ is equal to the equal sides. Every rhombus is a parallelogram so we conclude that radius $OC$ is parallel to side $AB$ and radius $OA$ is parallel to side $BC$ .

In $\triangle ABC$

, suppose that the radius through $C$

is parallel to side $AB$

, and the radius through $A$

is parallel to side $BC$

. PROVE that $\angle A=\angle C=30^{\circ}$

and $\angle B=120^{\circ}$

Suppose that a triangle $ABC$ with circumcenter $O$ has the property that radius $OA$ is parallel to side $BC$ and radius $OC$ is parallel to side $AB$ . Such a triangle is necessarily obtuse, so that the circumcenter $O$ lies outside the triangle and we again have a quadrilateral $OABC$ . Join $AC$ and let $\angle CAB=\alpha^{\circ}$ . Since $OC$ is parallel to $AB$ , we have that $\angle OCA=\alpha^{\circ}$ . Since $O$ is the circumcenter, triangle $AOC$ is isosceles and so $\angle OAC=\angle OCA=\alpha^{\circ}$ . The situation is shown below, with the circumcircle included:

Since $OA$ is parallel to $CB$ , we have that $\angle ACB=\angle OAC=\alpha^{\circ}$ . Thus, triangle $ABC$ is isosceles; moreover, $\angle ABC=180^{\circ}-2\alpha^{\circ}$ . By the inscribed angle theorem:

$180^{\circ}+2\alpha^{\circ}=2(180^{\circ}-2\alpha^{\circ})\implies \alpha=30$

Thus, the interior angles of triangle $ABC$ are $\angle A=\angle C=30^{\circ}$ and $\angle B=120^{\circ}$ .

Let $a,b,c$

be the side-lengths of triangle $ABC$

, whose area $K$

satisfies $K=\frac{1}{2}ac\cos A=\frac{1}{2}ac\cos C$

. PROVE that triangle $ABC$

is the $30^{\circ}-30^{\circ}-120^{\circ}$

isosceles triangle.

Since the area satisfies $\frac{1}{2}ac\cos A=\frac{1}{2}ac\cos C$ , we have that $A=C$ .

Normally, $K=\frac{1}{2}ac\sin B$ . And so we must have $\sin B=\cos A$ . In turn, this implies

$B=90^{\circ}-A\quad\text{or}\quad 180^{\circ}-B= 90^{\circ}-A$

$B=90^{\circ}-A$ is not possible due to the fact that $A=C$ and $A+B+C=180^{\circ}$ . So we take $180^{\circ}-B= 90^{\circ}-A$ , which yields $B=90^{\circ}+A$ . Using $A+B+C=180^{\circ}$ we get

$A+(90^{\circ}+A)+A=180^{\circ}\implies A=30^{\circ}=C$

Consequently, $B=120^{\circ}$ .

Let $O$

be the circumcenter of triangle $ABC$

. If the convex quadrilateral $OABC$

is a rhombus, PROVE that triangle $ABC$

is the $30^{\circ}-30^{\circ}-120^{\circ}$

isosceles triangle.

No matter how the vertices are arranged, we’ll end up with a rhombus in which a diagonal is equal to the side-lengths. The interior angles of such a rhombus are $60^{\circ},60^{\circ},120^{\circ},120^{\circ}$ . Now just consider the triangle that remains when $O$ is removed.

Let $O$

be the circumcenter of triangle $ABC$

. If the convex quadrilateral $OABC$

is a rhombus, PROVE that the rhombus cannot be a square.

Consider various arrangements of the rhombus $OABC$ , for example the one shown below:

Apply the inscribed angle theorem to see that it is not possible to have $OABC$ be a square.

Takeaway

In any triangle $ABC$ , let $a,b,c$ be the side-lengths, $H$ the orthocenter, $O$ the circumcenter, and $R$ the circumradius. Then the following statements are equivalent:

$a=c=R$
$\angle A=\angle C=30^{\circ}$
the reflection of $B$ over $AC$ is $O$
the area $K$ satisfies $K=\frac{1}{2}ac\cos A=\frac{1}{2}ac\cos C$
the reflection of $A$ over $BC$ is $H$ and the reflection of $C$ over $AB$ is $H$

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 area $K$ satisfies $K=\frac{1}{2}ac\cos A=\frac{1}{2}ac\cos C$
6. the circle with diameter $OH$ passes through vertices $A$ and $C$
7. radius $OA$ is parallel to side $CB$ and radius $OC$ is parallel to side $AB$
8. the reflection of $A$ over $BC$ is $H$ and the reflection of $C$ over $AB$ is $H$
9. $AF_a$ 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.)

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.)

Author: Slopefessor Fr1day

Nine-point center equals a vertex III

Takeaway

Task

On the geometric mean theorem III

Takeaway

Task