Nine-point center equals a vertex III

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.)