From reflection triangle to Euler line

If we begin with triangle $ABC$

in which $\angle A=\angle C=30^{\circ},\angle B=120^{\circ}$

and reflect vertex $A$

over side $BC$

, we obtain the orthocenter. If we reflect vertex $C$

over side $AB$

we obtain the orthocenter again.

If we now reflect vertex $B$ over side $AC$ we obtain the circumcenter.

It follows, in the case of the $30^{\circ}-30^{\circ}-120^{\circ}$ -triangle, that the reflection triangle is degenerate; in addition, it degenerates to a special line connecting the centroid, circumcenter and orthocenter, known as the Euler line.

Is the $30^{\circ}-30^{\circ}-120^{\circ}$ -triangle the only triangle whose reflection triangle degenerates to the Euler line?

In $\triangle ABC$

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

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

. PROVE that the reflection of vertex $A$

over side $BC$

coincides with the orthocenter.

To see how this holds, we’ll show that the length $h_a$ of the altitude through $A$ is half the distance from $A$ to the orthocenter $H$ . Indeed:

$h_a=\frac{bc}{2R}=\frac{\sqrt{3}a\times a}{2a}=\frac{\sqrt{3}a}{2}$

and

$AH=\sqrt{4R^2-a^2}=\sqrt{4a^2-a^2}=\sqrt{3}a$

This shows that the reflection of vertex $A$ over side $BC$ is the orthocenter $H$ .

In $\triangle ABC$

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

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

. PROVE that the reflection of vertex $C$

over side $AB$

coincides with the orthocenter.

Since $a=c$ , we have that $h_c=h_a$ and $CH=AH$ . Thus, the reflection of vertex $C$ over side $AB$ is also the orthocenter.

In $\triangle ABC$

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

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

. PROVE that the reflection of vertex $B$

over side $CA$

coincides with the circumcenter.

To see this, note that since triangle $ABC$ is isosceles with $B$ as the apex, the circumcenter lies on the axis of symmetry through $B$ . We just have to show that the radius through $B$ is twice the altitude through $B$ . The altitude through $B$ is:

$h_b=\frac{ac}{2R}=\frac{a\times a}{2a}=\frac{a}{2}\implies a=2\times h_b; \text{ that is, }R=2h_b$

PROVE that the $30^{\circ}-30^{\circ}-120^{\circ}$

-triangle has a degenerate reflection triangle.

We’ve already seen this by explicitly determining the reflection triangle from the preceding three examples. Alternatively, the necessary and sufficient condition for a degenerate reflection triangle is $\cos A\cos B\cos C=-\frac{3}{8}$ . This condition holds in the case of the $30^{\circ}-30^{\circ}-120^{\circ}$ -triangle:

$\cos 30^{\circ}\times \cos 120^{\circ}\times \cos 30^{\circ}=\frac{\sqrt{3}}{2}\times -\frac{1}{2}\times\frac{\sqrt{3}}{2}=-\frac{3}{8}$

If a triangle has the property that the reflections of two of its vertices over the corresponding opposite sides coincide with the orthocenter, PROVE that such a triangle must be the $30^{\circ}-30^{\circ}-120^{\circ}$

-triangle.

Let $R$ be the circumradius of the triangle. If the reflection of vertex $A$ over side $BC$ is the orthocenter, then we must have $b^2+c^2=4R^2$ , as per the equivalent statements here. Similarly, if the reflection of vertex $C$ over side $AB$ is the orthocenter, then we must have $a^2+b^2=4R^2$ . These two relations yield $a=c$ . Since $a^2+b^2=4R^2$ is also equivalent to $(b^2-a^2)^2=c^2(a^2+b^2)$ , we obtain $b^2=3a^2$ . Thus, the triangle is the $30^{\circ}-30^{\circ}-120^{\circ}$ -triangle.

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 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. radius $OA$ is parallel to side $CB$ and radius $OC$ is parallel to side $AB$
6. the reflection of $A$ over $BC$ is $H$ and the reflection of $C$ over $AB$ is $H$