Nine-point center coincides with vertex

Towards the end of this article, the author states that if the nine-point center of a triangle coincides with a vertex of the triangle, then the triangle will be obtuse isosceles.

True.

Today’s post will specify the obtuse isosceles triangle in question. In fact, it turns out to be just a special case of triangles that satisfy our modified Pythagorean identity:

(1) $\begin{equation*} (a^2-b^2)^2=(ac)^2+(cb)^2 \end{equation*}$

Take $a=c$ in equation (1) and that’s it. That’s what we seek.

Consider $\triangle ABC$

with vertices at $A(-4,0)$

, $B(0,0)$

, $C(2,2\sqrt{3})$

. Verify that its nine-point center is precisely vertex $B$

Observe that $a=c=4$ and $b=4\sqrt{3}$ . Further, equation (1) is satisfied:

$(b^2-a^2)^2=(48-16)^2=1024=16(16+48)=c^2(a^2+b^2)$

and so the triangle enjoys one (and all) of the equivalent statements in our previous post. In particular, its orthocenter $H$ is a reflection of vertex $C$ over side $AB$ , namely $H(2,-2\sqrt{3})$ , as shown below:

Also, the radius through vertex $C$ must be parallel to side $AB$ , giving the point $O(-2,2\sqrt{3})$ as the circumcenter. Notice how we obtained both the orthocenter $H$ and the circumcenter $O$ without a single calculation of altitude or right bisector.

The nine-point center is the midpoint of $O(-2,2\sqrt{3})$ and $H(2,-2\sqrt{3})$ , and this is $(0,0)=B$ .

$B$ for $B$ oom.

Easy theorem

There’s only one triangle whose orthocenter coincides with a vertex. In our next two examples, we also show that there’s only one triangle whose nine-point center coincides with a vertex. Fair enough.

In $\triangle ABC$

, suppose that the nine-point center $N$

coincides with vertex $B$

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

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

Let $B=N$ . Since the nine-point circle goes through the midpoint of $AB$ and has radius equal to half the circumradius of the parent triangle, we have that

$\frac{R}{2}=\frac{c}{2}\implies R=c$

But then $R=\frac{c}{2\sin C}$ , by the extended law of sines. So

$c=\frac{c}{2\sin C}\implies\sin C=\frac{1}{2}\implies \angle C=30^{\circ},150^{\circ}$

Similarly, the nine-point circle passes through the midpoint of $BC$ , so the radius from $B$ to this midpoint is $\frac{a}{2}$ :

$\frac{R}{2}=\frac{a}{2}\implies R=a\implies \frac{a}{2\sin A}=a\implies \angle A=30^{\circ},150^{\circ}$

The only permissible choice of $\angle A$ and $\angle C$ is $\angle A=\angle C=30^{\circ}$ . Then $\angle B=120^{\circ}$ .

$B$ for $B$ ig.

In any $\triangle ABC$

with $\angle A=\angle C=30^{\circ}$

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

, PROVE that the nine-point center $N$

coincides with vertex $B$

Consider the distance from $B$ to $N$ , given by

$BN^2=\frac{2BO^2+2BH^2-OH^2}{4}=\frac{2R^2+2a^2-(9R^2-a^2-b^2-c^2)}{4}$

Noting that $R=a=c$ and $b^2=3a^2$ then yields $BN=0$ .

Thus, there’s only one triangle with the property that its nine-point center coincides with one of the triangle’s vertices. And it’s the isosceles triangle with base angles of $30^{\circ}$ .

Having seen the main point, you can jump straight to the exercises at this point.

Equilateral triangles

Although the parent triangle in the preceding examples is not equilateral, there are a couple of equilateral triangles associated with it.

If the nine-point center of $\triangle ABC$

coincides with vertex $B$

, PROVE that the orthic triangle is equilateral.

By example 2 we have:

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

when the nine-point center coincides with $B$ . Since $\angle B$ is o $B$ tuse, the interior angles of the resulting orthic triangle are:

$\begin{equation*}\begin{split}&\boxed{2\times\angle A}, ~\boxed{2\times \angle B-180^{\circ}},~\boxed{2\times \angle C}\\ \implies & \boxed{2\times 30^{\circ}},~\boxed{2\times 120^{\circ}-180^{\circ}},~\boxed{2\times 30^{\circ}}\\ &60^{\circ},~60^{\circ},~60^{\circ} \end{split} \end{equation*}$

and so the orthic triangle is equilateral.

Let $O$

be the circumcenter of $\triangle ABC$

. If the nine-point center coincides with vertex $B$

, PROVE that $\triangle OAB$

and $\triangle OBC$

are both equilateral.

Being the circumcenter, $O$ is equidistant from $A$ and $B$ : $OA=OB=R$ . In addition, $AB=c=R$ . This shows that $\triangle OAB$ is equilateral. Similarly, $\triangle OBC$ is equilateral.

Let $H$

be the orthocenter of $\triangle ABC$

. If the nine-point center coincides with vertex $B$

, PROVE that $\triangle ACH$

is equilateral.

By the usual notation, we have $AC=b$ . By one of the conditions in our previous post, we have $AH=b$ . By example 2, $a=c=R$ . Then:

$CH^2=4R^2-c^2=4c^2-c^2=3c^2=3a^2=b^2\implies CH=b$

If $\triangle ACH$

is equilateral, PROVE that the nine-point center coincides with vertex $B$

Assume that $\triangle ACH$ is equilateral. Since $AC=b$ , we have that $AH=b$ and $CH=b$ as well.

$\therefore AH=CH\implies AH^2=CH^2\implies 4R^2-a^2=4R^2-c^2\implies a=c$

Using $a=c$ in equation (1) gives $b^2=3a^2$ . Alternatively, we get $\angle A=\angle C=30^{\circ}$ and $\angle B=120^{\circ}$ . Thus, by example 3, the nine-point center coincides with vertex $B$ .

Extra tangents

The circumcircle of any triangle that satisfies equation (1) comes naturally with one tangent: the segment from the orthocenter $H$ is a tangent to the circumcircle at vertex $C$ . In the event that the nine-point center coincides with vertex $B$ , we get two extra tangents.

If the side-lengths of $\triangle ABC$

satisfy equation (1), PROVE that both $CD$

and $HD$

are tangents to the nine-point circle at $D$

, where $D$

is the foot of the altitude from $C$

Look here:

If $B=N$

, PROVE that segment $AH$

is tangent to the nine-point circle.

Look here:

If $B=N$

, PROVE that segment $AC$

is tangent to the nine-point circle.

Look here:

Takeaway

Consider $\triangle ABC$ with side-lengths $a,b,c$ , circumradius $R$ , circumcenter $O$ , orthocenter $H$ , and nine-point center $N$ . If equation (1) is satisfied, then the following statements are equivalent:

$\triangle OAC$ and $\triangle OBC$ are both equilateral
$HA$ is tangent to the circumcircle at $A$
the orthic triangle is equilateral
$\triangle ACH$ is equilateral
$N$ coincides with $B$
$a:b:c=1:\sqrt{3}:1$
$a=c=R$

A triangle that satisfies these has to have $\angle A=\angle C=30^{\circ},\angle B=120^{\circ}$ .

Tasks

Let be the orthocenter, circumcenter, and nine-point center of . If , PROVE that:
- the points $A,O,C,H$ are concyclic
- the circumcenter of $\triangle OAH$ is $N$
- the circumcenter of $\triangle OCH$ is $N$
- the incenter of $\triangle ACH$ is $N$
- the nine-point circle of $\triangle ABC$ coincides with the incircle of $\triangle ACH$ .
In , let be the circumcenter and the circumradius. Let be the circumcenter of and let be its circumradius. If equation (1) is satisfied, PROVE that:
- $R_c=\frac{R^3}{ab}$
- $R^2=h_cD_c$ , where $h_c$ is the altitude from $C$ and $D_c$ is the diameter of the circumcircle of $\triangle AOB$ .
(Expanded list) Consider a non-right triangle with side-lengths , altitudes , circumradius , circumcenter , nine-point center , and orthocenter . PROVE that the following twenty eight statements are equivalent:
- $AH=b$
- $BH=a$
- $CH=2h_C$
- $R=\frac{|a^2-b^2|}{2c}$
- $h_C=R\cos C$
- $\cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)}$
- $\cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}$
- $\cos C=\frac{2ab}{a^2+b^2}$
- $\cos^2 A+\cos^2 B=1$
- $\sin^2 A+\sin^2 B=1$
- $a\cos A+b\cos B=0$
- $2\cos A\cos B+\cos C=0$
- $OH^2=5R^2-c^2$
- $h_A^2+h_B^2=AB^2$
- $\frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}$
- $a^2+b^2=4R^2$
- $A-B=\pm 90^{\circ}$
- $(a^2-b^2)^2=(ac)^2+(cb)^2$
- $AH^2+BH^2+CH^2=8R^2-c^2$
- $a=2R\sin A,~b=2R\cos A,~c=2R\cos 2A$
- radius $OC$ is parallel to side $AB$
- the nine-point center lies on $AB$
- the geometric mean theorem holds
- the orthic triangle is obtuse isosceles
- the bisector of $\angle C$ has length $l$ , where $l^2=\frac{2a^2b^2}{a^2+b^2}$
- the orthocenter is a reflection of vertex $C$ over side $AB$
- segment $HC$ is tangent to the circumcircle at point $C$
- segment $HD$ is tangent to the nine-point circle at $D$ , where $D$ is the foot of the altitude from vertex $C$ .
  (We’ll continue to update this list until we reach at least a certain limit).
(Euler line) Consider a right triangle with circumcenter , nine-point center , and . Let be the circumcenter of and let be the circumcenter of . Then the points are co-linear, as we saw earlier and will see later. PROVE that:
- $\triangle OO_aO_b$ is similar to $\triangle ABC$
- $N$ divides $O_aO_b$ in the ratio $a^2:b^2$ (or $b^2:a^2$ )
- the line segment $O_aNO_b$ is perpendicular to the Euler line
- the quadrilateral $COO_aO_b$ is a kite (don’t confuse $COO_aO_b$ with a certain functional group in organic chemistry).
(Euler line) If triangle satisfies equation (1), PROVE that:
- $CO_c$ is perpendicular to the Euler line
- $\triangle OO_aO_b$ is similar to the parent $\triangle ABC$
- $\triangle O_aO_bO_c$ is similar to the orthic triangle of $\triangle ABC$
- $N$ divides $O_aO_b$ in the ratio $a^2:b^2$ (or $b^2:a^2$ ).