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:

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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 Boom.

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}.

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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 Big.

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 oBtuse, 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:

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If B=N, PROVE that segment AH is tangent to the nine-point circle.

Look here:

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If B=N, PROVE that segment AC is tangent to the nine-point circle.

Look here:

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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:

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

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

Tasks

  1. Let H, O,N be the orthocenter, circumcenter, and nine-point center of \triangle ABC. If B=N, 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.
  2. In \triangle ABC, let O be the circumcenter and R the circumradius. Let O_c be the circumcenter of \triangle AOB and let R_c 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.
  3. (Expanded list) Consider a non-right triangle ABC with side-lengths a,b,c, altitudes h_A,h_B,h_C, circumradius R, circumcenter O, nine-point center N, and orthocenter H. 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).
  4. (Euler line) Consider a right triangle ABC with circumcenter O, nine-point center N, and \angle C=90^{\circ}. Let O_a be the circumcenter of \triangle BOC and let O_b be the circumcenter of \triangle AOC. Then the points N,O_a,O_b 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).
  5. (Euler line) If triangle ABC 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).