*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)

Take in equation (1) and that’s it. That’s what we *seek*.

*nine-point center*is precisely vertex .

Observe that and . Further, equation (1) is satisfied:

and so the triangle enjoys one (and all) of the equivalent statements in our previous post. In particular, its orthocenter is a reflection of vertex over side , namely , as shown below:

Also, the radius through vertex must be parallel to side , giving the point as the circumcenter. Notice how we obtained both the orthocenter and the circumcenter without a single calculation of altitude or right bisector.

The nine-point center is the midpoint of and , and this is .

for 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.

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

But then , by the extended law of sines. So

Similarly, the nine-point circle passes through the midpoint of , so the radius from to this midpoint is :

The only permissible choice of and is . Then .

for ig.

Consider the distance from to , given by

Noting that and then yields .

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 .

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.

*orthic triangle is equilateral*.

By example 2 we have:

when the nine-point center coincides with . Since is otuse, the interior angles of the resulting orthic triangle are:

and so the orthic triangle is equilateral.

Being the circumcenter, is equidistant from and : . In addition, . This shows that is equilateral. Similarly, is equilateral.

By the usual notation, we have . By one of the conditions in our previous post, we have . By example 2, . Then:

Assume that is equilateral. Since , we have that and as well.

Using in equation (1) gives . Alternatively, we get and . Thus, by example 3, the nine-point center coincides with vertex .

## Extra tangents

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

Look here:

Look here:

Look here:

## Takeaway

Consider with side-lengths , circumradius , circumcenter , orthocenter , and nine-point center . If equation (1) is satisfied, then the following statements are *equivalent*:

- and are both equilateral
- is tangent to the circumcircle at
- the orthic triangle is equilateral
- is equilateral
- coincides with

A triangle that satisfies these has to have .

## Tasks

- Let be the orthocenter, circumcenter, and nine-point center of . If , PROVE that:
- the points are concyclic
- the circumcenter of is
- the circumcenter of is
- the incenter of is
- the nine-point circle of coincides with the incircle of .

- 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:
- , where is the altitude from and is the diameter of the circumcircle of .

- (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*:- radius is parallel to side
- the nine-point center lies on
- the geometric mean theorem holds
- the orthic triangle is obtuse isosceles
- the bisector of has length , where
- the orthocenter is a reflection of vertex over side
- segment is tangent to the circumcircle at point
- segment is tangent to the nine-point circle at , where is the foot of the altitude from vertex .

(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:- is similar to
- divides in the ratio (or )
- the line segment is perpendicular to the Euler line
- the quadrilateral is a kite (don’t confuse with a certain functional group in organic chemistry).

- (Euler line) If triangle satisfies equation (1), PROVE that:
- is perpendicular to the Euler line
- is similar to the parent
- is similar to the orthic triangle of
- divides in the ratio (or ).