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