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.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(-4,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-bc3f15725e0a4923e8ba6e2df2ad7df3_l3.png)
![Rendered by QuickLaTeX.com B(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-7045877de833cb7e53ba1ef4ddfaeb24_l3.png)
![Rendered by QuickLaTeX.com C(2,2\sqrt{3})](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-51f4dd5600aa397f923da5141eeb676f_l3.png)
![Rendered by QuickLaTeX.com B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3f8c40a7f801883b444525b6eef00398_l3.png)
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.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com N](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-55a8b451f85dacca12e4dd6869a45cca_l3.png)
![Rendered by QuickLaTeX.com B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3f8c40a7f801883b444525b6eef00398_l3.png)
![Rendered by QuickLaTeX.com \angle A=\angle C=30^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-408e55b18738970715ba119fa33d44b9_l3.png)
![Rendered by QuickLaTeX.com \angle B=120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-09aa04cb2daf52ea1c2a9cb2b19f19f8_l3.png)
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.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com \angle A=\angle C=30^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-408e55b18738970715ba119fa33d44b9_l3.png)
![Rendered by QuickLaTeX.com \angle B=120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-09aa04cb2daf52ea1c2a9cb2b19f19f8_l3.png)
![Rendered by QuickLaTeX.com N](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-55a8b451f85dacca12e4dd6869a45cca_l3.png)
![Rendered by QuickLaTeX.com B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3f8c40a7f801883b444525b6eef00398_l3.png)
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.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3f8c40a7f801883b444525b6eef00398_l3.png)
By example 2 we have:
when the nine-point center coincides with . Since
is o
tuse, the interior angles of the resulting orthic triangle are:
and so the orthic triangle is equilateral.
![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3f8c40a7f801883b444525b6eef00398_l3.png)
![Rendered by QuickLaTeX.com \triangle OAB](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4b4bb4bda67aba63ef201e80402962b1_l3.png)
![Rendered by QuickLaTeX.com \triangle OBC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0f71cfa1120fb30c6f21d09fd09abfcd_l3.png)
Being the circumcenter, is equidistant from
and
:
. In addition,
. This shows that
is equilateral. Similarly,
is equilateral.
![Rendered by QuickLaTeX.com H](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d2121613062e686e294d3867f5da2955_l3.png)
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3f8c40a7f801883b444525b6eef00398_l3.png)
![Rendered by QuickLaTeX.com \triangle ACH](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c09051ea2fb08a3995cf7e62a859af3e_l3.png)
By the usual notation, we have . By one of the conditions in our previous post, we have
. By example 2,
. Then:
![Rendered by QuickLaTeX.com \triangle ACH](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c09051ea2fb08a3995cf7e62a859af3e_l3.png)
![Rendered by QuickLaTeX.com B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3f8c40a7f801883b444525b6eef00398_l3.png)
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.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com CD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4bc93c6cd404423a20a306998c2cc0cf_l3.png)
![Rendered by QuickLaTeX.com HD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-66661e5faa9cb279472889b3be74291a_l3.png)
![Rendered by QuickLaTeX.com D](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-e17a98724148be48d8f2415999c8e078_l3.png)
![Rendered by QuickLaTeX.com D](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-e17a98724148be48d8f2415999c8e078_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
Look here:
![Rendered by QuickLaTeX.com B=N](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-fe6301079a307f28114d596e9ea27ab2_l3.png)
![Rendered by QuickLaTeX.com AH](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-a434d48715d21e65d7e712e7a1dd02b2_l3.png)
Look here:
![Rendered by QuickLaTeX.com B=N](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-fe6301079a307f28114d596e9ea27ab2_l3.png)
![Rendered by QuickLaTeX.com AC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-2509ee19f0ee692f93e2b93071b678f5_l3.png)
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
.
- the points
- 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
).