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Problem statement
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
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![Rendered by QuickLaTeX.com h_A,h_B,h_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-acb86546a5e499bc6fc6a70f93a164a7_l3.png)
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- the orthic triangle is obtuse isosceles
- radius
is parallel to side
- the nine-point center lies on
- the geometric mean theorem holds
- the bisector of
has length
, where
- segment
is tangent to the circumcircle at point
- the orthocenter is a reflection of vertex
over side
.
The connection between the orthic triangle and the geometric mean theorem seems cool. The restriction to non-right triangles is crucal: some of the statements hold in right triangles, but not the entire chain of statements.
Our favorites: . Notice the last two.
Partial solutions
A complete proof will require about implications, and we’ve already seen some of these implications in previous iterations of our discussions: May 14, May 28, June 14, and June 28. Below we make some random selections.
![Rendered by QuickLaTeX.com A-B=\pm 90^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-90b6276720d28703c650aac802230805_l3.png)
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
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To be specific, let and set
. Then
and
. Consider the circumcircle shown below:
The angle which the major arc subtends at the center of the circle is twice the angle it subtends at the circumference, and so:
Since is isosceles, we have that
This shows that is parallel to
.
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![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com OC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-6bd1936408b65821e01b1d186f80594d_l3.png)
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![Rendered by QuickLaTeX.com A-B=\pm 90^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-90b6276720d28703c650aac802230805_l3.png)
Begin by drawing the circumcircle:
If , then
, as marked above. Reason: radius
is parallel to side
by assumption. Next,
is isosceles, and so
. In turn, this yields
. The major arc
now subtends an angle of
at the center of the circle. This means that it subtends an angle of
at the circumference. Thus,
. The difference between
and
is then
.
A different orientation having angle placed at
will also yield the same conclusion.
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![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A-B=\pm 90^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-90b6276720d28703c650aac802230805_l3.png)
![Rendered by QuickLaTeX.com OC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-6bd1936408b65821e01b1d186f80594d_l3.png)
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Let and set
. Then
and
. Draw the circumcircle, like so:
Since is obtuse, its orthocenter
is situated outside the triangle as shown above. Join
and
. Extend side
to meet
at
. Since
now becomes the foot of the altitude from
, we have that
. Since
is parallel to
, it follows that
.
Radius is perpendicular to tangent at the point of contact. Implies: is a tangent to the circumcircle at
.
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![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com CH](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-eb260559028e5f33a17b8b60463585e4_l3.png)
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Apply the Pythagorean theorem to the right triangle below:
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![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com CH](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-eb260559028e5f33a17b8b60463585e4_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com H](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d2121613062e686e294d3867f5da2955_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com AB](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0b7694a0a3a6ff6caddf255cb508f628_l3.png)
We had from the preceding example. This in turn implies
. So
below is isosceles with
:
Extend side to meet
at
. Since
now becomes the foot of the altitude from
, we have that
. Altitude
bisects the base, so
. This proves that
is a reflection of
over side
.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com (a^2-b^2)^2=(ac)^2+(cb)^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-6a663d57e999fe0e03e37ad5e103f29b_l3.png)
![Rendered by QuickLaTeX.com N](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-55a8b451f85dacca12e4dd6869a45cca_l3.png)
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We focus on the internal case and use the segment addition postulate. Note that in any triangle with orthocenter
, circumcenter
, and nine-point center
we have:
We proved before that when
. So
After some simplifications, we obtain
Similarly:
Let’s examine the absolute values. There are four cases to consider.
First, we can’t have and
simultaneously. Otherwise, their sum must be greater than zero as well; but their sum is
.
Next, suppose that and
. Then the sum is
, and so:
Because , this leads to
. This is a special case. If the points
aren’t co-linear, then in
, the median through
passes through the nine-point circle, and so the length of this median is a radius of the nine-point circle, namely
. We now have a triangle
in which the sum of two sides is
and a median has length
. This is impossible (see the exercises at the end). Indeed, the side-lengths of
have to be of the form
for sides
(or sides
). If we compute the cosine of the angle at
, we obtain
The third and fourth cases are the same. For example, and
. Then take
and obtain
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![Rendered by QuickLaTeX.com AB](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0b7694a0a3a6ff6caddf255cb508f628_l3.png)
![Rendered by QuickLaTeX.com (a^2-b^2)^2=(ac)^2+(cb)^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-6a663d57e999fe0e03e37ad5e103f29b_l3.png)
Using the fact that , we have:
After all the trouble of the previous example, we don’t want to bother you with another seemingly lengthy procedure, but note that results after simplifications.
![Rendered by QuickLaTeX.com \angle C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-6835a09db6bd08b5c057500d4e5e102c_l3.png)
![Rendered by QuickLaTeX.com l^2=\frac{2a^2b^2}{a^2+b^2}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-f2924b6045924e2c9a749284aafa5994_l3.png)
![Rendered by QuickLaTeX.com (a^2-b)^2=(ac)^2+(cb)^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-46c79db7ba448974adc2f49cefb22470_l3.png)
Normally, . So
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![Rendered by QuickLaTeX.com l](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b66a460035e996842441cf0d37f57e5f_l3.png)
![Rendered by QuickLaTeX.com \angle C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-6835a09db6bd08b5c057500d4e5e102c_l3.png)
![Rendered by QuickLaTeX.com l^2=\frac{2ab}{a^2+b^2}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-63864a3baf504e55ba37a68a44004ef6_l3.png)
First isolate from
as
and then use in the standard angle bisector formula:
Peculiar scenario
Below is a reward of the labour in example 6.
Take with
. Then
. The nine-point center is precisely
, since (according to example 6):
Our next post will show that no other triangle has this property.
Takeaway
Let be a non-right triangle with side-lengths
, altitudes
, circumradius
, circumcenter
, orthocenter
, and nine-point center
. Then the following statements are equivalent:
- radius
is parallel to side
- segment
is tangent to the circumcircle at point
- the nine-point center lies on
- the orthic triangle is obtuse isosceles
- the geometric mean theorem holds
- the bisector of
has length
, where
- the orthocenter is a reflection of vertex
over side
.
No need to wait until 2026 for twenty six equivalent statements.
Tasks
- (Identical traits) Consider a right triangle
with side-lengths
, circumradius
, circumcenter
, nine-point center
, and
. Let
be the circumcenter of
and let
be its circumradius. Similarly, let
be the circumcenter of
and let
be its circumradius.
- Show that
are co-linear.
- PROVE that
and
.
- Deduce that the ratio in which the bisector of
divides side
is the same as the circumradii ratio
.
- Show that
- (Identical traits) Consider a non-right triangle
with side-lengths
, circumradius
, circumcenter
, and nine-point center
. Let
be the circumcenter of
and let
be its circumradius. Similarly, let
be the circumcenter of
and let
be its circumradius. If
:
- Show that
are co-linear.
- PROVE that
and
.
- Deduce that the ratio in which the bisector of
divides side
is the same as the circumradii ratio
.
(There goes a non-right triangle that mimics a right triangle in many aspects.)
- Show that
- (Isosceles trapezium) Suppose that an obtuse
satisfies any of the equivalent statements considered in this post. Let
be a point on the circumcircle such that
is a diameter.
- PROVE that
.
- Deduce that quadrilateral
is an isosceles trapezium.
- PROVE that
- (Impossible triangle) In triangle
, suppose that
and that the length of the median from
is
, as per the special case encountered in the course of example 6.
- PROVE that the lengths of sides
and
must be
and
(or the other way).
- Using the triangle inequality, deduce that the three points
are co-linear.
- PROVE that the lengths of sides
- PROVE that the following two statements are equivalent for any triangle
:
- the length
of the bisector of
satisfies
.