(1)
then the following statements are equivalent:
is the golden ratio
Today’s post will now establish the above equivalence. (Note that the fourth condition, , which motivated today’s title, was coined by the author of this article.)
![Rendered by QuickLaTeX.com \frac{b}{a}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-9c07a6f9ef794812a52882f17301f73c_l3.png)
![Rendered by QuickLaTeX.com (ab)^2=(ac)^2+(cb)^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0a48a34f4eb028fcff5647206e66f636_l3.png)
Suppose that is the golden ratio, then
, and so
. This then gives
. In turn:
![Rendered by QuickLaTeX.com (ab)^2=(ac)^2+(cb)^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0a48a34f4eb028fcff5647206e66f636_l3.png)
![Rendered by QuickLaTeX.com \sin A\sin B=\sin C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-65490f2ae62f9f9eeee60ae63d9d7581_l3.png)
Let be the circumradius. Then, by the extended law of sines, we have:
The assumption implies
. (Henceforth we’ll frequently use the fact that
is equivalent to equation (1).) So:
![Rendered by QuickLaTeX.com \sin A\sin B=\sin C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-65490f2ae62f9f9eeee60ae63d9d7581_l3.png)
![Rendered by QuickLaTeX.com \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-6b8ca9f518299b6e98020c1572487628_l3.png)
Using the extended law of sines as in the previous example, the condition gives
. Since
, we obtain
So-called upside-down Pythagorean identity. Ours works under an added condition, namely equation (1).
![Rendered by QuickLaTeX.com \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-6b8ca9f518299b6e98020c1572487628_l3.png)
![Rendered by QuickLaTeX.com h_C=c](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-9ef04196bc7c6d8030cb2841aee63223_l3.png)
![Rendered by QuickLaTeX.com h_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-12f43127818fc8246bd1d6642e18ee7c_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com c=AB](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-ad4bb3e6c2a3ae2754000af62ae37ca0_l3.png)
Re-arrange the given condition as
and obtain . The length of the altitude from vertex
is given by
![Rendered by QuickLaTeX.com h_C=c](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-9ef04196bc7c6d8030cb2841aee63223_l3.png)
![Rendered by QuickLaTeX.com \frac{b}{a}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-9c07a6f9ef794812a52882f17301f73c_l3.png)
If , then
. Now use this in equation (1) to get
Thus, is the golden ratio.
Takeaway
Let be a triangle whose side-lengths
satisfy equation (1). Then the following statements are equivalent:
is the golden ratio
.
The last condition fixes the interior angles of the given triangle :
,
,
.
Task
- (Early sixties) In a non-right triangle
, let
be the side-lengths,
the altitudes,
the feet of the altitudes from the respective vertices,
the circumradius,
the circumcenter,
the nine-point center,
the orthocenter,
the midpoint of side
,
the reflection of
over side
,
the reflection of
over side
, and
the reflection of
over side
. PROVE that the following sixty-one statements are equivalent:
or
is congruent to
is congruent to
is isosceles with
is isosceles with
is right angled at
is the circumcenter of
is right-angled at
is right-angled at
- quadrilateral
is a rectangle
- the points
are concyclic with
as diameter
- the reflection
of
over
lies internally on
- the reflection
of
over
lies externally on
- radius
is parallel to side
is the reflection of
over side
- the nine-point center lies on
- the orthic triangle is isosceles with
- the geometric mean theorem holds
- the bisector of
has length
, where
- the orthocenter is a reflection of vertex
over side
- segment
is tangent to the circumcircle at point
- median
has the same length as the segment
- the bisector
of
is tangent to the nine-point circle at
is a convex kite with diagonals
and
- altitude
is tangent to the nine-point circle at
- segment
is tangent to the nine-point circle at
.
(short of the target.)
- (Extra feature) If
satisfies equation (??), PROVE that its nine-point center
divides
in the ratio
.