Earlier, we considered the “altitude version” of the the geometric mean theorem, which states that
(1)
as per the diagram above. There is also the “leg version” of the geometric mean theorem:
(2)
where is the hypotenuse.
If denotes the circumradius, then
in a right triangle and so the relations in (2) can be re-written:
(3)
Since and
are merely the distances from the foot of the altitude, nothing makes equation (3) exclusive to right triangles. As such, we’ll now derive similar relations for non-right triangles that satisfy our modified Pythagorean identity
(4)
(5)
(6)
You may jump straight to the exercises after the sixth example.
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_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 F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com a,b,c](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-57e6624e316682b226e631d7234f001d_l3.png)
![Rendered by QuickLaTeX.com a^2=(BF_C)(2R)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-79e3faa1100e637e1266c6637246a359_l3.png)
Apply Pythagorean theorem to and remember that equation (4) is equivalent to
. This gives:
Co-linearity
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_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 F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com a,b,c](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-57e6624e316682b226e631d7234f001d_l3.png)
![Rendered by QuickLaTeX.com b^2=(AF_C)(2R)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4347473688269a87936b6511e753eac1_l3.png)
As before apply Pythagorean theorem to , bearing in mind that equation (4) is equivalent to
. This gives:
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_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 F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com a^2=(BF_C)(2R)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-79e3faa1100e637e1266c6637246a359_l3.png)
![Rendered by QuickLaTeX.com a,b,c](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-57e6624e316682b226e631d7234f001d_l3.png)
As per the diagram below
we normally have
If in addition , then:
which we know to be equivalent to equation (4).
Coincidence
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_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 F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com b^2=(AF_C)(2R)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4347473688269a87936b6511e753eac1_l3.png)
![Rendered by QuickLaTeX.com a,b,c](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-57e6624e316682b226e631d7234f001d_l3.png)
As per the diagram below
we normally have
If in addition , then:
which we know to be equivalent to equation (4).
- we feel that the usual ten examples appear too much to digest at once, especially because we really want the reader to get the best out of every post;
- we will be returning to in-class teaching, after having taken full advantage of teaching from home — and saving some time — for the past few months.
Assuming the reader actually puts mind to the things we write, probably reducing the volume will aid better understanding of the material. We may end up reducing the “volume” and increasing the “pressure”, to obey Boyle’s law.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(-6,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-79c16eb208dc820695dfd9a30fa1c847_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(2,4)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-e2fb9eee47c72cdf6ab23e648c29df8c_l3.png)
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_l3.png)
![Rendered by QuickLaTeX.com F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com a^2=(BF_C)(2R)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-79e3faa1100e637e1266c6637246a359_l3.png)
From the diagram below
we get . Notice that equation (4) holds:
and so we can compute the circumradius using
, an equation that’s equivalent to (4).
Also, . Then:
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(-6,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-79c16eb208dc820695dfd9a30fa1c847_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(2,4)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-e2fb9eee47c72cdf6ab23e648c29df8c_l3.png)
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_l3.png)
![Rendered by QuickLaTeX.com F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com b^2=(AF_C)(2R)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4347473688269a87936b6511e753eac1_l3.png)
Here, the only modification to the previous example is that this time around. Then:
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(4,4)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4acc2a75b5891b8bc98456dcbc05edc4_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(1,-2)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c0508a321bf286d7f4b8683a94a1694c_l3.png)
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_l3.png)
![Rendered by QuickLaTeX.com F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com a^2=(BF_C)(2R)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-79e3faa1100e637e1266c6637246a359_l3.png)
From the diagram below
we get . Notice that equation (4) holds:
and so we can compute the circumradius using
, an equation that’s equivalent to (4).
Also, and
. Then:
Circumradii
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(4,4)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4acc2a75b5891b8bc98456dcbc05edc4_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(1,-2)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c0508a321bf286d7f4b8683a94a1694c_l3.png)
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_l3.png)
![Rendered by QuickLaTeX.com F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com b^2=(AF_C)(2R)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4347473688269a87936b6511e753eac1_l3.png)
Recall from the previous example that the foot of the altitude from is
. The only modification is the calculation of
, and it is
Then
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-a4c057a4f58741a4e3fc4d5121ae3540_l3.png)
![Rendered by QuickLaTeX.com B(3,6)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0b67631b2dbdedc7cc3d349be4e9a2d8_l3.png)
![Rendered by QuickLaTeX.com C(0,10)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-f5c0cc9017e2a098d03651147c899333_l3.png)
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_l3.png)
![Rendered by QuickLaTeX.com F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com a^2=(BF_C)(2R)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-79e3faa1100e637e1266c6637246a359_l3.png)
From the diagram below
we get . Notice that equation (4) holds:
and so we can compute the circumradius using
, an equation that’s equivalent to (4).
Also, , being the distance from
to
. Then:
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-a4c057a4f58741a4e3fc4d5121ae3540_l3.png)
![Rendered by QuickLaTeX.com B(3,6)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0b67631b2dbdedc7cc3d349be4e9a2d8_l3.png)
![Rendered by QuickLaTeX.com C(0,10)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-f5c0cc9017e2a098d03651147c899333_l3.png)
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_l3.png)
![Rendered by QuickLaTeX.com F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com b^2=(AF_C)(2R)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4347473688269a87936b6511e753eac1_l3.png)
As in the preceding example, the foot of the altitude from vertex is
. The distance
is then
. The circumradius is as before:
. Then:
Takeaway
In a non-right , let
be the side-lengths,
the circumradius, and
the foot of the altitude from vertex
. Then the following statements are equivalent:
satisfy equation (4)
.
And many more.
Task
- (Late thirties) 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, and
the orthocenter. PROVE that the following thirty seven statements are equivalent:
- radius
is parallel to side
- 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
- segment
is tangent to the circumcircle at point
is a convex kite with diagonals
and
- segment
is tangent to the nine-point circle at
.
(In terms of sheer numbers, the conditions above have now surpassed the number of statements in the Invertible Matrix Theorem. Next stop is to top the number of statements characterizing tangential quadrilaterals. Next stop after that? To stop pestering you with the same exercises over and over.)
- (Less tasks)