On the geometric mean theorem II

Rendered by QuickLaTeX.com

Earlier, we considered the “altitude version” of the the geometric mean theorem, which states that

(1)   \begin{equation*}h_C=\sqrt{pq} \end{equation*}

as per the diagram above. There is also the “leg version” of the geometric mean theorem:

(2)   \begin{equation*} a^2=qc\quad\& \quad b^2=pc \end{equation*}

where c=p+q is the hypotenuse.

If R denotes the circumradius, then c=2R in a right triangle and so the relations in (2) can be re-written:

(3)   \begin{equation*} a^2=q(2R)\quad\& \quad b^2=p(2R) \end{equation*}

Since p and q 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)   \begin{equation*} (a^2-b^2)^2=(ac)^2+(cb)^2 \end{equation*}

Rendered by QuickLaTeX.com

(5)   \begin{equation*} a^2=(BF_C)(2R) \end{equation*}

(6)   \begin{equation*} b^2=(AF_C)(2R) \end{equation*}

You may jump straight to the exercises after the sixth example.

Let R be the circumradius of a non-right \triangle ABC, and let F_C be the foot of the altitude from vertex C. If the side-lengths a,b,c satisfy equation (4), PROVE that a^2=(BF_C)(2R).

Rendered by QuickLaTeX.com

Apply Pythagorean theorem to \triangle BCF_C and remember that equation (4) is equivalent to a^2+b^2=4R^2. This gives:

    \begin{equation*} \begin{split} BF_C^2&=BC^2-CF_C^2\\ &=a^2-\left(\frac{ab}{2R}\right)^2\\ &=\left(\frac{a}{2R}\right)^2(4R^2-b^2)\\ &=\left(\frac{a}{2R}\right)^2(a^2)\\ \implies BF_C&=\frac{a^2}{2R}\\ \therefore a^2&=(BF_C)(2R) \end{split} \end{equation*}

Co-linearity

Let R be the circumradius of a non-right \triangle ABC, and let F_C be the foot of the altitude from vertex C. If the side-lengths a,b,c satisfy equation (4), PROVE that b^2=(AF_C)(2R).

Rendered by QuickLaTeX.com

As before apply Pythagorean theorem to \triangle ACF_C, bearing in mind that equation (4) is equivalent to a^2+b^2=4R^2. This gives:

    \begin{equation*} \begin{split} AF_C^2&=AC^2-CF_C^2\\ &=b^2-\left(\frac{ab}{2R}\right)^2\\ &=\left(\frac{b}{2R}\right)^2(4R^2-a^2)\\ &=\left(\frac{b}{2R}\right)^2(b^2)\\ \implies AF_C&=\frac{b^2}{2R}\\ \therefore b^2&=(AF_C)(2R) \end{split} \end{equation*}

Let R be the circumradius of a non-right \triangle ABC, and let F_C be the foot of the altitude from vertex C. If a^2=(BF_C)(2R), PROVE that the side-lengths a,b,c satisfy equation (4).

As per the diagram below

Rendered by QuickLaTeX.com

we normally have

    \[BF_C=\sqrt{a^2-\left(\frac{ab}{2R}\right)^2}=\frac{a}{2R}\sqrt{4R^2-b^2}.\]

If in addition a^2=(BF_C)(2R), then:

    \[a^2=\left(\frac{a}{2R}\sqrt{4R^2-b^2}\right)(2R)\implies a=\sqrt{4R^2-b^2}\implies a^2+b^2=4R^2,\]

which we know to be equivalent to equation (4).

Coincidence

Let R be the circumradius of a non-right \triangle ABC, and let F_C be the foot of the altitude from vertex C. If b^2=(AF_C)(2R), PROVE that the side-lengths a,b,c satisfy equation (4).

As per the diagram below

Rendered by QuickLaTeX.com

we normally have

    \[AF_C=\sqrt{b^2-\left(\frac{ab}{2R}\right)^2}=\frac{b}{2R}\sqrt{4R^2-a^2}.\]

If in addition b^2=(AF_C)(2R), then:

    \[b^2=\left(\frac{b}{2R}\sqrt{4R^2-a^2}\right)(2R)\implies b=\sqrt{4R^2-a^2}\implies a^2+b^2=4R^2,\]

which we know to be equivalent to equation (4).

From now on, we may be reducing the “volume” of each post (the number of examples per post) for two main reasons:

  • 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.

Consider \triangle ABC with vertices at A(-6,0), B(0,0), and C(2,4). If its circumradius is R, and F_C is the foot of the altitude from vertex C, PROVE that a^2=(BF_C)(2R).

From the diagram below

Rendered by QuickLaTeX.com

we get a^2=20, b^2=80, c^2=36. Notice that equation (4) holds:

    \[(b^2-a^2)^2=(80-20)^2=3600=36(20+80)=c^2(a^2+b^2)\]

and so we can compute the circumradius R using a^2+b^2=4R^2, an equation that’s equivalent to (4).

    \[\implies R=\frac{\sqrt{a^2+b^2}}{2}=\frac{\sqrt{100}}{2}=5\]

Also, BF_C=2. Then:

    \[a^2=20~\&~(BF_C)(2R)=(2)(2\times 5)=20\]

Consider \triangle ABC with vertices at A(-6,0), B(0,0), and C(2,4). If its circumradius is R, and F_C is the foot of the altitude from vertex C, PROVE that b^2=(AF_C)(2R).

Here, the only modification to the previous example is that AF_C=8 this time around. Then:

    \[b^2=80~\&~(AF_C)(2R)=(8)(2\times 5)=80\]

Consider \triangle ABC with vertices at A(4,4), B(0,0), and C(1,-2). If its circumradius is R, and F_C is the foot of the altitude from vertex C, PROVE that a^2=(BF_C)(2R).

From the diagram below

Rendered by QuickLaTeX.com

we get a^2=5, b^2=45, c^2=32. Notice that equation (4) holds:

    \[(b^2-a^2)^2=(45-5)^2=1600=32(50)=32(5+45)=c^2(a^2+b^2)\]

and so we can compute the circumradius R using a^2+b^2=4R^2, an equation that’s equivalent to (4).

    \[\implies R=\frac{\sqrt{a^2+b^2}}{2}=\frac{\sqrt{50}}{2}=\frac{5\sqrt{2}}{2}\]

Also, B(0,0) and F_C=\left(-\frac{1}{2},-\frac{1}{2}\right)\implies BF_C=\frac{1}{2}\sqrt{2}. Then:

    \[a^2=5~\&~(BF_C)(2R)=\left(\frac{1}{2}\sqrt{2}\right)\left(2\times \frac{5\sqrt{2}}{2}\right)=5\]

Circumradii

Consider \triangle ABC with vertices at A(4,4), B(0,0), and C(1,-2). If its circumradius is R, and F_C is the foot of the altitude from vertex C, PROVE that b^2=(AF_C)(2R).

Recall from the previous example that the foot of the altitude from C is F_C\left(-\frac{1}{2},-\frac{1}{2}\right). The only modification is the calculation of AF_C, and it is

    \[AF_C=\sqrt{\left(4+\frac{1}{2}\right)^2+\left(4+\frac{1}{2}\right)^2}=\frac{9}{2}\sqrt{2}.\]

Then

    \[b^2=45~\&~(AF_C)(2R)=\left(\frac{9}{2}\sqrt{2}\right)\left(2\times\frac{5\sqrt{2}}{2}\right)=45\]

Consider \triangle ABC with vertices at A(0,0), B(3,6), and C(0,10). If its circumradius is R, and F_C is the foot of the altitude from vertex C, PROVE that a^2=(BF_C)(2R).

From the diagram below

Rendered by QuickLaTeX.com

we get a^2=25, b^2=100, c^2=45. Notice that equation (4) holds:

    \[(b^2-a^2)^2=(100-25)^2=5625=45(25+100)=c^2(a^2+b^2)\]

and so we can compute the circumradius R using a^2+b^2=4R^2, an equation that’s equivalent to (4).

    \[\implies R=\frac{\sqrt{a^2+b^2}}{2}=\frac{\sqrt{125}}{2}=\frac{5\sqrt{5}}{2}\]

Also, BF_C=\sqrt{5}, being the distance from B(3,6) to F_C(4,8). Then:

    \[a^2=25~\&~(BF_C)(2R)=(\sqrt{5})\left(2\times \frac{5\sqrt{5}}{2}\right)=25\]

Consider \triangle ABC with vertices at A(0,0), B(3,6), and C(0,10). If its circumradius is R, and F_C is the foot of the altitude from vertex C, PROVE that b^2=(AF_C)(2R).

As in the preceding example, the foot of the altitude from vertex C is F_C(4,8). The distance AF_C is then 4\sqrt{5}. The circumradius is as before: R=\frac{5\sqrt{5}}{2}. Then:

    \[b^2=45~\&~(AF_C)(2R)=(4\sqrt{5})\left(2\times\frac{5\sqrt{5}}{2}\right)=45.\]

Takeaway

In a non-right \triangle ABC, let a,b,c be the side-lengths, R the circumradius, and F_C the foot of the altitude from vertex C. Then the following statements are equivalent:

  1. a,b,c satisfy equation (4)
  2. a^2=(BF_C)(2R)
  3. b^2=(AF_C)(2R)
  4. a^2+b^2=4R^2.

And many more.

Task

  • (Late thirties) In a non-right triangle ABC, let a,b,c be the side-lengths, h_A,h_B,h_C the altitudes, F_A,F_B, F_C the feet of the altitudes from the respective vertices, R the circumradius, O the circumcenter, N the nine-point center, and H the orthocenter. PROVE that the following thirty seven statements are equivalent:
    1. AH=b
    2. BH=a
    3. CH=2h_C
    4. h_A=AF_B
    5. h_B=BF_A
    6. AF_C=\frac{b^2}{2R}
    7. BF_C=\frac{a^2}{2R}
    8. \frac{a}{c} =\frac{h_C}{AF_B}
    9. \frac{b}{c}=\frac{h_C}{BF_A}
    10. \frac{a}{b}=\frac{BF_A}{AF_B}
    11. R=\frac{|a^2-b^2|}{2c}
    12. h_C=R\cos C
    13. \cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)}
    14. \cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}
    15. \cos C=\frac{2ab}{a^2+b^2}
    16. \cos^2 A+\cos^2 B=1
    17. \sin^2 A+\sin^2 B=1
    18. a\cos A+b\cos B=0
    19. 2\cos A\cos B+\cos C=0
    20. a\cos A-b\cos B=\sqrt{a^2+b^2}\cos C
    21. OH^2=5R^2-c^2
    22. h_A^2+h_B^2=AB^2
    23. \frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}
    24. a^2+b^2=4R^2
    25. A-B=\pm 90^{\circ}
    26. (a^2-b^2)^2=(ac)^2+(cb)^2
    27. AH^2+BH^2+CH^2=8R^2-c^2
    28. a=2R\sin A,~b=2R\cos A,~c=2R\cos 2A
    29. radius OC is parallel to side AB
    30. the nine-point center lies on AB
    31. the orthic triangle is obtuse isosceles
    32. the geometric mean theorem holds
    33. the bisector of \angle C has length l, where l^2=\frac{2a^2b^2}{a^2+b^2}
    34. the orthocenter is a reflection of vertex C over side AB
    35. segment HC is tangent to the circumcircle at point C
    36. AF_ABF_B is a convex kite with diagonals AB and F_AF_B
    37. segment HF_C is tangent to the nine-point circle at F_C.
      (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)

The above equivalence, which we will constantly update, will likely be our only task per post. After we reach a certain limit, we will return to other exercises.