This is a paragraph.

Category: General

Introductory; impromptu.

On the geometric mean theorem II

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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*}

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(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).

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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).

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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

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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

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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

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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

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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

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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.

Nine-point center coincides with vertex

Towards the end of this article, the author states that if the nine-point center of a triangle coincides with a vertex of the triangle, then the triangle will be obtuse isosceles.

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

Take a=c in equation (1) and that’s it. That’s what we seek.

Consider \triangle ABC with vertices at A(-4,0), B(0,0), C(2,2\sqrt{3}). Verify that its nine-point center is precisely vertex B.

Observe that a=c=4 and b=4\sqrt{3}. Further, equation (1) is satisfied:

    \[(b^2-a^2)^2=(48-16)^2=1024=16(16+48)=c^2(a^2+b^2)\]

and so the triangle enjoys one (and all) of the equivalent statements in our previous post. In particular, its orthocenter H is a reflection of vertex C over side AB, namely H(2,-2\sqrt{3}), as shown below:

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Also, the radius through vertex C must be parallel to side AB, giving the point O(-2,2\sqrt{3}) as the circumcenter. Notice how we obtained both the orthocenter H and the circumcenter O without a single calculation of altitude or right bisector.

The nine-point center is the midpoint of O(-2,2\sqrt{3}) and H(2,-2\sqrt{3}), and this is (0,0)=B.

B for Boom.

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.

In \triangle ABC, suppose that the nine-point center N coincides with vertex B. PROVE that \angle A=\angle C=30^{\circ} and \angle B=120^{\circ}.

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Let B=N. Since the nine-point circle goes through the midpoint of AB and has radius equal to half the circumradius of the parent triangle, we have that

    \[\frac{R}{2}=\frac{c}{2}\implies R=c\]

But then R=\frac{c}{2\sin C}, by the extended law of sines. So

    \[c=\frac{c}{2\sin C}\implies\sin C=\frac{1}{2}\implies \angle C=30^{\circ},150^{\circ}\]

Similarly, the nine-point circle passes through the midpoint of BC, so the radius from B to this midpoint is \frac{a}{2}:

    \[\frac{R}{2}=\frac{a}{2}\implies R=a\implies \frac{a}{2\sin A}=a\implies \angle A=30^{\circ},150^{\circ}\]

The only permissible choice of \angle A and \angle C is \angle A=\angle C=30^{\circ}. Then \angle B=120^{\circ}.

B for Big.

In any \triangle ABC with \angle A=\angle C=30^{\circ} and \angle B=120^{\circ}, PROVE that the nine-point center N coincides with vertex B.

Consider the distance from B to N, given by

    \[BN^2=\frac{2BO^2+2BH^2-OH^2}{4}=\frac{2R^2+2a^2-(9R^2-a^2-b^2-c^2)}{4}\]

Noting that R=a=c and b^2=3a^2 then yields BN=0.

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 30^{\circ}.

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.

If the nine-point center of \triangle ABC coincides with vertex B, PROVE that the orthic triangle is equilateral.

By example 2 we have:

    \[\angle A=\angle C=30^{\circ},~\angle B=120^{\circ}\]

when the nine-point center coincides with B. Since \angle B is oBtuse, the interior angles of the resulting orthic triangle are:

    \begin{equation*}\begin{split}&\boxed{2\times\angle A}, ~\boxed{2\times \angle B-180^{\circ}},~\boxed{2\times \angle C}\\ \implies & \boxed{2\times 30^{\circ}},~\boxed{2\times 120^{\circ}-180^{\circ}},~\boxed{2\times 30^{\circ}}\\ &60^{\circ},~60^{\circ},~60^{\circ} \end{split} \end{equation*}

and so the orthic triangle is equilateral.

Let O be the circumcenter of \triangle ABC. If the nine-point center coincides with vertex B, PROVE that \triangle OAB and \triangle OBC are both equilateral.

Being the circumcenter, O is equidistant from A and B: OA=OB=R. In addition, AB=c=R. This shows that \triangle OAB is equilateral. Similarly, \triangle OBC is equilateral.

Let H be the orthocenter of \triangle ABC. If the nine-point center coincides with vertex B, PROVE that \triangle ACH is equilateral.

By the usual notation, we have AC=b. By one of the conditions in our previous post, we have AH=b. By example 2, a=c=R. Then:

    \[CH^2=4R^2-c^2=4c^2-c^2=3c^2=3a^2=b^2\implies CH=b\]

If \triangle ACH is equilateral, PROVE that the nine-point center coincides with vertex B.

Assume that \triangle ACH is equilateral. Since AC=b, we have that AH=b and CH=b as well.

    \[\therefore AH=CH\implies AH^2=CH^2\implies 4R^2-a^2=4R^2-c^2\implies a=c\]

Using a=c in equation (1) gives b^2=3a^2. Alternatively, we get \angle A=\angle C=30^{\circ} and \angle B=120^{\circ}. Thus, by example 3, the nine-point center coincides with vertex B.

Extra tangents

The circumcircle of any triangle that satisfies equation (1) comes naturally with one tangent: the segment from the orthocenter H is a tangent to the circumcircle at vertex C. In the event that the nine-point center coincides with vertex B, we get two extra tangents.

If the side-lengths of \triangle ABC satisfy equation (1), PROVE that both CD and HD are tangents to the nine-point circle at D, where D is the foot of the altitude from C.

Look here:

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If B=N, PROVE that segment AH is tangent to the nine-point circle.

Look here:

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If B=N, PROVE that segment AC is tangent to the nine-point circle.

Look here:

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Takeaway

Consider \triangle ABC with side-lengths a,b,c, circumradius R, circumcenter O, orthocenter H, and nine-point center N. If equation (1) is satisfied, then the following statements are equivalent:

  1. \triangle OAC and \triangle OBC are both equilateral
  2. HA is tangent to the circumcircle at A
  3. the orthic triangle is equilateral
  4. \triangle ACH is equilateral
  5. N coincides with B
  6. a:b:c=1:\sqrt{3}:1
  7. a=c=R

A triangle that satisfies these has to have \angle A=\angle C=30^{\circ},\angle B=120^{\circ}.

Tasks

  1. Let H, O,N be the orthocenter, circumcenter, and nine-point center of \triangle ABC. If B=N, PROVE that:
    • the points A,O,C,H are concyclic
    • the circumcenter of \triangle OAH is N
    • the circumcenter of \triangle OCH is N
    • the incenter of \triangle ACH is N
    • the nine-point circle of \triangle ABC coincides with the incircle of \triangle ACH.
  2. In \triangle ABC, let O be the circumcenter and R the circumradius. Let O_c be the circumcenter of \triangle AOB and let R_c be its circumradius. If equation (1) is satisfied, PROVE that:
    • R_c=\frac{R^3}{ab}
    • R^2=h_cD_c, where h_c is the altitude from C and D_c is the diameter of the circumcircle of \triangle AOB.
  3. (Expanded list) Consider a non-right triangle ABC with side-lengths a,b,c, altitudes h_A,h_B,h_C, circumradius R, circumcenter O, nine-point center N, and orthocenter H. PROVE that the following twenty eight statements are equivalent:
    • AH=b
    • BH=a
    • CH=2h_C
    • R=\frac{|a^2-b^2|}{2c}
    • h_C=R\cos C
    • \cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)}
    • \cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}
    • \cos C=\frac{2ab}{a^2+b^2}
    • \cos^2 A+\cos^2 B=1
    • \sin^2 A+\sin^2 B=1
    • a\cos A+b\cos B=0
    • 2\cos A\cos B+\cos C=0
    • OH^2=5R^2-c^2
    • h_A^2+h_B^2=AB^2
    • \frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}
    • a^2+b^2=4R^2
    • A-B=\pm 90^{\circ}
    • (a^2-b^2)^2=(ac)^2+(cb)^2
    • AH^2+BH^2+CH^2=8R^2-c^2
    • a=2R\sin A,~b=2R\cos A,~c=2R\cos 2A
    • radius OC is parallel to side AB
    • the nine-point center lies on AB
    • the geometric mean theorem holds
    • the orthic triangle is obtuse isosceles
    • the bisector of \angle C has length l, where l^2=\frac{2a^2b^2}{a^2+b^2}
    • the orthocenter is a reflection of vertex C over side AB
    • segment HC is tangent to the circumcircle at point C
    • segment HD is tangent to the nine-point circle at D, where D is the foot of the altitude from vertex C.
      (We’ll continue to update this list until we reach at least a certain limit).
  4. (Euler line) Consider a right triangle ABC with circumcenter O, nine-point center N, and \angle C=90^{\circ}. Let O_a be the circumcenter of \triangle BOC and let O_b be the circumcenter of \triangle AOC. Then the points N,O_a,O_b are co-linear, as we saw earlier and will see later. PROVE that:
    • \triangle OO_aO_b is similar to \triangle ABC
    • N divides O_aO_b in the ratio a^2:b^2 (or b^2:a^2)
    • the line segment O_aNO_b is perpendicular to the Euler line
    • the quadrilateral COO_aO_b is a kite (don’t confuse COO_aO_b with a certain functional group in organic chemistry).
  5. (Euler line) If triangle ABC satisfies equation (1), PROVE that:
    • CO_c is perpendicular to the Euler line
    • \triangle OO_aO_b is similar to the parent \triangle ABC
    • \triangle O_aO_bO_c is similar to the orthic triangle of \triangle ABC
    • N divides O_aO_b in the ratio a^2:b^2 (or b^2:a^2).