On the geometric mean theorem

Given a right triangle with an altitude h_C as shown below:

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the geometric mean theorem states that

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

As shown here, equation (1) is equivalent to the Pythagorean identity:

(2)   \begin{equation*} c^2=a^2+b^2 \end{equation*}

However, the equivalence holds because the altitude is internal. In the case of an external altitude, we present an analogous geometric mean theorem:

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(3)   \begin{equation*}h_C=\sqrt{AF_C\times BF_C}\end{equation*}

Here, equation (3) is equivalent to:

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

Consider \triangle ABC with vertices at A(-6,0), B(0,0), C(2,4). Let F_C be the foot of the altitude from vertex C, and let h_C be the length of this altitude. Show that h_C=\sqrt{AF_C\times BF_C}.

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We have

    \[h_C=4,~AF_C=8,~BF_C=2,\boxed{4=\sqrt{2\times 8}}\implies h_C=\sqrt{AF_C\times BF_C}\]

External altitudes

In examples 2 and 3 below, we prove the equivalence between equations (3) and (4).

Establish an analogous geometric mean theorem for obtuse triangles.

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Apply the Pythagorean theorem to triangles ACF_C and BCF_C:

    \begin{equation*} \begin{split} h_C^2&=a^2-s^2\\ h_C^2&=b^2-(c+s)^2\\ \therefore a^2-s^2&=b^2-(c+s)^2\\ s&=\frac{(b^2-a^2)-c^2}{2c}\\ &=\frac{a^2}{\sqrt{a^2+b^2}}\\ AF_C&=c+s\\ &=\frac{b^2-a^2}{\sqrt{a^2+b^2}}+\frac{a^2}{\sqrt{a^2+b^2}}\\ &=\frac{b^2}{\sqrt{a^2+b^2}}\\ h_C&=\frac{ab}{\sqrt{a^2+b^2}}\\ AF_C\times BF_C&=(c+s)\times s\\ &=\frac{b^2}{\sqrt{a^2+b^2}}\times \frac{a^2}{\sqrt{a^2+b^2}}\\ \therefore h_C&=\sqrt{AF_C\times BF_C} \end{split} \end{equation*}

Note that \triangle ACF_C and \triangle BCF_C are similar, so one can actually use this fact to obtain an easier proof.

PROVE that if an obtuse triangle enjoys the geometric mean theorem, then a relation of the form (4) must hold.

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Assume that in the diagram above one has h_C=\sqrt{AF_C\times BF_C}. As before, h_C^2=a^2-s^2=b^2-(c+s)^2, giving

    \[s=\frac{b^2-a^2-c^2}{2c} \implies c+s=\frac{b^2-a^2+c^2}{2c}\]

    \begin{equation*} \begin{split} h_C^2&=AF_C\times BF_C\\ \therefore a^2-\left(\frac{b^2-a^2-c^2}{2c}\right)^2&=\scriptstyle\left(\frac{b^2-a^2+c^2}{2c}\right)\left(\frac{b^2-a^2-c^2}{2c}\right)\\ 4a^2c^2-\left[(b^2-a^2)^2+c^4-2c^2(b^2-a^2)\right]&=(b^2-a^2)^2-c^4\\ \therefore c^2(a^2+b^2)&=(b^2-a^2)^2\\ \end{split} \end{equation*}

Thus, we have a characterization of obtuse triangles that satisfy the geometric mean theorem.

Extra attributes

When a triangle satisfies equation (4), it is peculiar in its own right, but also shares some things in common with the right triangle.

If \triangle ABC satisfies equation (4), PROVE that AF_B=AF_A and BF_A=BF_B, where F_A,F_B are the feet of the altitudes from A,B.

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Note that (by definition) AF_A=h_A and BF_B=h_B, where h_A and h_B are the lengths of the altitudes from A and B. From the above diagram:

    \begin{equation*} \begin{split} AF_B^2&=AB^2-BF_B^2\\ &=c^2-h_B^2\\ &=c^2-\left(\frac{ac}{2R}\right)^2\\ &=\frac{c^2}{4R^2}(4R^2-a^2)\\ &=\frac{c^2}{4R^2}(b^2)\Longleftarrow a^2+b^2=4R^2\\ &=\left(\frac{bc}{2R}\right)^2\\ &=h_A^2\\ \therefore AF_B&=h_A=AF_A \end{split} \end{equation*}

Similarly, BF_A=BF_B.

Suppose that \triangle ABC satisfies equation (4). Using the diagram below, PROVE that ap=cs.

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From the previous example we had p=h_B and BF_B=h_B, so:

    \[ap=ah_B=a\frac{ac}{2R} =\frac{a^2c}{2R}.\]

From the diagram

    \[s=\sqrt{a^2-h_C^2}=\sqrt{a^2-\left(\frac{ab}{2R}\right)^2}=\frac{a^2}{2R}\implies cs=\frac{a^2c}{2R}.\]

Suppose that \triangle ABC satisfies equation (4). Using the diagram below, PROVE that p(p+a)=qr.

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Note that q=h_A (example 4) and r=\frac{a}{R}h_C, where R is the circumradius.

    \[qr=\frac{a}{R}h_Ah_C.\]

And

    \[p(p+a)=h_B(h_B+a)=\frac{ac}{2R}\left(\frac{ac}{2R}+a\right)=\frac{a^2c}{R}\frac{b^2}{4R^2}=\frac{a}{R}(h_Ah_C).\]

(Along the line we used equation (4) and its equivalent representation a^2+b^2=4R^2.)

Conclusion: p(p+a)=qr or BF_A\times CF_A=AF_B\times CF_B.

If \triangle ABC satisfies equation (4), PROVE that \cos^2A+\cos^2B=1.

The relation \cos^2A+\cos^2B=1 also holds for right triangles. Using equation (4) and the cosine formula, we obtain

    \[\cos A=-\frac{b}{\sqrt{a^2+b^2}},~\cos B=\frac{a}{\sqrt{a^2+b^2}}\implies \cos^2A+\cos^2B=1.\]

If \triangle ABC satisfies equation (4), PROVE that A-B=90^{\circ} or B-A=90^{\circ}.

This is probably the easiest way to see the relationship between right triangles and triangles satisfying equation (4).

From example 7 above we had \cos^2A+\cos^2B=1. But then \cos^2A+\sin^2A=1 also, giving \cos^2A+\cos^2B=\cos^2 A+\sin^2A.

    \[\cos B=\pm \sin A\implies A+B=90^{\circ}~\textrm{or}~B-A=90^{\circ}.\]

Our triangle is obtuse so we choose B-A=90^{\circ}.

If \cos^2A+\cos^2B=1, PROVE that \triangle ABC is either a right triangle or else satisfies equation (4).

Using \cos^2A+\cos^2B=1, the cosine formula, and skipping a couple of steps, we have:

    \begin{equation*} \begin{split} \left(\frac{b^2+c^2-a^2}{2bc}\right)^2+\left(\frac{a^2+c^2-b^2}{2bc}\right)^2&=1\\ (a^2+b^2)c^4-2(a^4+b^4)c^2+(a^2+b^2)(a^2-b^2)&=0\\ \frac{2(a^4+b^4)\pm\sqrt{4(a^4+b^4)^2-4(a^2+b^2)^2(a^2-b^2)^2}}{2(a^2+b^2)}&=c^2\\ \frac{2(a^4+b^4)\pm 4a^2b^2}{2(a^2+b^2)}&=c^2\\ a^2+b^2&=c^2\\ \textrm{or}~\frac{(a^2-b^2)^2}{a^2+b^2}&=c^2 \end{split} \end{equation*}

The second possibility \frac{(a^2-b^2)^2}{a^2+b^2}=c^2 is just equation (4) re-arranged.

Essential awareness

The preceding examples show that our newest arrival — any \triangle ABC satisfying equation (4) –“rivals” the right triangle in many aspects. More on that later. For now, any triangle whose side-slopes form a geometric progression is also comparable to a right triangle. There, a modified geometric mean theorem holds (see example 10).

In \triangle ABC, let the slopes of the sides AB,BC,CA be a,ar,ar^2. Let h_A be the length of the altitude from vertex A. Then

(5)   \begin{equation*}h_A=\frac{ar(r-1)}{\sqrt{|(a^2r^2+r)(a^2r^3+1)|}}\sqrt{pq}=k\sqrt{pq},~~\frac{p}{q}=\frac{a^2r^2+r}{a^2r^3+1} \end{equation*}

where p+q=BC and k=\frac{ar(r-1)}{\sqrt{|(a^2r^2+r)(a^2r^3+1)|}}.

Earlier aim

Our original plan was to focus on equation (5) throughout today, but we were redirected along the way. Kindly bear with us for not completely fulfilling the promise made in example 8 of our post on December 28, 2020.

Consider triangle ABC with vertices at A(1,4), B(3,6), and C(0,0). Verify that equation (5) is satisfied.

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The foot of the altitude from A is F_A(9/5,18/5), as shown above. By the distance formula:

    \[p=\frac{9}{5}\sqrt{5},~q=\frac{6}{5}\sqrt{5},~h_A=\frac{2}{5}\sqrt{5}.\]

The slopes of the sides of \triangle ABC form a geometric progression with a=1 and r=2, so we can now kalkulate k:

    \[k=\frac{ar(r-1)}{\sqrt{(a^2r^2+r)(a^2r^3+1)}}=\frac{2(2-1)}{\sqrt{(4+2)(8+1)}}=\frac{\sqrt{6}}{9}\]

Finally we compare both h_A and k\sqrt{pq}:

    \[h_A=\frac{2}{5}\sqrt{5},~k\sqrt{pq}=\frac{\sqrt{6}}{9}\sqrt{\frac{9}{5}\sqrt{5}\times \frac{6}{5}\sqrt{5}}=\frac{2}{5}\sqrt{5}.\]

Equal. Equation (5) is satisfied.

Takeaway

For any obtuse triangle ABC, the following statements are equivalent:

  • (a^2-b^2)^2=(ac)^2+(cb)^2
  • the orthic triangle is isosceles
  • the geometric mean theorem holds.

We have a longer chain of equivalence in the exercises below.

Tasks

  1. (Long chain) For any obtuse triangle ABC, PROVE that each of the following statements implies the others:
    • \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
    • a^2+b^2=4R^2
    • h_C=R\cos C
    • h_A^2+h_B^2=AB^2
    • \frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}
    • (a^2-b^2)^2=(ac)^2+(cb)^2
    • A-B=90^{\circ} or B-A=90^{\circ}
    • the orthic triangle is isosceles
    • the geometric mean theorem holds.
      (Some of the statements above are satisfied in a right triangle, but the entire chain of statements are no longer equivalent in that case.) For something different, some of the above statements are clearly redundant; notwithstanding, just admire the sheer quantity. The aim is to increase the numbers until as many statements as there are in the Invertible Matrix Theorem are reached (and “breached”).
  2. (Linear combination) Suppose that \triangle ABC satisfies equation (4).
    • PROVE that a\cos A-b\cos B-\sqrt{a^2+b^2}\cos C=0.
    • Write a\cos A-b\cos B in the form k\sin(B-A).
  3. Consider \triangle ABC with vertices at A(0,0), B(4,4), (1,-2). Denote its orthocenter and circumcenter by H and O as usual.
    • Verify that a^2=45,b^2=5,c^2=32. Hence, deduce that (a^2-b^2)^2=(ac)^2+(cb)^2 (in other words, equation (4) is satisfied)
    • PROVE that O\equiv \left(\frac{7}{2},\frac{1}{2}\right). Hence, deduce that radius OC is parallel to side AB
    • PROVE that H\equiv \left(-2,1\right). Deduce that the orthocenter is the reflection of vertex C over side AB.
  4. PROVE that the two statements below are equivalent for a triangle ABC:
    • it is a right triangle
    • its area is h_C\times m_C, the product of the altitude and median from the same vertex.
  5. Consider \triangle ABC with side-lengths a,b,c and altitudes h_A,h_B,h_C.
    • If \angle C=90^{\circ}, PROVE that \frac{a}{h_A}+\frac{b}{h_B}=\frac{c}{h_C} and \frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C} (notice the difference in the right members of both equations).
    • Deduce that \left(\frac{c}{h_C}\right)^2-\frac{a}{b}\frac{h_B}{h_A}-\frac{b}{a}\frac{h_A}{h_B}=2, under the condition above.
    • Suppose that the triangle satisfies equation (4) instead. PROVE that \frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}.
    • Deduce that \frac{h_A}{a}+\frac{h_B}{b}=\frac{h_C}{c} if, and only if, \frac{b}{a} is the golden ratio, for any triangle satisfying equation (4).
Bonus Question: \triangle ABC is a right triangle if A+B=90^{\circ}. What name would you give to a triangle in which A-B=90^{\circ}?

If you’ve been conversant with this poster’s style, you must have realized that he doesn’t hide from expressing gratitude to God publicly. The underlying reason dates back to a certain Thursday, June 14, 2018: beautiful, beautiful day. The poster’ll never forget that day.

By the way, if you’ve been observant, the poster often tries to post on the 14th and 28th (=14+14), and the reason for that should now be easy to understand.

All things being equal, the next iteration of the poster’s appreciation comes up on Thursday, October 14, 2021. Until then: stay safe, do math, and give thanks.