Geometric mean for angle bisectors

The three traditional triangle cevians are medians, altitudes, and angle bisectors. Of these three, only the altitude is usually considered in the geometric mean theorem. However, the median also has an associated, isolated geometric mean theorem (example 1). Since what is good for the goose is also good for the angle bisector, we thought it good to present a geometric mean theorem for this third triangle cevian.

Let a,b,c be the side-lengths of \triangle ABC, and let l be the length of the bisector of \angle C, as shown below:

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Then l is the geometric mean of m and n if, and only if:

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

We’ll see the simplifications that result when equation (1) restricts to right triangles

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

and the simplifications that result when (1) is considered in the context of triangles satisfying

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

Let’s set the ball rolling with what a right triangle enjoys solely.

PROVE that a median is the geometric mean of the two segments into which it divides the opposite side if, and only if, the parent triangle is a right triangle.

Suppose that \triangle ABC is a right triangle with \angle C=90^{\circ}. Then the median from vertex C is half of the length of the hypotenuse: m_c=\frac{1}{2}c and so m_c is the geometric mean of \frac{1}{2}c and \frac{1}{2}c. On the other hand, if a triangle has the property that one of its medians is the geometric mean of the two segments on the opposite side, for example, m_c^2=\frac{1}{2}c\times \frac{1}{2}c, we show that the triangle is a right triangle. Indeed:

    \begin{equation*} \begin{split} m_c^2&=\frac{1}{2}c\times \frac{1}{2}c\\ \frac{2a^2+2b^2-c^2}{4}&=\frac{1}{4}c^2\\ 2a^2+2b^2&=2c^2\\ \therefore a^2+b^2&=c^2\\ \end{split} \end{equation*}

So the triangle is a right triangle. Since all medians are internal, this property is unique to right triangles.

Formula derivation

Our main result is example 4, for which we may need the next two examples.

In the diagram below, CD is the bisector of \angle C. PROVE that m=\frac{b}{a+b}c and n=\frac{a}{a+b}c.

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We first have that m+n=c. Also:

    \[\angle ADC + \angle CDB=180^{\circ}\implies \sin \angle ADC=\sin \angle CDB\]

The sine rule applied to \triangles ADC and CDB then gives:

    \begin{equation*} \begin{split} \frac{m}{\sin\theta}&=\frac{b}{\sin \angle ADC}\\ \frac{n}{\sin\theta}&=\frac{a}{\sin \angle CDB}\\ \therefore \frac{m}{n}&=\frac{b}{a}\\ m&=\frac{b}{a}n\\ c-n&=\frac{b}{a}n\\ c&=\left(\frac{b}{a}+1\right)n\\ c&=\frac{a+b}{a}n\\ \therefore n&=\frac{a}{a+b}c\\ \therefore m&=\frac{b}{a+b}c\\ \end{split} \end{equation*}

Basically, m:n=b:a. This is very useful in finding the coordinates of the foot of an angle bisector, which in turn is useful in finding the coordinates of the incenter of a triangle.

PROVE that l^2=ab\left(1-\left(\frac{c}{a+b}\right)^2\right), where l is the length of the bisector of \angle C.

Let’s refer to the diagram in example 2 again:

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and then apply Stewart’s theorem to \triangle ABC to obtain

    \[a^2m+b^2n=c(l^2+mn)\implies l^2=\frac{a^2m+b^2n}{c}-mn\]

We now use the fact that m=\frac{b}{a+b}c and n=\frac{a}{a+b}c from example 2.

    \begin{equation*} \begin{split} l^2&=\frac{a^2m+b^2n}{c}-mn\\ &=\frac{a^2\times \frac{b}{a+b}c+b^2\times \frac{a}{a+b}c}{c}-\frac{b}{a+b}c\times \frac{a}{a+b}c\\ &=ab-\frac{abc^2}{(a+b)^2}\\ &=ab\left(1-\frac{c^2}{(a+b)^2}\right) \end{split} \end{equation*}

An alternative procedure is to use area arguments as well as the cosine formula and double angle identities for \sin 2\theta and \cos 2\theta. This way one avoids going through Stewart’s theorem and an explicit determination of m and n.

In the diagram below, PROVE that the length l of the bisector of \angle C is the geometric mean of the segments AD and DB if, and only if, (a+b)^2=2c^2.

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First suppose that l is the geometric mean of m and n. Then l^2=mn.

    \begin{equation*} \begin{split} \implies ab\left(1-\left(\frac{c}{a+b}\right)^2\right)&=\frac{a}{a+b}c\times \frac{b}{a+b}c\\ 1-\frac{c^2}{(a+b)^2}&=\frac{c^2}{(a+b)^2}\\ 1&=2\frac{c^2}{(a+b)^2}\\ \therefore (a+b)^2&=2c^2\\ \end{split} \end{equation*}

Conversely, suppose that (a+b)^2=2c^2.

    \begin{equation*} \begin{split} l^2&= ab\left(1-\left(\frac{c}{a+b}\right)^2\right)\\ &=ab\left(1-\frac{c^2}{2c^2}\right)\\ &=\frac{1}{2}(ab)\\ mn&=\frac{b}{a+b}c\times \frac{a}{a+b}c\\ &=\frac{abc^2}{(a+b)^2}\\ &=\frac{abc^2}{2c^2}\\ &=\frac{1}{2}ab\\ \therefore l^2&=mn \end{split} \end{equation*}

Notice how we obtain another equivalence: l^2=mn\iff (\sqrt{2}l)^2=ab. Translate into words: we killed two birds with one stone.

For a quick example consider triangle ABC with vertices at A(0,0), B\left(6\sqrt{2},0\right), and C\left(2\sqrt{2},\sqrt{17}\right), shown below:

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The foot D of the bisector of \angle C has coordinates \left(\frac{5}{2}\sqrt{2},0\right):=D. By the distance formula:

    \[l^2=\left(\frac{5}{2}\sqrt{2}-2\sqrt{2}\right)^2+\left(0-\sqrt{17}\right)^2=\frac{35}{2}.\]

Also:

    \[m=\frac{5}{2}\sqrt{2},~n=\frac{7}{2}\sqrt{2}\implies mn=\frac{35}{2}=l^2.\]

Thus, CD is the geometric mean of AD and DB.

Fifteen degrees

We now restrict equation (1) to right triangles and to triangles satisfying equation (3) . It turns out that the interior angles are all multiples of 15^{\circ}.

If the bisector of \angle C=90^{\circ} in a right triangle satisfies equation (1), PROVE that the triangle is isosceles.

Using the fact that a^2+b^2=c^2 and (a+b)^2=2c^2, we have:

    \[(a+b)^2=2(a^2+b^2)\implies (a-b)^2=0\implies a=b\]

The angles of the triangle will then be 45^{\circ}, 45^{\circ}, 90^{\circ}. All multiples of 15^{\circ}.

If the bisector of the apex \angle C in an isosceles triangle satisfies equation (1), PROVE that the triangle is a right triangle.

In this case a=b. Together with equation (1), we have

    \[(a+b)^2=2c^2\implies (2a)^2=2c^2\implies c^2=2a^2=a^2+b^2\]

Alternatively, the bisector is a median in this case, and so the conclusion follows from example 1 and example 4.

Suppose that \triangle ABC satisfies equation (3). If the bisector of \angle C satisfies equation (1), PROVE that a^2-4ab+b^2=0.

We have: (a+b)^2=2c^2\& (a^2-b^2)^2 =c^2(a^2+b^2). Isolate c^2 and simplify:

    \[(a+b)^2=2\left(\frac{(a^2-b^2)^2}{a^2+b^2}\right)\implies a^2+b^2=2(a-b)^2\implies \scriptstyle a^2-4ab+b^2=0\]

Suppose that \triangle ABC satisfies equation (3). If the bisector of \angle C satisfies equation (1), PROVE that \angle A=15^{\circ}, \angle B=105^{\circ}, \angle C=60^{\circ}.

Since \cos C=\frac{2ab}{a^2+b^2} (see here), we have (by example 7 above):

    \[\cos C=\frac{2ab}{4ab}=\frac{1}{2}\implies \angle C=60^{\circ}\]

Also, \cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)} and \cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)} yield \angle A=105^{\circ} and \angle B=15^{\circ} (or \angle A=15^{\circ} and \angle B=105^{\circ}).

Further developments

Here are two out of a few more consequences.

Let a,b,c be the side-lengths of \triangle ABC. If equations (1) and (3) are satisfied, PROVE that the circumradius R is the geometric mean of a and b.

We have a^2+b^2=4R^2 and a^2-4ab+b^2=0. Eliminate a^2+b^2 to obtain R^2=ab.

If \triangle ABC satisfies equations (1) and (3), PROVE that the squared lengths h_C^2,l^2,R^2 form a geometric progression. Here, l is the length of the bisector of \angle C, h_C is the length of the altitude from vertex C, and R is the circumradius.

In example 4 we had l^2=\frac{1}{2}(ab); in example 9 above we had R^2=ab. Now let’s calculate h_C:

    \[h_C=\frac{ab}{2R} =\frac{ab}{\sqrt{a^2+b^2}}=\frac{ab}{\sqrt{4ab}}\implies h_C^2=\frac{1}{4}(ab)\]

We see that h_C^2,l^2,R^2 form a geometric progression with common ratio r=2.

Takeaway

In triangle ABC, let D be the foot of the bisector of \angle C. Then the two statements below are equivalent:

  • CD is the geometric mean of AD and DB
  • \sqrt{2} CD is the geometric mean of AC and CB.

Further, for any triangle ABC whose side-lengths a,b,c satisfy equation (1), the following two statements are equivalent:

  • \triangle ABC is a right triangle
  • a^2-2ab+b^2=0.

Finally, for any triangle ABC whose side-lengths a,b,c satisfy equation (1), the two statements below are equivalent:

Tasks

  1. Using \triangle ABC with vertices at A(-6,0), B(4,4), C(2,4), verify that:
    • its incenter is I\left(-3+\sqrt{5},-1+\sqrt{5}\right).
    • the ratio of the coordinates (x,y) of the incenter is golden; that is, it satisfies x^2-xy-y^2=0\implies \left(\frac{x}{y}\right)^2-\frac{x}{y}-1=0.
    • the slope of the bisector of \angle B is -\frac{1+\sqrt{5}}{2}, the negative of the golden ratio.
  2. Let a,b,c be the side-lengths of triangle ABC.
    • PROVE that the harmonic mean of \frac{ac}{a+b} and \frac{bc}{a+b} is \frac{2abc}{(a+b)^2}.
    • Deduce that the bisector of \angle C is the harmonic mean of the two segments on the opposite, if (a+b)^2=c^2+2c.
  3. Suppose that the side-lengths a,b,c of triangle ABC satisfy equation (3), namely (a^2-b^2)^2=(ac)^2+(cb)^2.
    • PROVE that the length l of the bisector of \angle C satisfies l^2=\frac{2a^2b^2}{a^2+b^2}.
    • Deduce that l^2 is the harmonic mean of a^2 and b^2.
  4. Suppose that \triangle ABC satisfies the usual Pythagorean identity c^2=a^2+b^2.
    • If the bisector of \angle A satisfies equation (1) adapted as (b+c)^2=2a^2, PROVE that a^2=8b^2.
    • Under the above, deduce that \sin A=\frac{2\sqrt{2}}{3}, \sin B=\frac{1}{3}.
  5. In \triangle ABC, let the side-lengths be a,b,c, and let R,h_C,l denote the circumradius, altitude from C, and the length of the bisector of \angle C. If \triangle ABC satisfies equation (3), PROVE that the squared lengths h_C^2,l^2,R^2 form a geometric progression with common ratio r=2.
  6. In \triangle ABC, let R,h_C,m_C denote the circumradius, altitude from C, and the median from C. If \triangle ABC satisfies equations (1) and (3), PROVE that m_C^2=h_C^2+R^2.
    (Basically a right triangle can be formed with the median m_C as the hypotenuse.)

A pseudo right triangle

We explore more properties of triangles whose side-lengths a,b,c satisfy a modified Pythagorean identity:

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

Unsurprisingly, such triangles mirror many of the properties of right triangles, with minor differences. So, for the time being, we’re giving them the pseudonym pseudo right triangles.

In our calculations we’ll frequently utilize the fact that equation (1) is equivalent to

(2)   \begin{equation*}a^2+b^2=4R^2\end{equation*}

where R is the circumradius of the parent triangle ABC.

Identities

A random selection of simple properties, in addition to the ones we already explored before and maybe in preparation for subsequent discussions.

Let H be the orthocenter of \triangle ABC with side-lengths a,b,c. If equation (1) is satisfied, PROVE that AH=b and BH=a.

    \begin{equation*} \begin{split} AH^2&=4R^2-a^2\\ &=(a^2+b^2)-a^2\\ &=b^2\\ \therefore AH&=b\\ \end{split} \end{equation*}

Once one has AH=b, one also has BH=a automatically. They imply each other. Symbolically, AH=b\iff BH=a.

Let H be the orthocenter of \triangle ABC with side-lengths a,b,c. If equation (1) is satisfied, PROVE that CH=2h_C, where h_C is the length of the altitude from vertex C.

    \begin{equation*} \begin{split} CH^2&=4R^2-c^2\\ &=(a^2+b^2)-c^2\\ &=(a^2+b^2)-\frac{(a^2-b^2)^2}{a^2+b^2}\\ &=\frac{(a^2+b^2)^2-(a^2-b^2)^2}{a^2+b^2}\\ &=\frac{4a^2b^2}{a^2+b^2}\\ h_C&=\frac{ab}{2R}\\ &=\frac{ab}{\sqrt{a^2+b^2}}\\ \therefore CH^2&=4h_C^2\\ CH&=2h_C \end{split} \end{equation*}

For a right triangle this would be CH=0, if \angle C=90^{\circ}. In general, where a right triangle has a “degenerate” property, a triangle satisfying equation (1) doesn’t seem to. A case in point is the orthic triangle.

If \triangle ABC satisfies equation (1), PROVE that the orthocenter H is a reflection of vertex C over side AB.

This follows from example 2 above, because we had CH=2h_C, together with the fact that h_C is the length of the altitude from C. In other words, the distance from C to the foot of the altitude from C is equal to the distance from the foot of this altitude to the orthocenter.

If \triangle ABC satisfies equation (1), PROVE that OH^2=R^2+CH^2, where H is the orthocenter, O the circumcenter, and R the circumradius.

    \begin{equation*} \begin{split} OH^2&=9R^2-(a^2+b^2+c^2)\\ &=9R^2-(4R^2+c^2)\\ &=5R^2-c^2\\ &=R^2+(4R^2-c^2)\\ &=R^2+CH^2 \end{split} \end{equation*}

So we can form a right triangle with hypotenuse equal to the Euler segment OH having R and CH as legs.

Suppose that \triangle ABC satisfies equation (1). PROVE that bs=ah_C and bp=ch_C, as per the diagram below.

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Using the fact that a^2+b^2=4R^2 and the Pythagorean theorem

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

gives bs=ah_C, as desired. Next, multiply both sides by p and use the fact that ap=cs from our previous post:

    \[p(bs)=p(ah_C)\implies bp(s)=(pa)h_C=(cs)h_C\implies bp=ch_C.\]

Suppose that \triangle ABC satisfies equation (1). PROVE that \frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}.

    \begin{equation*} \begin{split} \frac{h_A}{a}+\frac{h_B}{b}&=\frac{bc}{2R}\left(\frac{1}{a}\right)+\frac{ac}{2R}\left(\frac{1}{b}\right)\\ &=\frac{c}{2R}\left(\frac{a^2+b^2}{ab}\right)\\ &=\frac{c}{\sqrt{a^2+b^2}}\left(\frac{a^2+b^2}{ab}\right)\\ &=\frac{c\sqrt{a^2+b^2}}{ab}\\ &=\frac{c(2R)}{ab}\\ \frac{c}{h_C}&=\frac{c}{ab/(2R)}\\ &=\frac{c(2R)}{ab}\\ \therefore \frac{c}{h_C}&=\frac{h_A}{a}+\frac{h_B}{b} \end{split} \end{equation*}

If \frac{b}{a} is the golden ratio, then \frac{h_A}{a}+\frac{h_B}{b}=\frac{h_C}{c}.

Inequalities

If \triangle ABC satisfies equation (1), PROVE that qr> pa and qr> cs, as per the diagram below.

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In our previous post we had qr=p(p+a)=p^2+pa. Re-arrange as p^2=qr-pa. Then

    \[p^2> 0\implies qr-pa> 0\implies qr> pa\]

We also had ap=cs in our last post, so qr> pa too.

Inverses

Two “upside-down” Pythagorean identities.

If \triangle ABC satisfies equation (1), PROVE that \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{h_C^2}.

Since h_C^2=\frac{a^2b^2}{a^2+b^2}, we have

    \[\frac{1}{h_C^2}=\frac{a^2+b^2}{a^2b^2}=\frac{1}{a^2}+\frac{1}{b^2}\]

(This happens in a right triangle too if \angle C=90^{\circ}.)

Suppose that \triangle ABC satisfies equation (1). PROVE that \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2} if \frac{a}{b} is the golden ratio.

Equation (1) says that c^2(a^2+b^2)=(a^2-b^2)^2. Now, if \frac{a}{b} is the golden ratio, then a^2-ab-b^2=0\implies a^2-b^2=ab.

    \begin{equation*} \begin{split} c^2(a^2+b^2)&=(a^2-b^2)^2\\ \frac{1}{c^2}&=\frac{a^2+b^2}{(a^2-b^2)^2}\\ &=\frac{a^2+b^2}{(ab)^2}\\ \implies \frac{1}{c^2}&=\frac{1}{a^2}+\frac{1}{b^2} \end{split} \end{equation*}

(Also happens in a Kepler triangle \cdots.)

Consider triangle ABC which satisfies equation (1). With reference to the diagram below, PROVE that \frac{b}{a} is the geometric mean of \frac{1}{s} and c+s and hence \frac{a}{b} is the geometric mean of s and \frac{1}{c+s}.

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In example 5 we had s=\frac{a}{b}h_C, so (assuming b> a) we have

    \[s=\frac{a}{b}\frac{ab}{2R}=\frac{a^2}{\sqrt{a^2+b^2}}=\frac{a^2c}{b^2-a^2}\]

Re-arrange:

    \[s(b^2-a^2)=a^2c\implies sb^2=a^2(c+s)\implies \left(\frac{b}{a}\right)^2=\frac{c+s}{s}\]

This would also imply \left(\frac{a}{b}\right)^2=\frac{s}{c+s}. And so \frac{b}{a} is the geometric mean of \frac{1}{s} and c+s, while \frac{a}{b} is the geometric mean of s and \frac{1}{c+s}.

Takeaway

For any obtuse triangle ABC satisfying equation (1), the following statements are equivalent:

  • \frac{a}{b} is the golden ratio
  • (ab)^2=(ac)^2+(cb)^2
  • \sin A\sin B=\sin C
  • \frac{h_A}{a}+\frac{h_B}{b}=\frac{h_C}{c}
  • \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}.

The golden ratio acts as a “bridge” between a Kepler triangle and any obtuse triangle satisfying equation (1).

Tasks

  1. (Keeping track) Let a,b,c be the side-lengths of an obtuse triangle ABC, R its circumradius, O its circumcenter, H its orthocenter, and h_A,h_B,h_C the altitudes. PROVE that each of the following statements implies the others:
    • AH=b
    • BH=a
    • CH=2h_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
    • a^2+b^2=4R^2
    • 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)^2=(ac)^2+(cb)^2
    • A-B=90^{\circ} or B-A=90^{\circ}
    • AH^2+BH^2+CH^2=8R^2-c^2
    • the orthic triangle is isosceles
    • the geometric mean theorem holds
    • 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
      (21 statements to help you remember 2021. Aside: the ratio of the length of the angle bisector of \angle C to the length of the altitude from vertex C is \sqrt{2}, under the above equivalence.)
  2. (Kepler triangle) Suppose that \triangle ABC satisfies the usual Pythagorean identity c^2=a^2+b^2. PROVE that the following four statements are equivalent:
    • b,a,c form a geometric progression
    • \frac{c}{b} is the golden ratio
    • (ac)^2=(ab)^2+(bc)^2
    • \frac{1}{a^2}=\frac{1}{b^2}+\frac{1}{c^2}.
  3. (Kosnita theorem) If \triangle ABC satisfies equation (1), PROVE that the Kosnita point coincides with vertex C. (Somewhat exciting property.)
  4. PROVE that the two statements below are equivalent for a triangle ABC:
    • R=\frac{a^2-b^2}{2c}
    • h_C=R\cos C
      (If these were added to the original 21 statements in the first exercise, we will obtain a total of 23. Cool. In terms of sheer numbers, we really wanted something that mirrors the Invertible Matrix Theorem. Reached.)
  5. Consider \triangle ABC with vertices at A(0,0), B(4,4), C(1,-2). Its orthocenter is H(-2,1), and its circumcenter is O\left(\frac{7}{2},\frac{1}{2}\right).
    • Verify that the foot of the altitude from vertex C is F_C\left(-\frac{1}{2},-\frac{1}{2}\right). Deduce that H is the reflection of vertex C over side AB.
    • PROVE that the midpoint of AB together with H,O,F_C form a parallelogram.
      (The midpoint of AB coincides with our favourite point W in this case.)