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.

Rendered by QuickLaTeX.com

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.

Rendered by QuickLaTeX.com

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}.

Rendered by QuickLaTeX.com

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.)