A pseudo right triangle – High School Geometry

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.

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.

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 $\frac{1}{c+s}$

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

(Keeping track) Let be the side-lengths of an obtuse triangle , its circumradius, its circumcenter, its orthocenter, and 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.)
(Kepler triangle) Suppose that satisfies the usual Pythagorean identity . PROVE that the following four statements are equivalent:
- $b,a,c$ form a geometric progression
- $\frac{c}{b}$ is the golden ratio
- $(ac)^2=(ab)^2+(bc)^2$
- $\frac{1}{a^2}=\frac{1}{b^2}+\frac{1}{c^2}$ .
(Kosnita theorem) If $\triangle ABC$ satisfies equation (1), PROVE that the Kosnita point coincides with vertex $C$ . (Somewhat exciting property.)
PROVE that the two statements below are equivalent for a triangle :
- $R=\frac{a^2-b^2}{2c}$
- $h_C=R\cos C$
  (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.)
Consider with vertices at , , . Its orthocenter is , and its circumcenter is .
- 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.)