A note on Kepler triangle – High School Geometry

A Kepler triangle is first of all a right triangle, and so it enjoys the Pythagorean identity:

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

In addition, it has the property that its side-lengths form a geometric progression, for example:

(2) $\begin{equation*} b,a,c\iff a^2=bc \end{equation*}$

Equations (1) and (2) force the ratio of the hypotenuse $c$ to one of the legs ( $a$ or $b$ ) to be golden:

(3) $\begin{equation*} \frac{c}{b}=\varphi,~~\varphi:=\textrm{golden ratio} \end{equation*}$

In this post, we show that a Kepler triangle also satisfies

(4) $\begin{equation*} (ac)^2=(ab)^2+(bc)^2 \end{equation*}$

which, in conjunction with equation (1), constitutes what one may call a double Pythagorean identity. We also show that if any of the four equations above hold, then the remaining three become equivalent.

Show that the Kepler triangle satisfies equation (4).

For the Kepler triangle we have $c=\varphi,~b=1,~a=\sqrt{\varphi}$ , where $\varphi$ is the golden ratio. By definition of $\varphi$ , we have $\varphi^2=\varphi+1$ and so $c^2=a^2+b^2$ .

$\begin{equation*} \begin{split} \varphi^2&=\varphi+1\\ \implies\varphi^3&=\varphi^2+\varphi\\ (ac)^2&=\left(\sqrt{\varphi}\times\varphi\right)^2\\ &=\varphi^3\\ (ab)^2+(bc)^2&=\left(\sqrt{\varphi}\times 1\right)^2+\left(1\times\varphi\right)^2\\ &=\varphi+\varphi^2\\ \therefore (ac)^2&=(ab)^2+(bc)^2 \end{split} \end{equation*}$

Note that a random right triangle need not satisfy equation (4). For example, the very familiar case with $c=5$ , $b=4$ , $a=3$ doesn’t.

Simple characterizations

A couple of trivial equivalent statements that offer alternate descriptions of the Kepler triangle.

Suppose that a triangle satisfies equation (1). PROVE that equations (2), (3), (4) become equivalent.

In other words, the following three statements are equivalent for any right triangle (in which $c^2=a^2+b^2$ ):

the side-lengths form a geometric progression $b,a,c$
the ratio $\frac{c}{b}$ is the golden ratio
the identity $(ac)^2=(ab)^2+(bc)^2$ holds.

The result follows if we show that $(1)\implies(2)\implies(3)\implies(1)$ .

So let’s suppose that $(1)$ holds, that is, the sequence $b,a,c$ is geometric. This gives $a^2=bc$ . Since we also have that $c^2=a^2+b^2$ , it means that

$c^2=bc+b^2\implies c^2-bc-b^2=0\implies \left(\frac{c}{b}\right)^2-\frac{c}{b}-1=0$

Thus, $\frac{c}{b}$ satisfies the defining equation for the golden ratio, and so $(2)$ holds.

Next, we show that $(2)\implies(3)$ . So suppose that $\frac{c}{b}$ is the golden ratio. Then

$\left(\frac{c}{b}\right)^2-\frac{c}{b}-1=0\implies c^2=bc+b^2\implies c^2-b^2=bc$

Multiply both sides of $c^2=a^2+b^2$ by $a^2$ to get

$a^2c^2=a^2a^2+a^2b^2\implies (ac)^2=(ab)^2+(bc)^2$

since $a^2=c^2-b^2$ from the Pythagorean identity and also $c^2-b^2=bc$ from the golden ratio. Thus, $(3)$ holds.

Finally, we show that $(3)\implies(1)$ . We have $(ac)^2=(ab)^2+(bc)^2$ and $c^2=a^2+b^2$ :

$a^2(a^2+b^2)=a^2b^2+b^2c^2\implies a^4=b^2c^2\implies a^2=bc$

So the sequence $b,a,c$ is geometric, and $(1)$ is proved.

Easy equivalence.

Suppose that a triangle satisfies equation (2). PROVE that equations (1), (3), (4) become equivalent.

In other words, the following three statements are equivalent for any triangle in which $a^2=bc$ :

the triangle is a right triangle with $c^2=a^2+b^2$
the ratio $\frac{c}{b}$ is the golden ratio
the identity $(ac)^2=(ab)^2+(bc)^2$ holds.

Follow the same argument as in example 2, bearing in mind that $a^2=bc$ holds throughout in this case.

Suppose that a triangle satisfies equation (3). PROVE that equations (1), (2), (4) become equivalent.

In other words, the following three statements are equivalent for any triangle in which $\frac{c}{b}=\varphi$ :

the triangle is a right triangle with $c^2=a^2+b^2$
the side-lengths form a geometric progression $b,a,c$
the identity $(ac)^2=(ab)^2+(bc)^2$ holds.

Follow the same argument as in example 2, bearing in mind that $\frac{c}{b}=\varphi$ holds throughout in this case.

Suppose that a triangle satisfies equation (4). PROVE that equations (1), (2), (3) become equivalent.

In other words, the following three statements are equivalent for any triangle in which $(ac)^2=(ab)^2+(bc)^2$ :

the triangle is a right triangle with $c^2=a^2+b^2$
the side-lengths form a geometric progression $b,a,c$
the ratio $\frac{c}{b}$ is the golden ratio $\varphi$ .

Follow the same argument as in example 2, bearing in mind that $(ac)^2=(ab)^2+(bc)^2$ holds throughout in this case.

Similar concept

Given $\triangle ABC$ in which $\angle C=90^{\circ}$ , the Pythagorean theorem states that

(5) $\begin{equation*} AC^2+CB^2=AB^2 \end{equation*}$

(Or the usual $a^2+b^2=c^2$ .) But then the left side of equation (5) is just the sum of the squares of two altitudes — the altitude from vertex $A$ and the altitude from vertex $B$ — and so we can re-write (5) as:

(6) $\begin{equation*} h_A^2+h_B^2=AB^2 \end{equation*}$

where $h_A$ and $h_B$ are the altitudes from $A$ and $B$ . The latter equation expresses the fact that in a right triangle, the sum of the squares of two altitudes is equal to the square of the third side (the hypotenuse). We’ll use this intrinsic characteristic of right triangles as the defining condition for the triangles under consideration in the second part of today’s discussion.

PROVE that $h_A^2+h_B^2=AB^2$

if, and only if, $a^2+b^2=4R^2$

, where $R$

is the circumradius of triangle $ABC$

First suppose that $h_A^2+h_B^2=AB^2$ . Since $h_A=\frac{bc}{2R}$ and $h_B=\frac{ac}{2R}$ , we get

$\left(\frac{bc}{2R}\right)^2+\left(\frac{ac}{2R}\right)^2=c^2\implies a^2+b^2=4R^2$

Conversely, suppose that $a^2+b^2=4R^2$ . Then:

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

QED: $h_A^2+h_B^2=AB^2\iff a^2+b^2=4R^2.$

PROVE that $\cos C=0$

or $\cos C=\frac{2ab}{a^2+b^2}$

, if $a^2+b^2=4R^2$

Begin with $R=\frac{c}{2\sin C}$ as per the extended law of sines. The rest is:

$\begin{equation*} \begin{split} a^2+b^2&=4\left(\frac{c}{2\sin C}\right)^2\\ (a^2+b^2)\sin^2C&=c^2\\ (a^2+b^2)(1-\cos^2C)&=a^2+b^2-2ab\cos C\\ \cos C\left((a^2+b^2)\cos C-2ab\right)&=0\\ \implies \cos C&=0,\frac{2ab}{a^2+b^2} \end{split} \end{equation*}$

PROVE that $(a^2-b^2)^2=(ac)^2+(bc)^2$

, if a non-right triangle satisfies $a^2+b^2=4R^2$

Since the given triangle is not a right triangle, we take $\cos C=\frac{2ab}{a^2+b^2}$ from the preceding example. However, since $\cos C$ is always equal to $\frac{a^2+b^2-c^2}{2ab}$ in any triangle, we have:

$\begin{equation*} \begin{split} \frac{2ab}{a^2+b^2}&=\frac{a^2+b^2-c^2}{2ab}\\ (a^2+b^2)\Big(a^2+b^2-c^2\Big)&=4a^2b^2\\ (a^2+b^2)^2-4a^2b^2&=(a^2+b^2)c^2\\ (a^2-b^2)^2&=(ac)^2+(bc)^2 \end{split} \end{equation*}$

Similar to equation (4) for the Kepler triangle. Let’s take a shot at making it exact.

Consider a non-right triangle that satisfies $a^2+b^2=4R^2$

. PROVE that $(ab)^2=(ac)^2+(cb)^2$

if, and only if, $\frac{a}{b}$

is the golden ratio.

First suppose that $(ab)^2=(ac)^2+(cb)^2$ . Since $(a^2-b^2)^2=(ac)^2+(bc)^2$ from the preceding example we must then have $(a^2-b^2)^2=(ab)^2$ . Positive square roots: $a^2-b^2=ab$ , or $\left(\frac{a}{b}\right)^2-\frac{a}{b}-1=0$ , and so $\frac{a}{b}$ is the golden ratio. The converse is also easy. So basically, we have non-right triangles that tend to behave like the Kepler triangle.

Special case

A sample triangle that satisfies equation (6) twice.

Consider an isosceles triangle $ABC$

in which $\angle A=120^{\circ}$

, $\angle B=30^{\circ}$

, and $\angle C=30^{\circ}.$

For this triangle we have:

$\begin{equation*} \begin{split} h_A^2+h_B^2&=AB^2\\ h_A^2+h_C^2&=AC^2\\ h_B^2+h_C^2&=\frac{1}{2}BC^2\\ h_B+h_C&=BC.\\ \end{split} \end{equation*}$

Takeaway

For any non-right triangle $ABC$ with side-lengths $a,b,c (a\neq b)$ , circumradius $R$ , and altitudes $h_A,h_B,h_C$ , the five statements below are equivalent:

$h_C=R\cos C$
$\cos C=\frac{2ab}{a^2+b^2}$
$a^2+b^2=4R^2$
$h_A^2+h_B^2=AB^2$
$(a^2-b^2)^2=(ac)^2+(bc)^2.$

In addition, if a non-right triangle satisfies any of the equivalent conditions above, then each of the following three statements implies the others:

$\frac{a}{b}$ is the golden ratio
$(ab)^2=(ac)^2+(cb)^2$
$a,\sqrt[4]{5}c,b$ is a geometric sequence.

In a nutshell, we have a non-right triangle version of the Kepler triangle.

Tasks

If a non-right triangle satisfies equation (6), PROVE that the length of the altitude from vertex $C$ can be given by $h_C=R\cos C$ .
Let be the side-lengths of a right triangle, where is the hypotenuse, and let be its circumradius. PROVE that:
- $a^2+b^2=4R^2$
- $a,c,b$ (in that order) cannot be a geometric sequence.
PROVE: For any non-right triangle with side-lengths , circumradius , and altitudes , the five statements below are equivalent:
- $h_C=R\cos C$
- $\cos C=\frac{2ab}{a^2+b^2}$
- $a^2+b^2=4R^2$
- $h_A^2+h_B^2=AB^2$
- $(a^2-b^2)^2=(ac)^2+(bc)^2.$
Suppose that a non-right triangle satisfies any of the equivalent conditions in the preceding exercise. PROVE that:
- $HO^2=5R^2-c^2$
- $AH^2+BH^2+CH^2=8R^2-c^2$
- $4(AN^2+BN^2+CN^2)=7R^2+c^2$ , where $H=$ orthocenter, $O=$ circumcenter, $N=$ nine-point center
Consider with vertices at , , and . Verify that:
- the circumradius is $R=5$
- $a^2+b^2=4R^2$ .