An extension of the extended sine law

In any triangle $ABC$

with circumcenter $O$

and circumradius $R$

, the extended sine law is:

(1) $\begin{equation*} \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R \end{equation*}$

If we now denote by $R_a,R_b,R_c$ the radii of the circumcircles of triangles $BCO,CAO,ABO$ in that order, then:

(2) $\begin{equation*} \scriptstyle R_a\cos A=R_b\cos B=R_c\cos C=\frac{R}{2}} \end{equation*}$

Extension confirmed

In one line of equations, the extension that arises from a combination of (1) and (2) is:

(3) $\begin{equation*} \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R=4R_a\cos A=4R_b\cos B=4R_c\cos C \end{equation*}$

Please don’t take equation (3) too seriously: there are many relations in Euclidean geometry that can be expressed in terms of the circumradius of a triangle, and equation (3) is just one of them. Meanwhile, note that:

one of the cosine terms in (2) (or (3)) requires an absolute value if the parent triangle is obtuse
only right triangles fail some of the equalities in (2) (or (3)).

Let $ABC$

be an acute triangle with circumcenter $O$

. Denote by $R_a,R_b,R_c$

the circumradii of the triangles $BCO,CAO,ABO$

, in that order. Verify equation (2), namely: $R_a\cos A=R_b\cos B=R_c\cos C=\frac{R}{2}$

, where $R$

is the circumradius of the parent triangle $ABC$

Whether the parent triangle is acute or obtuse, at least one of the associated triangles $BCO,CAO,ABO$ has to be obtuse. However, this won’t affect the calculations since all we have to do is use the extended sine law.

Here $\triangle ABC$ is acute, so it’s circumcenter $O$ is located inside the triangle as shown above. Consider $\triangle BCO$ . By the inscribed angle theorem, we have:

$\angle BOC=2\alpha\implies \angle OBC=\angle OCB=90-\alpha$

Now apply the extended sine law to triangle $BOC$ , noting that $BO=OC=R$ :

$2R_a=\frac{R}{\sin(90-\alpha)}=\frac{a}{\sin 2\alpha}$

and so $R=2R_a\cos \alpha$ . In the case where one of the triangles $BCO,CAO,ABO$ is obtuse, we also obtain the same result.

Let $ABC$

be an obtuse triangle with circumcenter $O$

. Denote by $R_a,R_b,R_c$

the circumradii of the triangles $BCO,CAO,ABO$

, in that order. Verify equation (2), namely: $R_a\cos A=R_b\cos B=R_c\cos C=\frac{R}{2}$

, where $R$

is the circumradius of the parent triangle $ABC$

In this case, the circumcenter is located outside the triangle. A sample diagram is shown below:

In $\triangle AOC$ , angle $AOC$ is equal to $360-2\beta$ , while angles $OAC$ and $OCA$ are both equal to $\beta-90$ . Apply the extended sine law again to get:

$2R_b=\frac{R}{\sin(\beta-90)}\implies R=2R_b\sin(\beta-90)=-2R_b\cos\beta$

This is where an absolute value is needed.

Exception circumvented

If the parent triangle $ABC$ is right-angled, let’s say at $C$ , then $\cos C=0$ and so we need modify equation (2) slightly.

Let $ABC$

be a right triangle in which $\angle C=90^{\circ}$

. PROVE that $aR_b=bR_a$

We would have $\cos A=\frac{b}{c}$ and $\cos B=\frac{a}{c}$ . Using equation (2) we have:

$R=2R_a\cos A=2R_b\cos B\implies R=2R_a\left(\frac{b}{c}\right)=2R_b\left(\frac{a}{c}\right)$

This gives $aR_b=bR_a$ .

Let $ABC$

be a right triangle in which $\angle C=90^{\circ}$

. PROVE that $R^2=aR_b$

and $R^2=bR_a$

Note that since triangle $ABC$ is right-angled at $C$ , we have $\cos A=\frac{b}{c}$ . From $R=2R_a\cos A$ isolate $R_a$ :

$R_a=\frac{cR}{2b}=\frac{(2R)R}{2b}=\frac{R^2}{b}\implies R^2=bR_a$

Similarly, $R^2=aR_b$ . Together we get $R^2=aR_b=bR_a$ . And

$\frac{R_a}{a}=\frac{R_b}{b}=\frac{\sqrt{R_a^2+R_b^2}}{\sqrt{a^2+b^2}}.$

Exploration continues

We’ll have need to use equation (2) in future. For now, consider what happens in the $30^{\circ}-30^{\circ}-120^{\circ}$ isosceles triangle.

In the $30^{\circ}-30^{\circ}-120^{\circ}$

isosceles triangle, PROVE that $\frac{R_a}{a}=\frac{R_b}{b}=\frac{R_c}{c}$

Suppose that $\angle B=120^{\circ}$ . Then $b^2=3a^2=3c^2$ . Using equation (2):

$\begin{equation*} \begin{split} R_a\cos 30^{\circ}&=R_b|\cos 120^{\circ}|=R_c\cos 30^{\circ}\\ R_a\left(\frac{\sqrt{3}}{2}\right)&=R_b\left(\frac{1}{2}\right)=R_c\left(\frac{\sqrt{3}}{2}\right)\\ R_a\left(\frac{\sqrt{3}}{2abc}\right)&=R_b\left(\frac{1}{2abc}\right)=R_c\left(\frac{\sqrt{3}}{2abc}\right)\\ \therefore\frac{R_a}{a}&=\frac{R_b}{b}=\frac{R_c}{c} \end{split} \end{equation*}$

Equation changes

An update about our ongoing hide-and-seek game with some hackers. As you can see below, we were initially hoping to use an exponential equation to model their menace, but the data suddenly changed after October 28, and so did our equation.

How wonderful would it be if the hackers spent all that time focusing on analytic geometry?

Takeaway

In an obtuse triangle $ABC$ , let $R_a,R_b,R_c$ be as mentioned in the post. Then the following statements are equivalent:

$(b^2-a^2)^2=(ac)^2+(cb)^2$
$aR_b=bR_a$ .

If instead triangle $ABC$ is acute, then the following statements are equivalent:

$c^2=a^2+b^2$
$aR_b=bR_a$ .

Tasks

(Similar triangles) Let be a non-right triangle with circumcenter . Denote by the circumcenters of the triangles , in that order.
1. PROVE that $\triangle O_aO_bO_c$ is similar to the orthic triangle
2. Find the similarity ratio.
(Similar triangles) Let be a right triangle with circumcenter and . Denote by the circumcenters of the triangles , in that order.
1. PROVE that $\triangle O_aO_bO$ is similar to the parent triangle $ABC$
2. Find the similarity ratio.