# Prove that $\Delta ABC$ is an equilateral triangle if and only if the complex coordinates $a$,$b$, and $c$ satisfy a relation.

Let $$\triangle ABC$$ be a triangle in the complex plane and let $$a$$,$$b$$ and $$c$$, respectively, be the complex coordinates of its vertices. Suppose that the tiangle is inscribed in the circle $$C(0,1)$$. Prove that $$\triangle ABC$$ is equilateral if and only if $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$$ and $$(a+b)(b+c)(c+a)\neq0\,.$$

I tried to convert $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$$ into a simpler equation by multipling it by $$(a+b)(b+c)(c+a)$$

In the end I obtained $$a^2+b^2+c^2+3ab+3bc+3ca=0$$ but I don't know how to continue from here. Can you help me, please?

• Interestingly the triangles $\{a,b,c\}$ and $\{a+b,b+c,c+a\}$ are always congruent. Mar 9 at 18:27

One direction (the direction assuming that $$ABC$$ is an equilateral triangle) is obvious. Here's a geometric proof of the converse.

Let $$k$$ be the given unit circle. If $$D$$, $$E$$, and $$F$$ are the midpoints of $$BC$$, $$CA$$, and $$AB$$, then note that the complex coordinates of the images $$D'$$, $$E'$$, and $$F'$$ of $$D$$, $$E$$, and $$F$$ under the inversion about $$k$$ are $$\dfrac{2}{\bar{b}+\bar{c}}$$, $$\dfrac{2}{\bar{c}+\bar{a}}$$, and $$\dfrac{2}{\bar{a}+\bar{b}}$$, respectively. Note that the origin $$O$$ is the incenter of the triangle $$D'E'F'$$. Your equation says that the centroid of the triangle $$D'E'F'$$ coincides with its incenter. This only happens when $$D'E'F'$$ is an equilateral triangle, whence $$ABC$$ is also an equilateral triangle.

• Very nice solution!Thank you! Mar 9 at 20:59

During all the proof I will call $$a=z_0$$, $$b=z_1$$ and $$c=z_2$$.

$$\Longrightarrow)$$

If $$z_0$$, $$z_1$$ and $$z_2$$form a equilateral triangle inscribed in $$C(0,1)$$, then we can write $$z_j$$ as $$z_j=m\cdot e^{i\frac{\pi j}{3}};\quad \text{with } |m|=1.$$ Then, since $$z_j$$ are rotations of the cubic roots of $$1$$, then $$\sum\limits_{j=0}^2 m\cdot e^{i\frac{\pi j}{3}}=m\cdot \underbrace{\sum\limits_{j=0}^2 e^{i\frac{\pi j}{3}}}_{=0}=0.$$

The other condition is easy to proof since none of the $$z_j$$ are diametrically opposite (ie, an equilateral triangle inscribed in a circumference can't have two vertices diametrically opposite).

$$\Longleftarrow)$$

The proof of this statement can be found (with this same notation) here.

• The first way of the problem is correct,thanks.But the other way around is not too clear.I couldn't find it on the link Mar 9 at 20:29

Given equilateral triangle, i.e.

$$a= e^{i\alpha}, \>\>\>\>\>b = e^{i(\alpha+\frac{2\pi} 3)}, \>\>\>\>\>c= e^{i(\alpha-\frac{2\pi} 3)}$$

it is straightforward to verify $$a^2+b^2+c^2+3ab+3bc+3ca=0$$.

Conversely, given

$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$$

rewrite the equation in term of $$b =a e^{i x}$$, $$c =a e^{ i y}$$

$$\frac{1}{1+ e^{i x}}+\frac{1}{1+ e^{i y}}+\frac{1}{e^{i x} + e^{i y}}=0\tag1$$

Suppose $$(x,y)$$ is a solution, then $$(y,x)$$ is also a solution due to party, as well as $$(-x,-y)$$ via conjugate. which implies $$x = -y$$. Substitute into (1) to obtain $$\cos x=-\frac12$$, or $$x= -y= \pm \frac {2\pi}3$$. Thus

$$b = a e^{\pm i\frac{2\pi}3}, \>\>\>\>\>c=a e^{\mp i \frac{2\pi} 3}$$

hence, equilateral triangles.

• Thank you, but I think that your solution is incomplete.You proved that if $\Delta ABC$ is equilateral,then the equation happens.But you haven't proved the opposite afirmation which is the hard part.This is an "if and only if" problem Mar 9 at 15:53
• Yes,that is both ways.I was talking about $a+b+c=0$.That is true only the first way,only if the triangle is equilateral Mar 9 at 15:58
• @alien2003 - let me think backwards Mar 9 at 16:02
• Ok.The backwards part is the harder part so it would be really good if you could do it.I failed to do it. Mar 9 at 16:05
• @alien2003 - see the edit for proving the reverse. Mar 9 at 22:58