2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Interestingly the triangles $\{a,b,c\}$ and $\{a+b,b+c,c+a\}$ are always congruent. $\endgroup$
    – Joffan
    Mar 9 at 18:27
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Very nice solution!Thank you! $\endgroup$
    – alien2003
    Mar 9 at 20:59
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 $\endgroup$
    – alien2003
    Mar 9 at 20:29
1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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 $\endgroup$
    – alien2003
    Mar 9 at 15:53
  • $\begingroup$ 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 $\endgroup$
    – alien2003
    Mar 9 at 15:58
  • $\begingroup$ @alien2003 - let me think backwards $\endgroup$
    – Quanto
    Mar 9 at 16:02
  • $\begingroup$ Ok.The backwards part is the harder part so it would be really good if you could do it.I failed to do it. $\endgroup$
    – alien2003
    Mar 9 at 16:05
  • $\begingroup$ @alien2003 - see the edit for proving the reverse. $\endgroup$
    – Quanto
    Mar 9 at 22:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.