TIFR 2011 complex number

Question

Three distinct points $x, y, z$ lie on a unit circle of the complex plane and satisfy $x+y+z=0$. Then $x, y, z$ form the vertices of.

An isosceles but not equilateral triangle.
An equilateral triangle.
A triangle of any shape.
A triangle whose shape can't be determined.
None of the above.

Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. — Martin R, Jun 12 '19 at 17:35
@MartinR I'm new in this topic, I'm unable to solve, please explain. — Lakshman Patel, Jun 12 '19 at 17:37
What does $x + y + z = 0$ mean? That if you treat them as vectors, their sum is the zero vector? — paulinho, Jun 12 '19 at 17:42
I just want to say that the answer is 2) An equilateral triangle using symmetry — xrfxlp, Jun 12 '19 at 17:51
https://math.stackexchange.com/questions/1397066/clarification-regarding-a-question — lab bhattacharjee, Jun 12 '19 at 21:07

Somos · Answer 1 · 2019-06-13T20:09:34.380

1

The points $0, x, y, x+y$ form a parallelogram but $x+y=-z$ which implies $|x+y|=|z|=|x|=|y|$. This means that both $0, x, x+y$ and $0, y, x+y$ form congruent equilateral triangles. Similarly, $x, -z, y, -x, z, -y$ form the vertices of a regular hexagon and hence $x, y, z$ form an equilateral triangle.

Another solution: Suppose $a, b, c$ are three points. Assume that thetriangle ($\triangle abc$) has centroid $d$ which is also the center of the circumcircle ($\bigcirc abc$). By using a complex linear transformation ($\, z \mapsto (z-d)/(c-d)\,$) we can assume (without loss of generality) that $$d=0,\quad c=1,\quad a+b=-1, |a|=|b|=1.$$ Now also $\,\bar{a}+\bar{b}=-1\,$ and this implies the diagonals of quadrilateral $\,a\bar{b}b\bar{a}\,$ bisect each other and hence is a parallelogram. This now implies that $\,a=\bar{b}\,$ and they are primitive curbe roots of unity. Hence $\triangle abc$ is equilateral.

edited Jun 13 '19 at 20:09

answered Jun 12 '19 at 20:14

Somos

37,457
3
35
85

+1 Another way along these lines: $0,x,y,x+y$ clearly forms a rhombus with side lengths $1$. Then $x,z,x+y$ is clearly a right triangle with hypotenuse 2 and short side 1, so it is a 30-60-90. The same is true for $y,z, x+y$. So the angle $\angle xzy$ is 60. By symmetry, so are the remaining angles in the triangle formed by $xyz$ – rschwieb Jun 13 '19 at 14:31
@rschwieb I don't see the reason why $x,z,x+y$ is a right triangle. It is not clear to me. What is the reason for it? – Somos Jun 13 '19 at 16:45
1

that's a classic theorem of geometry: a triangle inscribed in a circle such that the hypotenuse is the diameter is a right triangle. – rschwieb Jun 13 '19 at 16:46
@rschwieb You are right, of course. – Somos Jun 13 '19 at 16:48
Actually, since the focus of this question is complex numbers, I was really hoping to find an algebraic solution... but as usual it seems much easier to reason geometrically. – rschwieb Jun 13 '19 at 16:51

score 0 · Answer 2 · answered Jun 12 '19 at 18:04

Let's prove the answer is 2).

Write $x=e^{ia}$ etc. so $e^{ia}+e^{ib}+e^{ic}=0$ and $e^{i(b-a)}+e^{i(c-a)}=-1$. Equating imaginary parts, $\sin(b-a)=\sin(a-c)$; equating real ones, $\cos(b-a)+\cos(a-c)=-1$. The cosines are equal, rather than equal and opposite, and so each is $-\frac12$, whence $\sin(b-a)=\pm\frac{\sqrt{3}}{2}=\pm\sin\frac{2\pi}{3}$. Note in particular that $\cos(b-a)=-\frac12\implies b-a\ne\pm\frac{\pi}{3}$.

The angles between $x,\,y,\,z$ are $\frac{2\pi}{3}$ each, while the circumradii at the vertices split $\triangle xyz$ into isosceles triangles of base angles $\frac12\left(\pi-\frac{2\pi}{3}\right)=\frac{\pi}{6}$. Thus each internal angle of $\triangle xyz$ is $\frac{\pi}{3}$, as expected.

TIFR 2011 complex number

2 Answers2