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Let $A, B, C, D$ be points on a unit circle. Prove that if $A+B+C+D=0$, then $A,B,C,D$ make a rectangle. (Use complex numbers.)

How do I prove this? I tried to use the dot product of 2 adjacent sides, but I got an ugly trig expression.

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1  
Then don't convert the dot product into trig! If you post the details of what you have tried, people can give you suggestions as to what you've overlooked or what you're doing wrong, and what you can do next.... –  Hurkyl Apr 2 '13 at 6:44
    
You might consider the representation of the points as means of either two of them and their positive/negative deviations from the means. Then with simple algebra some properties pop up, for instance, that the real and the complex values of the means must be negative of each other. I didn't go through this completely, but it might be a nice approach avoiding the complicated trig-formulae. (Upps, after posting this I see the answer of @Ivan which seems to use the similar approach) –  Gottfried Helms Apr 2 '13 at 7:11
    
... I should add, that the formula for the unit-length of the variables $a,b,c,d$ (when taken as vectors in the complex plane, $a_r^2 + a_i^2=1$ with the representation for the complex numbers $a = a_r + a_i\cdot i$) is used. –  Gottfried Helms Apr 2 '13 at 7:17

6 Answers 6

up vote 2 down vote accepted

Well, if you really want a proof which uses complex numbers...

If $A+B \not =0, A+C \not =0$, $$(\frac{A+B}{2}) \cdot (A-B)=0$$ $$(\frac{A+B}{2}) \cdot (C-D)=(-\frac{C+D}{2}) \cdot (C-D)=0$$ (Here we are using dot product)

Since $A+B \not =0$, then the vector represented by $(\frac{A+B}{2})$ is perpendicular to $AB,$ and $CD$, so $AB//CD$. Similarly $AC//BD$, since $A+C \not =0$. Thus $ABDC$ is a paralellogram, so $A-B=C-D$ (since $A-B \not =D-C$), giving $A+D=B+C=0$.

Thus either $A+B=0, A+C=0,$ or $A+D=0$.

By symmetry it suffices to consider when $A+B=0$, then $C+D=0$. $(A-C) \cdot (B-C)=(A-C) \cdot (-A-C)=0$ so $AC \perp BC$. Similarly the other 3 angles are also right angles, so we get a rectangle.

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How do you derive your first equality ? –  Ewan Delanoy Apr 2 '13 at 7:48
    
@EwanDelanoy after removing the factor $\frac{1}{2}$, we get $A \cdot A-B \cdot B=0$, since $|A|=|B|=1$. –  Ivan Loh Apr 2 '13 at 7:55
    
nonsense. For example we have $|1|=|i|$ but $1\times1-i\times i$ is not zero. –  Ewan Delanoy Apr 2 '13 at 7:56
    
@EwanDelanoy dot product. –  Ivan Loh Apr 2 '13 at 7:57
    
ok, I got it now. –  Ewan Delanoy Apr 2 '13 at 7:58

The sum of two unit vectors lies on the line that bisects the angle between them, and the length of the sum determines the angle.

Having two equal and opposite such sums forces the existence of a symmetry relating one pair of summands to the other. Four points on a circle that can be divided into two pairs related by a symmetry, form a rectangle.

Maybe I am missing an extremely simple solution with complex numbers, but this seems to be a pure geometry problem where complex numbers do not help much. Of course you can prove the geometry statements using complex numbers, as an exercise.

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That should say: four points on a circle divided symmetrically into pairs, with an additional symmetry relating the pairs, form a rectangle. –  zyx Apr 3 '13 at 19:49

Let $A+B = 2x$. Then $C+D = -2x$. By rotating the four points (i.e. by multiplying on both sides of the two equations by $e^{-i\arg x}$), we may assume WLOG that $x$ is real. Hence $A,B,C,D$ must take the following forms: \begin{align*} A&=x+iu,\\ B&=x-iu,\\ C&=-x+iv,\\ D&=-x-iv, \end{align*} where $u$ and $v$ are real numbers. As $|A|=|C|=1$, it follows that $|u|=|v|=\sqrt{1-x^2}$. Hence $BACD$ is a rectangle if $v=u$, or $BADC$ is a rectangle if $v=-u$.

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This is the complex number implementation of the geometric solution. –  zyx Apr 2 '13 at 7:49
    
I didn't read your solution, but you're right. –  user1551 Apr 2 '13 at 8:13
    
+1: I don't really see why this can't be considered a pure complex number solution. –  Aryabhata Apr 3 '13 at 8:52

I suppose the assumption is that $A,B,C,D$ are all distinct, otherwise it is not necessarily true.

Here is a pure complex number only proof.

Assume that $A+B \ne 0$ and $A + D \ne 0$. We will show that this implies that $A + C = 0$.

Since $$A+B+C+D = 0 \quad \quad (1)$$ we must have that $$\overline{A} + \overline{B} + \overline{C} + \overline{D} = 0$$ where $\overline{z}$ is the conjugate of $z$ and thus

$$\frac{1}{A} + \frac{1}{B} + \frac{1}{C} +\frac{1}{D} = 0 \quad \quad \quad (2)$$

$(1)$ and $(2)$ imply that $$A + B = -(C+D) $$ and $$\frac{A+B}{AB} = -\frac{C+D}{CD}$$

and thus $$AB = CD\quad \quad \quad (3)$$ (because $A+B \neq 0$).

Similary because $A + D \ne 0$, we get $$AD = BC\quad \quad \quad (4)$$

Now $(3)$ and $(4)$ imply (just divide) that $B^2 = D^2$ and hence $B+D = -(A+C) = 0$.

Now rotate the plane around the origin so that $\overline{A} = D$. (This is always possible).

Since rotation is just multiplying by some non-zero $w$, we still have that $A+C = 0$

Thus we have that $D = \overline{A} $, $C = -A$ and $B = -\overline{A}$ and thus $A,B,C,D$ form a rectangle.

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+1. This shows that (1) and (2) imply that the set of four points is invariant under $z \to -z$. –  zyx Apr 3 '13 at 21:29
    
@zyx: Yes, we don't need them to be on the unit circle for that. Another way to look at it: they are roots of $z^4 + pz^2 + q = 0$. –  Aryabhata Apr 4 '13 at 4:48

another way is to say $A=cosQ1+isinQ1,B=cosQ2+isinQ2,C=cosQ3+isinQ3,D=cosQ4+isinQ4$, we assume $0 \leq Q1\leq Q2 \leq Q3\leq Q4 \leq 2\pi$, which sould not effect the final result. the our target is to proof $Q3-Q1=Q4-Q2=\pi$

so we can get:

$cosQ1+cosQ2+cosQ3+cosQ4=0,sinQ1+sinQ2+sinQ3+sinQ4=0$, that is:

$sinQ1+sinQ3=-(sinQ2+sinQ4)$......[1]

$cosQ1+cosQ3=-(cosQ2+cosQ4)$......[2]

if [1]=0 , we can get sinQ1=-sinQ3 and sinQ2=-sinQ4, according to our assumption, we can get $Q3=Q1+\pi$ and $Q4=Q2+\pi$,so ABCD is rectangle.

if [2]=0, we have $Q3=Q1+\pi $ or $Q3=\pi-Q1$ and $Q4=Q2+\pi $ or $Q4=\pi-Q2$.

if $Q3=Q1+\pi$ and $Q4=Q2+\pi $, then QED

if $Q3=\pi-Q1$ and $Q4=\pi-Q2 $, we put in [1] and get $sinQ1=-sinQ2$,$Q2=\pi+Q1$,that is $Q2 \geq Q3$,only when $Q1=0$ ,then $Q2=Q3=\pi,Q4=2\pi$,which is a very special case for the rectangle.

if $Q3=\pi-Q1$ and $Q4=Q2+\pi$, put in {1], we have $sinQ1=0$, then $Q1=0$ and $ Q3=\pi$, which also means $Q3-Q1=\pi$ QED

if $Q3=\pi+Q1$ and $Q4=Q2-\pi$, we have $Q1=Q2=0, Q3=Q4=\pi$ which is also a special case for the rectangle.

if [1]and [2] are both none zero,

[1] canbe $2sin\dfrac{Q1+Q3}{2}cos\dfrac{Q3-Q1}{2}=-2sin\dfrac{Q2+Q4}{2}cos\dfrac{Q4-Q2}{2}$ .....[3]

[2] can be $2cos\dfrac{Q1+Q3}{2}cos\dfrac{Q3-Q1}{2}=-2cos\dfrac{Q2+Q4}{2}cos\dfrac{Q4-Q2}{2}$ ......[4]

and $\dfrac{[3]}{[4]}$,we get $tan\dfrac{Q1+Q3}{2}=-tan\dfrac{Q2+Q4}{2}$

since $0 \leq \dfrac{Q1+Q3}{2} \leq \dfrac{Q2+Q4}{2} \leq 2\pi$,

then we must have

$\dfrac{Q2+Q4}{2}-\dfrac{Q1+Q3}{2}=\pi$......[5] or

$\dfrac{Q2+Q4}{2}+\dfrac{Q1+Q3}{2}=\pi$......[6]

in both case : put it in [3] and [4], then [3]+[4],we have $cos\dfrac{Q3-Q1}{2}=0$ which casue [1] and [2] be zero. so it is imposible. that is all.

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Say A+B!=0 and A+D!=0 and consider the quadrilateral with vertices 0,A,A+B,A+B+C=-D. It's a rhombus bc all sides have length 1. So A and C are parallel, as are B and D, forcing A=-C and B=-D. The angles in the original quadrilateral A,B,C,D are all 90 as the diagonals are diameters of the circle.

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