Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let ABCD be a rhombus, its interior angles are $\alpha<\Pi/2$ and $(\Pi-\alpha)$.

Let w, x, y, z four points located respectively in (A,B), (B,C), (C,D), (D,A).

Suppose we have as inputs the points w, z, y, z and the angle $\alpha$,

Is there a geometric method to find the points A, B, C, D such that $angle(x,C,y)=\alpha$ ?

Illustration:

Thank you.

share|improve this question
    
how did you draw and include the image? –  oks Mar 26 '13 at 23:57
1  
I drew the image using GeoGebra. –  user68645 Mar 26 '13 at 23:58
1  
Thankyou! GeoGebra looks good. –  oks Mar 27 '13 at 0:11

2 Answers 2

up vote 1 down vote accepted

No. For instance, suppose the rhombus were actually a square with coordinates at (0,0), (0,2), (2,2), (2,0). Suppose you are given $w,z,y,x$ as mid points of the lines of the square at (0,1), (1,2), (2,1), (1,0) and you are given that $\alpha = \pi/2$.

You might hit on the right square, but you might also fit a square by just joining up the given points.

Edit, suppose the given points were a bit closer to the true vertices. There is more than one square that will fit.

two squares centred on origin, off by a slight tilt

Edit 2 (for $\alpha \ne \frac{\pi}{2}$) You can also construct 2 congruent rhombuses (with $\alpha \ne \pi/2$) from the same points

2 rhombuses centred on origin (off by a quarter turn)

Edit 3 (preserving the position of $\alpha$)

enter image description here

share|improve this answer
    
@user68645 No, you can make 2 distinct non-square rhombuses from the same points as well –  oks Mar 27 '13 at 8:38
1  
Suppose the points are chosen in general position on the rhombus? I think this is what the OP wants to ask. In this case, I suspect there will be at most two solutions, and finding both of them should answer the OP's question. –  Peter Shor Mar 27 '13 at 21:34

I don't know about a geometric solution, but a numerical solution can be obtained like this.

Let $m$ be some arbitrary slope and let $\theta = tan^{-1}(m)$.

Let Line$_x$ be the straight line passing through $x$ with slope $m$. Line$_x$ is well-defined by the point it passes through and its slope.

Let Line$_y$ be the straight line passing through $y$ with slope $m' = tan(\theta + \alpha)$. Line$_x$ and Line$_y$ intersect at point $C'$, say, and by construction the angle $(x,C', y)$ is $\alpha$.

Let Line$_z$ be the straight line passing through $z$ with slope $m$. Line$_z$ and Line$_y$ intersect at point $D'$, say.

Let $A'$ be the point distance $C'D'$ from $D'$ (in the direction with slope $-m$). Let $B'$ be the point distance $C'D'$ from $C'$ (in the direction with slope $-m$).

Let $e$ be the distance of point $w$ from the line $A'B'$

Given the co-ordinates of $w$, $x$, $y$, $z$, and slope $m$, the coordinates of $A', B', C', D'$ and the distance $e$ can be calculated using linear equations and Pythagoras.

So $e$ is an explicit function of $m$, say $e = f(m)$. Thus some numerical solver can be used to find $m*: e* = f(m*) = 0$ which will be where $w$ is on the line $A'B'$.

In some special cases(as shown in the other answer) there will be more than one solution of $f(m)=0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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