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Let ABCD be a rhombus, its interior angles are $\alpha<\frac{\pi}{2}$ 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 $\text{angle}(x,C,y)=\alpha$ ?


Thank you.

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

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@user68645 No, you can make 2 distinct non-square rhombuses from the same points as well – oks Mar 27 '13 at 8:38
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 $f(m)=0$.

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