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Suppose we are given sides $a,b,c,d$. We need to construct a cyclic quadrilateral with the given sides. How can we do that?

Thank you very much in advance

Regards.

Asinomás
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    Parameshvara's Formula gives the circumradius in terms of the side lengths. – Blue Nov 27 '14 at 15:54
  • I don't know how I can use that, I mean. now I know what circle to inscribe it in. But how do I get the angles right so the sides actually fit in? – Asinomás Nov 27 '14 at 16:00
  • With the radius (call it $r$), you can construct an isosceles $\triangle OAB$ with $|AB|=a$ and $|OA|=|OB|=r$. This $O$ is the circumcenter. A radius-$b$ circle about $B$ meets the circumcircle at $C$; a radius-$c$ circle about $C$ meets the circumcircle at $D$; if $C$ and $D$ have been chosen properly, then $|DA|=d$, so that $\square ABCD$ is your quadrilateral. Alternatively, other formulas on the Wikipedia page give you the quad's angles. – Blue Nov 27 '14 at 16:20
  • I guess this answer works. But constructing a side with the required length is a really long construction in euclidean geometry. You would need like 50 constructions to get a length of the side given by parameshvara's formula. I think there is a better solution. – Asinomás Nov 28 '14 at 03:29
  • This is the shortest path I've seen so far: The angle $A$ between edges $a$ and $d$ satisfies $$\tan\frac{A}{2} = \sqrt{\frac{s-a}{s-b}\frac{s-d}{s-c}}$$ where $s := (a+b+c+d)/2$. Segments of length $s-a$, $s-b$, $s-c$, $s-d$ are easily constructed, as are segments of length $p:=(s-a)/(s-b)$ and $q:=(s-d)/(s-c)$. Then, the geometric mean $\sqrt{pq}$ is also straightforward, giving a segment of length $\tan(A/2)$. That effectively gives angle $A$, which, along with segments $a$ and $d$, determines a triangle with the desired circumcircle; the fourth point of the quadrilateral is easily found. – Blue Nov 28 '14 at 05:26

1 Answers1

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Here's a consolidated construction of the angle $\alpha$ between edges of length $a$ and $d$. As noted in Wikipedia's "Cyclic Quadrilateral" article, the angle satisfies $$\tan\frac{\alpha}{2} = \sqrt{\frac{s-a}{s-b}\frac{s-d}{s-c}} \tag{$\star$}$$ where $s := \frac{1}{2}(a+b+c+d)$ is the semi-perimeter.

enter image description here

In the diagram, the circle about $O$ through $U$ and $V$ has unit radius, and $S$ is easily constructed such that $|OS| = s$. Circles of radius $a$, $b$, $c$, $d$ about $S$ give points $S_a$, $S_b$, $S_c$, $S_d$ on $OS$ such that $S_x = s-x$.

We connect $S_b$ and $S_c$ (corresponding to the denominators of the fractions in $(\star)$) to $U$ and $V$, respectively. Parallels to the connecting lines, passing through $S_a$ and $S_d$ (corresponding to the numerators), determine points $P$ and $Q$ such that $$|OP| = p := \frac{s-a}{s-b} \qquad |OQ| = q := \frac{s-d}{s-c}$$

The semicircle with diameter $PQ$ meets $\overleftrightarrow{OS}$ at $T$ such that $$|OT| = \sqrt{pq} = \tan\frac{\alpha}{2} = \tan \angle OUT = \tan \angle OVT$$

Therefore, $\angle UTV = 180^\circ - \alpha$. From there, finding $\alpha$ itself is trivial. (Of course, $180^\circ - \alpha$ itself is the angle between edges of length $b$ and $c$ in the quadrilateral.) This is enough information to complete the construction of the entire quadrilateral.

Blue
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