I am given the lengths of 4 sides of a convex quadrilateral as $a$, $b$, $c$, $d$.

I am also given the sequence of the sides (i.e. it is $abcd$, not $abdc$, or $acbd$, or $acdb$,... etc).

I know that Bretschneider's_formula

$$ K=\sqrt{(s-a)(s-b)(s-c)(s-d)-a b c d \cdot \cos ^{2}\left(\frac{\alpha+\gamma}{2}\right)} $$ (where $s=\frac{a+b+c+d}2$ is the semi-perimeter) can give the area if the sum of either pair of opposite angles are known.

But I do not know the angles of any of the vertices (interior or exterior).

I can generate two quadrilateral equations by generating diagonal lines $e$, $f$ and using the cosine rule of the triangle...

$$ a^2 + b^2 - 2.a.b.\cos(\alpha) (= e^2) = c^2 + d^2 - 2.c.d.\cos(\gamma) $$

$$ b^2 + c^2 - 2.b.c.\cos(\beta) (= f^2) = a^2 + d^2 - 2.a.d.\cos(\delta) $$

These two equations have four unknowns (i.e. the interior angles: $\alpha$, $\beta$, $\gamma$, $\delta$). $\delta$ is given by:

$$ \delta = 360 - \alpha-\beta-\gamma $$

so this leave three unknowns but only two equations.

Where do I go from here?


I have accepted Ted Shifrin's answer, if opposite angles sum to $\pi$ then we can use a simplified form of Bretschneider's formula to get the (maximum) area K.

$$ K^2 = [(s-a)(s-b)(s-c)(s-d)] $$

For finding the maximum area we dont need to know what any of the four individual vertex angles actually are.


Ted indicates that it can be proved that a quadrilateral with opposite angles summing to $\pi$ must be concyclic and vice-versa. It is straightforward to prove that, in a concyclic quadrilateral, the opposite angles are supplementary, by drawing diagonals and using the rule that the angle subtended by a chord at the centre is twice the angle subtended at the perimeter.

Related question:- geometric-argument-as-to-why-the-cyclic-quadrilateral-has-the-maximal-area

This paper (PDF) by Thomas Peter (2003):- Maximising the area of a quadrilateral gives a proof that:- "For any quadrilateral with given edge lengths, there is a cyclic quadrilateral with the same edge lengths". It goes on to prove that:- "The cyclic quadrilateral Q has the largest area of all quadrilaterals with sides of the same length as those of Q".

Also I deduce that any angle of a cyclic quadrilateral can be obtained from:- $$ cos(\alpha) = (a^2 + b^2 - c^2 - d^2)/(2(a.b +c.d)) $$ where $\alpha$ is the interior angle between sides a and b. Which is nice. :)


2 Answers 2


You will find that the optimal solution is for the opposite angles to add to $\pi$ and, thus, for the quadrilateral to be concyclic (inscribed in a circle).

  • 1
    $\begingroup$ Hah, so much simpler than my solution. $\endgroup$
    – Lee Mosher
    Commented Jul 21, 2014 at 17:02
  • $\begingroup$ I read that a quadrilateral can only be concyclic if opposite angles sum to $\pi$. But is the reverse proven? i.e. "All quadrilaterals whose opposite angles sum to $\pi$ can be inscribed in a circle". $\endgroup$
    – steveOw
    Commented Jul 22, 2014 at 13:19
  • 1
    $\begingroup$ Yes, @steveOw. There is a unique circle through 3 of the 4 points. If the 4th point lies inside or outside the circle, then geometry tells us that the opposite angles will not sum to $\pi$. $\endgroup$ Commented Jul 22, 2014 at 15:08
  • $\begingroup$ Thanks @Ted Shifrin. I can see how a proof might be made using the relationships between a chord and peripheral angles and central angles in a circle. $\endgroup$
    – steveOw
    Commented Jul 22, 2014 at 15:38

Notice that on the right hand side of Bretschneider's formula, everything is constant except $\alpha$ and $\gamma$. So the area is maximized if and only if $\cos^2 (\frac{\alpha+\gamma}{2})$ is minimized. But $\alpha$ and $\gamma$ are not independent. From the figure one can see that $\alpha$ and $\gamma$ are constrained by applying the law of cosines to two of the triangles in the picture: $$(*) \qquad b^2 + c^2 - 2bc \cos(\gamma) = a^2 + d^2 - 2ad \cos(\alpha) $$ So now you have a constrained optimization problem, and you can try to solve it using Lagrange's method: find the minimum of $\cos^2 (\frac{\alpha+\gamma}{2})$ subject to the constraint $(*)$ (and restricted to $\alpha,\gamma \in [0,\pi]$).

You might be able to simplify the problem even further using some symmetry. For example: if you restrict to $\alpha,\gamma \in [0,\pi/2]$, so that $\cos(\alpha+\gamma) \ge 0$, then $\cos^2(\frac{\alpha+\gamma}{2})$ is minimized if and only if $\cos(\frac{\alpha+\gamma}{2})$ is also minimized.

  • $\begingroup$ thanks. Like you say:- minimizing the RHS of the Bretschneider's formula gives the maximum area which is all I need and so I just need to assume $\alpha + \gamma = \pi$...I dont need to know the actual values of the angles. I should have seen that myself. $\endgroup$
    – steveOw
    Commented Jul 22, 2014 at 18:52
  • $\begingroup$ @SteveOw I should also say that this depends on the fact that every set of quadrilaterals with sides abcd, and each side shorter than the sum of the other three sides, contains a quadrilateral which is concyclic and has supplementary opposite angles. $\endgroup$
    – steveOw
    Commented Nov 22, 2021 at 18:26

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