Congratulations to Katharina from the International School in Strasbourg and to Chris from Heckmondwike Grammar School for their solutions to Bendy Quad. Here is Katharina's solution :

To solve this problem I only used one formula which is called the Al Kashi theorem in the country I go to school (France). It is actually an extension of Pythagoras' theorem and is true for every triangle: $a^2 = b^2 + c^2 - 2bc \times \cos (k)$ where $a$ is the side opposite to the vertex $A$ and $k$ the angle at $A$; $b$ and $c$ are the two other sides of the triangle. By transforming it one obtains: $\cos(k) = (b^2 + c^2 - a^2)/(2bc)$.

Take angles $p$, $q$, $r$ and $s$ with $p$ formed by sides 4 and 5 (i.e. the sides with the lengths 4 and 5), $q$ formed by the sides 5 and 6, $r$ by the sides 6 and 3 and $s$ by the sides 3 and 4).

${\bf First}$ ${\bf part}$ ${\bf (convex}$ ${\bf quadrilateral}$ ${\bf with}$ ${\bf 60}$ ${\bf degree}$ ${\bf angle):}$

According to the Al Kashi theorem: $$3^2 + 4^2 -24\cos s = d^2 = 5^2 +6^2 - 60\cos q$$ which simplifies to give $$2\cos s + 3 = 5\cos q.$$ As $\cos s = 0.5$ this gives $\cos q = 0.8$ and $q = 36.9^o$

To calculate the other angles, use the Sine Rule to find the two parts of angle $p$ and of angle $r$ cut by the diagonal $d = \sqrt 13$. $ p =132.9^{\circ}$ and $r =130.2^{\circ}$ (to 1 decimal place).

${\bf Second}$ ${\bf Part}$ ${\bf (cyclic}$ ${\bf quadrilateral):}$

We have found $2\cos s + 3 = 5\cos q.$ For a cyclic quadrilateral $s + q = 180$ so $\cos 3s = - \cos q$. Hence $7\cos q = 3$ so that $q= 64.6^{\circ}$ and $s = 115.4^{\circ}$.

By drawing the other diagonal and using the same method we find that $45 - 36\cos r = 41 - 40\cos p$ which gives $$ 1 + 10\cos p = 9\cos r.$$ So $\cos r = 1/19$ giving $r = 87.0^{\circ}$ and $p = 93.0^{\circ}$.

${\bf Third}$ ${\bf Part}$ ${\bf (general}$ ${\bf convex}$ ${\bf quadrilateral):}$

We proved that $2\cos s + 3 = 5\cos q.$ Thus (because $-1\leq \cos \theta \leq 1 $ for all $\theta$) , we have $$-4 \leq \cos s \leq 1\ \rm{and} \ 1/5 \leq \cos q \leq 1.$$ This places no constraint on the value of angle $s$ but it gives $0^{\circ} \leq q\leq 78.5^{\circ}$ (to 1 decimal place).

Similarly $ 1 + 10\cos p = 9\cos r$ so we have $$-1 \leq \cos p \leq 0.8 \ \rm{and} \ -1 \leq \cos r \leq 11/9.$$ This places no constraint on the value of angle $r$ but it gives $36.9^{\circ} \leq p\leq 180^{\circ}$ (to 1 decimal place).

Once one angle is known then the opposite angle is determined from the relations: $2\cos s + 3 = 5\cos q$ and $ 1 + 10\cos p = 9\cos r$. Then using the same method as used for the case $s = 60$ the other two angles are also determined.