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 part (convex quadrilateral with 60 degree 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 $q = 36.9^o$ \par 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^o$ and $r =130.2^o$ (to 1 decimal place). quad 1

Second Part (cyclic quadrilateral):
We have found

2 \cos s + 3 = 5 \cos q .

For a cyclic quadrilateral

s + q = 180

\cos 3 s = - \cos q

. Hence

7 \cos q = 3

so that

q = 64 . 6^{o}

and

s = 115 . 4^{o}

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 $r = 87 . 0^{o}$ and $p = 93 . 0^{o}$ .

Third Part (general convex quadrilateral)
We proved that

2 \cos s + 3 = 5 \cos q .

Thus (because

- 1 \leq \cos θ \leq 1

for all

θ

) , we have

- 4 \leq \cos s \leq 1 and 1 / 5 \leq \cos q \leq 1 .

This places no constraint on the value of angle

s

but it gives

0^{o} \leq q \leq 78 . 5^{o}

(to 1 decimal place).

Similarly $1 + 10 \cos p = 9 \cos r$ so we have

$- 1 \leq \cos p \leq 0.8 and - 1 \leq \cos r \leq 11 / 9 .$

This places no constraint on the value of angle $r$ but it gives $36 . 9^{o} \leq p \leq 180^{o}$ (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.