B2. In a certain quadrilateral, the four angles are each two-digit numbers. These four numbers can be placed in the 2 by 2 grid shown, with one digit in each cell.
Find all the possibilities for the set of four angles.
SolUTION
Throughout this solution we will assume that $b \geq c$ because the problem remains unchanged if we were to swap $b$ and $c$.

It is useful to note that a two-digit number ' $x y$ ' can be written algebraically as $10 x+y$ and that the sum of the four interior angles of a quadrilateral is $360^{\circ}$.

Combining the previous two facts we see that $10 a+b+10 a+c+10 b+d+10 c+d=360$. This can be simplified to give:
$$
20 a+11 b+11 c+2 d=360
$$

If we assume that $a \leq 7$, then the maximum interior angle sum we can achieve is $20 \times 7+11 \times$ $9+11 \times 9+2 \times 9=356<360$ (where $a=7$ and $b=c=d=9$ ). As this is less than 360 we know that $a \geq 8$, and so $a$ (which must be a digit) may only equal 8 or 9 . We will consider each of these two cases separately.

If $a=8$ then it must be that $b=9$, or else the four interior angles are each less than $90^{\circ}$ and the interior angle sum would be less than $360^{\circ}$. Substituting $a=8$ and $b=9$ into (1) and then simplifying gives $11 c+2 d=101$. The largest possible value of $2 d$ is $2 \times 9=18$, and so $11 c$ must be at least $101-18=83$. This means that $c$ must be at least 8 . If $c=8$ then $d$ would not be a whole number and so not a digit. If $c=9$ then $d=1$.

If $a=9$ then $b$ must be equal to 8 or 9 , or else the maximum interior angle sum we can achieve is $20 \times 9+11 \times 7+11 \times 7+2 \times 9=352<360$ (where $a=9, b=c=7$ and $d=9$ ).
If $a=9$ and $b=8$, substituting into (1) and then simplifying gives $11 c+2 d=92$. This means that $11 c$ must be at least $92-18=74$ and so $c$ is at least 7 . If $c=7$ then $d$ is not a whole number and so not a digit. If $c=8$ then $d=2$. We know that $c$ cannot equal 9 because at the start we assumed it was less than or equal to $b$.
If $a=9$ and $b=9$, substituting into (1) and then simplifying gives $11 c+2 d=81$. This means that $11 c$ must be at least $81-18=63$ and so $c$ is at least 6 . If $c=6$ or 8 then $d$ is not a whole number and so not a digit. If $c=7$ then $d=2$. We know that $c$ cannot equal 9 because then $d$ would be negative.
There are three sets of four angles that satisfy the problem: $\{89,89,91,91\},\{98,98,82,82\}$ and $\{99,97,92,72\}$.