16. In $\triangle A B C, A B=33 \mathrm{~cm}, A C=21 \mathrm{~cm}$ and $B C=x \mathrm{~cm}$, where $x$ is an integer. $D$ is a point on the segment $A B$ and $E$ is a point on the segment $A C$ such that $A D=D E=E C=y \mathrm{~cm}$, where $y$ is again an integer. Find $x$.
(2 marks)
在 $\triangle A B C$ 中, $A B=33 \mathrm{~cm} 、 A C=21 \mathrm{~cm} 、 B C=x \mathrm{~cm}$, 其中 $x$ 是整數。 $D$ 是線段 $A B$上的一點, $E$ 則是線段 $A C$ 上的一點, 使得 $A D=D E=E C=y \mathrm{~cm}$, 其中 $y$ 亦是整數。求 $x$ 。
16. 30
16. Note that $y21-y$, or $y>7$. It follows that $y$ is between 8 and 20 inclusive.
Now, applying cosine law in $\triangle A D E$ and $A B C$, we get
$$
\cos A=\frac{y^{2}+(21-y)^{2}-y^{2}}{2 y(21-y)}
$$
and
$$
\cos A=\frac{33^{2}+21^{2}-x^{2}}{2(33)(21)}
$$
respectively. Equating and simplifying, we get
$$
y\left(2223-x^{2}\right)=14553=3^{3} \times 7^{2} \times 11 .
$$

Since $8 \leq y \leq 20$, the only possible values of $y$ are 9 and 11. These correspond to $x^{2}=606$ and 900 respectively. As the former is not a perfect square, we take the latter which corresponds to $x=30$.