16. On the plane there are two triangles, each with side lengths 18,24 and 30 . If the two triangles do not completely overlap, but share the same circumcircle as well as the same inscribed circle, find the area of the region common to both triangles.
(2 marks)
平面上有兩個三角形，邊長都是 18、24、30。若兩個三角形並非完全重疊，
但它們的外接圓相同, 內切圓也相同, 求兩個三角形重疊部分的面積。
16. 132
16. First of all, note that a 18-24-30 triangle is a 6time magnification of a 3-4-5 triangle, and so is right-angled. Thus the midpoint of the hypotenuse of each triangle is the centre of their common circumcircle, and the in-radius is $(18+24-30) \div 2=6$. Let one of the triangles be $\triangle A B C$ where $\angle A<\angle B<\angle C=90^{\circ}$, with circumcentre $O$.
Let $\ell$ be the diameter of the circumcircle passing through the mid-point of $A C$. Since $O$ is the mid-point of $A B, \ell$ is parallel to $B C$, and since $A C=24$, it is 12 units from $B C$. As the diameter of the inscribed circle is also 12 units, $\ell$ is tangent to the inscribed circle.

Therefore $\ell$ must be the hypotenuse of the other triangle. It is then easy to see that the regions which belong to exactly one of the two triangles (i.e. the six shaded triangular regions in the figure) are all similar to $\triangle A B C$.

The area of $\triangle A B C$ is $18 \times 24 \div 2=216$. We focus on the three shaded regions at $A, B$ and $C$. At $A$, the area of the shaded region is clearly $216 \div 4=54$ (in fact, it is a 9-12-15 triangle). The region at $B$ is a 6-8-10 triangle with area $6 \times 8 \div 2=24$. (To see this, we note that the line segment $B O=15$ is split into two parts, with the second part being of length 9 as can be seen by comparing with the 9-12-15 shaded triangle at $A$.) Similarly, the region at $C$ is a 3-4-5 triangle with area $3 \times 4 \div 2=6$. It follows that the answer is $216-54-24-6=132$.