HMMT February

Notice, first of all, that $18-24-30$ is 6 times $3-4-5$, so the triangles are right. Thus, the midpoint of the hypotenuse of each is the center of their common circumcircle, and the inradius is $\frac{1}{2}(18+24-30)=6$. Let one of the triangles be $A B C$, where $\angle A<\angle B<\angle C=90^{\circ}$. Now the line $\ell$ joining the midpoints of sides $A B$ and $A C$ is tangent to the incircle, because it is the right distance (12) from line $B C$. So, the hypotenuse of the other triangle lies along $\ell$. We may formulate this thus: The hypotenuse of each triangle is parallel to the shorter leg, and therefore perpendicular to the longer leg, of the other. Now it is not hard to see, as a result of these parallel and perpendicularisms, that the other triangle "cuts off" at each vertex of $\triangle A B C$ a smaller, similar right triangle. If we compute the dimensions of these smaller triangles, we find that they are as follows: 9-12-15 at $A, 6-8-10$ at $B$, and 3-4-5 at $C$. The total area chopped off of $\triangle A B C$ is thus $$\frac{9 \cdot 12}{2}+\frac{6 \cdot 8}{2}+\frac{3 \cdot 4}{2}=54+24+6=84$$ The area of $\triangle A B C$ is $18 \cdot 24 / 2=216$. The area of the region common to both the original triangles is thus $216-84=132$.