2019 AMC 10B Problem 15

Let $T_1$ have legs $a, b$ and hypotenuse $c$, with $ab = 2$.

If the two shared sides were both legs of both triangles, the areas would match, contradiction. If $c$ were the hypotenuse of $T_2$ as well and one leg were shared, the other legs would be equal too, so the triangles would be congruent, again contradicting the areas. Hence the hypotenuse $c$ of $T_1$ is a leg of $T_2$. The second shared side is a leg of $T_1$, say $b$; it is shorter than $c$, so it cannot be the hypotenuse of $T_2$ and must be the other leg.

So $T_2$ has legs $b$ and $c$ with $bc = 4$. Dividing by $ab = 2$ gives $c = 2a$. Then $c^2 = a^2 + b^2$ becomes $4a^2 = a^2 + b^2$, so $b^2 = 3a^2$ and $b = a\sqrt3$. From $ab = 2$: $a^2\sqrt3 = 2$, so $a^2 = \frac{2}{\sqrt3}$.

The third sides are $a$ (for $T_1$) and the hypotenuse of $T_2$, $\sqrt{b^2 + c^2} = \sqrt{3a^2 + 4a^2} = a\sqrt7$. Their product is $a^2\sqrt7$, and its square is
$$7a^4 = 7 \cdot \frac{4}{3} = \frac{28}{3}.$$

The answer is $\boxed{\textbf{(A)}\ \frac{28}{3}}$.

Area and side-sharing constraints interact through which roles the shared sides play. Eliminating the impossible pairings leaves one configuration, and the area ratio $2$ then translates into a length ratio $c = 2a$, which forces $T_1$ to be $30$-$60$-$90$. Asking for the square of the product is a hint that the lengths themselves involve nested radicals; work with $a^2$ throughout.

• Assuming the two triangles share both legs (impossible with different areas) or that the shared sides are both hypotenuses.
• Reporting $a^2\sqrt7 \approx 3.06$ or the unsquared product, instead of its square $\frac{28}{3}$.