2019 AMC 10B Problem 15

First work out which sides can be the shared ones. Then the areas give the lengths.

Call the legs of $T_1$ $a$ and $b$ and its hypotenuse $c$. Its area is $1$, so $ab = 2$.

Step 1: rule out the easy guesses.

1. If the two shared sides were the two legs of both triangles, the areas would be equal. But they are $1$ and $2$. Impossible.
2. If one shared side were the hypotenuse of both, and the other a leg of both, then by the Pythagorean theorem the third sides would match too. The triangles would be identical, with equal areas. Impossible.
3. So some shared length is a hypotenuse in one triangle and a leg in the other.

Step 2: decide which way round.

1. Could the hypotenuse of $T_2$ be a leg of $T_1$, say $b$? Then the other shared side is a leg of $T_2$, so it is shorter than $b$. It must be $a$ (not $c$, which is longer than $b$). Then $T_2$ has one leg $a$ and another leg shorter than $b$, so its area is less than $\frac12 ab = 1$. But its area is $2$. Impossible.
2. So the hypotenuse $c$ of $T_1$ is a leg of $T_2$. The other shared side is a leg of $T_1$, say $b$. It is shorter than $c$, and the hypotenuse of $T_2$ is longer than $c$, so $b$ is the other leg of $T_2$.

Step 3: find the shape.

1. $T_2$ has legs $b$ and $c$ and area $2$, so $bc = 4$. Compare with $ab = 2$: dividing gives $c = 2a$.
2. In $T_1$: $a^2 + b^2 = c^2 = 4a^2$, so $b^2 = 3a^2$ and $b = a\sqrt3$. (So $T_1$ is a $30$-$60$-$90$ triangle.)
3. From $ab = 2$: $a \cdot a\sqrt3 = 2$, so $a^2 = \frac{2}{\sqrt3}$ and $a^4 = \frac{4}{3}$.

Step 4: the third sides.

1. The side of $T_1$ that is not shared is $a$.
2. The side of $T_2$ that is not shared is its hypotenuse: $\sqrt{b^2 + c^2} = \sqrt{3a^2 + 4a^2} = a\sqrt7$.
3. Their product is $a \cdot a\sqrt7 = a^2\sqrt7$. Its square is $7a^4 = 7 \cdot \frac43 = \frac{28}{3}$.

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

The hard part is not the algebra but deciding which side of one triangle matches which side of the other. Listing the possibilities and throwing out the ones that force equal areas leaves just one picture. Also notice that the question asks for the square of the product: that is a hint that the lengths themselves are messy, so work with $a^2$ and $a^4$ instead of $a$.