2022 AMC 10B Problem 16

Place the rectangle with corners $(0,0)$ and $(8,4)$. Call the square's vertices $P$ (on the bottom side), $Q$ (on the right side), $R$ (on the top side) and $T$ (outside), with $PQ$ and $PR$ the two sides meeting at $P$.

Going from $P$ to $Q$ moves $(u, v)$ with $u^2 + v^2 = 25$; since $PR$ is $PQ$ rotated $90^\circ$, going from $P$ to $R$ moves $(-v, u)$. $R$ is on the top side, $4$ units above $P$, so $u = 4$ and then $v = 3$. Thus $Q$ is $4$ right and $3$ up from $P$; since $Q$ is on the right side, $P = (4, 0)$, $Q = (8, 3)$, $R = (1, 4)$ and $T = Q + (R - P) = (5, 7)$.

The only part of the square outside the rectangle is the piece above the top side $y = 4$: triangle $RTU$, where $U$ is where side $QT$ crosses $y = 4$. This triangle has a right angle at $T$ (a corner of the square) and $RT = 5$. Side $RT$ has slope $\frac34$, so $\angle TRU$ has tangent $\frac34$ and
$$TU = RT \cdot \tan\angle TRU = 5 \cdot \tfrac34 = \tfrac{15}{4}.$$
Its area is $\frac12 \cdot 5 \cdot \frac{15}{4} = \frac{75}{8}$.

The overlap is therefore $25 - \frac{75}{8} = \frac{125}{8} = 15\frac58$.

The answer is $\boxed{\textbf{(D)}\ 15\dfrac{5}{8}}$.

A square with one vertex on the bottom and one on the top of a height-$4$ rectangle, with side $5$, must be tilted along a $3$-$4$-$5$ triangle; the perpendicular sides are $90^\circ$ rotations of each other, so one right-triangle relation determines the whole figure. Once the geometry is pinned down, "area of the overlap" is just "square minus the protruding triangle," and that triangle is similar to the $3$-$4$-$5$ triangle as well.

Shoelace on the overlap quadrilateral $P(4,0)$, $Q(8,3)$, $U\left(\tfrac{29}{4}, 4\right)$, $R(1,4)$: the cross terms are $12 + \frac{41}{4} + 25 - 16 = \frac{125}{4}$, and half of that is $\frac{125}{8}$.

• Assuming the square's vertices on the rectangle sit at 'nice' positions (like the midpoint) or that only a small corner of the square pokes out, then guessing from the choices.
• Computing the protruding triangle with legs $3$ and $4$ (area $6$) rather than $5$ and $\frac{15}{4}$, giving $19$, which is not even a choice.