2022 AMC 10B Problem 16

The overlap is the whole square minus the small triangle that sticks out of the top of the rectangle. So first pin down exactly where the square sits.

1. Set up coordinates. Put the rectangle's bottom-left corner at $(0, 0)$ and its top-right corner at $(8, 4)$. Call the square's corners $P$ (on the bottom side), $Q$ (on the right side), $R$ (on the top side) and $T$ (the corner outside, above the rectangle).
2. The side $PR$. It is $5$ long and goes from the bottom side up to the top side, so it rises $4$. By the Pythagorean theorem it moves $\sqrt{5^2 - 4^2} = 3$ sideways. From the figure it goes left, so $R$ is $3$ left and $4$ up from $P$. (The figure draws this right triangle, with a vertical line from $R$ down to the bottom side.)
3. The side $PQ$. It is the side $PR$ turned a quarter turn, so the $3$ and the $4$ swap: $Q$ is $4$ right and $3$ up from $P$.
4. Find the points. $Q$ is on the right side, where $x = 8$, so $P$ is at $x = 8 - 4 = 4$. That gives
$$P = (4, 0),\quad Q = (8, 3),\quad R = (1, 4).$$
$T$ is $3$ left and $4$ up from $Q$, so $T = (5, 7)$.
5. The piece sticking out. Only the part above the line $y = 4$ is outside the rectangle. It is a triangle with corners $R$, $T$, and the point $U$ where side $QT$ crosses the top of the rectangle. It has a right angle at $T$, because $T$ is a corner of the square, and the side $RT$ is $5$.
6. Its other leg, $TU$. The side $RT$ goes from $(1, 4)$ to $(5, 7)$: $4$ across and $3$ up. So triangle $RTU$ has the same shape as a $3$-$4$-$5$ triangle lying on its long side: its right angle is at $T$, and the leg $TU$ is to the leg $RT$ as $3$ is to $4$. So $TU = \frac34 \times RT = \frac34 \times 5 = \frac{15}{4}$.
7. Area of the piece. $\frac12 \times 5 \times \frac{15}{4} = \frac{75}{8}$.
8. Overlap. $25 - \frac{75}{8} = \frac{200}{8} - \frac{75}{8} = \frac{125}{8} = 15\frac58$.

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

A square of side $5$ that stretches across a rectangle of height $4$ has to be tilted like a $3$-$4$-$5$ triangle. Because neighbouring sides of a square are the same length and at right angles, knowing one side's "across and up" tells you the next side's too (the numbers swap). Once the square is placed, the overlap is simply the square minus the triangle poking out, and that triangle has the same $3 : 4$ shape.

Check step 6 with coordinates. Side $QT$ goes from $(8, 3)$ to $(5, 7)$: $3$ left for every $4$ up. To rise from $y = 3$ to $y = 4$ it goes $\frac34$ left, so $U = \left(\frac{29}{4}, 4\right)$. Then the triangle $RTU$ has a base $RU = \frac{29}{4} - 1 = \frac{25}{4}$ along the top edge and a height of $7 - 4 = 3$ above it, so its area is $\frac12 \times \frac{25}{4} \times 3 = \frac{75}{8}$, the same as before.

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