2021 AMC Fall 10B Problem 25

Step 1: pin down the tilt. The $1 \times 3$ rectangle is tilted. Say its short side (length $1$) goes $p$ across and $r$ up, so $p^2 + r^2 = 1$. The long side is at a right angle to it, so it goes $3r$ across and $3p$ up.

1. Width of the big square. The $1 \times 3$ rectangle touches the left, bottom and right sides. Across the bottom it spans one long side plus one short side: width $= 3r + p$.
2. Height of the big square. Go up the left side: the long side rises $3p$ to reach the left wall, then the short side rises $r$ to the rectangle's top corner. The unit square sits on that corner: its side to the left wall has length $1$ and runs $p$ across (from the corner back to the wall), so, just like the rectangle's short side, it rises $r$; and its next side rises $p$ to reach the top. Height $= 3p + r + r + p = 4p + 2r$.
3. Square, so width = height. $3r + p = 4p + 2r$, which gives $r = 3p$. Then $p^2 + 9p^2 = 1$, so $p = \frac{1}{\sqrt{10}}$ and $r = \frac{3}{\sqrt{10}}$. The big square's side is $3r + p = \frac{10}{\sqrt{10}} = \sqrt{10}$.

Step 2: nicer units. Measure lengths in units of $\frac{1}{\sqrt{10}}$. Then $p = 1$, $r = 3$, and the big square is $10 \times 10$ with corner $(0, 0)$. Areas in these units are $10$ times too big, so divide by $10$ at the end. The pieces now have whole-number corners:
- $1 \times 3$ rectangle: $(9, 0)$, $(10, 3)$, $(1, 6)$, $(0, 3)$. Its top side runs from $(1, 6)$ to $(10, 3)$ along the line $x + 3y = 19$.
- Unit square: $(1, 6)$, $(0, 9)$, $(3, 10)$, $(4, 7)$. Rectangle $R$ has its corner at $P = (4, 7)$.

Step 3: rectangle $R$. $R$ has corner $P = (4, 7)$, one corner $U$ on the top side $y = 10$, one on the right side $x = 10$, and one corner $Q$ on the line $x + 3y = 19$.

1. Going from $P$ to $U$ climbs $3$ (from $7$ to $10$). Call the sideways distance $k$, so $P \to U$ is "$k$ across, $3$ up."
2. Side $PQ$ is at a right angle to $PU$, so it goes in the direction "$3$ across, $k$ down." Say $PQ$ is $\lambda$ times that: $Q = (4 + 3\lambda,\ 7 - k\lambda)$.
3. Fourth corner on the right side. The fourth corner is $U$ moved the same way as $P \to Q$, so its $x$-coordinate is $4 + k + 3\lambda = 10$. So $k + 3\lambda = 6$.
4. $Q$ on the line. $(4 + 3\lambda) + 3(7 - k\lambda) = 19$, which simplifies to $\lambda(k - 1) = 2$.
5. Solve. Put $k = 6 - 3\lambda$ into step 4: $\lambda(5 - 3\lambda) = 2$, so $3\lambda^2 - 5\lambda + 2 = 0$, which factors as $(3\lambda - 2)(\lambda - 1) = 0$.
- $\lambda = 1$, $k = 3$: $PU$ is $(3, 3)$, length $3\sqrt2$, and $PQ$ is $(3, -3)$, length $3\sqrt2$. Area $18$.
- $\lambda = \frac23$, $k = 4$: $PU$ is $(4, 3)$, length $5$, and $PQ$ is $(2, -\frac83)$, length $\sqrt{4 + \frac{64}{9}} = \frac{10}{3}$. Area $\frac{50}{3}$.
6. Both fit. For $\lambda = 1$ the corners are $(4,7), (7,10), (10,7), (7,4)$. For $\lambda = \frac23$ they are $(4,7), (8,10), (10, \frac{22}{3}), (6, \frac{13}{3})$. In both, $Q$ lands on the rectangle's top side between $x = 1$ and $x = 10$.

Step 4: real areas. Divide by $10$: $\frac{18}{10} = \frac95$ and $\frac{50}{3} \div 10 = \frac53$. The sum is
$$\frac95 + \frac53 = \frac{27}{15} + \frac{25}{15} = \frac{52}{15},$$
so $m + n = 52 + 15 = 67$.

The answer is $\boxed{\textbf{(E)}\ 67}$.

Two of the three pieces are rigid enough to fix the big square completely, while $R$ has some freedom, which is why the problem says "all possible values." Three touching conditions on $R$ give equations with a quadratic at the end, so two rectangles work. Choosing units that put every known corner at whole-number coordinates keeps the algebra small.