2021 AMC Fall 10B Problem 25

Step 1: the fixed pieces. Let the $1 \times 3$ rectangle's long side make angle $\theta$ with the horizontal, its corners on the left, bottom and right sides of the big square. Its horizontal span is the big side, $s = 3\cos\theta + \sin\theta$; its top corner $T$ has height $3\sin\theta + \cos\theta$ and is $\sin\theta$ from the left side.

The unit square has a vertex at $T$, one on the left side and one on the top. Its side from $T$ to the left side has run $\sin\theta$, hence rise $\cos\theta$; the next side rises $\sin\theta$ more to reach the top, so $s = 4\sin\theta + 2\cos\theta$. Equating: $\tan\theta = \frac13$ and $s = \sqrt{10}$.

Step 2: coordinates. Scale by $\sqrt{10}$ so the big square is $[0,10]^2$. The rectangle's top side is the segment of $x + 3y = 19$ from $(1,6)$ to $(10,3)$, and the unit square's free vertex is $P = (4,7)$.

Step 3: rectangle $R$. $R$ has vertex $P$, vertex $Q$ on $x + 3y = 19$, a vertex on $x = 10$, and vertex $U$ on $y = 10$. Let its sides from $P$ be $u(\cos\alpha, \sin\alpha)$ toward $U$ and $q(\sin\alpha, -\cos\alpha)$ toward $Q$. The conditions are $u\sin\alpha = 3$, $q(3\cos\alpha - \sin\alpha) = 6$, and $q\sin\alpha + u\cos\alpha = 6$. With $t = \tan\alpha$ they combine to
$$\frac{6t}{3 - t} + \frac{3}{t} = 6 \quad\Longrightarrow\quad 4t^2 - 7t + 3 = 0,$$
so $t = \frac34$ or $t = 1$.

For $t = \frac34$: $u = 5$, $q = \frac{10}{3}$, area $\frac{50}{3}$. For $t = 1$: $u = q = 3\sqrt2$, area $18$. Both fit in the square. Areas scale by $10$, so the true areas sum to $\frac53 + \frac95 = \frac{52}{15}$, giving $m + n = 67$.

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

Two of the three inscribed pieces are rigid enough to fix the big square completely, and the phrase "all possible values" signals that the last piece has a free parameter with finitely many valid settings. Three incidence conditions on three unknowns (tilt and two side lengths) produce a quadratic, hence two configurations. Choosing coordinates that make the picture integral ($10 \times 10$) keeps the algebra clean.

• Assuming R must be parallel to the 1x3 rectangle or to the unit square, which finds only one of the two configurations.
• Forgetting to convert areas back after scaling the picture by $\sqrt{10}$ (areas scale by $10$), or reporting one area instead of the sum.