2022 AMC 10A Problem 25

Step 1: counting lattice points. A square of side $n$ with lattice-point corners has $n + 1$ points in each row and $n + 1$ rows, so $(n + 1)^2$ points. For example, side $2$ gives $9$ points.

Step 2: squares $R$ and $S$. Let the sides be $r$, $s$, $t$. We are told $(r + 1)^2 = \frac94 (s + 1)^2$, so $r + 1 = \frac32 (s + 1)$. That means $s + 1$ is even. Write
$$s + 1 = 2k, \qquad r + 1 = 3k.$$
$R$ sits to the right of the $y$-axis and $S$ to the left. They share only the points on the $y$-axis from $0$ up to $s$, which is $s + 1 = 2k$ points. So $R$ and $S$ together contain
$$9k^2 + 4k^2 - 2k = 13k^2 - 2k \text{ points}.$$

Step 3: square $T$. $T$ has a quarter of those points:
$$(t + 1)^2 = \frac{13k^2 - 2k}{4}.$$

Step 4: where $T$ sits. $T$ must cross the $y$-axis (if it were entirely on one side, one of the two fractions would be $0$ and the other not). Say $T$ runs from $x = -u$ to $x = t - u$. Its top corners are in $R \cup S$, so $T$ is no taller than $S$. Counting columns times rows:
- $S$ and $T$ share the columns $x = -u, \ldots, 0$, which is $u + 1$ columns, each with $t + 1$ points.
- $R$ and $T$ share the columns $x = 0, \ldots, t - u$, which is $t - u + 1$ columns, each with $t + 1$ points.

Step 5: the "27 times" condition. $S$ has $4k^2$ points and $R$ has $9k^2$, so
$$\frac{(u + 1)(t + 1)}{4k^2} = 27 \cdot \frac{(t - u + 1)(t + 1)}{9k^2}.$$
Cancel $t + 1$ and multiply both sides by $36k^2$: $9(u + 1) = 108(t - u + 1)$, so
$$u + 1 = 12(t - u + 1).$$

Step 6: what this says about $t$. Let $v = t - u + 1$, the number of columns of $T$ on the $R$ side. Then $u + 1 = 12v$, and $T$ has $t + 1 = (u + 1) + v - 1 = 13v - 1$ columns (the $y$-axis column was counted in both). So $t + 1$ is one less than a multiple of $13$.

Step 7: combine with Step 3. Write $N = t + 1$. We need $4N^2 = 13k^2 - 2k$ with $N = 13v - 1$.
- Since $N$ is one less than a multiple of $13$, $N^2$ is one more than a multiple of $13$, so $4N^2$ is $4$ more than a multiple of $13$.
- $13k^2 - 2k$ is a multiple of $13$ minus $2k$.
- So $2k + 4$ must be a multiple of $13$, which means $k + 2$ is a multiple of $13$: $k = 11, 24, 37, 50, \ldots$
- Also $k$ must be even, since otherwise $13k^2 - 2k$ is odd while $4N^2$ is even. That leaves $k = 24, 50, 76, \ldots$

Step 8: try them in order.
- $k = 24$: $\frac{13 \cdot 576 - 48}{4} = \frac{7440}{4} = 1860$. Not a perfect square ($43^2 = 1849$, $44^2 = 1936$).
- $k = 50$: $\frac{13 \cdot 2500 - 100}{4} = \frac{32400}{4} = 8100 = 90^2$. And $90 = 13 \cdot 7 - 1$, so $v = 7$ works.

Step 9: read off the sides and check. $k = 50$ gives $r = 149$, $s = 99$, $t = 89$, and $u = 12 \cdot 7 - 1 = 83$. $T$ runs from $x = -83$ to $x = 6$ with height $89$. Its top corners $(-83, 89)$ and $(6, 89)$ lie in $S$ and $R$. Every larger $k$ makes all three sides bigger, so this is the smallest total:
$$149 + 99 + 89 = 337.$$

The answer is $\boxed{\textbf{(B)}\ 337}$.

Every statement about the squares is really a lattice-point count, and each count is either a square $(n+1)^2$ or columns times rows. Translating them gives one equation in whole numbers plus a condition on remainders. Remainder and even/odd checks shrink the search to a short list, and the geometry is checked at the end.

A lattice square of side $n$ contains $(n+1)^2$ lattice points. Let the sides be $r, s, t$. From $(r+1)^2 = \frac94 (s+1)^2$ put $r + 1 = 3k$ and $s + 1 = 2k$. Squares $R$ and $S$ overlap only in the $s+1$ points on the $y$-axis, so $R \cup S$ has $9k^2 + 4k^2 - 2k = 13k^2 - 2k$ points, and $(t+1)^2 = \frac{13k^2 - 2k}{4}$.

Let $T$ occupy $-u \le x \le t - u$ with $0 \le u \le t$; its top vertices lie in $S$ and $R$, so also $t \le s$. Counting columns times rows, $S \cap T$ has $(u+1)(t+1)$ points and $R \cap T$ has $(t-u+1)(t+1)$. The stated condition is
$$\frac{(u+1)(t+1)}{4k^2} = 27 \cdot \frac{(t-u+1)(t+1)}{9k^2} \quad\Longrightarrow\quad u + 1 = 12(t - u + 1).$$
Writing $v = t - u + 1 \ge 1$ gives $u + 1 = 12v$ and $n := t + 1 = 13v - 1$, so $n \equiv -1 \pmod{13}$.

Now solve $4n^2 = 13k^2 - 2k = k(13k - 2)$. Modulo $13$: $4 \equiv -2k$, so $k \equiv 11 \pmod{13}$. The product $k(13k-2)$ must be even, so $k$ is even; together, $k \equiv 24 \pmod{26}$.

• $k = 24$: $\frac{24 \cdot 310}{4} = 1860$, not a square.
• $k = 50$: $\frac{50 \cdot 648}{4} = 8100 = 90^2$, and $90 = 13 \cdot 7 - 1$ has the required form ($v = 7$, $u = 83$).

Hence $r = 149$, $s = 99$, $t = 89$; indeed $t \le s$ and $u \le s$, so $T$'s top vertices lie in $R \cup S$. Larger $k$ only enlarges every side, so the minimum is $149 + 99 + 89 = 337$.

The answer is $\boxed{\textbf{(B)}\ 337}$.