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2022 AMC 10A Problem 25

2022 AMC 10A Problem 25AMC 10 Step by Step

Name: Date: September 13, 2026

Coordinate Geometry · difficulty 5 of 5 · about 15 min · position P21-25

Let $R$ , $S$ , and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the $x$ -axis. The left edge of $R$ and the right edge of $S$ are on the $y$ -axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$ . The top two vertices of $T$ are in $R \cup S$ , and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S.$ See the figure (not drawn to scale). The fraction of lattice points in $S$ that are in $S \cap T$ is $27$ times the fraction of lattice points in $R$ that are in $R \cap T$ . What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$ ?

A)
$336$
B)
$337$
C)
$338$
D)
$339$
E)
$340$

Answer

Key insight

Side n has (n+1)^2 lattice points; set r+1 = 3k, s+1 = 2k, and the 27-times condition forces t+1 congruent to -1 mod 13, then 4(t+1)^2 = 13k^2 - 2k.

Solution

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 $\boxed{\textbf{(B)}\ 337}$ .

Why this works

Every geometric statement here is a lattice-point count, and each count is a product (columns times rows) or a square. Translating them yields one Diophantine equation, $4n^2 = 13k^2 - 2k$ , plus the side condition $n \equiv -1 \pmod{13}$ . Congruence and parity filters shrink the search to a short list, and the geometric constraints ( $u \le s$ , $t \le s$ ) are checked at the end.

The trap

Counting lattice points as the area n^2 instead of (n+1)^2, or double counting the shared y-axis column when finding the points of R union S.

Common mistakes

Counting lattice points as the area n^2 instead of (n+1)^2, or double counting the shared y-axis column when finding the points of R union S.
Stopping at the first $k$ that makes $13k^2 - 2k$ four times a square without checking $t + 1 \equiv -1 \pmod{13}$ , or forgetting that $T$ may straddle the $y$ -axis.

Techniques

Organized listing / direct enumeration · Substitute to simplify (u = x+1/x, shifting, scaling)