AMC10 2022 A

Let $r$ be the number of lattice points on the side length of square $R$, $s$ be the number of lattice points on the side length of square $S$, and $t$ be the number of lattice points on the side length of square $T$. Note that the actual lengths of the side lengths are the number of lattice points minus $1$, so we can work in terms of $r, s, t$ and subtract $3$ to get the actual answer at the end. Furthermore, note that the number of lattice points inside a rectangular region is equal to the number of lattice points in its width times the number of lattice points along its length. Using this fact, the number of lattice points in $R$ is $r^2$, the number of lattice points in $S$ is $s^2$, and the number of lattice points in $T$ is $t^2$. Now, by the first condition, we have \[r^2=\frac{9}{4}\cdot s^2 \implies r = \frac{3}{2}s \quad \quad \quad \quad \quad (1)\] The second condition, the number of lattice points contained in $T$ is a fourth of the number of lattice points contained in $R \cup S$. The number of lattice points in $R \cup S$ is equal to the sum of the lattice points in their individually bounded regions, but the lattice points along the y-axis for the full length of square $S$ is shared by both of them, so we need to subtract that out. In all, this condition yields us $t^2 = \frac{1}{4}\cdot(r^2 + s^2 - s )\implies t^2 = \frac{1}{4}\cdot\left(\frac{9}{4}\cdot s^2 + s^2 - s \right)$ $\implies t^2=\frac{1}{4}\cdot\frac{13s^2-4s}{4} \implies 16t^2= s(13s-4)$ Note from $(1)$ that $s$ is a multiple of $2$. We can write $s=2j$ and substitute: $16t^2=2j(26j-4) \implies 4t^2=j(13j-2)$. Note that $j$ must be divisible by two for the product to be divisible by 4. Thus we make another substitution, $j=2k$: \[4t^2=2k(26k-2) \implies t^2 = k(13k-1) \quad \quad \quad \quad \quad (2)\] Finally we look at the last condition; that the fraction of the lattice points inside $S$ that are inside $S \cap T$ is $27$ times the fraction of lattice points inside $r$ that are inside $R \cap T$. Let $x$ be the number of lattice points along the bottom of the rectangle formed by $S \cap T$, and $y$ be the number of lattice points along the bottom of the the rectangle formed by $R \cap T$. Therefore, the number of lattice points in $S\cap T$ is $xt$ and the number of lattice points in $R \cap T$ is $yt$. Thus by this condition, $\frac{xt}{s^2} = 27 \cdot \frac{yt}{r^2} \implies \frac{x}{s^2} = 27 \cdot \frac{y}{\frac{9}{4}\cdot s^2} \implies x= 12y$ Finally, notice that $t=x+y-1=12y+y-1$ (subtracting overlap), and so we have \[t=13y-1 \quad \quad \quad \quad \quad (3)\] Now notice that by $(3)$ , $t\equiv -1 \pmod{13}\implies t^2 \equiv 1 \pmod{13}$. However, by $(2)$ , $t^2 \equiv k \cdot -1 \pmod{13}$. Therefore, $-k \equiv 1 \pmod{13} \implies k \equiv -1 \pmod{13}$ Also, by $(2)$ , we know $k$ must be a perfect square since $k$ is relatively prime to $13k-1$ (Euclids algorithm) and the two must multiply to a perfect square. Hence we know two conditions on $k$, and we can now guess and check to find the smallest that satisfies both. We check $k=12$ first since its one less than a multiple of $13$, but this does not work. Next, we have $k=25$ which works because $25$ is a perfect square. Thus we have found the smallest $k$, and therefore the smallest $r, s, t$. Now we just work backwards: $j= 2k = 50$ and $s=2j=100$. Then $r=\frac{3}{2}\cdot 100 = 150$. Finally, from $(2)$ , $t^2=25(13\cdot25-1) \implies t^2 = 25 \cdot 324 \implies t=5\cdot 18=90$. Finally, the sum of each square’s side lengths is $r+s+t-3=340-3=337=\boxed{\textbf{(B) }337}$.