2022 AMC 10A Problem 25

Let $r$ be the number of lattice points on the side length of $R.$ Similarly define $s$ for $S$ and $t$ for $T.$ Note that the number of lattice points in a rectangle is the product of the number of lattice points along its width and the number of lattice points along its length. The first condition gives us that $$r^2 = \dfrac{9}{4} \cdot s^2$$$$r = \dfrac{3}{2} \cdot s \tag*{(1)}$$ The number of lattice points in $R \cup S$ is the sum of the lattice points in each of the regions, but there is overlap along the $y$-axis where $S$ touches it. The second condition, therefore, yields $$t^2 = \dfrac{1}{4}(r^2 + s^2 - s)$$ $$t^2 = \dfrac{1}{4}(\dfrac{9}{4} \cdot s^2 + s^2 - s)$$ $$t^2 = \dfrac{1}{4} \cdot \dfrac{13s^2 - 4s}{4}$$ $$16t^2 = s(13s - 4).$$ From $(1),$ we get that $s$ is a multiple of $2.$ We can substitute $s$ with $2j$ to get $$16t^2 = 2j(26j - 4)$$ $$4t^2 = j(13j - 2).$$ For the product to be divisible by $4,$ $j$ must be divisible by $2.$ We can again substitute $j$ with $2k$ to get $$4t^2 = 2k(26k - 2)$$ $$t^2 = k(13k - 1) \tag*{(2)}$$ 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 rectangle formed by $R \cap T.$ Using these variables, we get that the number of lattice points in $S \cap T$ is $xt$ and in $R \cap T$ is $yt.$ The third condition gives us that $$\dfrac{xt}{s^2} = 27 \cdot \dfrac{yt}{r^2}$$ $$\dfrac{x}{s^2} = 27 \cdot \dfrac{y}{\dfrac{9}{4} s^2}$$ $$x = 12y.$$ We also know that $t = x + y - 1$ (accounting for overlap), and this yields $$t = 13y - 1 \tag*{(3)}$$ $(3)$ gives us that $$\begin{gathered} t \equiv -1 \pmod{13}, \\ t^2 \equiv 1 \pmod{13}. \end{gathered}$$ However, by $(2),$ we get that $$t^2 \equiv -k \pmod{13}$$ $$k \equiv -1 \pmod{13}.$$ By $(2),$ we also get that $k$ is a perfect square since it is relatively prime to $13k - 1,$ and they must multiply to a perfect square. Thus $k$ must be a perfect square satisfying $k\equiv-1\pmod{13}.$ The smaller positive squares $1,4,9,16$ do not satisfy this congruence, while $k=25$ does, so $25$ is the least possible value. From this value of $k,$ we get that $j = 2 \cdot 25 = 50,$ $s = 2 \cdot 50 = 100,$ and $r = \dfrac{3}{2} \cdot 100 = 150.$ We can also find that $$t^2 = 25(13 \cdot 25 - 1) = 25 \cdot 324$$ $$t = 5 \cdot 18 = 90.$$ Therefore, $$r + s + t = 340.$$ The question, however, asked for the sum of the side lengths. The side lengths of the squares are $1$ less than the number of lattice points on the side, so we have to subtract $3.$ This value is attainable: equation $(3)$ gives $y=7$ and hence $x=84.$ Since $x\le s=100$ and $y\le r=150,$ a square $T$ with $90$ lattice points per side can straddle the $y$-axis with the required overlaps, and its top vertices lie in $R\cup S.$ Therefore, the desired answer is $340 - 3 = 337.$ Thus, B is the correct answer.