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Difference between revisions of "2022 AMC 10A Problems/Problem 25"

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<math>16t^2=2j(26j-4) \implies 4t^2=j(13j-2)</math>.
 
<math>16t^2=2j(26j-4) \implies 4t^2=j(13j-2)</math>.
 
Note that <math>j</math> must be divisible by two for the product to be divisible by 4. Thus we make another substitution, <math>j=2k</math>:
 
Note that <math>j</math> must be divisible by two for the product to be divisible by 4. Thus we make another substitution, <math>j=2k</math>:
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<math>4t^2=2k(26k-2) \implies t^2 = k(13k-1)</math>.
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<math>4t^2=2k(26k-2) \implies </math> <cmath>t^2 = k(13k-1) \quad \quad \quad (2)</cmath>
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Revision as of 00:16, 13 November 2022

Problem 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).

P252022A.png

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 $T$?

$\textbf{(A) }336\qquad\textbf{(B) }337\qquad\textbf{(C) }338\qquad\textbf{(D) }339\qquad\textbf{(E) }340$

Solution

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 $r^2=\frac{9}{4}\cdot s^2 \implies r = \frac{3}{2}s$ $\quad \quad \quad (1)$

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 $t^2 = \frac{1}{4}\cdot(r^2 + s^2 - s )\implies t^2 = \frac{1}{4}\cdot(\frac{9}{4}\cdot s^2 + s^2 - s )$ $\implies t^2=\frac{1}{4}\cdot\frac{13s^2-4s}{4} \implies 16t^2= s(13s-4)$

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, $j=2k$: $4t^2=2k(26k-2) \implies$ \[t^2 = k(13k-1) \quad \quad \quad (2)\]



Solution in progress

~KingRavi

See Also

2022 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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