AMC 10 · 2022 · #25

Grade 8 geometry-2d

coordinate-geometrydivisibility-rulesmulti-digit-arithmetic convert-to-algebraidentify-subproblemsguess-and-check ↑ Prerequisites: divisibility-rules

📏 Long solution 💡 3 insights 📊 Diagram

Problem

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 $T$ ?

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

Pick an answer.

(A)

336

(B)

337

(C)

338

(D)

339

(E)

340

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Toolkit + CCSS Solution

Understand

Restated: Three axis-aligned lattice squares — $R$ (first quadrant, left edge on $y$-axis), $S$ (second quadrant, right edge on $y$-axis), and $T$ (bottom edge on $x$-axis, top two vertices in $R \cup S$, straddling the $y$-axis) — satisfy: $R$ has $\tfrac{9}{4}$ as many lattice points as $S$; $T$ has $\tfrac{1}{4}$ as many lattice points as $R \cup S$; the fraction of $S$'s lattice points lying in $S \cap T$ is $27$ times the fraction of $R$'s lattice points lying in $R \cap T$. Find the minimum possible value of $\ell_R + \ell_S + \ell_T$ (edge lengths).

Givens: $R$, $S$, $T$ are axis-aligned squares with vertices at lattice points; bottom edges on the $x$-axis; Left edge of $R$ on the $y$-axis; right edge of $S$ on the $y$-axis; $T$ straddles the $y$-axis; its top two vertices lie inside $R \cup S$; $\# R = \tfrac{9}{4}\#S$ (lattice-point counts); $\#T = \tfrac{1}{4}\#(R \cup S)$; $\tfrac{\#(S \cap T)}{\#S} = 27 \cdot \tfrac{\#(R \cap T)}{\#R}$; Answer choices: (A) $336$, (B) $337$, (C) $338$, (D) $339$, (E) $340$

Unknowns: Minimum value of $\ell_R + \ell_S + \ell_T$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #13 Convert to Algebra, #7 Identify Subproblems, #6 Guess and Check, #3 Eliminate Possibilities

Tool #1 (Diagram) — sketch $R$ and $S$ meeting at the $y$-axis with $T$ straddling, label edge lengths $\ell_R, \ell_S, \ell_T$ and lattice-point counts $r = \ell_R + 1, s = \ell_S + 1, t = \ell_T + 1$. Tool #7 (Subproblems) — the three given conditions translate, one by one, into three equations in $r, s, t$ (plus a horizontal split of $T$ across the axis). Tool #13 (Algebra) — combine them to a single Diophantine equation $t^2 = k(13k - 1)$ where $s = 4k$. Tool #6 (Guess & Check) — the coprime factors $k$ and $13k - 1$ must each be a perfect square, so test $k = 1^2, 2^2, 3^2, \ldots$ until $13k - 1$ is also a perfect square. Tool #3 (Eliminate) — the answer choices $336$ to $340$ confirm the magnitude.

Execute — Answer: B

#1 Draw a Diagram 5.G.A.2 Step 1

Sketch and label.
Let $r = \ell_R + 1, s = \ell_S + 1, t = \ell_T + 1$ (lattice points per edge).
Then $\#R = r^2, \#S = s^2, \#T = t^2$.
The first condition $\#R = \tfrac{9}{4} \#S$ gives $r^2 = \tfrac{9}{4} s^2$, so $r = \tfrac{3}{2} s$, forcing $s$ even (so $r$ is an integer) and $r > s$, hence $\ell_R > \ell_S$.

$$r = \tfrac{3}{2} s, \quad s \text{ even}$$

💡 Grade 5 — set up coordinate sketch and count lattice points on a square edge.

#7 Identify Subproblems 7.SP.C.8 Step 2

Lattice points in $R \cup S$.
The squares share only their common vertical edge on the $y$-axis.
Since $\ell_R > \ell_S$, the shared edge has length $\ell_S$ with $s = \ell_S + 1$ lattice points.
By inclusion-exclusion: $\#(R \cup S) = r^2 + s^2 - s$.
The second condition $\#T = \tfrac{1}{4}\#(R \cup S)$ becomes $4 t^2 = r^2 + s^2 - s$.
Substituting $r^2 = \tfrac{9}{4} s^2$ and multiplying by $4$ gives $16 t^2 = 9 s^2 + 4 s^2 - 4 s = 13 s^2 - 4 s = s(13 s - 4)$.

$$16 t^2 = s(13 s - 4)$$

💡 Grade 7 inclusion-exclusion on lattice point sets, then algebraic simplification.

#7 Identify Subproblems 6.RP.A.3 Step 3

Decompose $T$ along the $y$-axis.
Let $T$ extend across the $y$-axis with $y_p$ lattice points on its bottom edge to the right of (or on) the axis and $x_p$ to the left of (or on), with the $y$-axis column counted in both, so $t = x_p + y_p - 1$.
Then $R \cap T$ is a $y_p$-by-$t$ rectangular block of lattice points (sharing $T$'s full height $t$ since $T$'s top is inside $R \cup S$ and the entire top sits within the squares' vertical extents).
So $\#(R \cap T) = y_p \cdot t$ and $\#(S \cap T) = x_p \cdot t$.
The third condition $\tfrac{\#(S \cap T)}{s^2} = 27 \cdot \tfrac{\#(R \cap T)}{r^2}$ becomes $\tfrac{x_p t}{s^2} = 27 \cdot \tfrac{y_p t}{r^2}$.
Using $r^2 = \tfrac{9}{4} s^2$: $x_p = 27 \cdot \tfrac{4}{9} y_p = 12 y_p$.
So $t = x_p + y_p - 1 = 13 y_p - 1$ and $t \equiv -1 \pmod{13}$, giving $t^2 \equiv 1 \pmod{13}$.

$$x_p = 12 y_p, \quad t = 13 y_p - 1, \quad t^2 \equiv 1 \pmod{13}$$

💡 Grade 6 ratio reasoning — the $27$ folds neatly into the $\tfrac{9}{4}$ ratio between $R$ and $S$.

#13 Convert to Algebra 8.EE.C.7 Step 4

Simplify the Diophantine equation.
Since $s$ is even, write $s = 4k$ (we'll check the $\bmod 4$ count): with $s = 2j$, $16 t^2 = 2j(26 j - 4) = 4 j (13 j - 2)$, so $4 t^2 = j(13 j - 2)$, which forces $j$ even (since the right side must be a multiple of $4$ and $13j - 2$ has opposite parity from $j$).
Writing $j = 2k$ gives $s = 4 k$ and $4 t^2 = 2k(26 k - 2) = 4 k (13 k - 1)$, i.e.
$t^2 = k(13 k - 1)$.

$$s = 4k, \quad t^2 = k(13 k - 1)$$

💡 Grade 8 — substitute $s = 4k$ to clean up the equation; now both factors on the right are coprime.

#13 Convert to Algebra 8.EE.A.2 Step 5

$k$ and $13 k - 1$ are coprime ($\gcd(k, 13 k - 1) = \gcd(k, -1) = 1$), so for their product to be a perfect square BOTH must be perfect squares.
Let $k = m^2$.
Then we need (i) $13 m^2 - 1$ is a perfect square, and (ii) the modular condition $t^2 \equiv 1 \pmod{13}$ becomes $k(13k - 1) \equiv 1 \pmod{13}$, i.e.
$k \cdot (-1) \equiv 1$, so $k \equiv -1 \equiv 12 \pmod{13}$, i.e.
$m^2 \equiv 12 \pmod{13}$.

$$k = m^2, \quad 13 m^2 - 1 \text{ a perfect square}, \quad m^2 \equiv 12 \pmod{13}$$

💡 Grade 8 — perfect-square reasoning plus a quick modular check to find the smallest $m$.

#6 Guess and Check 8.EE.A.2 Step 6

Guess and check small $m$.
Squares mod $13$: $m = 1, 2, 3, 4, 5$ give $m^2 \equiv 1, 4, 9, 3, 12$.
The first $m$ with $m^2 \equiv 12 \pmod{13}$ is $m = 5$.
Check: $k = 25$, $13 k - 1 = 325 - 1 = 324 = 18^2$ — a perfect square.
So the minimum is $m = 5, k = 25$.

$$m = 5, \;k = 25, \;13 k - 1 = 324 = 18^2$$

💡 Grade 8 — test $m = 1, 2, 3, 4, 5$ by hand; $5^2 = 25$ leaves remainder $12$ on division by $13$, and $324 = 18^2$ is the perfect square we need.

#13 Convert to Algebra 5.NBT.B.5 Step 7

Read off the answers.
$s = 4 k = 100$, $r = \tfrac{3}{2} s = 150$, $t^2 = 25 \cdot 324 = 8100$, $t = 90$.
So $\ell_R = r - 1 = 149$, $\ell_S = s - 1 = 99$, $\ell_T = t - 1 = 89$, and the sum is $149 + 99 + 89 = 337$, choice (B).

$$\ell_R + \ell_S + \ell_T = 149 + 99 + 89 = 337 \;\Rightarrow\; \textbf{(B)}$$

💡 Grade 5 — basic multi-digit arithmetic to finish.

[1] #1 5.G.A.2 Sketch and label. Let $r = \ell_R + 1, s = \ell_S + 1, t = \ell_T + 1$ (lattice

[2] #7 7.SP.C.8 Lattice points in $R \cup S$. The squares share only their common vertical edge

[3] #7 6.RP.A.3 Decompose $T$ along the $y$-axis. Let $T$ extend across the $y$-axis with $y_p$

[4] #13 8.EE.C.7 Simplify the Diophantine equation. Since $s$ is even, write $s = 4k$ (we'll chec

[5] #13 8.EE.A.2 $k$ and $13 k - 1$ are coprime ($\gcd(k, 13 k - 1) = \gcd(k, -1) = 1$), so for t

[6] #6 8.EE.A.2 Guess and check small $m$. Squares mod $13$: $m = 1, 2, 3, 4, 5$ give $m^2 \equi

[7] #13 5.NBT.B.5 Read off the answers. $s = 4 k = 100$, $r = \tfrac{3}{2} s = 150$, $t^2 = 25 \cd

Review

Reasonableness: Sanity. The three lattice-point ratios ($9/4$, $1/4$, $27$) all involve the same prime factor pattern ($2$ and $3$), so a tight solution exists with $r : s : t = 150 : 100 : 90 = 15 : 10 : 9$ — nice ratio. The corresponding $y_p$ from $t = 13 y_p - 1$ is $y_p = 7$ and $x_p = 12 \cdot 7 = 84$, so $T$ has $7$ lattice columns inside $R$ and $84$ columns inside $S$ along its bottom — both are $\le s = 100$ and $\le r = 150$, so $T$ truly fits with top vertices inside $R \cup S$. The sum $149 + 99 + 89 = 337$ matches choice (B) exactly. Choices $336$ to $340$ confirm the answer band; $337$ is precisely the unique value the system forces.

Alternative: Tool #3 (Eliminate Possibilities): given the structure $r : s : t = 3 : 2 : t/k$ and the requirement $t = 13 y_p - 1$, the edge sum $r + s + t - 3$ must be $1 \pmod{?}$ — equivalent to checking which of $336, 337, 338, 339, 340$ can decompose as $(150 - 1) + (100 - 1) + (\text{lattice-edge}_T - 1)$ for an integer $\ell_T$. Only $\ell_T = 89$ yields the modular and perfect-square pattern, ruling out the neighbors.

CCSS standards used (min grade 8)

5.G.A.2 Represent real-world and mathematical problems by graphing points in the first quadrant of the coordinate plane (Sketching $R, S, T$ in the coordinate plane and counting lattice points $(l+1)^2$ on each square.)
5.NBT.B.5 Fluently multiply multi-digit whole numbers using the standard algorithm (Computing $25 \cdot 324 = 8100$, $\sqrt{8100} = 90$, and the final sum $149 + 99 + 89 = 337$.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Reducing the third condition $\tfrac{x_p}{s^2} = 27 \cdot \tfrac{y_p}{r^2}$ to $x_p = 12 y_p$.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Inclusion-exclusion $\#(R \cup S) = r^2 + s^2 - s$ on the lattice-point sets.)
8.EE.A.2 Use square root and cube root symbols to represent solutions to equations; evaluate square roots of small perfect squares (Recognizing $k$ and $13 k - 1$ as coprime perfect squares, and checking $324 = 18^2$ and $8100 = 90^2$.)
8.EE.C.7 Solve linear equations in one variable (Substituting $s = 4 k$ and simplifying $16 t^2 = s(13 s - 4)$ to $t^2 = k(13 k - 1)$.)

⭐ This AMC 10 problem only needs Grade 8 algebra and square roots you already know — read each lattice-point ratio as a clean equation in $r, s, t$, simplify to $t^2 = k(13k - 1)$ with $s = 4k$, and notice $k$ and $13k - 1$ must each be a perfect square. Testing $k = 1, 4, 9, 16, 25$ by hand shows $k = 25$ works (since $13 \cdot 25 - 1 = 324 = 18^2$), giving $r = 150, s = 100, t = 90$ and $\ell_R + \ell_S + \ell_T = 149 + 99 + 89 = 337$.