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

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 ∪ S, straddling the y-axis) — satisfy: R has

\frac{9}{4}

as many lattice points as S; T has

\frac{1}{4}

as many lattice points as R ∪ S; the fraction of S's lattice points lying in S ∩ T is 27 times the fraction of R's lattice points lying in R ∩ T. Find the minimum possible value of ℓ_R + ℓ_S + ℓ_T (edge lengths).

Pick an answer.

(A)

336

(B)

337

(C)

338

(D)

339

(E)

340

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram) — sketch R and S meeting at the y-axis with T straddling, label edge lengths ℓ_R, ℓ_S, ℓ_T and lattice-point counts r = ℓ_R + 1, s = ℓ_S + 1, t = ℓ_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 t² = 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 = 1², 2², 3², … until 13k - 1 is also a perfect square. Tool #3 (Eliminate) — the answer choices 336 to 340 confirm the magnitude.

1STEP 1

Let r, s, t be the lattice points per edge, so #R = r², #S = s², #T = t². The ratio #R = $\frac{9}{4}$ #S gives r = $\frac{3}{2}$ s, forcing s even and r ＞ s.

r =

\frac{3}{2}

s, s even

2STEP 2

R and S share only their y-axis edge (s points), so #(R ∪ S) = r² + s² - s; then #T = $\frac{1}{4}$ #(R ∪ S) becomes 16 t² = s(13 s - 4).

16 t² = s(13 s - 4)

3STEP 3

Split T at the y-axis: R ∩ T, S ∩ T are t-tall blocks, so the 27-ratio gives x_p = 12 y_p and t = 13 y_p - 1, so t² ≡ 1 (mod 13).

x_p = 12 y_p, t = 13 y_p - 1, t² ≡ 1 (mod 13)

4STEP 4

Since s is even, substitute s = 4k; the equation reduces cleanly to t² = k(13 k - 1), with the two factors coprime.

s = 4k, t² = k(13 k - 1)

5STEP 5

Since k and 13 k - 1 are coprime, both must be perfect squares: set k = m²; reducing mod 13 forces m² ≡ 12 (mod 13).

k = m², 13 m² - 1 a perfect square, m² ≡ 12 (mod 13)

6STEP 6

Test m = 1…5: the first with m² ≡ 12 (mod 13) is m = 5, and k = 25 makes 13 k - 1 = 324 = 18², a perfect square.

m = 5, k = 25, 13 k - 1 = 324 = 18²

7STEP 7

Back-substitute: s = 100, r = 150, t = 90, so ℓ_R + ℓ_S + ℓ_T = 149 + 99 + 89 = 337, choice (B).

ℓ_R + ℓ_S + ℓ_T = 149 + 99 + 89 = 337 → (B)

Answer

337

Sanity. The three lattice-point ratios (

\frac{9}{4}

\frac{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 · 7 = 84, so T has 7 lattice columns inside R and 84 columns inside S along its bottom — both are ≤ s = 100 and ≤ r = 150, so T truly fits with top vertices inside R ∪ 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.

💡Key takeaway

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 t² = 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 · 25 - 1 = 324 = 18²), giving r = 150, s = 100, t = 90 and ℓ_R + ℓ_S + ℓ_T = 149 + 99 + 89 = 337.