Competition · AMC preparation · step 4 of 4

AMC 10 · 2021A · #24

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Lattice or Non-Aligned Polygon Area

coordinate-geometryarea-rectanglesabsolute-value convert-to-algebraidentify-subproblems ↑ Prerequisites: coordinate-geometry

📏 Long solution 💡 2 insights

Problem

The region enclosed by the graphs of (x + a y)² = 4 a² and (a x - y)² = a² (with a ＞ 0) is a quadrilateral. Find its area as a function of a, valid for all a ＞ 0.

Pick an answer.

(A)

$\frac{8a^2}{(a+1)^2}$

(B)

$\frac{4a}{a+1}$

(C)

$\frac{8a}{a+1}$

(D)

$\frac{8a^2}{a^2+1}$

(E)

$\frac{8a}{a^2+1}$

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram) — sketch the four lines; once you see two parallel pairs with perpendicular slopes, the region is obviously a rectangle. Tool #7 (Subproblems) — separately compute the rectangle's length (distance between x + a y = ± 2a) and width (distance between a x - y = ± a), then multiply. Tool #9 (Easier Problem) — plug a = 1 to get a concrete square, verify the answer-choice candidates collapse to the same number to spot-check the formula. Tool #3 (Eliminate) — a = 1 instantly knocks out wrong choices. Tool #13 (Algebra) — the distance-between-parallel-lines formula (|c₁ - c₂|)/(√(A² + B²)) for A x + B y = c gives the cleanest derivation.

1STEP 1

Factor into two line pairs

Each squared equation splits into two parallel lines: x + a y = ± 2 a and a x - y = ± a — four lines, two parallel pairs.

x + a y = ± 2 a; a x - y = ± a

2STEP 2

Check the slopes are perpendicular

Solved for y, the slopes are - $\frac{1}{a}$ and a; their product a · (- $\frac{1}{a}$ ) = -1, so the pairs are perpendicular and the region is a rectangle.

slope₁ = -1/a, slope₂ = a, slope₁ · slope₂ = -1

3STEP 3

Find the rectangle's length

Distance between x + a y = 2a and x + a y = -2 a via $\frac{|c₁ - c₂|}{\sqrt{A² + B²}}$ gives length = $\frac{4a}{\sqrt{a² + 1}}$ .

length = (4 a)/(√(a² + 1))

4STEP 4

Find the rectangle's width

Same formula for a x - y = ± a: width = $\frac{|a - (-a)|}{\sqrt{a² + 1}}$ = $\frac{2a}{\sqrt{a² + 1}}$ .

width = (2 a)/(√(a² + 1))

5STEP 5

Multiply for the area

Multiply: area = length × width = $\frac{4a}{\sqrt{a² + 1}}$ · $\frac{2a}{\sqrt{a² + 1}}$ = $\frac{8a²}{a² + 1}$ .

Area = (8 a²)/(a² + 1)

6STEP 6

Spot-check the simplest case

Spot-check: at a = 1 the area is 4, and at a = 2 it is $\frac{32}{5}$ — only (D) matches both, since (C) and (E) diverge at a = 2.

a = 1: 4; a = 2: 32/5; only (D) consistent

Answer

8a²/(a²+1)

Behavior as a → 0^+: length → 0 (lines x + 0 · y = 0 collapse to x = 0), width → 0, area → 0. Our formula:

\frac{8 a²}{a² + 1}

→ 0. ✓. As a → ∞: length ≈

\frac{4 a}{a}

= 4, width ≈

\frac{2 a}{a}

= 2, area → 8. Our formula:

\frac{8 a²}{a² + 1}

→ 8. ✓. Maximum at a = ?: derivative shows monotonic increase from 0 to 8, never exceeding 8. Everything geometric is consistent with choice (D).

💡Key takeaway

This AMC 10 problem only needs Grade 8 lines and the Pythagorean distance-between-parallel-lines formula you already know — factor (X)² = c² into two parallel lines per equation, notice the two pairs of slopes - $\frac{1}{a}$ and a are perpendicular (so the region is a rectangle), then multiply the two distances $\frac{4a}{\sqrt{a² + 1}}$ · $\frac{2a}{\sqrt{a² + 1}}$ = $\frac{8a²}{a² + 1}$ .

Factor into two line pairs
Check the slopes are perpendicular
Find the rectangle's length
Find the rectangle's width
Multiply for the area
Spot-check the simplest case

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