AMC 10 · 2002 · #25

Grade 9 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid

completing-the-squarecircle-equationdifference-of-squarescircular-sector symmetry-argumentconvert-to-algebra ↑ Prerequisites: completing-the-squarecircle-equation

📏 Long solution 💡 3 insights

Problem

Let f(x) = x² + 6x + 1, and let R be the set of points (x,y) in the plane that satisfy both f(x) + f(y) ≤ 0 and f(x) - f(y) ≤ 0. Which of the listed numbers is closest to the area of R?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Organize Information in More Ways

As written, both conditions are quadratic inequalities in two variables, which say nothing to the eye. Tool #4 (Introduce a Variable) is used twice: completing the square rewrites f, and then u = x+3, v = y+3 slides the origin. Tool #15 (Organize Information in More Ways) is the main move — after that slide, one condition is a disk centred at the origin and the other is simply |u| ≤ |v|, so both conditions share the same centre and the same symmetry. Tool #1 (Draw a Diagram) turns that pair into one picture: a circle cut by two perpendicular lines through its own centre. Tool #16 (Change Focus) finishes it — nothing needs to be integrated, because the region is a whole number of congruent slices of the circle.

1STEP 1

Complete the square in f

Completing the square turns the first condition into (x+3)² + (y+3)² ≤ 16.

f(t) = (t+3)² - 8 → f(x)+f(y) ≤ 0 ⇔ (x+3)² + (y+3)² ≤ 16

2STEP 2

Read the first condition as a disk

That is a disk of radius 4 centred at (-3,-3), with area 16π.

(x+3)² + (y+3)² ≤ 4²: disk, centre (-3,-3), radius 4, area 16π

3STEP 3

Factor the second condition

The second condition factors into (x-y)(x+y+6) ≤ 0 — two straight lines, not a curve.

f(x) - f(y) = (x-y)(x+y+6) ≤ 0

4STEP 4

Slide the origin to the centre

Shifting the origin to the centre makes both conditions read u² + v² ≤ 16 and |u| ≤ |v|.

u = x+3, v = y+3 → u² + v² ≤ 16 and (u-v)(u+v) = u² - v² ≤ 0 ⇔ |u| ≤ |v|

5STEP 5

Count quarters of the disk

The two perpendicular lines through the centre cut the disk into four congruent quarters.

four congruent quarters of area 4π each; |u| ≤ |v| keeps the top and bottom ones

6STEP 6

Add the two pieces and compare

Keeping two of four gives half the disk, 8π ≈ 25.13, closest to 25, choice (E).

[R] = 1/2 · 16π = 8π ≈ 25.13 → (E)

Answer

Test specific points to confirm the region is the top-and-bottom one and not the left-and-right one. At the centre (-3,-3): f(-3) = 9 - 18 + 1 = -8, so the sum is -16 ≤ 0 and the difference is 0 ≤ 0; the centre is in R, as it must be since it sits on both boundary lines. At (-3,-7), the bottom of the disk: f(-7) = 49 - 42 + 1 = 8, so the sum is -8 + 8 = 0 (on the circle) and the difference is -16 ≤ 0; in R. At its mirror image (-7,-3), the left of the disk: the difference is +16 ＞ 0; not in R. Bottom in, left out — exactly the claim. A numerical check agrees: counting the two conditions on a fine grid over [-8,2]×[-8,2] gives an area of about 25.14, against 8π = 25.133. One more guard against a common slip — forgetting to halve would give 16π ≈ 50.3, which is nowhere near any choice, so the halving step is genuinely load-bearing rather than decorative.

💡Key takeaway

Slide the whole picture until the circle's centre sits at the origin — the two cutting lines land on that centre too, and the region you want is simply half the circle.

Complete the square in f
Read the first condition as a disk
Factor the second condition
Slide the origin to the centre
Count quarters of the disk
Add the two pieces and compare