AMC 10 · 2006 · #11

Grade 8 geometry-2d
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
coordinate-geometrypolynomial-factoring convert-to-algebra ↑ Prerequisites: polynomial-factoring
📏 Medium solution 💡 2 insights
Problem
The equation (x+y)² = x² + y² is graphed in the coordinate plane, where x and y range over all real numbers. The graph is the set of all points (x, y) that make the equation true. Which of these describes that set: the empty set, one point, two lines, a circle, or the entire plane?

Pick an answer.

(A)
the\,empty\,set
(B)
$\textrm{one\,point}$
(C)
$\textrm{two\,lines}$
(D)
$\textrm{a\,circle}$
(E)
$\textrm{the\,entire\,plane}$

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Organize Information in More Ways

The equation looks tangled, but Tool #15 (Organize Information in More Ways) says to rewrite it into a form you already recognize. Expanding the squared side and cancelling the matching pieces turns the whole thing into a tiny statement, xy = 0, whose graph is easy to name. Tool #3 (Eliminate Possibilities) then uses the multiple-choice setup: once you know the graph is unbounded, a circle and a single point are out, and a quick test point rules out the empty set and the whole plane. Tool #1 (Draw a Diagram) confirms the survivors as two familiar lines. The whole difficulty melts once the equation is rewritten.

1STEP 1

Expand the squared side

The only messy piece is (x+y)²; expanding it to x² + 2xy + y² turns the equation into x² + 2xy + y² = x² + y².

(x+y)² = x² + 2xy + y² → x² + 2xy + y² = x² + y²
2STEP 2

Cancel and simplify

Both sides carry an x² and a y²; subtract them and only 2xy = 0 is left, so dividing by 2 gives xy = 0.

x² + 2xy + y² = x² + y² → 2xy = 0 → xy = 0
3STEP 3

Use the zero-product rule

A product is zero only if one factor is zero, so xy = 0 holds exactly when x = 0 or y = 0.

xy = 0 ⇔ x = 0 or y = 0
4STEP 4

Read off the graph

x = 0 draws the vertical y-axis and y = 0 draws the horizontal x-axis, so the graph is those two axis lines — choice (C).

x = 0 (the y-axis) and y = 0 (the x-axis) → two lines → (C)
Answer
two lines
Test some points against the original equation. At (2, 0): left side (2+0)² = 4, right side 2² + 0² = 4 — equal, and this point is on the x-axis. At (0, 3): left side 9, right side 9 — equal, and it sits on the y-axis. At (1, 1), an off-axis point: left side (1+1)² = 4, right side 1 + 1 = 2 — not equal, so it is correctly excluded. Points work exactly when they lie on an axis, so the graph is the two axes: unbounded, so not a circle (choice D) and not a single point (choice B); a test point works, so not empty (choice A); and (1,1) fails, so not the whole plane (choice E). Choice (C), two lines, is the only fit.
💡Key takeaway

Expand the square, cancel the matching pieces, and the whole equation shrinks to xy = 0 — a product is zero only when a factor is zero, so the graph is the two axis lines.

  • Expand the squared side
  • Cancel and simplify
  • Use the zero-product rule
  • Read off the graph

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