AMC 10 Step by Step

2006 AMC 10A Problem 11

P11-15~1 minPrint

Which of the following describes the graph of the equation (x+y)2=x2+y2(x+y)^2=x^2+y^2 ?

Log in to save your progress.
Show solution without answering

Key insight

Expand the left side. The x^2 and y^2 cancel, leaving xy = 0, which is true when x = 0 or when y = 0: two lines, the two axes.

Solution

Simplify the equation first, then describe every point that satisfies it.

1. Expand: (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2.
2. The equation becomes x2+2xy+y2=x2+y2x^2 + 2xy + y^2 = x^2 + y^2. Take x2+y2x^2 + y^2 away from both sides:
2xy=0,xy=0.2xy = 0, \qquad xy = 0.
3. A product is zero exactly when one of the factors is zero. So a point works if x=0x = 0 or if y=0y = 0.
4. The points with x=0x = 0 form the yy-axis, like (0,5)(0, 5). The points with y=0y = 0 form the xx-axis, like (3,0)(3, 0). Check (3,0)(3, 0) in the original: (3+0)2=9(3 + 0)^2 = 9 and 32+02=93^2 + 0^2 = 9. It works.

So the graph is two lines that cross at the origin.

The answer is (C) two lines\boxed{\textbf{(C)}\ \text{two lines}}.

Why this works

The graph of an equation is the set of all points that make it true. Simplify as much as you can, then ask "which points work?" When the simplified equation says a product equals zero, the graph is everything that makes the first factor zero together with everything that makes the second factor zero.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Expand the left side: (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2. Cancelling x2+y2x^2 + y^2 from both sides leaves
2xy=0⟹xy=0.2xy = 0 \quad\Longrightarrow\quad xy = 0.
A product is zero exactly when a factor is zero, so the solutions are all points with x=0x = 0 (the yy-axis) together with all points with y=0y = 0 (the xx-axis). That is two lines crossing at the origin.

The answer is (C) two lines\boxed{\textbf{(C)}\ \text{two lines}}.

Review in the AoPS books

Similar problems

Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.