2016 AMC 10B Problem 21

Replacing $x$ by $-x$ or $y$ by $-y$ leaves the equation unchanged, so the graph is symmetric across both axes. Compute the area in the first quadrant and multiply by $4$.

For $x, y \ge 0$ the equation is $x^2 - x + y^2 - y = 0$. Completing the square,
$$\left(x - \tfrac12\right)^2 + \left(y - \tfrac12\right)^2 = \tfrac12,$$
a circle with center $\left(\tfrac12, \tfrac12\right)$ and radius $\tfrac{\sqrt2}{2}$. It passes through $(0,0)$, $(1,0)$ and $(0,1)$.

The segment from $(1,0)$ to $(0,1)$ has length $\sqrt2$, twice the radius, so it is a diameter: it splits the circle into two semicircles. The half on the far side from the origin lies in the first quadrant; the half containing the origin is cut off by the axes, and the only part of the enclosed region there is the triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$.

First-quadrant area $=$ triangle $+$ semicircle $= \tfrac12 + \tfrac12 \pi \left(\tfrac{\sqrt2}{2}\right)^2 = \tfrac12 + \tfrac{\pi}{4}$.

Multiplying by $4$: total area $= 2 + \pi$.

The answer is $\boxed{\textbf{(B)}\ \pi + 2}$.

Absolute values split the plane into four congruent pieces, and in each piece the equation is an honest circle after completing the square. The region is then a square of side $\sqrt2$ (the diamond with vertices $(\pm1, 0)$, $(0, \pm1)$) with a semicircular cap on each of its four sides; four semicircles make two full circles of area $\tfrac{\pi}{2}$ each, plus the square's area $2$.

Sketch the curve: a rounded four-petal shape passing through $(\pm1, 0)$ and $(0, \pm1)$. The inner diamond has area $2$, so the answer must be $2$ plus something involving $\pi$; only (B) has the form $\pi + 2$, and $2\pi + 2$ is not offered.

• Treating the first-quadrant piece as the full circle of radius $\tfrac{\sqrt2}{2}$, which overcounts the semicircle lying outside the quadrant.
• Using the diamond's side length $\sqrt2$ as if it were an area contribution, arriving at choices with $\sqrt2$ in them.