2018 AMC 10B Problem 22

Since $x, y \le 1$, the side of length $1$ is the longest, so it is opposite the largest angle. Two conditions are needed:

1. A triangle exists: the two shorter sides must exceed the longest, $x + y > 1$ (the other two inequalities hold automatically).
2. The angle opposite the side $1$ is obtuse: $x^2 + y^2 < 1^2$.

In the unit square of all $(x, y)$, the second condition is the interior of the quarter disk of radius $1$ centered at the origin, area $\frac{\pi}{4}$. The first condition removes the triangle below the line $x + y = 1$, which has area $\frac{1}{2}$ and lies entirely inside that quarter disk. The favorable region is therefore the quarter disk with the triangle cut out:
$$P = \frac{\pi}{4} - \frac{1}{2} \approx 0.7854 - 0.5 = 0.2854.$$

The closest choice is $0.29$, so the answer is $\boxed{\textbf{(C)}\ 0.29}$.

Uniform random pairs turn probability into area in the unit square, and each geometric condition becomes a curve: the triangle inequality is a line, the obtuse condition is a circle. Identifying the longest side first is what keeps the problem to a single case; then the region is a standard shape minus another.

The choices are spread out, so a rough estimate suffices: the quarter disk has area about $0.785$; the region below $x + y = 1$ is exactly half the square, $0.5$, and it lies inside the quarter disk, so the answer is just under $0.29$. Any answer near $0.28$ rules out everything but (C).

• Dropping the triangle inequality and answering pi/4 = 0.79, or reversing the obtuse condition to x^2 + y^2 > 1 and getting 0.21.
• Treating $x$ or $y$ as a possible longest side; with both at most $1$, the side $1$ is always longest, so only one obtuse condition applies.