2018 AMC 10B Problem 22

Picking $x$ and $y$ at random from $0$ to $1$ is like throwing a dart at a $1$ by $1$ square, where the point hit is $(x, y)$. The probability is the area of the good part of the square.

1. The longest side. Both $x$ and $y$ are at most $1$, so the side of length $1$ is the longest. The largest angle is across from it.
2. Condition 1: it is a triangle. The two shorter sides must add to more than the longest:
$$x + y > 1.$$
(The other two triangle rules, like $x + 1 > y$, are automatic.)
3. Condition 2: it is obtuse. A triangle has an obtuse angle across from its longest side exactly when the square of that side is bigger than the sum of the squares of the other two. (Equal would be a right angle, as in the Pythagorean theorem.) So
$$x^2 + y^2 < 1.$$
4. Draw the regions.
- $x^2 + y^2 < 1$ is the inside of a quarter circle of radius $1$ in the corner at $(0,0)$. Its area is $\frac14 \pi \cdot 1^2 = \frac{\pi}{4}$.
- $x + y > 1$ is the part of the square above the line from $(1,0)$ to $(0,1)$.
- The triangle below that line, with corners $(0,0), (1,0), (0,1)$, has area $\frac12$ and sits completely inside the quarter circle.
5. The good region is the quarter circle with that triangle removed:
$$\frac{\pi}{4} - \frac12 \approx 0.785 - 0.5 = 0.285.$$

The closest choice is $0.29$.

The answer is $\boxed{\textbf{(C)}\ 0.29}$.

When two numbers are picked at random from $0$ to $1$, probabilities become areas in the unit square. Each condition draws a boundary: the triangle rule gives a straight line, and the obtuse rule gives a circle. Knowing which side is longest keeps it to just these two conditions.

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 $\boxed{\textbf{(C)}\ 0.29}$.