2018 AMC 10B Problems/Problem 22

Real numbers $x$ and $y$ are chosen independently and uniformly at random from the interval $[0,1]$ . Which of the following numbers is closest to the probability that $x,y,$ and $1$ are the side lengths of an obtuse triangle?

$\textbf{(A)} \text{ 0.21} \qquad \textbf{(B)} \text{ 0.25} \qquad \textbf{(C)} \text{ 0.29} \qquad \textbf{(D)} \text{ 0.50} \qquad \textbf{(E)} \text{ 0.79}$

Solution

The Pythagorean Inequality tells us that in an obtuse triangle, $a^{2} + b^{2} < c^{2}$ . The triangle inequality tells us that $a + b > c$ . So, we have two inequalities: $x^2 + y^2 < 1$ $x + y > 1$ The first equation is $\frac14$ of a circle with radius $1$ , and the second equation is a line from $(0, 1)$ to $(1, 0)$ . So, the area is $\frac{\pi}{4} - \frac12$ which is approximately $0.29$ .

Solution 2

Note that the obtuse angle in the triangle has to be opposite the side that is always length 1. This is because the largest angle is always opposite the largest side, and if 2 sides of the triangle were 1, the last side would have to be greater than 1 to make an obtuse triangle. Using this observation, we can set up a law of cosines where the angle is opposite 1:

$1^2=x^2+y^2-2xy\cos(\theta)$

where $x$ and $y$ are the sides that go from $[0,1]$ and $\theta$ is the angle opposite the side of length 1.

By isolating $\cos(\theta)$ , we get:

$\frac{1-x^2-y^2}{-2xy} = \cos(\theta)$

For $\theta$ to be obtuse, $\cos(\theta)$ must be negative. Therefore, $\frac{1-x^2-y^2}{-2xy}$ is negative. Since $x$ and $y$ must be positive, $-2xy$ must be negative, so we must make $1-x^2-y^2$ positive. From here, we can set up the inequality $x^2+y^2<1$ Additionally, to satisfy the definition of a triangle, we need: $x+y>1$ The solution should be the overlap between the two equations in the 1st quadrant.

By observing that $x^2+y^2<1$ is the equation for a circle, the amount that is in the 1st quadrant is $\frac{\pi}{4}$ . The line can also be seen as a chord that goes from $(0, 1)$ to $(1, 0)$ . By cutting off the triangle of area $\frac{1}{2}$ that is not part of the overlap, we get $\frac{\pi}{4} - \frac{1}{2} \approx \boxed{0.29}$ .

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2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2018 AMC 10B Problems/Problem 22

Solution

Solution 2

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