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Difference between revisions of "2018 AMC 10B Problems/Problem 22"

(Created page with "Real numbers <math>x</math> and <math>y</math> are chosen independently and uniformly at random from the interval <math>[0,1]</math>. Which of the following numbers is closest...")
 
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<math>\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}</math>
 
<math>\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}</math>
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==Solution 1==
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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:
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<cmath>1^2=x^2+y^2-2xy\cos(\theta)</cmath>
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where <math>x</math> and <math>y</math> are the sides that go from <math>[0,1]</math> and <math>\theta</math> is the angle opposite the side of length 1.
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By isolating <math>\cos(\theta)</math>, we get:
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<cmath>\frac{1-x^2-y^2}{-2xy} = \cos(\theta)</cmath>
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For <math>\theta</math> to be obtuse, <math>\cos(\theta)</math> must be negative. Therefore, <math>\frac{1-x^2-y^2}{-2xy}</math> is negative. Since <math>x</math> and <math>y</math> must be negative, <math>1-x^2-y^2</math> must be positive. From here, we can set up the inequality
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<cmath>x^2+y^2<1</cmath>
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Additionally, to satisfy the definition of a triangle, we need:
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<cmath>x+y>1</cmath>
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The solution should be the overlap between the two equations in the 1st quadrant.
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By observing that <math>x^2+y^2<1</math> is the equation for a circle, the amount that is in the 1st quadrant is <math>\frac{\pi}{4}</math>. The line can also be seen as a chord that goes from <math>(0, 1)</math> to <math>(1, 0)</math>. By cutting off the triangle of area <math>\frac{1}{2}</math> that is not part of the overlap, we get <math>\frac{\pi}{4} - \frac{1}{2} \approx \boxed{0.29}</math>.
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-allenle873

Revision as of 15:32, 16 February 2018

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 1

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 negative, $1-x^2-y^2$ must be 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 $\frac{\pi}{4} - \frac{1}{2} \approx \boxed{0.29}$.

-allenle873

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