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

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==Problem==
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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 to the probability that <math>x,y,</math> and <math>1</math> are the side lengths of an obtuse triangle?
 
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 to the probability that <math>x,y,</math> and <math>1</math> are the side lengths of an obtuse triangle?
  
 
<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 ==  
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== Solution 1==  
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The Pythagorean Inequality tells us that in an obtuse triangle, <math>a^{2} + b^{2} < c^{2}</math>. The triangle inequality tells us that <math>a + b > c</math>. Thus we have two inequalities:
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The Pythagorean Inequality tells us that in an obtuse triangle, <math>a^{2} + b^{2} < c^{2}</math>. The triangle inequality tells us that <math>a + b > c</math>. So, we have two inequalities:
 
<cmath>x^2 + y^2 < 1</cmath>
 
<cmath>x^2 + y^2 < 1</cmath>
 
<cmath>x + y > 1</cmath>
 
<cmath>x + y > 1</cmath>
 
The first equation is <math>\frac14</math> of a circle with radius <math>1</math>, and the second equation is a line from <math>(0, 1)</math> to <math>(1, 0)</math>.
 
The first equation is <math>\frac14</math> of a circle with radius <math>1</math>, and the second equation is a line from <math>(0, 1)</math> to <math>(1, 0)</math>.
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So, the area is <math>\frac{\pi}{4} - \frac12</math> which is approximately <math>0.29</math>.
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So, the area is <math>\frac{\pi}{4} - \frac12</math> which is approximately <math>\boxed{\textbf{(C)} ~0.29}</math>
  
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==Solution 2==
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==Solution 2 (Trig)==
  
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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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Note that the obtuse angle in the triangle has to be opposite the side that is always length <math>1</math>. This is because the largest angle is always opposite the largest side, and if two sides of the triangle were <math>1</math>, the last side would have to be greater than <math>1</math> to make an obtuse triangle. Using this observation, we can set up a law of cosines where the angle is opposite <math>1</math>:
  
 
<cmath>1^2=x^2+y^2-2xy\cos(\theta)</cmath>
 
<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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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 <math>1</math>.
  
 
By isolating <math>\cos(\theta)</math>, we get:
 
By isolating <math>\cos(\theta)</math>, we get:
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Additionally, to satisfy the definition of a triangle, we need:
 
Additionally, to satisfy the definition of a triangle, we need:
 
<cmath>x+y>1</cmath>
 
<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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The solution should be the overlap between the two equations in the first 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 first 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{\textbf{(C)} ~0.29}</math>.
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..why would you do this? for what purpose? its much more complicated and this is the AMC 10! -Orion 2010
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==Video Solution & More by MegaMath==
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https://www.youtube.com/watch?v=d6oFfN5N_70
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== Video Solution by OmegaLearn ==
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https://youtu.be/LwtoLiBwO-E?t=316
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 +
~ pi_is_3.14
  
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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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==Video Solution==
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https://youtu.be/tWkE_c3Fa3I -- Geometric Probability and Inequalities!
  
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-allenle873
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https://www.youtube.com/watch?v=GHAMU60rI5c
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2018|ab=B|num-b=21|num-a=23}}
 
{{AMC10 box|year=2018|ab=B|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:56, 25 September 2024

Problem

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

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

Solution 2 (Trig)

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 two 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 first quadrant.

By observing that $x^2+y^2<1$ is the equation for a circle, the amount that is in the first 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 $\frac{\pi}{4} - \frac{1}{2} \approx \boxed{\textbf{(C)} ~0.29}$.

..why would you do this? for what purpose? its much more complicated and this is the AMC 10! -Orion 2010

Video Solution & More by MegaMath

https://www.youtube.com/watch?v=d6oFfN5N_70

Video Solution by OmegaLearn

https://youtu.be/LwtoLiBwO-E?t=316

~ pi_is_3.14

Video Solution

https://youtu.be/tWkE_c3Fa3I -- Geometric Probability and Inequalities!

https://www.youtube.com/watch?v=GHAMU60rI5c

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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All AMC 10 Problems and Solutions

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