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

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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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==Video Solution & More by MegaMath==
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https://www.youtube.com/watch?v=d6oFfN5N_70
  
 
== Solution 1==  
 
== Solution 1==  
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<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>\boxed{C 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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latex edits - srisainandan6
 
  
 
==Solution 2 (Trig)==
 
==Solution 2 (Trig)==
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The solution should be the overlap between the two equations in the first quadrant.
 
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{0.29}</math>.
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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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-allenle873
 
  
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==Solution 3 (Bogus, not legitimate solution)==
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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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Similarly to Solution 1, note that The Pythagorean Inequality states that in an obtuse triangle, <math>a^{2} + b^{2} < c^{2}</math>.
 
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We can now complementary count to find the probability by reversing the inequality into:
 
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<cmath>a^{2} + b^{2} \geq c^{2}</cmath>
 
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Since it is given that one side is equal to <math>1</math>, and the closed interval is from <math>[0,1]</math>, we can say without loss of generality that <math>c=1</math>.
 
  
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The probability that <math>x^{2}</math> and <math>y^{2}</math> sum to <math>1</math> is equal to when both <math>x^{2}</math> and <math>y^{2}</math> are <math>0.5</math> (Edit: this is not true, as all the points (x,y) which lie on the unit circle centered at the origin satisfy <math>x^2+y^2=1</math>). We can estimate <math>\sqrt{0.5}</math> to be <math>\approx 0.707</math>.
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== Video Solution by OmegaLearn ==
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Now we know the probability that <math>a^{2} + b^{2} > 1</math> is just when <math>x</math> and/or <math>y</math> equal any value between <math>0.707</math> and <math>1</math>.
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https://youtu.be/LwtoLiBwO-E?t=316
  
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The probability that <math>x</math> or <math>y</math> lie between <math>0.707</math> and <math>1</math> is <math>0.293</math>.
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~ pi_is_3.14
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This gives us <math>\approx \boxed{C \ 0.29}</math>.
 
  
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-Dynosol
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==Video Solution==
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https://youtu.be/tWkE_c3Fa3I -- Geometric Probability and Inequalities!
  
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==Solution through video==
 
 
https://www.youtube.com/watch?v=GHAMU60rI5c
 
https://www.youtube.com/watch?v=GHAMU60rI5c
  

Revision as of 21:14, 2 July 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?

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

Video Solution & More by MegaMath

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

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 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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