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

Revision as of 10:04, 4 April 2021

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 $\boxed{0.29}$ , which is $\boxed{C}$

latex edits - srisainandan6

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 $\frac{\pi}{4} - \frac{1}{2} \approx \boxed{0.29}$ .

-allenle873

Video Solution

https://youtu.be/tWkE_c3Fa3I -- Geometric Probability and Inequalities! https://www.youtube.com/watch?v=GHAMU60rI5c

@@ Line 1: / Line 1: @@
+==Problem==
 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>
+== Solution 1==
+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 + 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>.
+So, the area is <math>\frac{\pi}{4} - \frac12</math> which is approximately <math>\boxed{0.29}</math>, which is <math>\boxed{C}</math>
+latex edits - srisainandan6
+==Solution 2 (Trig)==
+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>
+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:
+<cmath>\frac{1-x^2-y^2}{-2xy} = \cos(\theta)</cmath>
+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 positive, <math>-2xy</math> must be negative, so we must make <math>1-x^2-y^2</math> positive. From here, we can set up the inequality
+<cmath>x^2+y^2<1</cmath>
+Additionally, to satisfy the definition of a triangle, we need:
+<cmath>x+y>1</cmath>
+The solution should be the overlap between the two equations in the first quadrant.
+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>.
+-allenle873
+==Video Solution==
+https://youtu.be/tWkE_c3Fa3I -- Geometric Probability and Inequalities!
+https://www.youtube.com/watch?v=GHAMU60rI5c
+==See Also==
+{{AMC10 box|year=2018|ab=B|num-b=21|num-a=23}}
+{{MAA Notice}}

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
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