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

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

@@ Line 4: / Line 4: @@
 <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>
-==Video Solution & More by MegaMath==
-https://www.youtube.com/watch?v=d6oFfN5N_70
 == Solution 1==
@@ Line 33: / Line 31: @@
 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>.
+..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 ==

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