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

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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

Solution 3 (Bogus, not legitimate solution)

Similarly to Solution 1, note that The Pythagorean Inequality states that in an obtuse triangle, $a^{2} + b^{2} < c^{2}$ . We can now complementary count to find the probability by reversing the inequality into: $a^{2} + b^{2} \geq c^{2}$ Since it is given that one side is equal to $1$ , and the closed interval is from $[0,1]$ , we can say without loss of generality that $c=1$ .

The probability that $x^{2}$ and $y^{2}$ sum to $1$ is equal to when both $x^{2}$ and $y^{2}$ are $0.5$ (Edit: this is not true, as all the points (x,y) which lie on the unit circle centered at the origin satisfy $x^2+y^2=1$ ). We can estimate $\sqrt{0.5}$ to be $\approx 0.707$ . Now we know the probability that $a^{2} + b^{2} > 1$ is just when $x$ and/or $y$ equal any value between $0.707$ and $1$ .

The probability that $x$ or $y$ lie between $0.707$ and $1$ is $0.293$ . This gives us $\approx \boxed{C \ 0.29}$ .

-Dynosol

Video Solution

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

@@ Line 50: / Line 50: @@
 -Dynosol
-==Solution through video==
+==Video Solution==
 https://www.youtube.com/watch?v=GHAMU60rI5c

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