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Difference between revisions of "2018 AMC 10B Problems/Problem 22"
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== Solution ==  
 
== Solution ==  
  
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(Can someone please add a diagram and format)
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It's the same as the one below, but without the cosines.
  
 
Pythagorean Inequality tells us that in an obtuse triangle, <math>x^{2} + y^{2} < 1^{2}</math>.
 
Pythagorean Inequality tells us that in an obtuse triangle, <math>x^{2} + y^{2} < 1^{2}</math>.

Revision as of 16:39, 16 February 2018

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

It's the same as the one below, but without the cosines.

Pythagorean Inequality tells us that in an obtuse triangle, $x^{2} + y^{2} < 1^{2}$.

Triangle inequality tells us that $x + y > 1$

Geometric probability

The first equation is 1/4 of a circle with radius 1.

The second equation is a line from (0,1) to (1,0).

So, the area is (pi/4 - 1*1/2)/1 which is around 0.29

Solution 1

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:

\[1^2=x^2+y^2-2xy\cos(\theta)\]

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 1st quadrant.

By observing that $x^2+y^2<1$ is the equation for a circle, the amount that is in the 1st 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

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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