Difference between revisions of "2018 AMC 10B Problems/Problem 22"
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The second equation is a line from (0,1) to (1,0). | The second equation is a line from (0,1) to (1,0). | ||
− | So, the area is (pi | + | So, the area is <math>(\frac{\pi}{4} - 1\cdot \frac{1}{2})/1</math> which is around 0.29 |
==Solution 1== | ==Solution 1== |
Revision as of 20:41, 17 February 2018
Real numbers and
are chosen independently and uniformly at random from the interval
. Which of the following numbers is closest to the probability that
and
are the side lengths of an obtuse triangle?
Solution
It's the same as the one below, but without the cosines.
Pythagorean Inequality tells us that in an obtuse triangle, .
Triangle inequality tells us that
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 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:
where and
are the sides that go from
and
is the angle opposite the side of length 1.
By isolating , we get:
For to be obtuse,
must be negative. Therefore,
is negative. Since
and
must be positive,
must be negative, so we must make
positive. From here, we can set up the inequality
Additionally, to satisfy the definition of a triangle, we need:
The solution should be the overlap between the two equations in the 1st quadrant.
By observing that is the equation for a circle, the amount that is in the 1st quadrant is
. The line can also be seen as a chord that goes from
to
. By cutting off the triangle of area
that is not part of the overlap, we get
.
-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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