Difference between revisions of "2018 AMC 10B Problems/Problem 22"
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<cmath>x + y > 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>. | 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>0.29</math>. | + | So, the area is <math>\frac{\pi}{4} - \frac12</math> which is approximately <math>\boxed{0.29}</math>. |
+ | |||
+ | latex edits - srisainandan6 | ||
==Solution 2 (Trig)== | ==Solution 2 (Trig)== |
Revision as of 13:22, 14 April 2020
Contents
Problem
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 1
The Pythagorean Inequality tells us that in an obtuse triangle, . The triangle inequality tells us that
. So, we have two inequalities:
The first equation is
of a circle with radius
, and the second equation is a line from
to
.
So, the area is
which is approximately
.
latex edits - srisainandan6
Solution 2 (Trig)
Note that the obtuse angle in the triangle has to be opposite the side that is always length . This is because the largest angle is always opposite the largest side, and if two sides of the triangle were
, the last side would have to be greater than
to make an obtuse triangle. Using this observation, we can set up a law of cosines where the angle is opposite
:
where and
are the sides that go from
and
is the angle opposite the side of length
.
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 first quadrant.
By observing that is the equation for a circle, the amount that is in the first 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
Solution 3 (Bogus, not legitimate solution)
Similarly to Solution 1, note that The Pythagorean Inequality states that in an obtuse triangle, .
We can now complementary count to find the probability by reversing the inequality into:
Since it is given that one side is equal to
, and the closed interval is from
, we can say without loss of generality that
.
The probability that and
sum to
is equal to when both
and
are
(Edit: this is not true, as all the points (x,y) which lie on the unit circle centered at the origin satisfy
). We can estimate
to be
.
Now we know the probability that
is just when
and/or
equal any value between
and
.
The probability that or
lie between
and
is
.
This gives us
.
-Dynosol
Solution through video
https://www.youtube.com/watch?v=GHAMU60rI5c
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.