Difference between revisions of "2018 AMC 10B Problems/Problem 22"
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<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> | <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 == | + | == Solution 1== |
The Pythagorean Inequality tells us that in an obtuse triangle, <math>a^{2} + b^{2} < c^{2}</math>. The triangle inequality tells us that <math>a + b > c</math>. So, we have two inequalities: | The Pythagorean Inequality tells us that in an obtuse triangle, <math>a^{2} + b^{2} < c^{2}</math>. The triangle inequality tells us that <math>a + b > c</math>. So, we have two inequalities: | ||
<cmath>x^2 + y^2 < 1</cmath> | <cmath>x^2 + y^2 < 1</cmath> | ||
<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{\textbf{(C)} ~0.29}</math> |
− | ==Solution 2== | + | ==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 | + | Note that the obtuse angle in the triangle has to be opposite the side that is always length <math>1</math>. This is because the largest angle is always opposite the largest side, and if two sides of the triangle were <math>1</math>, the last side would have to be greater than <math>1</math> to make an obtuse triangle. Using this observation, we can set up a law of cosines where the angle is opposite <math>1</math>: |
<cmath>1^2=x^2+y^2-2xy\cos(\theta)</cmath> | <cmath>1^2=x^2+y^2-2xy\cos(\theta)</cmath> | ||
− | where <math>x</math> and <math>y</math> are the sides that go from <math>[0,1]</math> and <math>\theta</math> is the angle opposite the side of length 1. | + | where <math>x</math> and <math>y</math> are the sides that go from <math>[0,1]</math> and <math>\theta</math> is the angle opposite the side of length <math>1</math>. |
By isolating <math>\cos(\theta)</math>, we get: | By isolating <math>\cos(\theta)</math>, we get: | ||
Line 28: | Line 30: | ||
Additionally, to satisfy the definition of a triangle, we need: | Additionally, to satisfy the definition of a triangle, we need: | ||
<cmath>x+y>1</cmath> | <cmath>x+y>1</cmath> | ||
− | The solution should be the overlap between the two equations in the | + | The solution should be the overlap between the two equations in the first quadrant. |
− | By observing that <math>x^2+y^2<1</math> is the equation for a circle, the amount that is in the | + | 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 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 | https://www.youtube.com/watch?v=GHAMU60rI5c | ||
Latest revision as of 22:14, 2 July 2024
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?
Video Solution & More by MegaMath
https://www.youtube.com/watch?v=d6oFfN5N_70
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
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
.
..why would you do this? for what purpose? its much more complicated and this is the AMC 10! -Orion 2010
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
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.