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

Latest revision as of 15:56, 25 September 2024

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 $\boxed{\textbf{(C)} ~0.29}$

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 $\frac{\pi}{4} - \frac{1}{2} \approx \boxed{\textbf{(C)} ~0.29}$ .

..why would you do this? for what purpose? its much more complicated and this is the AMC 10! -Orion 2010

Video Solution & More by MegaMath

https://www.youtube.com/watch?v=d6oFfN5N_70

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

@@ Line 10: / Line 10: @@
 <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>.
-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 (Trig)==
@@ Line 30: / Line 30: @@
 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 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{0.29}</math>.
+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>.
--allenle873
+..why would you do this? for what purpose? its much more complicated and this is the AMC 10! -Orion 2010
-==Solution 3 (Bogus, not legitimate solution)==
+==Video Solution & More by MegaMath==
-Similarly to Solution 1, note that The Pythagorean Inequality states that in an obtuse triangle, <math>a^{2} + b^{2} < c^{2}</math>.
+https://www.youtube.com/watch?v=d6oFfN5N_70
-We can now complementary count to find the probability by reversing the inequality into:
-<cmath>a^{2} + b^{2} \geq c^{2}</cmath>
-Since it is given that one side is equal to <math>1</math>, and the closed interval is from <math>[0,1]</math>, we can say without loss of generality that <math>c=1</math>.
-The probability that <math>x^{2}</math> and <math>y^{2}</math> sum to <math>1</math> is equal to when both <math>x^{2}</math> and <math>y^{2}</math> are <math>0.5</math>(Edit: this is not true). We can estimate <math>\sqrt{0.5}</math> to be <math>\approx 0.707</math>.
+== Video Solution by OmegaLearn ==
-Now we know the probability that <math>a^{2} + b^{2} > 1</math> is just when <math>x</math> and/or <math>y</math> equal any value between <math>0.707</math> and <math>1</math>.
+https://youtu.be/LwtoLiBwO-E?t=316
-The probability that <math>x</math> or <math>y</math> lie between <math>0.707</math> and <math>1</math> is <math>0.293</math>.
+~ pi_is_3.14
-This gives us <math>\approx \boxed{C \ 0.29}</math>.
--Dynosol
+==Video Solution==
+https://youtu.be/tWkE_c3Fa3I -- Geometric Probability and Inequalities!
-==Solution through video==
 https://www.youtube.com/watch?v=GHAMU60rI5c

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
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