Difference between revisions of "2003 AMC 12B Problems/Problem 21"

Latest revision as of 09:23, 20 September 2024

Problem

An object moves $8$ cm in a straight line from $A$ to $B$ , turns at an angle $\alpha$ , measured in radians and chosen at random from the interval $(0,\pi)$ , and moves $5$ cm in a straight line to $C$ . What is the probability that $AC < 7$ ?

$\mathrm{(A)}\ \frac{1}{6} \qquad\mathrm{(B)}\ \frac{1}{5} \qquad\mathrm{(C)}\ \frac{1}{4} \qquad\mathrm{(D)}\ \frac{1}{3} \qquad\mathrm{(E)}\ \frac{1}{2}$

Solution 1 (Trigonometry)

By the Law of Cosines, $\begin{align*} AB^2 + BC^2 - 2 AB \cdot BC \cos \alpha = 89 - 80 \cos \alpha = AC^2 &< 49\\ \cos \alpha &> \frac 12\\ \end{align*}$

It follows that $0 < \alpha < \frac {\pi}3$ , and the probability is $\frac{\pi/3}{\pi} = \boxed{\textbf{(D) } \frac13 }$ .

Solution 2 (Analytic Geometry)

$WLOG$ , let the object turn clockwise.

Let $B = (0, 0)$ , $A = (0, -8)$ .

Note that the possible points of $C$ create a semi-circle of radius $5$ and center $B$ . The area where $AC < 7$ is enclosed by a circle of radius $7$ and center $A$ . The probability that $AC < 7$ is $\frac{\angle ABO}{180 ^\circ}$ .

The function of $\odot B$ is $x^2 + y^2 = 25$ , the function of $\odot A$ is $x^2 + (y+8)^2 = 49$ .

$O$ is the point that satisfies the system of equations: $\begin{cases} x^2 + y^2 = 25 \\ x^2 + (y+8)^2 = 49 \end{cases}$

$x^2 + (y+8)^2 - x^2 - y^2 = 49 - 25$ , $64 + 16y =24$ , $y = - \frac52$ , $x = \frac{5 \sqrt{3}}{2}$ , $O = (\frac{5 \sqrt{3}}{2}, - \frac52)$

Note that $\triangle BDO$ is a $30-60-90$ triangle, as $BO = 5$ , $BD = \frac{5 \sqrt{3}}{2}$ , $DO = \frac52$ . As a result $\angle CBO = 30 ^\circ$ , $\angle ABO = 60 ^\circ$ .

Therefore the probability that $AC < 7$ is $\frac{\angle ABO}{180 ^\circ} = \frac{60 ^\circ}{180 ^\circ} = \boxed{\textbf{(D) } \frac13 }$

~isabelchen

Solution 3 (Geometric Probability)

Setting $A = (0,0)$ we get that $B = (8,0)$ , after assuming segment AB to be straight in the x-direction relative to our coordinate system (in other words, due to symmetrically we can set $x = 8$ for point B). This gives $C = (8 + 5cos(\alpha), 5sin(\alpha))$ . Using the distance formula we get $sqrt((8 + 5cos(\alpha))^2 + (5sin(\alpha))^2) < 7$ . After algebra, this simplifies to $cos(\alpha) < -\frac{1}{2}$ . After evaluating the constraints of the problem, we land on option (D).

~PeterDoesPhysics

@@ Line 8: / Line 8: @@
 \qquad\mathrm{(E)}\ \frac{1}{2}</math>
-== Solution ==
+== Solution 1 (Trigonometry) ==
 By the [[Law of Cosines]],
 <cmath>\begin{align*}
-AB^2 + AC^2 - 2 AB \cdot AC \cos \alpha = 89 - 80 \cos \alpha = AC^2 &< 49\\
+AB^2 + BC^2 - 2 AB \cdot BC \cos \alpha = 89 - 80 \cos \alpha = AC^2 &< 49\\
-\cos \alpha &< \frac 12\\
+\cos \alpha &> \frac 12\\
 \end{align*}</cmath>
-It follows that <math>0 < \alpha < \frac {\pi}3</math>, and the probability is <math>\frac{\pi/3}{\pi} = \frac 13 \Rightarrow \mathrm{(D)}</math>.
+It follows that <math>0 < \alpha < \frac {\pi}3</math>, and the probability is <math>\frac{\pi/3}{\pi} = \boxed{\textbf{(D) } \frac13 }</math>.
+==Solution 2 (Analytic Geometry)==
+[[File:2003AMC12BP21.png|center|500px]]
+<math>WLOG</math>, let the object turn clockwise.
+Let <math>B = (0, 0)</math>, <math>A = (0, -8)</math>.
+Note that the possible points of <math>C</math> create a semi-circle of radius <math>5</math> and center <math>B</math>. The area where <math>AC < 7</math> is enclosed by a circle of radius <math>7</math> and center <math>A</math>. The probability that <math>AC < 7</math> is <math>\frac{\angle ABO}{180 ^\circ}</math>.
+The function of <math>\odot B</math> is <math>x^2 + y^2 = 25</math>, the function of <math>\odot A</math> is <math>x^2 + (y+8)^2 = 49</math>.
+<math>O</math> is the point that satisfies the system of equations: <math>\begin{cases} x^2 + y^2 = 25 \\ x^2 + (y+8)^2 = 49 \end{cases}</math>
+<math>x^2 + (y+8)^2 - x^2 - y^2 = 49 - 25</math>, <math>64 + 16y =24</math>, <math>y = - \frac52</math>, <math>x = \frac{5 \sqrt{3}}{2}</math>, <math>O = (\frac{5 \sqrt{3}}{2}, - \frac52)</math>
+Note that <math>\triangle BDO</math> is a <math>30-60-90</math> triangle, as <math>BO = 5</math>, <math>BD = \frac{5 \sqrt{3}}{2}</math>, <math>DO = \frac52</math>. As a result <math>\angle CBO = 30 ^\circ</math>, <math>\angle ABO = 60 ^\circ</math>.
+Therefore the probability that <math>AC < 7</math> is <math>\frac{\angle ABO}{180 ^\circ} = \frac{60 ^\circ}{180 ^\circ} = \boxed{\textbf{(D) } \frac13 }</math>
+~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
+==Solution 3 (Geometric Probability)==
+Setting <math>A = (0,0)</math> we get that <math>B = (8,0)</math>, after assuming segment AB to be straight in the x-direction relative to our coordinate system (in other words, due to symmetrically we can set <math>x = 8</math> for point B). This gives <math>C = (8 + 5cos(\alpha), 5sin(\alpha))</math>. Using the distance formula we get <math>sqrt((8 + 5cos(\alpha))^2 + (5sin(\alpha))^2) < 7</math>. After algebra, this simplifies to <math>cos(\alpha) < -\frac{1}{2}</math>. After evaluating the constraints of the problem, we land on option (D).
+~PeterDoesPhysics
 == See also ==

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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 12 Problems and Solutions

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