Difference between revisions of "2003 AMC 12B Problems/Problem 21"
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\qquad\mathrm{(E)}\ \frac{1}{2}</math> | \qquad\mathrm{(E)}\ \frac{1}{2}</math> | ||
− | == Solution == | + | == Solution 1 (Trigonometry) == |
By the [[Law of Cosines]], | By the [[Law of Cosines]], | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
AB^2 + BC^2 - 2 AB \cdot BC \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 & | + | \cos \alpha &> \frac 12\\ |
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | It follows that <math>0 < \alpha < \frac {\pi}3</math>, and the probability is <math>\frac{\pi/3}{\pi} = \frac 13 \ | + | 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 | ||
+ | ==Solution 4 (Triangle Inequality)== | ||
+ | |||
+ | |||
+ | Note that we can treat <math>\text{ABC}</math> as a triangle with side lengths <math>5</math>, <math>8</math> and <math>AC=x.</math> Because <math>0</math> and <math>\pi</math> are not pat of the interval of valid <math>\alpha</math> values, <math>\triangle \text{ABC}</math> is a non-degenerate triangle. Then, by the Triangle Inequality, <math>5+8>x,</math> <math>5+x>8,</math> and <math>8+x>5.</math> These reduce to <math>x<13,</math> <math>x>3,</math> and <math>x>-3.</math> Thus, the possible values of x are <math>3<x<13,</math> or <math>x=\text{[}4,5,6,7,8,9,10,11,12\text{]}.</math> Of these <math>9</math> possible <math>x,</math> <math>3</math> of them are less than <math>7,</math> so the probability that <math>x<7</math> is <math>\frac39=\frac13=\boxed{\text{(D)}}.</math> | ||
+ | |||
+ | |||
+ | ~~AndrewZhong2012~~ | ||
== See also == | == See also == |
Latest revision as of 10:01, 8 August 2025
Contents
Problem
An object moves cm in a straight line from
to
, turns at an angle
, measured in radians and chosen at random from the interval
, and moves
cm in a straight line to
. What is the probability that
?
Solution 1 (Trigonometry)
By the Law of Cosines,
It follows that , and the probability is
.
Solution 2 (Analytic Geometry)
, let the object turn clockwise.
Let ,
.
Note that the possible points of create a semi-circle of radius
and center
. The area where
is enclosed by a circle of radius
and center
. The probability that
is
.
The function of is
, the function of
is
.
is the point that satisfies the system of equations:
,
,
,
,
Note that is a
triangle, as
,
,
. As a result
,
.
Therefore the probability that is
Solution 3 (Geometric Probability)
Setting we get that
, 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
for point B). This gives
. Using the distance formula we get
. After algebra, this simplifies to
. After evaluating the constraints of the problem, we land on option (D).
~PeterDoesPhysics
Solution 4 (Triangle Inequality)
Note that we can treat as a triangle with side lengths
,
and
Because
and
are not pat of the interval of valid
values,
is a non-degenerate triangle. Then, by the Triangle Inequality,
and
These reduce to
and
Thus, the possible values of x are
or
Of these
possible
of them are less than
so the probability that
is
~~AndrewZhong2012~~
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.