Difference between revisions of "2015 AMC 10A Problems/Problem 25"
m (→Solution 3) |
(→Problem 25) |
||
(16 intermediate revisions by 12 users not shown) | |||
Line 4: | Line 4: | ||
<math>\textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63</math> | <math>\textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63</math> | ||
− | ==Solution 1== | + | ==Solution 1 (Calculus)== |
Divide the boundary of the square into halves, thereby forming <math>8</math> segments. Without loss of generality, let the first point <math>A</math> be in the bottom-left segment. Then, it is easy to see that any point in the <math>5</math> segments not bordering the bottom-left segment will be distance at least <math>\dfrac{1}{2}</math> apart from <math>A</math>. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least <math>0.5</math> apart from <math>A</math> is <math>\dfrac{0 + 1}{2} = \dfrac{1}{2}</math> because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.) | Divide the boundary of the square into halves, thereby forming <math>8</math> segments. Without loss of generality, let the first point <math>A</math> be in the bottom-left segment. Then, it is easy to see that any point in the <math>5</math> segments not bordering the bottom-left segment will be distance at least <math>\dfrac{1}{2}</math> apart from <math>A</math>. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least <math>0.5</math> apart from <math>A</math> is <math>\dfrac{0 + 1}{2} = \dfrac{1}{2}</math> because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.) | ||
Line 15: | Line 15: | ||
<cmath>P = \frac{1}{8} \left( 5 + \frac{1}{2} + 1 - \frac{\pi}{4} \right) = \frac{1}{32} \left( 26 - \pi \right).</cmath> | <cmath>P = \frac{1}{8} \left( 5 + \frac{1}{2} + 1 - \frac{\pi}{4} \right) = \frac{1}{32} \left( 26 - \pi \right).</cmath> | ||
− | + | Thus our answer is <math>\boxed{\textbf{(A) } 59}</math>. | |
+ | |||
+ | ~minor edit by [[User: Yiyj1|Yiyj1]] | ||
== Solution 2 == | == Solution 2 == | ||
Line 28: | Line 30: | ||
Thus the probability that the probability that the distance between the two points is at least <math>1/2</math> is given by | Thus the probability that the probability that the distance between the two points is at least <math>1/2</math> is given by | ||
<cmath> | <cmath> | ||
− | \frac{1}{4} \cdot \frac{1}{4}+ \frac{1}{2}(1 - \frac{\pi}{16}) + \frac{1}{4} =\frac{26-\pi}{32}. | + | \frac{1}{4} \cdot \frac{1}{4}+ \frac{1}{2}\left(1 - \frac{\pi}{16}\right) + \frac{1}{4} =\frac{26-\pi}{32}. |
</cmath> | </cmath> | ||
Therefore <math>a=26</math>, <math>b=1</math>, and <math>c=32</math>. Thus, <math>a+b+c=59</math> and the answer is <math>\textbf{(A).}</math> | Therefore <math>a=26</math>, <math>b=1</math>, and <math>c=32</math>. Thus, <math>a+b+c=59</math> and the answer is <math>\textbf{(A).}</math> | ||
+ | |||
==Solution 3== | ==Solution 3== | ||
Let our points be called Point A and Point B. Let us first choose Point A to be on some side of the square. We have three cases: | Let our points be called Point A and Point B. Let us first choose Point A to be on some side of the square. We have three cases: | ||
Line 45: | Line 48: | ||
− | Now taking the probabilities of the setups into account, our final probability is <cmath>\frac{1}{4}\cdot \left( \frac{1}{4} \right)+ \frac{1}{2}\cdot \left(\frac{16-\pi}{16}\right) + \frac{1}{4}\cdot \left(1\right)=\frac{26-\pi}{32}.</cmath>Thus <math>(a,b,c)=(26,1,32),</math> so <math>a+b+c= (A) \; 59.</math> | + | Now taking the probabilities of the setups into account, our final probability is <cmath>\frac{1}{4}\cdot \left( \frac{1}{4} \right)+ \frac{1}{2}\cdot \left(\frac{16-\pi}{16}\right) + \frac{1}{4}\cdot \left(1\right)=\frac{26-\pi}{32}.</cmath>Thus <math>(a,b,c)=(26,1,32),</math> so <math>a+b+c= \boxed{\mathbf{(A)}\;59}</math> |
+ | |||
+ | ==Geometric way to solve case 2== | ||
+ | The probability of case 2 (if the two points fall on adjacent sides) can be evaluated geometrically. Let the square have vertices at <math>(0,0)</math>, <math>(1,0)</math>, <math>(0,1)</math> and <math>(1,1)</math>, and WLOG, let point <math>A</math> be on the <math>x</math> axis and let point <math>B</math> be on the <math>y</math> axis. Let the midpoint of <math>AB</math> be <math>M</math>. We can draw the following conclusions: | ||
+ | |||
+ | 1. <math>M</math> must fall inside the square with vertices at <math>(0,0)</math>, <math>(0.5,0)</math> and <math>(0,0.5)</math> and <math>(0.5,0.5)</math>. | ||
+ | |||
+ | 2. <math>M</math> will fall inside the square described above randomly, with a uniform probability of landing anywhere. | ||
+ | |||
+ | 3. If and only if <math>M</math> is more than <math>0.25</math> units from the origin will <math>AB</math> be more than <math>0.5</math> units long. | ||
+ | |||
+ | The proof of each of these statements is left as an exercise to the reader. | ||
+ | |||
+ | Thus, the probability of case 2 can be mapped to the probability that a randomly chosen point inside the square with vertices at <math>(0,0)</math>, <math>(0.5,0)</math> and <math>(0,0.5)</math> and <math>(0.5,0.5)</math> is more than <math>0.25</math> units from the origin. By calculating areas, this is <math>\frac{16 - \pi}{16}</math>. | ||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2015amc10a/399 | ||
+ | |||
+ | ~naren_pr | ||
==See Also== | ==See Also== |
Latest revision as of 13:00, 18 August 2024
Contents
Problem 25
Let be a square of side length
. Two points are chosen independently at random on the sides of
. The probability that the straight-line distance between the points is at least
is
, where
,
, and
are positive integers with
. What is
?
Solution 1 (Calculus)
Divide the boundary of the square into halves, thereby forming segments. Without loss of generality, let the first point
be in the bottom-left segment. Then, it is easy to see that any point in the
segments not bordering the bottom-left segment will be distance at least
apart from
. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least
apart from
is
because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)
If the second point is on the left-bottom segment, then if
is distance
away from the left-bottom vertex, then
must be up to
away from the left-middle point. Thus, using an averaging argument we find that the probability in this case is
(Alternatively, one can equate the problem to finding all valid with
such that
, i.e.
is outside the unit circle with radius
)
Thus, averaging the probabilities gives
Thus our answer is .
~minor edit by Yiyj1
Solution 2
Let one point be chosen on a fixed side. Then the probability that the second point is chosen on the same side is , on an adjacent side is
, and on the opposite side is
. We discuss these three cases.
Case 1: Two points are on the same side. Let the first point be and the second point be
in the
-axis with
. Consider
a point on the unit square
on the Cartesian plane. The region
has the area of
. Therefore, the probability that
is
.
Case 2: Two points are on two adjacent sides. Let the two sides be on the x-axis and
on the y-axis and let one point be
and the other point be
. Then
and the distance between the two points is
. As in Case 1,
is a point on the unit square
. The area of the region
is
and the area of its complementary set inside the square (i.e.
) is
. Therefore, the probability that the distance between
and
is at least
is
.
Case 3: Two points are on two opposite sides. In this case, the probability that the distance between the two points is at least is obviously
.
Thus the probability that the probability that the distance between the two points is at least is given by
Therefore
,
, and
. Thus,
and the answer is
Solution 3
Let our points be called Point A and Point B. Let us first choose Point A to be on some side of the square. We have three cases:
Case 1: Point B is on the same side as Point A: This setup occurs with probability
This is the standard geometric probability problem. Since Point A and Point B can be anywhere on the side, we can't really "count" all of the possibilities. Hence, we translate the problem to a problem involving areas, which can be a finite value while still "containing" infinitely many points. Let side of a square be a number line from
to
and
and
be the values representing the positions of Point A and Point B respectively. Our problem now asks for the probability that
Graphing the inequality on a coordinate plane with
and
as the
and
-axes gives us a "good area" of
out of a "total area" of
Hence, the probability the inequality is satisfied is
Case 2: Point B is on a side adjacent to the side with Point A: This setup occurs with probability
This is a slight deviation of the same geometric probability principle. This time, let the common vertex of the two sides be
and the sides have side length
Again, let
and
be the values representing the positions of Point A and Point B respectively, so the Pythagorean Theorem yields
graphing into a quarter circle of radius
and a total area that is
Hence, our probability is
Case 3: Point B is on a side opposite to the side with Point A: This setup occurs with probability
Clearly, the distance between Point A and Point B are at least 1, so it must be at least
The probability in this case is
Now taking the probabilities of the setups into account, our final probability is Thus
so
Geometric way to solve case 2
The probability of case 2 (if the two points fall on adjacent sides) can be evaluated geometrically. Let the square have vertices at ,
,
and
, and WLOG, let point
be on the
axis and let point
be on the
axis. Let the midpoint of
be
. We can draw the following conclusions:
1. must fall inside the square with vertices at
,
and
and
.
2. will fall inside the square described above randomly, with a uniform probability of landing anywhere.
3. If and only if is more than
units from the origin will
be more than
units long.
The proof of each of these statements is left as an exercise to the reader.
Thus, the probability of case 2 can be mapped to the probability that a randomly chosen point inside the square with vertices at ,
and
and
is more than
units from the origin. By calculating areas, this is
.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2015amc10a/399
~naren_pr
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.