Difference between revisions of "2020 AMC 12B Problems/Problem 25"
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− | ==Problem | + | ==Problem== |
For each real number <math>a</math> with <math>0 \leq a \leq 1</math>, let numbers <math>x</math> and <math>y</math> be chosen independently at random from the intervals <math>[0, a]</math> and <math>[0, 1]</math>, respectively, and let <math>P(a)</math> be the probability that | For each real number <math>a</math> with <math>0 \leq a \leq 1</math>, let numbers <math>x</math> and <math>y</math> be chosen independently at random from the intervals <math>[0, a]</math> and <math>[0, 1]</math>, respectively, and let <math>P(a)</math> be the probability that | ||
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<math>\textbf{(A)}\ \frac{7}{12} \qquad\textbf{(B)}\ 2 - \sqrt{2} \qquad\textbf{(C)}\ \frac{1+\sqrt{2}}{4} \qquad\textbf{(D)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(E)}\ \frac{5}{8}</math> | <math>\textbf{(A)}\ \frac{7}{12} \qquad\textbf{(B)}\ 2 - \sqrt{2} \qquad\textbf{(C)}\ \frac{1+\sqrt{2}}{4} \qquad\textbf{(D)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(E)}\ \frac{5}{8}</math> | ||
+ | |||
+ | ==Solution 1 == | ||
+ | |||
+ | Let's start first by manipulating the given inequality. | ||
+ | |||
+ | <cmath>\sin^{2}{(\pi x)}+\sin^{2}{(\pi y)}>1</cmath> | ||
+ | <cmath>\sin^{2}{(\pi x)}>1-\sin^{2}{(\pi y)}=\cos^{2}{(\pi y)}</cmath> | ||
+ | |||
+ | Let's consider the boundary cases: <math>\sin{(\pi x)}=\pm \cos{(\pi y)}</math>. | ||
+ | |||
+ | <cmath>\sin{(\pi x)}=\pm \cos{(\pi y)}=\sin{(\tfrac 12 {\pi}\pm \pi y)}</cmath> | ||
+ | |||
+ | Solving the first case gives us <cmath>y=\tfrac{1}{2}-x \quad \textrm{and} \quad y=x-\tfrac{1}{2}.</cmath> Solving the second case gives us <cmath>y=x+\tfrac{1}{2}\quad \textrm{and} \quad y=\tfrac{3}{2}-x.</cmath> If we graph these equations in <math>[0,1]\times[0,1]</math>, we get a rhombus shape. | ||
+ | <asy> | ||
+ | defaultpen(fontsize(9)+0.8); size(200); | ||
+ | pen p=fontsize(10); | ||
+ | pair A,B,C,D,A1,A2,B1,B2,C1,C2,D1,D2,I,L; | ||
+ | A=MP("(0,0)",origin,down+left,p); B=MP("(1,0)",right,down+right,p); C=MP("(1,1)",right+up,up+right,p); D=MP("(0,1)",up,up+left,p); | ||
+ | A1=MP("",extension(A,B,(0.5,0),(0,0.5)),2*down,p); dot(A1); | ||
+ | A2=MP("",extension(A,D,(0.5,0),(0,0.5)),2*left,p); dot(A2); | ||
+ | B1=MP("",extension(B,C,(0.5,0),(0,-0.5)),2*right,p); dot(B1); | ||
+ | B2=MP("",extension(C,D,(0.5,1),(0,0.5)),2*up,p); dot(B2); | ||
+ | real a=0.7; | ||
+ | draw(A1--B1--B2--A2--cycle, gray+0.6); | ||
+ | draw(a*right--a*right+up, royalblue); | ||
+ | draw(A1--B2, royalblue+dashed); | ||
+ | draw(A--B--C--D--A, black+1.2); | ||
+ | dot("$(a,0)$",(a,0),down); dot("$(a,1)$",(a,1),up); | ||
+ | </asy> | ||
+ | Testing points in each section tells us that the inside of the rhombus satisfies the inequality in the problem statement. | ||
+ | |||
+ | From the region graph, notice that in order to maximize <math>P(a)</math>, <math>a\geq\tfrac{1}{2}</math>. We can solve the rest with geometric probability. | ||
+ | |||
+ | Instead of maximizing <math>P(a)</math>, we minimize <math>Q(a)=1-P(a)</math>. <math>Q(a)</math> consists of two squares (each broken into two triangles), one of area <math>\tfrac{1}{4}</math> and another of area <math>(a-\tfrac 12)^2</math>. To calculate <math>Q(a)</math>, we divide this area by <math>a</math>, so <cmath>Q(a) = \frac{1}{a}\left(\frac{1}{4}+(a-\tfrac 12)^2\right) = \frac{1}{a}\left(\frac{1}{2}+a^2-a\right)= a+\frac 1{2a}-1.</cmath> | ||
+ | |||
+ | By AM-GM, <math>a+\frac{1}{2a}\geq 2\sqrt{\frac{a}{2a}}=\sqrt{2}</math>, which we can achieve by setting <math>a=\frac{\sqrt{2}}{2}</math>. | ||
+ | |||
+ | Therefore, the maximum value of <math>P(a)</math> is <math>1-\min(Q(a))</math>, which is <math>1-(\sqrt{2}-1) =\boxed{\textbf{(B)}\ 2 - \sqrt{2}}</math> | ||
+ | |||
+ | ==Solution 2 (Trigonometry Identities)== | ||
+ | |||
+ | <cmath>\sin^2{(\pi x)} + \sin^2{(\pi y)} > 1, \quad \sin^2{(\pi x)} - (1 - \sin^2{(\pi y)} ) > 0, \quad \sin^2{(\pi x)} - \cos^2{(\pi y)} > 0</cmath> | ||
+ | |||
+ | <cmath>( \sin{(\pi x)} + \cos{(\pi y)} )(\sin{(\pi x)} - \cos{(\pi y)})> 0, \quad \left( \sin{(\pi x)} + \sin{( \frac{\pi}{2} - \pi y)} \right) \left( \sin{(\pi x)} - \sin{(\frac{\pi}{2} - \pi y)} \right)> 0</cmath> | ||
+ | |||
+ | <cmath>2\sin{\left( \frac{ \pi x + \frac{\pi}{2} - \pi y }{2} \right)} \cos{\left( \frac{ \pi x - \frac{\pi}{2} + \pi y }{2} \right)} 2 \sin{\left( \frac{ \pi x - \frac{\pi}{2} + \pi y }{2} \right)} \cos{\left( \frac{ \pi x + \frac{\pi}{2} - \pi y }{2} \right)} > 0 </cmath> | ||
+ | |||
+ | <cmath>2\sin{\left( \frac{ \pi x + \frac{\pi}{2} - \pi y }{2} \right)} \cos{\left( \frac{ \pi x + \frac{\pi}{2} - \pi y }{2} \right)} 2 \sin{\left( \frac{ \pi x - \frac{\pi}{2} + \pi y }{2} \right)} \cos{\left( \frac{ \pi x - \frac{\pi}{2} + \pi y }{2} \right)} > 0 </cmath> | ||
+ | |||
+ | <cmath>\sin{\left(\pi x + \frac{\pi}{2} - \pi y\right)} \sin{\left( \pi x - \frac{\pi}{2} + \pi y \right)} > 0 </cmath> | ||
+ | |||
+ | If <math>\sin{\left(\pi x + \frac{\pi}{2} - \pi y\right)} > 0</math> and <math> \sin{\left( \pi x - \frac{\pi}{2} + \pi y \right)} > 0 </math> | ||
+ | |||
+ | <cmath>0 < \pi x + \frac{\pi}{2} - \pi y < \pi \quad \text{and} \quad 0 < \pi x - \frac{\pi}{2} + \pi y < \pi</cmath> | ||
+ | |||
+ | <cmath>0 < x + \frac{1}{2} - y < 1 \quad \text{and} \quad 0 < x - \frac{1}{2} + y < 1</cmath> | ||
+ | |||
+ | <cmath>x - \frac{1}{2} < y < x + \frac{1}{2} \quad \text{and} \quad -x + \frac{1}{2} < y < -x + \frac{3}{2}</cmath> | ||
+ | |||
+ | If <math>\sin{\left(\pi x + \frac{\pi}{2} - \pi y\right)} < 0</math> and <math> \sin{\left( \pi x - \frac{\pi}{2} + \pi y \right)} < 0 </math> | ||
+ | |||
+ | <cmath>\pi < \pi x + \frac{\pi}{2} - \pi y < 2\pi \quad \text{and} \quad \pi < \pi x - \frac{\pi}{2} + \pi y < 2\pi</cmath> | ||
+ | |||
+ | <cmath>1 < x + \frac{1}{2} - y < 2 \quad \text{and} \quad 1 < x - \frac{1}{2} + y < 2</cmath> | ||
+ | |||
+ | <cmath> x - \frac{3}{2} < y < x - \frac{1}{2} \quad \text{and} \quad -x + \frac{3}{2} < y < -x + \frac{5}{2}</cmath> | ||
+ | |||
+ | Notice that <math>\sin{\left(\pi x + \frac{\pi}{2} - \pi y\right)} < 0</math> and <math> \sin{\left( \pi x - \frac{\pi}{2} + \pi y \right)} < 0 </math> cannot be true, because that means <math>x</math> is in the interval <math>[1, 2]</math>. Thus, <math>y</math> is in the the boundaries <math>x - \frac{1}{2} < y < x + \frac{1}{2}</math>, and <math>-x + \frac{1}{2} < y < -x + \frac{3}{2}</math>. | ||
+ | |||
+ | Finishing like Solution 1, 3, or 4 gives <math>P(a) =\boxed{\textbf{(B)}\ 2 - \sqrt{2}}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 3 (Calculus finish)== | ||
+ | |||
+ | We find the same region as in the first solution or the second solution, and again notice we must have <math>a \geq \frac{1}{2}</math>. | ||
+ | |||
+ | We now express <math>P</math> as a function of <math>b=(1-a)</math>. The triangle on the right of the line <math>x=b</math> is an isosceles right triangle with altitude <math>b</math>, so it has area <math>b^2</math>. The total area of the region to the left of <math>x=b</math> has area <math>1-b</math>. So | ||
+ | <cmath>P(b) = \frac{1/2 - b^2}{1-b}</cmath> | ||
+ | Differentiating using the quotient rule, we find <math>P</math> has local extrema at | ||
+ | <cmath>P'(b) = \frac{(1-h)(-2h)-(-1/2 - h^2)(-1)}{(1-h)^2} = 0</cmath> | ||
+ | Setting the numerator equal to <math>0</math> and solving the quadratic, we find <math>P</math> has extrema at <math>b = 1 \pm \frac{\sqrt{2}}{2}</math>. Only <math>b=1-\frac{\sqrt 2}{2}</math> is within the desired region. Plugging in, we get <math>P(1-\frac{\sqrt 2}{2}) = \boxed{\textbf{(B)}\ 2 - \sqrt{2}}</math> as our solution. We also need to check <math>P(b=0) = 1/2</math>. But <math>1/2 < 2 - \sqrt{2}</math>, and if this isn't immediately obvious, <math>1/2</math> isn't an answer choice anyways. | ||
+ | |||
+ | ~jd9 (AMGM scary) | ||
+ | |||
+ | ==Solution 4 (Calculus Finish)== | ||
+ | |||
+ | We find the same region as in the first solution or the second solution, and again notice <math>a \geq \frac{1}{2}</math>. | ||
+ | |||
+ | The numerator of <math>P(a)</math> is the area of the triangle with vertices <math>(0, \frac12), (\frac12, 0), (\frac12, 1)</math> plus the area of the rectangle with vertices <math>(\frac12,0), (a,0), (a,1), (\frac12,1)</math> subtract the area of the isosceles right triangle between <math>x=a, y=0,x - \frac{1}{2}</math> and the isosceles right triangle between <math>x=a, y=1, -x + \frac{3}{2}</math>. The denominator of <math>P(a)</math> is the area of rectangle with vertices <math>(0,0), (a,0), (a,1), (0,1)</math>. | ||
+ | |||
+ | <cmath>P(a) = \frac{ \frac14 + a - \frac12 -(a - \frac12)^2 }{a} = \frac{a - \frac14 -a^2 + a - \frac14 }{a} = \frac{-a^2 + 2a - \frac12 }{a} = -a + 2 - \frac{1}{2a}</cmath> | ||
+ | |||
+ | By the power rule, <math>P'(a) = -1 - \frac12(-1)\frac{1}{a^2} = -1 + \frac{1}{2a^2}</math>. Solving <math>P'(a) = 0</math>, we find that <math>a = \frac{ \sqrt{2} }{2}</math> | ||
+ | |||
+ | <math>\because P'(a) = 0</math> when <math>a = \frac{ \sqrt{2} }{2}</math>, <math>P'(a)</math> is positive when <math>a < \frac{ \sqrt{2} }{2}</math>, and <math>P'(a)</math> is negative when <math>a > \frac{ \sqrt{2} }{2}</math> | ||
+ | |||
+ | <math>\therefore P(a)</math> has a local maximum when <math>a = \frac{ \sqrt{2} }{2}</math>. | ||
+ | |||
+ | <cmath>P(\frac{ \sqrt{2} }{2}) = - \frac{\sqrt{2} }{2} + 2 - \frac{1}{2 \cdot \frac{\sqrt{2} }{2}} =\boxed{\textbf{(B)}\ 2 - \sqrt{2}}</cmath> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 5 (Calculus)== | ||
+ | |||
+ | Comparing the diagram of <math>\sin^{2}{(\pi x)}</math> and <math>\cos^{2}{(\pi y)}</math>, we can see if we let <math>x=k</math>, then <math>y\in (0+\lvert \dfrac{1}{2}-k\rvert,1-\lvert\dfrac{1}{2}-k\rvert)</math> will satisfy the above inequality. | ||
+ | |||
+ | Thus, we can write that | ||
+ | <cmath>P(a) = \frac{1}{a} (\int_0^{a}1-2\lvert\frac{1}{2}-x\rvert{\rm d}x)</cmath> | ||
+ | <cmath>P(a) = \frac{1}{a} (\int_0^{\frac{1}{2}}2x{\rm d}x + \int_{\frac{1}{2}}^{a}2-2x{\rm d}x) = -a+2-\frac{1}{2a}</cmath> | ||
+ | |||
+ | Then it is easy to find that <math>P(a)</math> is maximized when <math>a=\dfrac{\sqrt{2}}{2}</math>, which give us <math>P(\dfrac{\sqrt{2}}{2})=\boxed{\textbf{(B)}\ 2 - \sqrt{2}}</math> | ||
+ | |||
+ | ~ERiccc | ||
+ | |||
+ | ==Video Solution by On The Spot STEM== | ||
+ | https://www.youtube.com/watch?v=5goLUdObBrY | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC12 box|year=2020|ab=B|num-b=24|after=Last Problem}} | ||
+ | {{MAA Notice}} |
Latest revision as of 04:29, 24 July 2024
Contents
Problem
For each real number with
, let numbers
and
be chosen independently at random from the intervals
and
, respectively, and let
be the probability that
What is the maximum value of
Solution 1
Let's start first by manipulating the given inequality.
Let's consider the boundary cases: .
Solving the first case gives us Solving the second case gives us
If we graph these equations in
, we get a rhombus shape.
Testing points in each section tells us that the inside of the rhombus satisfies the inequality in the problem statement.
From the region graph, notice that in order to maximize ,
. We can solve the rest with geometric probability.
Instead of maximizing , we minimize
.
consists of two squares (each broken into two triangles), one of area
and another of area
. To calculate
, we divide this area by
, so
By AM-GM, , which we can achieve by setting
.
Therefore, the maximum value of is
, which is
Solution 2 (Trigonometry Identities)
If and
If and
Notice that and
cannot be true, because that means
is in the interval
. Thus,
is in the the boundaries
, and
.
Finishing like Solution 1, 3, or 4 gives
Solution 3 (Calculus finish)
We find the same region as in the first solution or the second solution, and again notice we must have .
We now express as a function of
. The triangle on the right of the line
is an isosceles right triangle with altitude
, so it has area
. The total area of the region to the left of
has area
. So
Differentiating using the quotient rule, we find
has local extrema at
Setting the numerator equal to
and solving the quadratic, we find
has extrema at
. Only
is within the desired region. Plugging in, we get
as our solution. We also need to check
. But
, and if this isn't immediately obvious,
isn't an answer choice anyways.
~jd9 (AMGM scary)
Solution 4 (Calculus Finish)
We find the same region as in the first solution or the second solution, and again notice .
The numerator of is the area of the triangle with vertices
plus the area of the rectangle with vertices
subtract the area of the isosceles right triangle between
and the isosceles right triangle between
. The denominator of
is the area of rectangle with vertices
.
By the power rule, . Solving
, we find that
when
,
is positive when
, and
is negative when
has a local maximum when
.
Solution 5 (Calculus)
Comparing the diagram of and
, we can see if we let
, then
will satisfy the above inequality.
Thus, we can write that
Then it is easy to find that is maximized when
, which give us
~ERiccc
Video Solution by On The Spot STEM
https://www.youtube.com/watch?v=5goLUdObBrY
See Also
2020 AMC 12B (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.