Difference between revisions of "2020 AMC 12B Problems/Problem 25"
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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== | + | ==Solution 1 == |
Let's start first by manipulating the given inequality. | Let's start first by manipulating the given inequality. | ||
| Line 15: | Line 15: | ||
<cmath>\sin^{2}{(\pi x)}>1-\sin^{2}{(\pi y)}=\cos^{2}{(\pi y)}</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)}=\ | + | Let's consider the boundary cases: <math>\sin{(\pi x)}=\pm \cos{(\pi y)}</math>. |
| − | <cmath>\sin{(\pi x)}=\cos{(\pi y)}=\sin{(\ | + | <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. | 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. | ||
| Line 34: | Line 34: | ||
draw(A1--B2, royalblue+dashed); | draw(A1--B2, royalblue+dashed); | ||
draw(A--B--C--D--A, black+1.2); | draw(A--B--C--D--A, black+1.2); | ||
| + | dot("$(a,0)$",(a,0),down); dot("$(a,1)$",(a,1),up); | ||
</asy> | </asy> | ||
Testing points in each section tells us that the inside of the rhombus satisfies the inequality in the problem statement. | Testing points in each section tells us that the inside of the rhombus satisfies the inequality in the problem statement. | ||
| Line 39: | Line 40: | ||
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. | 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] | |
| − | ==Video Solution | + | ==Video Solution by On The Spot STEM== |
| − | On The Spot STEM | ||
https://www.youtube.com/watch?v=5goLUdObBrY | https://www.youtube.com/watch?v=5goLUdObBrY | ||
Latest revision as of 06:17, 30 December 2023
Contents
Problem
For each real number 
with 
, let numbers 
and 
be chosen independently at random from the intervals ![$[0, a]$](http://latex.artofproblemsolving.com/c/e/7/ce7d05d4da0d50abf77ee166ae0c6474bee5a1cf.png)
and ![$[0, 1]$](http://latex.artofproblemsolving.com/a/b/1/ab178d831a786b92cb4c9ddc2d33578223036f98.png)
, respectively, and let 
be the probability that
![\[\sin^2{(\pi x)} + \sin^2{(\pi y)} > 1\]](http://latex.artofproblemsolving.com/4/d/d/4dda018877a7b217408e61c488287f3bb9d31582.png)
What is the maximum value of 
Solution 1
Let's start first by manipulating the given inequality.
![\[\sin^{2}{(\pi x)}+\sin^{2}{(\pi y)}>1\]](http://latex.artofproblemsolving.com/c/b/8/cb8efe9949fa008c546e52b562015960b29b76f7.png)
![\[\sin^{2}{(\pi x)}>1-\sin^{2}{(\pi y)}=\cos^{2}{(\pi y)}\]](http://latex.artofproblemsolving.com/a/3/e/a3e20e036a736e493cf3d41f17f106a79f6ebc86.png)
Let's consider the boundary cases: 
.
![\[\sin{(\pi x)}=\pm \cos{(\pi y)}=\sin{(\tfrac 12 {\pi}\pm \pi y)}\]](http://latex.artofproblemsolving.com/c/9/3/c93ef44e6b1df372b5c977cd09b9ef7f5d1ae6ac.png)
Solving the first case gives us ![\[y=\tfrac{1}{2}-x \quad \textrm{and} \quad y=x-\tfrac{1}{2}.\]](http://latex.artofproblemsolving.com/5/b/2/5b25d7cd56a6d0c43ef832495aab8ab53a5f7e05.png)
Solving the second case gives us ![\[y=x+\tfrac{1}{2}\quad \textrm{and} \quad y=\tfrac{3}{2}-x.\]](http://latex.artofproblemsolving.com/e/2/8/e2834ef3bae5c13dd5fb8845ce37722f9b42436e.png)
If we graph these equations in ![$[0,1]\times[0,1]$](http://latex.artofproblemsolving.com/a/6/a/a6aa56acfcb95ab5be4a1c4210c59219c425167b.png)
, 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]](http://latex.artofproblemsolving.com/4/3/d/43d922c21d526814bbd9d43648760262218cf640.png)
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 ![\[Q(a) = \frac{1}{a}\left(\frac{1}{4}+(a-\tfrac 12)^2\right) = \frac{1}{a}\left(\frac{1}{2}+a^2-a\right)= a+\frac 1{2a}-1.\]](http://latex.artofproblemsolving.com/0/8/1/081895533609872b95d9f66fc1fc82b496c8a642.png)
By AM-GM, 
, which we can achieve by setting 
.
Therefore, the maximum value of is 
, which is 
Solution 2 (Trigonometry Identities)
![\[\sin^2{(\pi x)} + \sin^2{(\pi y)} > 1, \quad \sin^2{(\pi x)} - (1 - \sin^2{(\pi y)} ) > 0, \quad \sin^2{(\pi x)} - \cos^2{(\pi y)} > 0\]](http://latex.artofproblemsolving.com/5/d/9/5d9cf8dcfaddae79ba1760b160e9959d26ba1d86.png)
![\[( \sin{(\pi x)} + \cos{(\pi y)} )(\sin{(\pi x)} - \cos{(\pi y)})> 0, \quad \left( \sin{(\pi x)} + \sin{( \frac{\pi}{2} - \pi y)} \right) \left( \sin{(\pi x)} - \sin{(\frac{\pi}{2} - \pi y)} \right)> 0\]](http://latex.artofproblemsolving.com/7/2/7/7275135d8f7ab2a4c8dedc9f3888d2515e75d9bd.png)
![\[2\sin{\left( \frac{ \pi x + \frac{\pi}{2} - \pi y }{2} \right)} \cos{\left( \frac{ \pi x - \frac{\pi}{2} + \pi y }{2} \right)} 2 \sin{\left( \frac{ \pi x - \frac{\pi}{2} + \pi y }{2} \right)} \cos{\left( \frac{ \pi x + \frac{\pi}{2} - \pi y }{2} \right)} > 0\]](http://latex.artofproblemsolving.com/3/2/8/3286fae5b4128443231db6fbb5f24930b3ea7a83.png)
![\[2\sin{\left( \frac{ \pi x + \frac{\pi}{2} - \pi y }{2} \right)} \cos{\left( \frac{ \pi x + \frac{\pi}{2} - \pi y }{2} \right)} 2 \sin{\left( \frac{ \pi x - \frac{\pi}{2} + \pi y }{2} \right)} \cos{\left( \frac{ \pi x - \frac{\pi}{2} + \pi y }{2} \right)} > 0\]](http://latex.artofproblemsolving.com/7/3/4/7345d267acc54c55f633516d9d9a729495af81ba.png)
![\[\sin{\left(\pi x + \frac{\pi}{2} - \pi y\right)} \sin{\left( \pi x - \frac{\pi}{2} + \pi y \right)} > 0\]](http://latex.artofproblemsolving.com/d/f/b/dfba9b81fab4e347abff00699e4024b58b973107.png)
If 
and 
![\[0 < \pi x + \frac{\pi}{2} - \pi y < \pi \quad \text{and} \quad 0 < \pi x - \frac{\pi}{2} + \pi y < \pi\]](http://latex.artofproblemsolving.com/e/8/0/e8046a5f5ffcfbf1c13345f2cbfb5c2857a4d64a.png)
![\[0 < x + \frac{1}{2} - y < 1 \quad \text{and} \quad 0 < x - \frac{1}{2} + y < 1\]](http://latex.artofproblemsolving.com/8/3/2/832219f3ed06bb4a5dd13b9744d09ff809303faf.png)
![\[x - \frac{1}{2} < y < x + \frac{1}{2} \quad \text{and} \quad -x + \frac{1}{2} < y < -x + \frac{3}{2}\]](http://latex.artofproblemsolving.com/0/c/5/0c5573c94c0d263e397afb95a4ada87844e84ea1.png)
If 
and 
![\[\pi < \pi x + \frac{\pi}{2} - \pi y < 2\pi \quad \text{and} \quad \pi < \pi x - \frac{\pi}{2} + \pi y < 2\pi\]](http://latex.artofproblemsolving.com/0/2/a/02a3486edea33802e586e5b0ec74c49acf155542.png)
![\[1 < x + \frac{1}{2} - y < 2 \quad \text{and} \quad 1 < x - \frac{1}{2} + y < 2\]](http://latex.artofproblemsolving.com/6/0/6/6069390d10ab9f07821b1a42ab7a545643928a6c.png)
![\[x - \frac{3}{2} < y < x - \frac{1}{2} \quad \text{and} \quad -x + \frac{3}{2} < y < -x + \frac{5}{2}\]](http://latex.artofproblemsolving.com/7/b/1/7b1f158598120149d96093309cc83c457a53a7d3.png)
Notice that and cannot be true, because that means is in the interval ![$[1, 2]$](http://latex.artofproblemsolving.com/f/6/e/f6e0fb93486b514b62f25bbf3c4724b4e53b3d36.png)
. 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
![\[P(b) = \frac{1/2 - b^2}{1-b}\]](http://latex.artofproblemsolving.com/6/a/f/6af3c54128994fb71522cff8f4346148b84c28bc.png)
Differentiating using the quotient rule, we find has local extrema at
![\[P'(b) = \frac{(1-h)(-2h)-(-1/2 - h^2)(-1)}{(1-h)^2} = 0\]](http://latex.artofproblemsolving.com/a/d/d/addfc1f25ea2531f7e8cf9649bb5339cb9e39045.png)
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 
.
![\[P(a) = \frac{ \frac14 + a - \frac12 -(a - \frac12)^2 }{a} = \frac{a - \frac14 -a^2 + a - \frac14 }{a} = \frac{-a^2 + 2a - \frac12 }{a} = -a + 2 - \frac{1}{2a}\]](http://latex.artofproblemsolving.com/b/4/f/b4f910c982b74cd2e0db8d7ef707e5a0b4aceea1.png)
By the power rule, 
. Solving 
, we find that 
when , 
is positive when 
, and is negative when 
has a local maximum when .
![\[P(\frac{ \sqrt{2} }{2}) = - \frac{\sqrt{2} }{2} + 2 - \frac{1}{2 \cdot \frac{\sqrt{2} }{2}} =\boxed{\textbf{(B)}\ 2 - \sqrt{2}}\]](http://latex.artofproblemsolving.com/d/1/2/d129c6e3d18e6c6dc065e0498bad01093419fbb7.png)
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. 
