Difference between revisions of "2020 AMC 12B Problems/Problem 25"
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Let's consider the boundary cases: <math>\sin{(\pi x)}=\pm \cos{(\pi y)}</math>. | Let's consider the boundary cases: <math>\sin{(\pi x)}=\pm \cos{(\pi y)}</math>. | ||
− | <cmath>\sin{(\pi x)}=\cos{(\pi y)}=\sin{(\tfrac 12 {\pi}\pm \pi y)}</cmath> | + | <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. |
Revision as of 16:15, 13 July 2022
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
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
Video Solution
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 |
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