2020 AMC 12B Problems/Problem 25
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 .
Video Solution by On The Spot STEM
https://www.youtube.com/watch?v=5goLUdObBrY
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
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