Problem:
For each real number a with 0≤a≤1, let numbers x and y be chosen independently at random from the intervals [0,a] and [0,1], respectively, and let P(a) be the probability that
sin2(πx)+sin2(πy)>1
What is the maximum value of P(a) ?
Answer Choices:
A. 127
B. 2−2
C. 41+2
D. 25−1
E. 85 Solution:
Let's start first by manipulating the given inequality.
sin2(πx)+sin2(πy)>1 sin2(πx)>1−sin2(πy)=cos2(πy)
Let's consider the boundary cases: sin(πx)=±cos(πy).
sin(πx)=±cos(πy)=sin(21π±πy)
Solving the first case gives us
y=21−x and y=x−21
Solving the second case gives us
y=x+21 and y=23−x
If we graph these equations in [0,1]×[0,1], 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 P(a),a≥21. We can solve the rest with geometric probability.
Instead of maximizing P(a), we minimize Q(a)=1−P(a).Q(a) consists of two squares (each broken into two triangles), one of area 41 and another of area (a−21)2. To calculate Q(a), we divide this area by a, so
Q(a)=a1(41+(a−21)2)=a1(21+a2−a)=a+2a1−1
By AM-GM, a+2a1≥22aa=2, which we can achieve by setting a=22.
Therefore, the maximum value of P(a) is 1−min(Q(a)), which is 1−(2−1)=(B)2−2
OR
We find the same region as in the first solution, and again notice we must have a≥21.
We now express P as a function of b=(1−a). The triangle on the right of the line x=b is an isosceles right triangle with altitude b, so it has area b2. The total area of the region to the left of x=b has area 1−b. So
P(b)=1−b1/2−b2
Differentiating using the quotient rule, we find P has local extrema at
P′(b)=(1−h)2(1−h)(−2h)−(−1/2−h2)(−1)=0
Setting the numerator equal to 0 and solving the quadratic, we find P has extrema at b=1±22. Only b=1−22 is within the desired region. Plugging in, we get P(1−22)= (B) 2−2 as our solution. We also need to check P(b=0)=1/2. But 1/2<2−2, and if this isn't immediately obvious, 1/2 isn't an answer choice anyways.
