Problem:
For each real number a with 0β€aβ€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(Οx)+sin2(Οy)>1
What is the maximum value of P(a) ?
Answer Choices:
A. 127β
B. 2β2β
C. 41+2ββ
D. 25ββ1β
E. 85β Solution:
Let's start first by manipulating the given inequality.
sin2(Οx)+sin2(Οy)>1 sin2(Οx)>1βsin2(Οy)=cos2(Οy)
Let's consider the boundary cases: sin(Οx)=Β±cos(Οy).
sin(Οx)=Β±cos(Οy)=sin(21βΟΒ±Οy)
Solving the first case gives us
y=21ββx and y=xβ21β
Solving the second case gives us
y=x+21β and y=23ββx
If we graph these equations in [0,1]Γ[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),aβ₯21β. We can solve the rest with geometric probability.
Instead of maximizing P(a), we minimize Q(a)=1βP(a).Q(a) consists of two squares (each broken into two triangles), one of area 41β and another of area (aβ21β)2. To calculate Q(a), we divide this area by a, so
By AM-GM, a+2a1ββ₯22aaββ=2β, which we can achieve by setting a=22ββ.
Therefore, the maximum value of P(a) is 1βmin(Q(a)), which is 1β(2ββ1)=(B)2β2ββ
OR
We find the same region as in the first solution, and again notice we must have aβ₯21β.
We now express P as a function of b=(1βa). 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 1βb. So
P(b)=1βb1/2βb2β
Differentiating using the quotient rule, we find P has local extrema at
Setting the numerator equal to 0 and solving the quadratic, we find P has extrema at b=1Β±22ββ. Only b=1β22ββ is within the desired region. Plugging in, we get P(1β22ββ)=(B)2β2ββ as our solution. We also need to check P(b=0)=1/2. But 1/2<2β2β, and if this isn't immediately obvious, 1/2 isn't an answer choice anyways.
