Consider a right triangle which has a leg that is uniformly distributed on [ 0 , 5 ] and hypotenuse has a fixed length of 5.
What is the expected area of the triangle? If the answer is A , then input ⌊ 1 0 0 0 A ⌋ .
Notation:
⌊
⋅
⌋
denotes the
floor function
.
Bonus: Generalize this for any hypotenuse k .
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If the hypotenuse of such triangle is 5 , then its legs are x and 2 5 − x 2 for some length x such that 0 < x < 5 .
The area of this triangle is 2 1 x 2 5 − x 2 .
Now, to find its expected area over the inteval ( 0 , 5 ) , we do
A = 5 − 0 ∫ 0 5 2 1 x 2 5 − x 2 d x
A = − 1 0 1 ⋅ 3 1 ( 2 5 − x 2 ) 2 3 ∣ ∣ ∣ ∣ ∣ 0 5
A = 6 2 5 ≈ 4 . 1 6 6 6 6 6 6 7
Hence, ⌊ 1 0 0 0 A ⌋ = 4 1 6 6 .
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Let the length of one of the legs be x , then the probability of uniformly distributed random variable X is given by P ( a ≤ X ≤ b ) = ∫ a b k d x .
Since the area of the right triangle with hypotenuse of k is A △ = 2 x k 2 − x 2 , then the expected value of A △ is as follows:
E ( A △ ) = ∫ 0 k 2 k x k 2 − x 2 d x = 2 k ∫ 0 k k x 1 − ( k x ) 2 d x = 2 k 2 ∫ 0 2 π sin θ cos 2 θ d θ = − 2 k 2 ∫ 1 0 cos 2 θ d cos θ = − 6 k 2 cos 3 θ ∣ ∣ ∣ ∣ 1 0 = 6 k 2 Let k x = sin θ , d x = k cos θ d θ
For k = 5 , E ( A △ ) = A = 6 2 5 ≈ 4 . 1 6 6 6 7 , ⟹ ⌊ 1 0 0 0 A ⌋ = 4 1 6 6 .