Just No Ordinary Triangle - 2

Geometry Level 4

Consider a right triangle which has a leg that is uniformly distributed on [ 0 , 5 ] [ 0, 5 ] and hypotenuse has a fixed length of 5.

What is the expected area of the triangle? If the answer is A A , then input 1000 A \lfloor 1000A \rfloor .


Notation: \lfloor \cdot \rfloor denotes the floor function .


Bonus: Generalize this for any hypotenuse k k .

Another version awaits here. .


The answer is 4166.

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2 solutions

Chew-Seong Cheong
May 21, 2017

Let the length of one of the legs be x x , then the probability of uniformly distributed random variable X X is given by P ( a X b ) = a b d x k \displaystyle P\left(a \le X \le b \right) = \int_a^b \dfrac {dx}k .

Since the area of the right triangle with hypotenuse of k k is A = x k 2 x 2 2 A_\triangle = \dfrac {x\sqrt{k^2-x^2}}2 , then the expected value of A A_\triangle is as follows:

E ( A ) = 0 k x k 2 x 2 2 k d x = k 2 0 k x k 1 ( x k ) 2 d x Let x k = sin θ , d x = k cos θ d θ = k 2 2 0 π 2 sin θ cos 2 θ d θ = k 2 2 1 0 cos 2 θ d cos θ = k 2 6 cos 3 θ 1 0 = k 2 6 \begin{aligned} E\left(A_\triangle \right) & = \int_0^k \frac {x\sqrt{k^2-x^2}}{2k} dx \\ & = \frac k2 \int_0^k \frac xk \sqrt{1-\left(\frac xk\right)^2 } \ dx & \small \color{#3D99F6} \text{Let }\frac xk = \sin \theta, \ dx = k \cos \theta \ d\theta \\ & = \frac {k^2}2 \int_0^\frac \pi 2 \sin \theta \cos^2 \theta \ d\theta \\ & = - \frac {k^2}2 \int_1^0 \cos^2 \theta \ d \cos \theta \\ & = - \frac {k^2}6 \cos^3 \theta \ \bigg|_1^0 \\ & = \frac {k^2}6 \end{aligned}

For k = 5 k=5 , E ( A ) = A = 25 6 4.16667 E\left(A_\triangle \right) = A = \dfrac {25}6 \approx 4.16667 , 1000 A = 4166 \implies \lfloor 1000A \rfloor = \boxed{4166} .

Efren Medallo
May 20, 2017

If the hypotenuse of such triangle is 5 5 , then its legs are x x and 25 x 2 \sqrt{25-x^2} for some length x x such that 0 < x < 5 0 < x < 5 .

The area of this triangle is 1 2 x 25 x 2 \frac{1}{2} x \sqrt{25 - x^2} .

Now, to find its expected area over the inteval ( 0 , 5 ) (0, 5) , we do

A = 0 5 1 2 x 25 x 2 d x 5 0 A = \frac { \displaystyle \int_0^{5} \frac{1}{2} x \sqrt{25 - x^2} \mathrm{d}x }{ 5 - 0 }

A = 1 10 1 3 ( 25 x 2 ) 3 2 0 5 A = - \frac{1}{10} \cdot \frac{1}{3} (25 - x^2)^{\frac{3}{2}} \Bigg|_0^{5}

A = 25 6 4.16666667 A = \frac{25}{6} \approx 4.16666667

Hence, 1000 A = 4166 \lfloor 1000A \rfloor = 4166 .

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