Expected Area of a Triangle Paradox

There have been several recent problems about expected values and unit circles. I came up with another problem like this, but I noticed something interesting. Here's the problem.

It is a well known fact that the triangle with vertices on opposite ends of the diameter of a circle and a third point on the circle is a right triangle. In a circle of radius 1,1, find the expected area of such a triangle.

Seems pretty simple, right? Actually, it has two different answers, depending on how you interpret the problem. Consider the following.

It is a well known fact that the triangle with vertices on opposite ends of the diameter of a circle and a third point on the circle is a right triangle. In a circle of radius 1,1, a diameter and a random point not on the diameter are randomly picked and connected to form a triangle. Find the expected area of this triangle.

And

It is a well known fact that the triangle with vertices on opposite ends of the diameter of a circle and a third point on the circle is a right triangle. In a circle of radius 1,1, an x-valuex\text{-value} is randomly selected from the diameter and its corresponding y-valuey\text{-value} is plotted on the circle. Find the expected area of this triangle.

Assume WLOG that the diameters of these circles are on the x-axisx\text{-axis}

These seem like the exact same problem. But there's two different ways to solve them, given the wording of the problem.


Solving the first problem\textbf{Solving the first problem}

The points on the diameter are (1,0)(-1,0) and (1,0).(1,0). These are the base of the triangle. We just have to find the average height of the circle. Any point on the unit circle can be represented by (cosθ,sinθ),(\cos\theta,\sin\theta), where θ\theta is the angle formed by drawing the ray starting at the origin and passing through (1,0)(1,0) and the ray starting at the origin and passing through the point. The height of the triangle is going to be sinθ.\sin\theta. The angle will be anywhere in the range [0,2π).[0,2\pi). We can just find the average of sinθ|\sin\theta| over the range [0,2π)[0,2\pi) to find the expected height. This will be the same as the average value of sinθ\sin\theta over [0,π).[0,\pi). 1π0πsinθ dθ=1πcosθ0π=1π(cosπcos0)=1π(11)=2π \begin{aligned} \dfrac{1}{\pi}\int_0^\pi\sin\theta\text{ }d\theta&=\dfrac{-1}{\pi}\left.\cos\theta\right|^\pi_0\\ &=\dfrac{-1}{\pi}(\cos\pi-\cos0)\\ &=\dfrac{-1}{\pi}(-1-1)\\ &=\dfrac{2}{\pi} \end{aligned} The height of this triangle has an average value of 2π\frac{2}{\pi} and the base is a constant 2,2, so the expected area of the triangle is 12×2×2π=2π\frac{1}{2}\times2\times\frac{2}{\pi}=\boxed{\frac{2}{\pi}}


Solving the second problem\textbf{Solving the second problem}

In this problem, you are randomly picking an x-valuex\text{-value} in the range (1,1)(-1,1) and plotting it on the circle, which we can again remove the parts of the circle in the third and fourth quadrants because of symmetry. We can model the semicircle by y=1x2.y=\sqrt{1-x^2}. The average value of this function is the area of the semicircle divided by 2,2, which is π22=π4.\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}. The base of the triangle is 22 and the average height is π4,\frac{\pi}{4}, so the expected area of the triangle is 12×2×π4=π4.\frac{1}{2}\times2\times\frac{\pi}{4}=\boxed{\frac{\pi}{4}}.


Wait, we've gotten two different answers to the same problem. Why is that? Consider the length of the arc we are picking a point from. In the first problem, we are picking any point from an arc of length π.\pi. In the second problem, the arc has length 2.2. But these points are still going to result in the same triangles. The variable arc in the first problem (A1)(A_1) and the variable arc in the second problem (A2)(A_2) are different; A1A2=π2.\frac{A_1}{A_2}=\frac{\pi}{2}. So we can then expect to see a difference in area.

See if you can try to explain the difference mathematically!

Note by Trevor B.
5 years, 4 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

The difference in the two cases is that the probability distributions are different. Hence, the expected value is different in each case.

In the first case, you are picking the third point with uniform distribution on the circular arc, whereas in second case, the uniform distribution in on the x-axis (diameter). In the first case, the distribution over x-axis would be a bowl shaped distribution, i.e. it would produce higher probability for smaller-area triangles, whereas in second case, as stated earlier, the distribution is uniform over x-axis. Hence, you get lower expected area in first case.

Prakhar Jaiswal - 5 years, 4 months ago

Log in to reply

Can i ask you one thing....how much time did it take to write such a long note with latex

Max B - 5 years, 4 months ago

Log in to reply

This took maybe about 1515 minutes. LaTeX isn't that hard if you've used it a lot.

Do you have anything to say about the paradox?

Trevor B. - 5 years, 4 months ago

Log in to reply

In the first case θ \theta is uniformly distributed over [0,π \pi] so by transform of variable distribution function of x=cosθx= \cos \theta will be= 1π1x2 \frac {1}{\pi \sqrt{1-x^2}} over [-1,1]

In the second case x is distributed uniformly over [-1,1] .

Jayanta Mandi - 5 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...