Skip to main content

Random chord on circle

Problem: 

Given a circle of radius 1, pick a chord at random. What is the chance that the length of the chord is less than the radius?


Solution:

Depending on how you pick it at random, the probability can be different!


1. If you pick it by randomly choosing 2 points on the circle's circumference, then the chance would be 1/3.

2. Every chord has a midpoint which is also a perpendicular bisector. If we pick this midpoint randomly within the circle, its chord length will be less than radius as long as it is outside of a smaller circle with radius $\frac{\sqrt{3}}{2}$. So the chance is 1/4 based on the ratio of area.

3. From the centre, randomly choose a radial direction and randomly pick a point along this radius, then draw a radially perpendicular line through this point, we'll get a chord. Then the chance would be \( 1 - \frac{\sqrt{3}}{2} \).

Surprising, huh. In fact, you can compute the distribution of the chord length given the picking method. Is there a way to pick a chord "uniformly"?

Another interesting and related problem: how do you uniformly pick a point within a circle? Can you prove that this method of picking a point is not uniform: first uniformly pick $\theta$ and then uniformly pick $r$, we get $x = r \sin \theta$, $y = r \cos \theta$?


Comments

Popular posts from this blog

641 | 2^32 + 1

Problem: Show that \( 641 | 2^{32} + 1 \). Solution: (From Problem Solving Strategies, Arthur Engel) \( 641 = 625 + 16 = 5^4 + 2^4 \). So \( 641 | 2^{32} + 2^{28} \cdot 5^4 \). Also, \( 641 = 640 + 1 = 2^7 \cdot 5 + 1\). So \( 641 | (2^7 \cdot 5)^4 - 1 = 2^{28}\cdot 5^4 - 1 \). Hence \( 641 | 2^{32} + 2^{28} \cdot 5^4 -(2^{28}\cdot 5^4 - 1) \). QED

2n+1 3n+1 squares then 5n+3 not prime

Problem: Prove that if \( 2n+1 \) and \( 3n+1 \) are both squares, then \( 5n+3 \) is not a prime. Solution: Proof by contradiction. Suppose that there is n such that 5n+3 is prime. Note that \( 4(2n+1) - (3n+1) = 5n+3 \). Let \( 2n+1 = p^2 \) and \( 3n+1 = q^2 \), where \( p, q > 0 \). We have \( (2p-q)(2p+q) = 5n+3 \). Since RHS is a prime, we must have \(2p - q = 1 \) and \( 2p + q = 5n+3 \). Solving for \( q \) we get \( 2q = 5n + 2 \). Substituting, we get \( 2q = q^2 + 2n + 1 \), or \( -2n = (q-1)^2 \). Since RHS is \( \geq 0 \), we can only have equality when \( n= 0 \) and \( q = 1 \). In this case, we have \( 5n + 3 = 8 \) not a prime. QED.

12 coins

Problem: There are 12 identical coins. There might be one that is a counterfeit, which is either heavier or lighter than the rest. Can you identify the counterfeit (if any) using a balance at most 3 times? Solution: I think this is one of the most difficult variations of the coins weighing problem. The main idea is to "mark" the coins after weighing them. Step 1. Divide the coins into groups of 4. Weigh the first two. If they have the same weight, go to step 4. Otherwise, mark the coins belonging to the heavier group 'H', lighter group 'L', and the unweighted ones 'S'. We know that the counterfeit is either in L or H. Step 2. Now we have 4 Hs, 4 Ls, and 4 Ss. We now form 3 groups: HHL, LLH, and HLS. Step 3. Weigh HHL against HLS. Case 3.1: HHL is lighter than HLS. Then either the L in HHL or the H in HLS is the counterfeit. Simply weigh one of them against S and conclude accordingly. Case 3.2: HHL is heavier than HLS. Then the counterfeit ...