Skip to main content

Geometry Puzzle: Center of Square in Circle

Source: Asked to me by a lot of people from IIT Bombay on email and posted by a few on "Post a Question" page.

What is the probability that two uniform random points (to be precise: i.i.d. with respect to Lebesgue measure) in the square are such that center of the square lies in the circle formed by taking the points as diameter.

Edit: Problem wording made more mathematical.

Update: (24/12/2012)
Correct solutions by Barun Kumar Kejriwal, Akhil Kumar, Aastha Airan, Arpit Goyal and Yashoteja in comments! Thanks a ton.


  1. Mark an arbitrary point P1 inside the square and draw a line L1 through P1 and the centre of the circle.
    Draw another line L2 perpendicular to L1 and passing through the centre of the square, O.

    Now, it becomes easy to see that any point in the portion of the square on the opposite side of L2 makes an obtuse angle with OP1.

    Thus, prob = 1/2

  2. (1/2)

    Hint for people who want to try: Diametrically opposite points form obtuse angle at any point inside the circle, right angle at points on the circle and acute angle at the points outside the circle

  3. The angle it the two points make with the center should be >= 90 (obtuse angle)
    With half probability the angle will be obtuse
    Probability = 1/2

  4. Take one point inside the square. Join it to the centre of square. Now draw a line perpendicular to the given line. The perpendicular line divides the square into parts with equal area.

    Now if I take the second point in the part which contains the first point, out points will subtend an acute angle at the centre.

    If I take second point in the other half, our points will subtend an obtuse angle at the centre.

    A point lies is circle obtained by two points as diameter if the two points subtend an obtuse angle.

    So probability that centre lies inside the circle is same as probability that it lies outside the circle.

    Ans: 0.5

  5. Probability is 0.5

    If 'O' is the center of the square,
    and 'A' and 'B' are 2 points inside the square
    chosen uniformly at random,

    then O lies inside circle sitting on AB iff angle AOB is obtuse.

    Now for any random point A inside the square,
    placing B in some parts of square leads to acute angle, others to obtuse angle. By symmetry it is easy to see that both areas are equal, meaning probability of selecting A and B such that O lies inside the circle sitting on them is 0.5

    (Draw lines 0A', OA", OC', OC" where,
    A', A", C', C" lie on square and C'C" perp to A'A", and A lies on OA'. then area on one side of C'C" corresponds to acute angle, other to obtuse angle, both areas being equal)

  6. Correct solutions by Barun Kumar Kejriwal, Akhil Kumar, Aastha Airan, Arpit Goyal and Yashoteja in comments! Thanks a ton.

  7. One more variation might be what is the probability that circle formed lies completely inside the square :)


Post a Comment

Popular posts from this blog

Asking a girl out

This is not a puzzle. So, for those of you who follow this puzzle blog, please bear with me for just one post. Interesting Math in this article though :P

Most of my friends already read an article that I wrote more than an year back - "Speak Up"

Here, inspired by the movie, The Beautiful Mind, I give a mathematical analysis of asking a girl out. Nice time it is. Feb 10. No plans for Feb 14 and I am sure this article makes me look even more geekier and all the more reason for me to believe that I will be alone, yet again. But what the hell, lets do it!

Note: This is not an independent analysis. There are many "mathematics sites" which does "similar" analysis.

@Consultants, correct me if I am wrong in my estimates. :P

Why is there a need to be selective?

From the age of 15, I guess there are approximately 3,600 girls I have liked (On average days, I don't see new girls. But going outside, I like about 30 girls. Saying that I go out once every week right …

Consecutive Heads

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.

1) Calculate P{X>Y}
2) What's the expected value of X
3) What's the expected value of Y

This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment.

(Solved by me finally after 13 months :))

Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently.

So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0"

f(State) = P(X>Y | "State" configuration initially)

f(0,0) = 1/4[f(…

Fraction Brainteaser

Sent to me by Gaurav Sinha

Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20 out of 28 Geometry questions. In total, Siddhant scores 25 out of 34. 

Vaibhav writes another Maths test and correctly answers 20 out of 25 Arithmetic questions and 6 out of 9 Geometry questions. in total, Vaibhav scores 26 out of 34.

Note that
a) Vaibhav scores more than Siddhant
b) Siddhant score better than Vaibhav in both individual topics - 5/6 > 20/25 and 20/28 > 6/9

How is it possible?