## Poll

 I love math! 16 votes (50%) Math is great. 12 votes (37.5%) My religion is mathology. 5 votes (15.62%) Women didn't speak to me until I was 30. 2 votes (6.25%) Total eclipse reminder -- 04/08/2024 10 votes (31.25%) I steal cutlery from restaurants. 3 votes (9.37%) I should just say what's on my mind. 6 votes (18.75%) Who makes up these awful names for pandas? 5 votes (15.62%) I like to touch my face. 10 votes (31.25%) Pork chops and apple sauce. 7 votes (21.87%)

32 members have voted

Gialmere Joined: Nov 26, 2018
• Posts: 2048
April 14th, 2020 at 7:11:02 PM permalink
Four points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points? (It is understood that each point is independently chosen relative to a uniform distribution on the sphere.)

I posted this problem on the song thread as a joke. It's a level 6 question on the Putnam test meaning it's really hard. The reason I'm placing it here is that it contains a shortcut. Using logic and visual intuition, it's possible solve the problem in your head without picking up a pencil. So, if you decide to have a go, there's no need to show your work. Just post the answer. Have you tried 22 tonight? I said 22.
Gialmere Joined: Nov 26, 2018
• Posts: 2048
April 30th, 2020 at 7:38:25 PM permalink
In the unlikely event someone is still contemplating this, try reducing the problem down by one dimension and consider a triangle inside of a circle.
Have you tried 22 tonight? I said 22.
gordonm888 Joined: Feb 18, 2015
• Posts: 3186
Thanks for this post from: April 30th, 2020 at 8:56:48 PM permalink
Its a beautiful graphic.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
unJon Joined: Jul 1, 2018
• Posts: 2844
Thanks for this post from: April 30th, 2020 at 9:01:34 PM permalink
Quote: Gialmere

In the unlikely event someone is still contemplating this, try reducing the problem down by one dimension and consider a triangle inside of a circle.

My guess for a triangle in a square is 1/3.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Gialmere Joined: Nov 26, 2018
• Posts: 2048
May 1st, 2020 at 6:56:21 PM permalink
Quote: unJon

Quote: Gialmere

In the unlikely event someone is still contemplating this, try reducing the problem down by one dimension and consider a triangle inside of a circle.

My guess for a triangle in a square is 1/3.

I'm not sure where the "square" comes from but, your answer (although incorrect) is in the ballpark.
Have you tried 22 tonight? I said 22.
ChesterDog Joined: Jul 26, 2010
• Posts: 964
Thanks for this post from: May 1st, 2020 at 7:21:26 PM permalink
Quote: Gialmere

Quote: unJon

Quote: Gialmere

In the unlikely event someone is still contemplating this, try reducing the problem down by one dimension and consider a triangle inside of a circle.

My guess for a triangle in a square is 1/3.

I'm not sure where the "square" comes from but, your answer (although incorrect) is in the ballpark.

For the 2-dimensional problem, calculus gives 1/4 as the probability of three random points on a circle enclosing the circle's center.

Now, I have to figure how to use this result to answer the 3-dimensional problem.
ChesterDog Joined: Jul 26, 2010
• Posts: 964
Thanks for this post from: May 2nd, 2020 at 5:02:04 AM permalink
Quote: Gialmere

In the unlikely event someone is still contemplating this, try reducing the problem down by one dimension and consider a triangle inside of a circle.

3-dimensional problem: The probability that the tetrahedron made by 4 random points on a sphere enclose the center of the center is 1/8?

2-dimensional problem: The probability that the triangle made by 3 random points on a circle enclose the center of the circle is 1/4.

1-dimensional problem: The probability that the segment made by 2 random points on a line segment enclose the center of the line segment is 1/2.

0-dimensional problem: The probability that the point made by 1 random point on a point encloses the center of the point is 1.

unJon Joined: Jul 1, 2018
• Posts: 2844
Thanks for this post from:  May 2nd, 2020 at 5:58:17 AM permalink
Quote: Gialmere

Quote: unJon

Quote: Gialmere

In the unlikely event someone is still contemplating this, try reducing the problem down by one dimension and consider a triangle inside of a circle.

My guess for a triangle in a square is 1/3.

I'm not sure where the "square" comes from but, your answer (although incorrect) is in the ballpark.

Oops circle.

I think the trick is that the problem is equal to the average % surface area encompassed by X-1 points.

So for a two dimensional circle, what is the average % circumference enclosed by two random points on the circle (use the smaller circumference segment)?

Three dimensional: what is the average % surface area covered by the shape enclosed by three random points (use the smaller surface area)?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Gialmere Joined: Nov 26, 2018
• Posts: 2048
Thanks for this post from: May 2nd, 2020 at 11:38:27 AM permalink Quote: ChesterDog

3-dimensional problem: The probability that the tetrahedron made by 4 random points on a sphere enclose the center of the center is 1/8?

Ding! Ding! Ding!

You are absolutely correct.

I'm not a big math wiz but I do enjoy logic problems. Since math and logic problems often overlap, YouTube suggested I watch this video and I thought it was pretty cool...

Have you tried 22 tonight? I said 22.
charliepatrick Joined: Jun 17, 2011
• Posts: 2424
May 2nd, 2020 at 11:52:37 AM permalink
Quote: unJon

...for a two dimensional circle, what is the average % circumference enclosed by two random points on the circle (use the smaller circumference segment)?...

Thanks for the thought - here's a simple proof. (Now to worry about the sphere issue!)

If the first point picked is assumed to be at the South (Pole), then the secondwill be somewhere (for simplicity assume the RHS) between North and South - in the diagram Point A.

For the triangle to contain the centre the point would have to be between B (where AB crosses the centre) and N (where NS crosses the centre). Thus the probability is length of arc NB / length of circumference. Length(NB)=Length(AS)=length of arc.

Length of arc is uniformly between 1/2 (A=N) and 0 (A=S), so the average is 1/4. Note it took some time for me to create this diagram so I landed up posting this after the above video. btw thanks for the video it's very instructive.
Last edited by: charliepatrick on May 2, 2020