Way back I watched a video by 3b1b on the tetrahedron in a sphere Putnam problem and tried to solve it on my own (having forgotten the entire video). Having no experience with the Putnam, and only an easy discrete math class in terms of proof-writing, I'm not sure how rigorous--if at all--my solution is. I want to take the Putnam this year, but am wondering if I even can begin to solve the questions properly. So here is my solution. I'm wondering, in your opinion, would this be a 0, 1, 8, or even a 10? Any opinions/advice would be greatly appreciated.
Prompt: What is the probability that four randomly chosen points on the surface of a sphere, when connected, form a shape that contains the center of the sphere?
First, let us solve some easier versions of this question.
Take the 1D version of this question, where we are tasked with finding the probability that two randomly chosen points on a line, when connected, pass through the center of this line. For this to occur, there must be a point to the right and to the left of the center. Pick any first point on that line, the probability of the second point being on the other side is 1⁄2, so the probability that their connection passes through the center is simply 1⁄2 as well.
Next, let us look at the 2D version, where we are tasked with finding the probability that three randomly chosen points on a circle, when connected, pass through the center. First, you can randomly choose two points on the circle, and let us call the angle of the arc connecting these two points alpha. If we connect these two points to the center of the circle with a straight line and extend to the other side of the circle, vertical angles are formed with a vertex at the circle’s center with measure alpha. In order for the third point to form a shape that passes through the center, the point must lie on the arc made from the vertical angle on the opposite side of the first two points. The probability that this happens for any angle alpha is alpha/360 degrees. Since the possible measures for the arc angle are uniformly distributed from 0 to 180 degrees, the average measure is 90 degrees. Therefore the probability that three randomly chosen points form a shape whose center passes through the center of the triangle is 90 degrees/360 degrees = ¼.
Returning back to the original question, again we notice symmetry. If we choose any three points on the surface of the sphere, the region for the fourth point to be in so that the connected shape passes through the center is identical to the region formed by the first three points and is on the opposite side made through extending lines from the first three points through the center to the other side of the sphere. So, the proportional area of that region of the sphere relative to the entire sphere’s area will determine the probability of the randomly chosen four points to contain the center. To determine this area, start with one point. To find the average surface area we can find what the average region formed by three random points looks like . So, given the sphere of radius R, the euclidean distance between the second point to the first can be from 0 to 2R. However this distribution is not uniform, as there is only 1 point to be 0 and 2R away yet an entire circular band of radius R of points to be R√2 away. This can be visualized by orienting the sphere so that the first point is northernmost, and the band of distance R√2 is the equator. However, the distribution is symmetric at R√2 as the northern and southern hemispheres of the sphere are identical and so for every point at a distance greater than R√2 there is a point at a distance below R√2. Therefore, the average euclidean distance from the first to the second point is R√2. Note that the arc length connecting these two points is a quarter of the circumference. Next we look at adding the third point. If we align the first two points so the straight line connecting them is parallel with the ground, we notice another symmetry. For any point on the left hemisphere, there is a mirrored version along the connecting line on the right side of the sphere. This means that the surface area will be the same regardless of which side the third point is on, therefore we only need to look at one hemisphere. Again, we see the same symmetry. For the first point, the third point can be from 0 to 2R away with a symmetric distribution. Same for the distance between the second and third point. And since the distribution is once again symmetric due to the symmetry of the sphere, on average the third point will be R√2 away from both the first and second point (in terms of Euclidean distance). This created a shape taking up a quarter of the hemisphere and therefore an eighth of the sphere. Thus, the probability that four randomly chosen points, when connected, form a shape that passes through the center of the sphere, is ⅛.
