bonanova

You're right. I didn't read to the end of your solution, and I think you've solved it.

HBTY, HBTY, HBDR, HBTY! Cake may arrive later in the day.

It's your birthday, and a cake has been brought to your table, frosted on the top and sides. The cake is 3" high and has a base in the shape of a regular pentagon, 3" on a side. It's your task to cut the cake into pieces for yourself and your six guests. You want each piece to have the same volume and the same amount of frosting. The cutting is conventional: as seen from above, the cuts are like spokes radiating from the pentagon's center. The cutting planes are perpendicular to the cake's base. You realize that if you had invited two fewer guests (5 people in all) your job would be much easier. But after some thought you devise a plan that gets the job done. How do you cut your cake to provide seven pieces with equal volume and frosting?

OK, got it. I did mis-read the solution. I ran across this puzzle some time ago, and wondered whether it could break the deterministic nature of 3x3 tic tac toe by giving freedom of choosing which marker to place, and found that it doesn't. Maybe ANY 3x3 game allows Player 1 to either win or at least avoid a loss.

@BMAD my pentagon solution seems inaccurate.I found out the side length point at (+) being off . I would like to request your better solution for im having trouble with finding the last point of side length.

Excellent point. "Within a circle" or "On a disk" are appropriate references to the location of geometrically interior points. Most likely what Phil meant. We would also use the complementary term "Ball" (solid) with "Sphere" (hollow.) In a single dimension, "Line segment" (having an interior) and "Point pairs" (hollow) would be used. Only in zero dimensions does a single term "Point" suffice. What leads us astray at times is the fact that colloquial usage is not so strict. A manhole cover can be called circular, for example, rather "disk-like," and everyone understands. Hmmm... I wonder what interior-inclusive, "disk-like" term we should use with "Square" and "Rectangle"? Not to mention Pentagon, ...

Now, that question 1 is solved, I am inclined to yield the opportunity of solving question 2 to others. I think the proof is fine as it is, but I believe there are a few typos. See below Question for bonanova: about question 1- the probability that the 4th point falls within the triangle, Tardy answering your question ... better late than never. Sextuple integrals are indeed the solution. Nasty, but thankfully someone has already done it.

A few puzzles posted in this forum have related to random triangles inside a circle. By evaluating nasty integrals, or by my preferred method, simulation, it can be shown, perhaps surprisingly, that triangles constructed from sets of three uniformly chosen points within a circle cover only about 7.388% of the circle's area on average. After looking at on the subject, I simulated 1 million triangles to determine the median area. It turns out to be about 5.335% of the circle's area. Read: a random triangle has a 50% chance of being smaller. If the distribution of random-triangle areas has a mean of about 7.4% and a median of about 5.3%, what value might you expect for the mode?

Yes, it's called the isoperimetric problem, generally stated, what is the curve of constant length that encloses the greatest area? Equivalently, what shape of constant area has the smallest perimeter? The answer is a circle. A 45o arc can be reflected 7 times into a full circle. Symmetry demands this answer, although rigorous proofs abound. Here is a general discussion. In three dimensions (think soap bubble) a sphere encloses more volume than any other closed surface of the same area. isoperimetric = iso (same) perimetric (perimeter).

Uniform density of points on the circle. Envision choosing x and y randomly along the sides of the enclosing square, and discarding the point if it lies outside the circle. Note that choosing a random point on a random radius favors picking points near the center. There is controversy (generally resolved now) about what constitutes a random chord (Bertrand paradox.) The usual resolution is to pick a random point on a random radius and draw the perpendicular chord. Interestingly, what fails for getting a random point succeeds for getting a random chord.

<tongue in cheek> Since it has long been accepted that the proof is in the pudding, not in the pi, perhaps the puzzle is not solvable. <remove tongue from cheek>

Yes, I see your point. I think the OP was not well formed. I edited it.

Kudos to both TSLF and SP.

I can do it with 3 multiplications and 4 additions. Still looking.