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# Yoruichi-san

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## Posts posted by Yoruichi-san

1. ### An Antiquated Math Challenge

well the shortest distance between any two points would be a straight line, so I'll go with that.

unless we are trying to find a curve with the minimum friction.

Yes, but the particle is not moving at constant speed . It starts out with no momentum (perhaps I should have specified that, but I thought it was implied), so its momentum is dictated by gravity.

A vertical line would allow free fall if it touches it initially, does that constitute sliding?

Lol...sorry it's the path it slides along, not by ;P.

2. ### LGBD R5 Vanguard Battle

Blank.

Panther's turn to shoot or pass.

3. ### LGBD R5 Vanguard Battle

BANG!

Standings:

Framm (East): 53

Panther (West): 247

Framm's turn to shoot or pass.

4. ### An Antiquated Math Challenge

You can use g to represent the gravitational constant, the answer should be in terms of g. I'm not sure what you mean by the other question, it's particle sliding down the path you choose without friction.

5. ### LGBD R5 Vanguard Battle

BANG!

Standings:

Framm (East): 3

Panther (West): 297

Panther's turn to shoot or pass.

6. ### Isosceles triangle inscribed in red-blue-green circle

Time for bed...got up way too early this morning...I'll leave it to plasmid .

7. ### LGBD R5 Vanguard Battle

Blank. There was 1 chip on the table from Framm.

Standings:

Framm (East): 53

Panther (West): 247

Framm's turn to shoot or pass.

8. ### The Chromatic Witch, Ep 2: The Power of Green

"Aha!" Agent Flamebirde had an idea. He created a 100 m high box underneath the door. Constructing a structure in this manner would allow the agents to float to the top on the will generated water of one agent, based on the idea of Agent Aaryan. Now they would just need some help getting the water the rest of the way up...

Captain Wraith smiled to himself as he studied Orin. He will make a good captain someday, if something were to happen to me...

9. ### Isosceles triangle inscribed in red-blue-green circle

If we divide the circumference into 10 equal portions with 10 equally spaced points, then at least 4 of the points must be the same color, and there is no way to pick 4 of the points without 3 of them forming an isosceles triangle.

So I see a general principle emerging here. I'm not sure I can completely describe it yet, but to start from the beginning:

Any triangle whose points lie on a circle is comprised of three inscribed angles. These angles have corresponding arc lengths, so a necessary and sufficient condition for a inscribed triangle to be isosceles is for two of the arc lengths to be equal.

If we put n points on the circle, we can figure out the minimum number of points that must be of the same color, call it k. We can think of 'painting' k of those points the same color as dividing the circle into k pieces. Picking any three points to form a triangle is summing adjacent pieces into three groups. If there is a way of forming three groups such that two groups have the same arc length total, then we have an isosceles triangle. So if there is no way of dividing the circumference into k pieces along the n points such that there is no way of forming three groups in which two of the groups have equal arc length, then there must be an isosceles triangle.

For the 2 color case, dividing the circumference into 5 pieces of 2pi/5 rad (72 deg) guaranteed that there was at least 3 points of one color, and there was no way to group the pieces into three groups so that no two groups had the same arc length.

For the 3 color case, for uniformly placed n points, n is 10, and k is 4. I'm not sure if there is a way of doing it with non-uniform placement with a smaller n, though...

I think this may be generalized further, but I'm not sure...it leaves an interesting exercise

10. ### LGBD R5 Vanguard Battle

Blank

Framm's turn to shoot or pass.

11. ### LGBD R5 Vanguard Battle

BANG!

Table contents and 50 chips from Framm go to Panther.

Standings:

Framm (East): 54

Panther (West): 246

Panther's turn to shoot or pass.

12. ### Line through a circle

Yeah, I agree with k-man that different methods of drawing give different answers, but overall I think I agree with JAWF on the answer...

The way I draw a chord is usually to pick a point on the circumference and then pick another point and connect them, so using that method and assuming each point on the circumference is equally likely, the second point must be >120 deg away from the first point for the chord to have length >sqrt(3), so the probability would be 1/3.

13. ### Boys and Girls Line Up

2nm/(m+n)

Don't have time to write the full derivation, but I calculated the probability that, taking any pair in the lineup, one would be a boy and one would be a girl, then calculated the expectation value by multiplying by the number of pairs (m+n-1).

14. ### Black and white plane

Oh, and, btw, TSLF, when I said I was "confident that you'd see", I meant I was confident in your intelligence to be able to find 7-move win, lol...

15. ### The Chromatic Witch, Ep 2: The Power of Green

The agents worked together to create water to float to the top. Then Agent Flamebirde had a great idea...create a wall to lessen the space needed, to lessen the amount of water they would need to create. However, the maximum dimensions of the wall were 100m, so it would fully encompass the 150m room...

"Wait..." Orin murmured slowly, as an idea came to him. "Who said we were confined to two dimensions? Remember the first thing the Witch created, when we got here..."

16. ### Rectangle on red-blue plane

Nice, vista, efficient and simple solution...I was trying to draw grids of right triangles

(Just thought it was about time someone actually said something nice about someone else's solution ;P)

Edit: Oh, and thanks for the puzzles, btw, witzar, they've been fun and interesting .

17. ### Black and white plane

Lol...sorry, read the 'p' as an 'f'...damb phone screens

...you're just not trying hard enough ;P

Seriously though, I know you're smart enough to see the point that 2 and 5 form an equilateral triangle with .

The point is (again, no pun intended) that the minimum number of points needed to prove the null hypothesis false is 7, but you can find an infinite number of points that don't if you select the wrong points...it's like a game you're playing against the null hypothesis, you can win in 7 moves, but you can stall the game for as long as you like...

Have you ever played gomoku? (We call it wu zhi lian, or "connect five") I love the game, I'm very good at it, if I do say so myself *whistling*, and the winning strategy is usually a forcing strategy that makes your opponent put pieces at certain spots to prevent you from connecting five (or an open-ended four). So I guess I like proofs that are the same way *shrugs*.

You're free to have your opinion about which kind of proof is better, I don't really care about getting 'best answer' or w/e so if witzar wants to give it to you, he can be my guest, but in the end, it's his call what he likes better .

18. ### LGBD R5 Vanguard Battle

Blank.

Framm's turn to shoot or pass.

19. ### LGBD R5 Vanguard Battle

Blank.

Panther's turn to shoot or pass.

20. ### Black and white plane

Lol...TSLF, you're not very good at paint by numbers, are you? (I'm just kidding )

I don't think you're following my proof...it's ordered in such a way for a reason...

The fact that you can paint the arrangement of 7 points a way in which there are no equilateral triangles does not refute my argument, since my argument was not that there is no way to paint such an arrangement as such. My argument was that if you start with the assumption/null hypothesis "there exists no equilateral triangles with points all the same color" then you would eventually reach a contradiction, which I was able to show in 7 points.

I.e. I did not start with a set of 7 points and try to paint them, I started with 2 points that were the same color (which it's pretty trivial to prove must exist), and found the corresponding points that I could make logical conclusions about based on the null hypothesis, and showed the null hypothesis was eventually cornered into a contradiction, hence must be false.

In your diagram, points 1 and 2 are different colors, which kind of misses the point (no pun intended ). Start with any two points that are the same color, say your 3 and 6, or your 1 and 3 (everything would just be rotated), or your 2 and 7 (everything would be rotated and scaled up), or any two points of the same color, and I'm confident you will see how each step of the argument is proven true by the previous steps and geometry .

21. ### LGBD R5 Vanguard Battle

The shot was blank.

Table contents go to Panther.

Standings:

Framm (East): 109

Panther (West): 191

Framm's turn to shoot or pass.

22. ### RGB plane

Here's the simplest, probably most elegant solution...that I feel dumb for not seeing earlier:

Choose a point of one color, say red, call it the origin. Draw a unit circle around it of points that cannot be red. Each set of two points on this circle that are a distance 1 from each other must be different colors, one blue and one green.

Now draw a circle of radius sqrt(3) with the red dot as the origin. Each point on this circle forms an equilateral triangle with two points on the unit circle, so all the points on this circle must be red. But there are infinite number of points on this circle that are distance 1 from each other, hence there exists two points with the same color that are distance 1 from each other.

23. ### An Antiquated Math Challenge

One of the most famous problems in the history of mathematics, posed by Johann Bernoulli in 1696 (I won't state the official name of it, in order to make it not too google-able ;P):

Find the curve along which a particle will slide without friction in the minimum time from one given point P to another point Q, the second point being lower than the first but not directly beneath it.

...Aquaman?

Crabs, crawfish...or like the god poseidon or sth?
25. ### RGB plane

Choose a point, color it one color, say blue. Choose a point a distance of one from the first point, it must be a different color, say green. Now find the points that are one unit away from both B and G, they must be the third color, red.

Each point has a unit circle around which represents the points that cannot be the same color as the point. Consider the ones for the two red points (colored in light turquoise in the diagram).

Pick two points on one of the unit circles that are a distance one from each other, call them 1 and 2. They are one unit away from the center of the circle as well as an outside point, call it 3. Since there are an infinite number of points of types 1 and 2, there are a corresponding infinite number of points of type 3, and they draw out a circle of radius sqrt(3) around R.

This circle (in light purple) intersects with the other unit circle. Hence there exists an equilateral unit triangle (four actually) for which all three points lie on the unit circles of points that cannot be red, designated 1', 2', and 3'. Since there are three points and only two colors left, two of these points must be the same color, hence two points of the same color a distance of one from each other must exist.

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