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Posts posted by bonanova


SpoilerYou arrange with a friend to tell you the sum of three numbers, one each of which will be told him by the group. No mention is made to the friend of the meaning of the numbers. You each then tell the friend your salary, and you divide the sum the friend tells you by 3. You thus determine the average without telling one another your salaries.

SpoilerThe Brooklynbound train arrives, each hour, exactly 54 minutes after the Bronxbound train. The time intervals determine the destination probabilities, and they are in the ratio of 54:6 = 9:1, favoring Brooklyn.

SpoilerYou tilt the glass as far as you can without spilling any water. If water still touches any part of the bottom of the glass, the glass is more than half full. Otherwise less. If the water just touches the edge of the bottom, it's exactly half full.

Maybe, maybe not.
SpoilerTaking the natural log of both sides yields
(1/x) ln(x) = (x) ln(x)
ln(x) [1/x + x] = 0
This has solutions
ln(x) = 0 ==> x = 1
And
1/x + x = 0 ==> x^{2} + 1 = 0 ==> x = (+/) i
So there are three solutions, one of which is real.

On 9/13/2017 at 2:21 AM, bonanova said:I, as well, stand by my answer.
Spoiler OP does not say "on" an ellipse.
 Nor a particular ellipse.
To finish my proof, fasten a string to the foci that loops a pencil at the ellipse point.
Draw an ellipse.Caveat:
SpoilerThe OP could be more definitive.
Take, for example, the phrase "that you know a foci is located at (0,0)."
My interpretation would be rock solid if OP said
How many points would you need to have to uniquely determine an ellipse "that has a focus located at (0,0)"?

It would be dead wrong if OP said
There is an ellipse that has a focus at (0,0).
How many points on that ellipse do you need to draw that ellipse?

To clarify. You don't have any of these abilities:
 Make a line segment of equal length to another segment, or equivalently make equallyspaced pairs of points. (You don't have a pair of dividers.)
 Make a line segment that is parallel to another line (segment).
 Make an arc. (You don't have a compass.)
But you do have the equivalent of a straightedge.

Don't see how it matters, unless ...
SpoilerYou do more than "take a shot."
If you keep shooting until someone is killed, then just don't be the first to shoot.
Why? Because a bullet in either of chambers 1 and 6 will kill the first shooter.
The other four shooters have a 1/6 death probability. 
SpoilerAs I understand these things, he can't escape a black hole that way.
The reason is that as objects enter the vicinity of a black hole (approach it, but before they cross the event horizon) they get spaghettified. They lose their structural integrity. If Superman had crossed the event horizon, the "front" of the spaceship would already have already been ripped from its "back" apart. Sadly, our super hero would also have suffered the same fate. Even if his super powers allow him to stay in one piece, his space ship would offer him no assistance in exiting the event horizon. He would have to do that on his own.
And, the OP asks whether the space ship would help him exit. The answer is, it wouldn't.
As I understand what happens, to ordinary matter, is the difference in gravitational attraction experienced by the near and far portions of objects is so extreme that they are pulled apart, like so much taffy, but worse  torn apart, almost at the molecular level. That damage would be irreparable. The space ship would disintegrate.

Thanks to araver for the game.
With a full (original) compliment of players the game would have unfolded very differently, even with a lucky lynch.

I'm headed to bed (US EDT). I'll read my obit when I wake up.
I hope something kind is said ...
I'm wondering ... has a player ever been BOTH of the first two kills before? Poor Aura

3 hours ago, nana77 said:Is there time to get the new actions in?
I am suspicious of Flame for pushing the tie d1. Even though the results were nice. And I am wary of Gavin. Plas scores some goodie points for his plan. idk about Bona.
I'm in. I'm changing my block now. (said the soldier willing to die for the cause.)

56 minutes ago, flamebirde said:Bona, who did you claim?
You mean which Goodie? I didn't.
But if the list of names were alphabetized I'd be at the tail end.
Thus I'm a non factor at this point. 
So ... if Dolores or Hector is still around ... then ... grab the gun and get it done?
Gav and I have claimed goodie roles. I don't know what (1 of 2 roles) Nana has claimed.
Flame and plas: just need one of you to claim and it's done. Wow.

11 hours ago, plasmid said:Considering that the only role with a color in its Role Description is the Man in Black, you might want to try another hint.
bonanova < (goodie color)

59 minutes ago, gavinksong said:On the other hand, I would really like to hear from bonanova.
Well other than the reference to my color, it's so early in the game I don't know what else to allude to.
That said, there are so many goodies, it seems likely one of us will swing tomorrow morning. Sad.

Agree Phil should post.
Also the other analyses are good reading; I'd like to see more.As for myself: I like my color.
1. Gavinksong  voting for Flamebirde
2. Flamebirde  voting for bonanova
3. Nana77  voting for plasmid
4. plasmid  voting for phil
5. phil1882
6. bonanova  voting for phil 
SpoilerEllipse is the locus of points the sum of whose distances from the foci is constant.
So, we just need (a) the other locus and (b) any point on the ellipse to determine the sum: two more points.

BTW I'm here, stumbling around in the dark ... Hi y'all.

And rotated?

SpoilerIf we can infer (OP does not say so, but ...) the town's population doubles every 30 years,
then it had population (1/2) x 30 years before it had population x. 
 What is his initial bet (stake)?
 Did you mean to say "Every time he wins, he raises his stake by 1/4 of his bankroll"?

Signups:
1. Gavinksong
2. Flamebirde
3. Nana77
4. plasmid
5. phil1882
6. bonanova
7. aura
8.
9.
10.
11.
12I've just moved, to Ohio, from NY and drowning in the details of it.
Reluctant to pledging my attention here. But,I'll give it a shot ... Should be able to check in at least daily.
 1

Red area: alternate (and harder) method
SpoilerWolfram gives the filling factor of hexagonal circle packing as f = [pi sqrt(3)]/6 = 0.9068996821.
Spoiler(A circle plus two red areas) tessellates the plane.
The circles have unit radii and pi areas.f(circle + 2 red) = circle
f(pi + 2 red) = pi.red = pi (1f)/2f = 0.1612544807739810

Proof that CaptainEd's answer is minimal.
SpoilerStart with any interior 2x2 region and color the 4 squares differently.
Without loss of generality,
x x x x x
x x 2 3 x
x x 1 4 x
x x x x x
x x x x xSquare(3,2), to the left of color 1 can only be 3 or 4
Case 1: color(3,2) = 3
SpoilerThis choice forces the coloring of 15 squares:
x 3 1 4 x
x 4 2 3 x
x 3 1 4 x
x 4 2 3 x
x 3 1 4 xwhere all the x squares receive (only) colors 1 or 2.
There are already 5 squares colored 3 and 4.
Colors 1 and 2 must each occur at least 4 more times,
making at least 7 squares with oolor 1
and at least 6 squares with color 2.Case 2: color(3,2) = 4
SpoilerThis choice forces only one other square
x x x x x
x 3 2 3 x
x 4 1 4 x
x x x x x
x x x x xSquare (4,2) can only be color 2 or color 3.
Case 2a: color(4,2) = 2
SpoilerThis choice forces the color of 15 squares:
x x x x x
2 3 2 3 2
1 4 1 4 1
3 2 3 2 3
x x x x xwhere all the x squares receive (only) colors 1 or 4.
There are already 5 squares colored 2 and 3
Colors 1 and 4 must each occur at least 4 more times,
making at least 7 squares with color 1
and at least 6 squares with color 4.Case 2b: color(4,2) = 3
SpoilerThis choice forces the color of 9 squares:
x x x x x
x 3 2 3 x
x 4 1 4 x
x 3 2 3 x
x x x x xwhere all the x squares receive (only) colors 1, 2 or 4.
Colors 1, 2 and 4 must each occur at least 4 more times,
making at least 5 squares with color 1,
and at least 6 squares with colors 2 and 4.Solution: There are exactly 4 squares with color 3.
Make color 3 Yellow
With hindsight,
SpoilerSquares (2,2) (2,4) (4,2) and (4,4)collectively touch ALL the other squares and and do not touch each other. So one solution is 4 squares. Enumeration of all cases proves that 4 is minimal.
The halfway glass
in New Logic/Math Puzzles
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You're right. I botched it. rodomac's wordless pics were in my mind, but they never made it to the keyboard.
"any part of" should be "all of."