Assuming that those are three bananas in the third statement, and one banana in the fourth, 21.
I am still the wizard that did it.
"On my business card, I am a corporate president. In my mind, I am a game developer. But in my heart, I am a gamer." -- Satoru Iwata
<scrimpy> at least I now know where every map, energy and save room in this game is
Also note that the grape bunches are different, as they contain 3+5+4=12 in the first picture, and 3+4+4=11 in the second. Using that
A=7
12G=5+A=5+7=12, G=1
A=1+3B 7=1+3b 6=3B 2=B
A+11G+B=7+11+2 =20
Look who's failing at counting. Good eye, Flip.
Edit: In before someone notices that the first apple is 99% the size of the third apple. <_<
I am still the wizard that did it.
"On my business card, I am a corporate president. In my mind, I am a game developer. But in my heart, I am a gamer." -- Satoru Iwata
<scrimpy> at least I now know where every map, energy and save room in this game is
I was thinking of flat earth bullshit and this problem came to my mind. The answer that I got was pretty big and complicated for a seemingly simple problem, let's see if some of you finds out.
Click to expand.
I was thinking of flat earth bullshit and this problem came to my mind. The answer that I got was pretty big and complicated for a seemingly simple problem, let's see if some of you finds out.
Click to expand.
Are those two lines both tangent to the circle?
Also, are we given theta? It seems very likely that the answer depends on the angle subtended by the line segments (the "towers"?).
Edit: After a few minutes of work, I've applied the Pythagorean theorem twice and the law of cosines once to obtain
c_1^2 = (r+h+a)^2 - r^2
c_2^2 = (r+h-b)^2 - r^2
(c_1 + c_2)^2 = (r+h+a)^2 + (r+h-b)^2 - 2*(r+h-a)*(r+h-b)*cos(theta).
The first two equations can be derived by considering a line from the center of the circle to the point where the red line intersects. If we are given theta, this constitutes three equations and three variables (c_1, c_2, and b) and so the rest is "trivial". I'll do the algebra anyway. Expect another edit in five or ten minutes.
Edit 2: So here's my answer, hastily derived in Notepad:
v = u*[cos(theta) +/- sqrt(cos^2(theta) - 1 + u^2*sin^2(theta))]/(1 - u^2*sin^2(theta))
where
u = (r+h+a)/r,
v = (r+h-b)/r
and so the above expression for v is written entirely in terms of known quantities. To find b explicitly, simply solve for v above, then take
b = r+h-r*v.
I may do a little more algebra to see if there's any more simplification to be done.
Edit 3: Last version of my solution. I hope this one is more readable than the previous one.
v = u*(cos(theta) +/- sqrt(2*w + w2))/(cos2(theta) - (2*w + w2)*sin2(theta))
where
u = (r+h+a)/r,
v = (r+h-b)/r,
w = (r+a)/r.
As before, solve for u and w (both written in terms of known quantities), then plug both into the right hand side of our expression for v. After finding v, take
b = r+h-r*v.
By the way, I'm not sure whether to take the positive or negative square root of the expression. It's very likely that one solution is non-physical.
Edit 4: I should have the correct answer now. It's the same as before, except with
theta = 2*cos-1(r/(r+h))
There is likely a way to remove all trigonometric dependence. I guess I'll work on that...
Quick reminder that we have [sup][/sup] and [sub][/sub] for neat superscript and subscript for anyone that likes to polish up their replies.
I've used them before, but I'm in kind of a lazy mood. This problem has too much going on in it for me to put in subscripts and superscripts everywhere, although I may include them in my final answer.
Joined: 1/16/2008
Posts: 358
Location: The Netherlands
Amaraticando wrote:
The lines are tangent and the angle is not given, as it is determined by R and h. The "towers" are perpendicular to the "ground".
I'm not sure if I'm interpreting this correctly... I have the impression that if the two 'towers' are the same height (h) and the lines are tangent to the circle then a=b=0
The lines are tangent and the angle is not given, as it is determined by R and h. The "towers" are perpendicular to the "ground".
I'm not sure if I'm interpreting this correctly... I have the impression that if the two 'towers' are the same height (h) and the lines are tangent to the circle then a=b=0
If you don't have trouble imagining it in three dimensions, the question is basically asking...
Suppose you are standing at the top of a tall tower and can barely see the top of an identical tower some miles away. If you were to climb a spire on the top of your tower, how far down the other tower would you be able to see?Amaraticando wrote:
The lines are tangent and the angle is not given, as it is determined by R and h. The "towers" are perpendicular to the "ground".
I see that now. I'll add one more line to my solution above.
I was thinking of flat earth bullshit and this problem came to my mind. The answer that I got was pretty big and complicated for a seemingly simple problem, let's see if some of you finds out.
Click to expand.
Let h1=h+a and h2=h-b. Then
cos-1 (r/(r+h1)) + cos-1 (r/(r+h2)) = 2 cos-1 (r/(r+h))
based on the angle subtended by the line segment between the tower and the horizon from the center of the earth, for each of the towers h (twice), h1, and h2. Solving for h2:
h2 = (r/D)-r
where D=cos( 2 cos-1 (r/(r+h)) - cos-1 (r/(r+h1)) )
Then take b=h-h2.
If you want D in terms of algebraic elements, then using that cos(cos-1(n))=n when -1≤n≤1 and sin(cos-1(p/q))=sqrt(q2-p2)/q when 0≤p≤q gives:
D= {2(r/(r+h))2-1}{r/(r+h1)} + 2{(sqrt(h2+2hr)/(r+h))(r/(r+h))}{sqrt(h12+2h1r)/(r+h1)}
All this problem is, is what Bobo the King said above with the ratio tending to 3:2:1, and the negative binomial distribution applied to the expected number of total drops required to get one of each gem.
If probability of success is p and probability of failure is q=1-p, then the expected number of trials required to obtain success is 1(p)+2(qp)+3(q2p)+... = p(1+2q+3q2+...) . Using the identity 1/(1-x)2 = 1+2x+3x2+... , the expected number is p(1/(1-q)2)=p/p2=1/p. Ex. The expected number of rolls of a 6-sided die required to roll a 1 is 1/(1/6)=6.
To calculate the expected number of total drops required to get one of each gem, condition on the order in which the types of gems are first encountered. That is, we calculate:
1 + P(Common)*{E(# of drops to get Uncommon or Rare) + P(Uncommon|Uncommon or Rare)*E(# of drops to get Rare) + P(Rare|Uncommon or Rare)*E(# of drops to get Uncommon)}
+ P(Uncommon)*{E(# of drops to get Common or Rare) + P(Common|Common or Rare)*E(# of drops to get Rare) + P(Rare|Common or Rare)*E(# of drops to get Common)}
+ P(Rare)*{E(# of drops to get Common or Uncommon) + P(Common|Common or Uncommon)*E(# of drops to get Uncommon) + P(Uncommon|Common or Uncommon)*E(# of drops to get Common)}
=
1 + (1/2)*{2 + (2/3)*6 + (1/3)*3} + (1/3)*{3/2 + (3/4)*6 + (1/4)*2} + (1/6)*{6/5 + (3/5)*3 + (2/5)*2}
=
1 + (1/2)*7 + (1/3)*(13/2) + (1/6)*(19/5) = 1 + 7/2 + 13/6 + 19/30 = 30/30 + 105/30 + 65/30 + 19/30 = 219/30 = 7.3
Since the ratio of the types of gems tends to 3:2:1, the expected number of common, uncommon, and rare gems, respectively is 7.3/2 = 3.65, 7.3/3 = 2.43333... , and 7.3/6 = 1.21666... . This is consistent with the numbers given by OmnipotentEntity's program.
Bravo, FractalFusion! I knew it had to do with the negative binomial distribution but I couldn't quite put all the pieces together. Your explanation is clear and it appears to be correct.
I used a very complicated tree-branching model, looking at the chances that a sequence of a certain type of gem would be terminated by each of the other gems, and then I got 4.7 common gems, 103/36 (about 2.861) uncommon gems, and 299/225 (about 1.329) rare gems; I wasn't confident in the model, and I had tinkered with it from the original one, which kept saying less than one rare gem on average.
The idea I had was to look at the average length of a sequence of one type of gem, then add on the average length of a sequence of that first type and a second type, which begins with that second type, and look at the average proportions of the gems in each two-type sequence, and I ended up using six cases: CU, CR, UC, UR, RC, and RU.
Suppose there's a round-robin tournament of 21 players, where each player plays each other player once (ie. 210 games played in total). Each game can end in win, loss or draw. Each player gets one point for a win, and half a point for a draw.
If one of the players draws every single one of his games (with all the other players winning/losing at least one of their games), this player will end up exactly at position 11 (ie. exactly in the middle). True or false?
False. By having every other player win nine games and tie a second game, the player you specify (who ties all of their games) can be in a 21-way tie for first place.
I am still the wizard that did it.
"On my business card, I am a corporate president. In my mind, I am a game developer. But in my heart, I am a gamer." -- Satoru Iwata
<scrimpy> at least I now know where every map, energy and save room in this game is
Another way to show this is false, without the distinguished player being in a tied place:
one player beats every other player except our distinguished player. All other matches are ties.
The player who wins gets 19.5 points, taking first place.
Our hero gets 10 points, taking second place.
The 19 other players gets 9.5 points each, taking third equal.
Each game rewards a total of 1 point amongst the players, thus 210 points total. A person who draws all will score 20x0.5=10 points, somebody who has 9 wins, 10 losses, and 1 draw scores 9x1 + 10x0 + 1x0.5=9+0+0.5=9.5 points. Clearly that's not the same score, nor can everybody else have that same value either. That would give total score 19x9.5 + 1x10 =190.5=/=210.
For now, start with every game being a draw, thus with scores {10,10,10,10,....,10}. In order for player P to not be in the middle, we need redistribute points such that at least 11 people have higher scores. Obviously just subtract from those lower than P, add to those higher than P. EG
{8.5, 9.5, 9.5, ... , P=10, 10.5, 10.5, ... , 10.5}
Remember that we cannot uniformly distribute these points, as there's only 9 people left of P, 11 right of P, thus we need 2 additional reductions in order to elevate 11 people above P. This would give 8.5 + 8x9.5 + 1x10 + 11x10.5=210 as required. This gives 11 people beating P, thus player P being in 12th place.
Intuitively one might think that drawing all games would end up squarely in the middle (especially if using a tie breaking system like the Sonneborn-Berger score), but that is indeed not so. Here is a results crosstable of a situation where player T has drawn all of his games, yet is in position 20:
Tagging for spoilers because I don't want to ruin other people's fun. FYI, I've marked the spoilers paragraph by paragraph, so you can follow along or get hints as need be.
Let the bread be 2 units by 2 units. Place the origin of the coordinate system at the bread's center and consider only the first quadrant. In fact, I'll only consider the portion of the first quadrant below the line y=x.Use polar coordinates to define the boundary between the eaten and uneaten bread. Because we are below y=x, the right edge is clearly going to be closer to a point on this boundary than the top edge. The boundary is such that the distance to the origin (r) is equal to the distance to the right edge (1 - r*cos(theta)).We have r = 1 - r*cos(theta). This can be rearranged to obtain r = 1/(1+cos(theta)).One-eighth of the area that's eaten is equal to the integral of this curve with respect to theta from 0 to pi/4. Because we're finding the area in polar coordinates, the integrand is 1/2*r*r*dtheta = 1/2*r^2*dtheta.We have the integral from 0 to pi/4 of 1/2*r^2*dtheta which is 1/2 times the integral from 0 to pi/4 of 1/(1+cos(theta))^2 dtheta.Plug this integral into Wolfram Alpha. What? Did you seriously expect me to do that integral myself? See for yourself. It's a monster and it requires an obscure u substitution of tan(theta/2) to put it into a solvable form.The definite integral (excluding the factor of 1/2) is (4*sqrt(2) - 5)/3.To find the fraction of the area that's eaten, multiply this by 1/2, then multiply by 8 because we only found one-eighth the area eaten, then divide by the total area of 4. This is 1/2*8/4 = 1, so just multiply by 1.The fraction eaten is (4*sqrt(2) - 5)/3 which is approximately 0.21895.
I would've thought the bread you eat ends up looking like this, in which case the answer is 1/4.
But that only considers distance in the direction to the origin, not how the sides of the square are closer than the corners are, so the corners need to be rounded off too.
Edit, here's my attempt with Cartesian coordinates rather than Polars. Starting with the upper right quadrant as before, calculating the lower half, thus an eighth of total area. For a given (x,y), we have the distance to the origin being r=(x2+y2)1/2, and the distance to the right hand edge being (1-x). Thus solving for r=1-x gives y=(1-2x)1/2. This is valid only until the top edge becomes closer, IE until y=x. Therefore this intersects with the line y=x at x=(1-2x)1/2, x=-1 +- 21/2, so that's where this eighth of the total area ends. Call this location L.
We therefore have the area just being under the y=x curve until point x=L, where it then turns into the weird curve y=(1-2x)1/2 until x=1/2. Thus we have total area
A= Int{x, x=0, L} + Int{(1-2x)1/2,x=L,1/2}
= L2/2 + ( -1/3.(1-2x)3/2 , x=L, 1/2 )
=(3-2.21/2)/2 + (0 + 1/3.(3-2.21/2)3/2)
=(3-2.21/2)/2 + (5.21/2-7)/3
=(9-6.21/2 )/6 + (10.21/2-14 )/6
=(-5 + 4.21/2 )/6 =~ 0.11.
Thus total area 8A~=0.88, thus ratio 8A/4=2A=0.218951416, as before.
But the question says (ambiguously, I suppose) "You hate the crust so much that you’ll only eat the portion of the sandwich that is closer to its center than to its edges", which suggests to me that your graphical interpretation is correct, Flip. Anything closer to the crust than (r/2, r/2) is "closer to an edge than the center" so you get that smaller square that is 1/4 the area of the sandwich.
Which suggests to me that the more general case is that you get to eat 1/n of an n-sided sandwich.
I am still the wizard that did it.
"On my business card, I am a corporate president. In my mind, I am a game developer. But in my heart, I am a gamer." -- Satoru Iwata
<scrimpy> at least I now know where every map, energy and save room in this game is
But the question says (ambiguously, I suppose) "You hate the crust so much that you’ll only eat the portion of the sandwich that is closer to its center than to its edges", which suggests to me that your graphical interpretation is correct, Flip. Anything closer to the crust than (r/2, r/2) is "closer to an edge than the center" so you get that smaller square that is 1/4 the area of the sandwich.
Which suggests to me that the more general case is that you get to eat 1/n of an n-sided sandwich.
I don't get it. Taking the width of the bread to be 2 units, the corners of the inner boundary are a distance of 1 unit from the outer edge and a distance of sqrt(2) units from the origin. I came up with a picture similar to Flip's when I started this problem, but quickly saw that it couldn't work. In what reasonable sense are those edges equidistant from the center and the crust?
It's early, and reading it again, I think you're right about this.
(I really should stop posting things at 5 AM.)
I am still the wizard that did it.
"On my business card, I am a corporate president. In my mind, I am a game developer. But in my heart, I am a gamer." -- Satoru Iwata
<scrimpy> at least I now know where every map, energy and save room in this game is
The extension to arbitrary regular polygons is obvious, and the limit as the polygons approach circles is also obvious.What is a little less obvious is the question of which shape is most efficient: Something to note, however, is that if, for any point on the edge of the eaten portion, the segment from the center to the edge of the sandwich passing through this point is not bisected by this point, that means the sandwich is not at the most efficient shape.For example, in the polygonal case, the corners could be rounded off with circular arcs so that the eaten region is the same, but the uneaten region is smaller.Also, it seems like you don't need to have the initial requirement that distances are measured with respect to the centroid, to end up with the most efficient shape as if you do have this initial requirement.I think a rigorous proof of this would require a similar strategy to the isoperimetric problem.
Click to expand.
-- WolframAlpha
But that only considers distance in the direction to the origin, not how the sides of the square are closer than the corners are, so the corners need to be rounded off too.
Edit, here's my attempt with Cartesian coordinates rather than Polars. Starting with the upper right quadrant as before, calculating the lower half, thus an eighth of total area. For a given (x,y), we have the distance to the origin being r=(x2+y2)1/2, and the distance to the right hand edge being (1-x). Thus solving for r=1-x gives y=(1-2x)1/2. This is valid only until the top edge becomes closer, IE until y=x. Therefore this intersects with the line y=x at x=(1-2x)1/2, x=-1 +- 21/2, so that's where this eighth of the total area ends. Call this location L.
We therefore have the area just being under the y=x curve until point x=L, where it then turns into the weird curve y=(1-2x)1/2 until x=1/2. Thus we have total area
A= Int{x, x=0, L} + Int{(1-2x)1/2,x=L,1/2}
= L2/2 + ( -1/3.(1-2x)3/2 , x=L, 1/2 )
=(3-2.21/2)/2 + (0 + 1/3.(3-2.21/2)3/2)
=(3-2.21/2)/2 + (5.21/2-7)/3
=(9-6.21/2 )/6 + (10.21/2-14 )/6
=(-5 + 4.21/2 )/6 =~ 0.11.
Thus total area 8A~=0.88, thus ratio 8A/4=2A=0.218951416, as before.