538 Riddler Express
April 11th, 2020 at 9:11:58 PM
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I suggest we have a separate thread for the 538.com math puzzles. Here is the one for this week:
Link: Can You Catch The Free T-Shirt?
Here is my answer. If some of the other math geniuses on the forum agree, I'll explain my solution.
Quote: 538.comThe rows of seats in the audience are all on the same level (i.e., there is no incline), they are numbered 1, 2, 3, etc., and the T-shirts are being launched from directly in front of Row 1. Assume also that there is no air resistance (yes, I know, that’s a big assumption). You also happen to know quite a bit about the particular model of T-shirt cannon being used — with no air resistance, it can launch T-shirts to the very back of Row 100 in the audience, but no farther.
The crew member aiming in your direction is still figuring out the angle for the launch, which you figure will be a random angle between zero degrees (straight at the unfortunate person seated in Row 1) and 90 degrees (straight up). Which row should you be sitting in to maximize your chances of nabbing the T-shirt?
Link: Can You Catch The Free T-Shirt?
Here is my answer. If some of the other math geniuses on the forum agree, I'll explain my solution.
Row 100. I figure the shirt will land there if the angle is between about 41.4 and 48.6. This represents about a 7.2% chance. It would take some integration to get the exact probability, but I'm pretty sure row 100 is right.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 11th, 2020 at 11:42:55 PM
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Quote: WizardI suggest we have a separate thread for the 538.com math puzzles. Here is the one for this week:
Link: Can You Catch The Free T-Shirt?
Here is my answer. If some of the other math geniuses on the forum agree, I'll explain my solution.Row 100. I figure the shirt will land there if the angle is between about 41.4 and 48.6. This represents about a 7.2% chance. It would take some integration to get the exact probability, but I'm pretty sure row 100 is right.
It all depends on the initial speed of the T-shirt and the shooting angle, and then the trajectory, Y = Xtanθ - gX^2/(2v^2)/(cosθ)^2 (it can be defined easily) . . .
When Y =0, X = 2 * v^2 * cosθ * sinθ/g = v^2 * sin2θ/g
Xmax = v^2 /g, when θ = 45 degree
If the initial speed of T-shirt is very high, definitely it will reach row 100.....
If no air resistance, then 1/2mv^2 = mgh....
AM I MISSING SOMETHING ? Please explain further. Thnaks
Last edited by: ssho88 on Apr 12, 2020
April 12th, 2020 at 9:06:01 AM
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Quote: ssho88It all depends on the initial speed of the T-shirt and the shooting angle, and then the trajectory, Y = Xtanθ - gX^2/(2v^2)/(cosθ)^2 (it can be defined easily) . . .
We're told we can ignore air resistance and the maximum row the t-shirt can reach is 100. That should be enough to fill in the unknowns.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 12th, 2020 at 10:45:42 AM
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Assuming each row is 1m apart, so Xmax = 100, then X = 100sin2θ, if we plot the graph, the max distance(row 100) area will cover "WIDEEST" range of angle near θ=45 degree(peak of the graph).
1) scenario 1, audience sit at row 100 will get the T-shirt if 99.5 <X<100, so 42.13 degree < θ < 45 degree, the θ range = 2.87 degree.
2) scenario 2, audience sit at row 99 will get the T-shirt if 98.5 <X<99, so 40.03 degree < θ < 40.95 degree, the θ range = 0.92 degree.
Therefore, audience sit at row 100 have more than 3 times higher chance to catch the T-shirt compare to audience sit at row 99.
So row 100 should have highest θ range compare to any other row.
EDITED.
After some thoughts, I think it can be proved this way, X = 100sin2θ, dX/dθ = 2cos2θ, for a fix value of dX, the higher the dθ, the lower the dX/dθ, but we know that the hgiher the dθ, the higher the probability to catch the T-shirt. Therefore, probability become maximum when dX/dθ is minimum, from the graph, the minimum value of dX/dθ = 0, 2cos2θ=0, θ=45 degree, X = 100 * sin90 = 100. So audience sit at row 100 have highest probability.
Last edited by: ssho88 on Apr 12, 2020
April 12th, 2020 at 11:18:34 AM
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Assume the rows are equally spaced apart, and the cannon is at "row 0".
Let V be the velocity of the cannon, and X be the angle.
The time that the T-shirt is in the air is the solution to 0 = (V sin X) t - 1/2 g t^2
t = 2V/g sin X
The horizontal distance = t cos X = 2V/g sin X cos X = V/g sin 2X
The maximum horizontal distance is where X = 45 degrees -> sin X = cos X = sqrt(2)/2
Express distances in terms of rows:
100 = V/g sin 2X = V/g -> V = 100g
The horizontal distance = 100 sin 2X
It will land in row N if the horizontal distance is between N - 1/2 and N + 1/2
N - 1/2 < 100 sin 2X < N + 1/2
(N - 1/2) / 100 < sin 2x < (N + 1/2) / 100
arcsin ((N - 1/2) / 100) < 2X < arcsin((N + 1/2) / 100)
arcsin ((N - 1/2) / 100) / 2 < X < arcsin((N + 1/2) / 100) / 2
For what N in {1, 2, ..., 99} is (arcsin((N + 1/2) / 100) - arcsin((N + 1/2) / 100)) / 2 a maximum, and are any of them less than (arcsin 1 - arcsin 0.995)?
I needed a little help with Excel for this one, but the maximum is at N = 100, so sit in row 100.
April 13th, 2020 at 1:55:51 AM
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Another interesting finding, audience at Row 100 have 20 times higher chance to catch the T-shirt compare to Row 1 !
April 13th, 2020 at 9:39:08 PM
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No secret insights here to hide behind spoilers, just a practical consideration. If (a) the canon has a launch speed capable of reaching row #100, (b) I have a seat in row #1, and (c) the shooter aims at very nearly zero loft, then I suspect that the in-flight time until the projectile passes my row will be so short that I would have no hope of having a reaction speed capable of snagging the shirt. My only hope of being a winner would be for the shooter to aim almost vertical so I get to watch the straight-up-and-down trajectory and wait for it to land in my lap.
I think some aspect of this reaction time factor may influence the optimal seat selection and not just calculations of how far a shirt will fly.
I think some aspect of this reaction time factor may influence the optimal seat selection and not just calculations of how far a shirt will fly.
April 14th, 2020 at 4:46:51 AM
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We are not talking about reaction of snagging but landing infront of the audience, and equal chance for any shooting angle from 0 to 90 degrees.
April 14th, 2020 at 6:02:59 AM
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I've assumed the shooter is on the 0 foot line and the furthest the cannon can fire is to land on the 100 foot line (I accept that chairs being 1ft apart is unlikely, but then a stadium having 100 rows is!).
As has been said the angle that shoots the furthest is 45o and the distance is K sin(2*45).
Thus if the value of sin(x) is between 0 and 0.01, seat #1 gets the t-shirt. As the derivative of sin is cos, then when the angle is small x increases fairly fast for a change in angle. Thus earlier seats have a smaller angle range that covers their seat compared to later seats. Thus earlier seats have less chance that seats "behind" them.
As has been said the angle that shoots the furthest is 45o and the distance is K sin(2*45).
Thus if the value of sin(x) is between 0 and 0.01, seat #1 gets the t-shirt. As the derivative of sin is cos, then when the angle is small x increases fairly fast for a change in angle. Thus earlier seats have a smaller angle range that covers their seat compared to later seats. Thus earlier seats have less chance that seats "behind" them.
1 | 0.637% |
2 | 0.637% |
3 | 0.637% |
4 | 0.637% |
5 | 0.637% |
6 | 0.638% |
7 | 0.638% |
8 | 0.638% |
9 | 0.639% |
10 | 0.640% |
11 | 0.640% |
12 | 0.641% |
13 | 0.642% |
14 | 0.643% |
15 | 0.643% |
16 | 0.644% |
17 | 0.645% |
18 | 0.647% |
19 | 0.648% |
20 | 0.649% |
21 | 0.650% |
22 | 0.652% |
23 | 0.653% |
24 | 0.655% |
25 | 0.657% |
26 | 0.658% |
27 | 0.660% |
28 | 0.662% |
29 | 0.664% |
30 | 0.666% |
31 | 0.668% |
32 | 0.671% |
33 | 0.673% |
34 | 0.676% |
35 | 0.678% |
36 | 0.681% |
37 | 0.684% |
38 | 0.687% |
39 | 0.690% |
40 | 0.693% |
41 | 0.696% |
42 | 0.700% |
43 | 0.703% |
44 | 0.707% |
45 | 0.711% |
46 | 0.715% |
47 | 0.719% |
48 | 0.723% |
49 | 0.728% |
50 | 0.733% |
51 | 0.738% |
52 | 0.743% |
53 | 0.748% |
54 | 0.754% |
55 | 0.759% |
56 | 0.765% |
57 | 0.772% |
58 | 0.778% |
59 | 0.785% |
60 | 0.792% |
61 | 0.800% |
62 | 0.807% |
63 | 0.816% |
64 | 0.824% |
65 | 0.833% |
66 | 0.843% |
67 | 0.852% |
68 | 0.863% |
69 | 0.874% |
70 | 0.885% |
71 | 0.898% |
72 | 0.911% |
73 | 0.924% |
74 | 0.939% |
75 | 0.954% |
76 | 0.971% |
77 | 0.989% |
78 | 1.007% |
79 | 1.028% |
80 | 1.050% |
81 | 1.073% |
82 | 1.099% |
83 | 1.127% |
84 | 1.157% |
85 | 1.191% |
86 | 1.228% |
87 | 1.269% |
88 | 1.315% |
89 | 1.368% |
90 | 1.428% |
91 | 1.497% |
92 | 1.579% |
93 | 1.676% |
94 | 1.796% |
95 | 1.948% |
96 | 2.150% |
97 | 2.434% |
98 | 2.879% |
99 | 3.743% |
100 | 9.011% |
April 14th, 2020 at 6:07:34 AM
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You have to choose your seat before the cannon is fired and must stay in that seat. In other words, no running after the cannon is fired to another seat. Reaction time should not play into it.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 14th, 2020 at 6:46:16 AM
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Shooting angle to reach seat 100 is 45 degree while shooting angle to reach seat 99, X = 100sin2θ, sin2θ = 99/100, θ =40.945.
dθ = 45 - 40.945 = 4.055, prob of seat 100 = 4.055/45 *100 = 9.01%.
dθ = 45 - 40.945 = 4.055, prob of seat 100 = 4.055/45 *100 = 9.01%.
April 15th, 2020 at 7:53:28 AM
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Assume the rows are equally spaced apart, and the cannon is at "row 0".
Let V be the velocity of the cannon, and X be the angle.
The time that the T-shirt is in the air is the solution to 0 = (V sin X) t - 1/2 g t^2
t = 2V/g sin X
The horizontal distance = t cos X = 2V2/g sin X cos X = V2/g sin 2X
Let y = 2X; the first derivative of V2/g sin y = V2/g cos y, which, for 0 < y < 180 (since 0 < X < 90), is 0 at y = 90 ->X = 45
Note that this is a maximum point, so the maximum horizontal distance = 100 = V2/g sin (2 * 45) = V2/g
For angle X, the horizontal distance = V2/g sin 2X = 100 sin 2X
Assume the T-shirt lands in row N if the distance is between N - 1/2 and N + 1/2 for N in {1, 2, 3, ..., 99}, and the distance is between 99.5 and 100 for N = 100
For N <= 99:
N - 1/2 < 100 sin 2X < N + 1/2
(N - 1/2) / 100 < sin 2X < (N + 1/2)
Since arcsin > 0 for all 2X in the possible values, taking the arcsin of all three values does not change the inequality signs:
arcsin ((N - 1/2) / 100) / 2 < X < arcsin((N + 1/2) / 100) / 2
The range of X for a given N is (arcsin((N + 1/2) / 100) - arcsin ((N - 1/2) / 100)) / 2
The first derivative of arcsin x is 1 / sqrt(1 - x2) dx
Let f(N) = (arcsin((N + 1/2) / 100) - arcsin ((N - 1/2) / 100)) / 2
f'(N) = (1 / 100 * 1 / sqrt(1 - (N + 1/2)2 / 10,000) - 1 / 100 * 1 / sqrt(1 - (N - 1/2)2 / 10,000)) / 2
= (100 / sqrt(10,000 - (N + 1/2)2) - 100 / sqrt(10,000 - (N - 1/2)2)) / 200
= (sqrt(10,000 - (N + 1/2)2) - sqrt(10,000 - (N - 1/2)2)) / 2
This is strictly increasing as N increases, so for rows 1-99, the largest range is at N = 99
The question becomes, which is larger - the angle range for N = 99, or for N = 100?
For N = 99, it is (arcsin 199/200 - arcsin 197/200) / 2
For N = 100, it is 2 (arcsin 1 - arcsin 199/200) / 2 = arcsin 1 - arcsin 199/200
(it is multiplied by 2 to cover the ranges < 45 degrees and > 45 degrees)
I can't find a way to determine this off the top of my head other than calculating the values directly
N = 100 is 90 - 84.268 = 5.732 degrees
N = 99 is (84.268 - 80.064) / 2 = 2.102 degrees
Thus, N = 100 is the most likely result
April 15th, 2020 at 8:55:27 AM
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Shouldn't there be three variables?: speed of the shirt when it leaves the cannon, angle of the cannon, and height of the cannon. A 5' gymnast holding it at her hip is going to shoot the t-shirt a different distance than 7' center holding it over his head, even with the same launch angle. We can assume it's from a fixed height, but wouldn't you need to know what that height is? Or is my intersection of physics and calculus all messed up?
April 15th, 2020 at 8:58:56 AM
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Quote: TomGShouldn't there be three variables?: speed of the shirt when it leaves the cannon, angle of the cannon, and height of the cannon. A 5' gymnast holding it at her hip is going to shoot the t-shirt a different distance than 7' center holding it over his head, even with the same launch angle. We can assume it's from a fixed height, but wouldn't you need to know what that height is? Or is my intersection of physics and calculus all messed up?
Height of cannon = height of seats.
May 22nd, 2020 at 5:41:57 PM
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Official write up of the T-Shirt problem.. (new problem down below): You were seated in an audience, when T-shirts were being launched via cannon in your direction. The rows of seats in the audience were all on the same level, they were numbered 1, 2, 3, etc., and the T-shirts were being launched from directly in front of Row 1.
Additionally, there was no air resistance, and the particular model of T-shirt cannon being used was able to launch T-shirts to the very back of Row 100 in the audience, but no farther.
If the T-shirt was launched at a random angle between zero degrees (straight at the unfortunate person seated in Row 1) and 90 degrees (straight up), which row should you have been sitting in to maximize your chances of nabbing the T-shirt?
While some solvers took an analytical approach, there was no shortage of simulations. For example, if you were to launch a T-shirt at whole number values of degrees between zero and 90, and you recorded which rows they landed in, here’s what you’d find: (sorry, graph didn't copy over)
Each blue dot represents a T-shirt fired at a different angle, and the red bars show how many T-shirts landed in each row. (Some rows didn’t get any T-shirts because this simulation only used whole number values of the angles.)
Already, it appears that the last few rows received more T-shirts than the others. This result was confirmed by Eli Luberoff, Jason Shaw, Ravi Chandrasekaran and Angelos Tzelepis, the last of whom sampled the angles from zero to 90 degrees at 0.01-degree steps. Every single one of them found that Row 100 had the greatest chance of receiving a T-shirt.
So what’s going on here? As solver J. D. Roaden explained, it helped If we looked at a graph of launch angle θ versus row number R, which physics tells us are related by the equation R = 100·sin(2θ), shown by the blue curve in the graph below:
range of angles that will land in Row 100
Meanwhile, the vertical red bar shows the range in angles that will launch a T-shirt to Row 100. Because the blue curve is flattest at the top, that means larger variations in angle have relatively little effect on row number. And so sure enough, Row 100 corresponds to the widest swath of angles. Solver Jonah Majumder specifically found that the chances of nabbing the T-shirt in Row 100 were just over 9 percent. Not bad!
Finally, as an aside, this phenomenon (whereby the maximum possible value is more likely because that’s where it’s changing the least), also explains why rainbows exist. 🌈Seriously! 🌈
Here is this week's puzzle:
To share a cylindrical muffin equally with his two toddlers, Robert makes three vertical cuts in a “Y” pattern, producing three equal pieces.
The next morning, his wife wants in on the fun. But before he can cut the muffin into quarters with an “X” pattern, one of his children suggests using an “A” pattern. If Robert were to produce equal fourths in this manner, what would be the ratio of length of the A’s middle bar to the radius of the muffin?
I noodled around and got pi/2, but I could easily be wrong.
Additionally, there was no air resistance, and the particular model of T-shirt cannon being used was able to launch T-shirts to the very back of Row 100 in the audience, but no farther.
If the T-shirt was launched at a random angle between zero degrees (straight at the unfortunate person seated in Row 1) and 90 degrees (straight up), which row should you have been sitting in to maximize your chances of nabbing the T-shirt?
While some solvers took an analytical approach, there was no shortage of simulations. For example, if you were to launch a T-shirt at whole number values of degrees between zero and 90, and you recorded which rows they landed in, here’s what you’d find: (sorry, graph didn't copy over)
Each blue dot represents a T-shirt fired at a different angle, and the red bars show how many T-shirts landed in each row. (Some rows didn’t get any T-shirts because this simulation only used whole number values of the angles.)
Already, it appears that the last few rows received more T-shirts than the others. This result was confirmed by Eli Luberoff, Jason Shaw, Ravi Chandrasekaran and Angelos Tzelepis, the last of whom sampled the angles from zero to 90 degrees at 0.01-degree steps. Every single one of them found that Row 100 had the greatest chance of receiving a T-shirt.
So what’s going on here? As solver J. D. Roaden explained, it helped If we looked at a graph of launch angle θ versus row number R, which physics tells us are related by the equation R = 100·sin(2θ), shown by the blue curve in the graph below:
range of angles that will land in Row 100
Meanwhile, the vertical red bar shows the range in angles that will launch a T-shirt to Row 100. Because the blue curve is flattest at the top, that means larger variations in angle have relatively little effect on row number. And so sure enough, Row 100 corresponds to the widest swath of angles. Solver Jonah Majumder specifically found that the chances of nabbing the T-shirt in Row 100 were just over 9 percent. Not bad!
Finally, as an aside, this phenomenon (whereby the maximum possible value is more likely because that’s where it’s changing the least), also explains why rainbows exist. 🌈Seriously! 🌈
Here is this week's puzzle:
To share a cylindrical muffin equally with his two toddlers, Robert makes three vertical cuts in a “Y” pattern, producing three equal pieces.
The next morning, his wife wants in on the fun. But before he can cut the muffin into quarters with an “X” pattern, one of his children suggests using an “A” pattern. If Robert were to produce equal fourths in this manner, what would be the ratio of length of the A’s middle bar to the radius of the muffin?
I noodled around and got pi/2, but I could easily be wrong.
Last edited by: rsactuary on May 22, 2020
May 22nd, 2020 at 7:17:43 PM
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Here is this week's puzzle:
To share a cylindrical muffin equally with his two toddlers, Robert makes three vertical cuts in a “Y” pattern, producing three equal pieces.
The next morning, his wife wants in on the fun. But before he can cut the muffin into quarters with an “X” pattern, one of his children suggests using an “A” pattern. If Robert were to produce equal fourths in this manner, what would be the ratio of length of the A’s middle bar to the radius of the muffin?
I noodled around and got pi/2, but I could easily be wrong.
sin(2θ) = (1/2 - θ/90) *pi, where θ is in degree -------Eq1
ratio of middle bar length to the radius = (pi * tanθ)^0.5 ------Eq2
solve Eq1 to find θ, then replace it into Eq2, my estimate answer is 1.178
To share a cylindrical muffin equally with his two toddlers, Robert makes three vertical cuts in a “Y” pattern, producing three equal pieces.
The next morning, his wife wants in on the fun. But before he can cut the muffin into quarters with an “X” pattern, one of his children suggests using an “A” pattern. If Robert were to produce equal fourths in this manner, what would be the ratio of length of the A’s middle bar to the radius of the muffin?
I noodled around and got pi/2, but I could easily be wrong.
sin(2θ) = (1/2 - θ/90) *pi, where θ is in degree -------Eq1
ratio of middle bar length to the radius = (pi * tanθ)^0.5 ------Eq2
solve Eq1 to find θ, then replace it into Eq2, my estimate answer is 1.178
May 22nd, 2020 at 7:21:51 PM
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Quote: rsactuary...Here is this week's puzzle:
To share a cylindrical muffin equally with his two toddlers, Robert makes three vertical cuts in a “Y” pattern, producing three equal pieces.
The next morning, his wife wants in on the fun. But before he can cut the muffin into quarters with an “X” pattern, one of his children suggests using an “A” pattern. If Robert were to produce equal fourths in this manner, what would be the ratio of length of the A’s middle bar to the radius of the muffin?
I noodled around and got pi/2, but I could easily be wrong.
I get approximately pi / 2.667197. This agrees with ssho88.
May 22nd, 2020 at 7:34:49 PM
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Quote: ChesterDogI get approximately pi / 2.667197. This agrees with ssho88.
Could you(or anyone) please teach me how to solve the below equation ?
sin(2θ) = (1/2 - θ/90) *pi, where θ is in degrees
May 22nd, 2020 at 8:21:10 PM
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Quote: ssho88...Could you(or anyone) please teach me how to solve the below equation ?...
Yours is a "transcendental" equation. Here's a link to a Wikipedia page: https://en.wikipedia.org/wiki/Transcendental_equation
The page says, "...most equations where the variable appears both as an argument to a transcendental function and elsewhere in the equation are not solvable in closed form..."
When I put your equation into WolframAlpha, it gives a numerical solution, so probably there is no closed-form solution.
I like radians more than degrees, so here is your equation and solution using radians: your equation.
By the way, how did you make the symbol for "theta" in your post?
May 22nd, 2020 at 9:25:15 PM
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Quote: ChesterDogYours is a "transcendental" equation. Here's a link to a Wikipedia page: https://en.wikipedia.org/wiki/Transcendental_equation
The page says, "...most equations where the variable appears both as an argument to a transcendental function and elsewhere in the equation are not solvable in closed form..."
When I put your equation into WolframAlpha, it gives a numerical solution, so probably there is no closed-form solution.
I like radians more than degrees, so here is your equation and solution using radians: your equation.
By the way, how did you make the symbol for "theta" in your post?
Thanks
symbol for "theta" :-
https://www.google.com/search?q=theta+symbol+in+word&oq=theta&aqs=chrome.5.69i57j0l6j5.3125j0j7&sourceid=chrome&ie=UTF-8
Just copy and paste
May 23rd, 2020 at 8:08:05 AM
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Assume the radius is 1
Let M be the midpoint of AC
AOM and COM are side-side-side congruent, so angle AOM = angle COM;
AOM + COM = AM = 180, so AOM and COM are both right angles
Let x be the measure of angle AOM (and COM) in degrees (I couldn't figure out how to enter lowercase theta in Photoshop Elements)
AM = sin x; OM = cos x
The area of the "wedge" of the circle bounded by AO and CO is PI * 2x / 360 = PI x / 180
The area of triangle AOC is 1/2 (2 sin x) (cos x) = 1/2 sin 2x
The area of the region bounded by the circle and AC is PI x / 180 - 1/2 sin 2x
Since this is a piece of the cake, it must be 1/4 of the total area, or PI / 4:
PI x / 180 - 1/2 sin 2x = PI / 4
PI (x - 45) / 90 = sin 2x
Using Excel approximation, I get x = 66.17322941 degrees
Let h be the height of triangle DAE measured from A
tan x = h / (DE / 2)
h = (DE tan x) / 2
The area of DAE = 1/2 DE h = 1/4 (DE)^2 tan x
This is also a piece of the cake, so its area is also PI / 4:
1/4 (DE)^2 tan x = PI / 4
(DE)^2 = PI cot x
DE = about 1.177863
Since the radius is 1, this is the ratio of the bar of the "A" cut to the radius
June 6th, 2020 at 7:18:25 AM
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This week's easier puzzle: While normally I at least have an idea how to proceed, even if I can't get to an answer, this week I have no idea.....
The astronomers on planet Xiddler have made several remarkable discoveries. After inventing the telescope, they quickly discovered a new planet in their solar system!
Xiddler is very much like Earth. The planet orbits its star in a nearly circular path, with an average distance of 150 million kilometers, a period of one Earth year and a day that lasts 24 hours. But unlike Earth, there weren’t any other known planets in the solar system…until now.
Moments after the Xiddlerian sun set below the horizon, three astronomers happened to focus their telescopes at the zenith of the evening sky, all seeing the same new planet. In their excitement, the astronomers race to Xiddler’s Grand Minister to deliver the momentous news.
The first astronomer says that, by her calculations, the newly discovered planet orbits their sun with a radius of 50 million kilometers. The second astronomer says that, by her calculations, the planet in fact orbits their sun with a radius of 300 million kilometers. The third astronomer disagrees with the other two — by her calculations, the planet has a very similar orbit to Xiddler, with a radius of 150 million kilometers.
Which astronomer should the Grand Minister believe?
The astronomers on planet Xiddler have made several remarkable discoveries. After inventing the telescope, they quickly discovered a new planet in their solar system!
Xiddler is very much like Earth. The planet orbits its star in a nearly circular path, with an average distance of 150 million kilometers, a period of one Earth year and a day that lasts 24 hours. But unlike Earth, there weren’t any other known planets in the solar system…until now.
Moments after the Xiddlerian sun set below the horizon, three astronomers happened to focus their telescopes at the zenith of the evening sky, all seeing the same new planet. In their excitement, the astronomers race to Xiddler’s Grand Minister to deliver the momentous news.
The first astronomer says that, by her calculations, the newly discovered planet orbits their sun with a radius of 50 million kilometers. The second astronomer says that, by her calculations, the planet in fact orbits their sun with a radius of 300 million kilometers. The third astronomer disagrees with the other two — by her calculations, the planet has a very similar orbit to Xiddler, with a radius of 150 million kilometers.
Which astronomer should the Grand Minister believe?
June 6th, 2020 at 7:19:33 AM
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My guess is that it would have to be 300M radius. I can't seem to come up with a scenario for the other two where they would have never seen the planet before. But that's not very mathematical.
June 6th, 2020 at 8:15:55 AM
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I guess 150M
Closer to the sun/star would be visible only during the day, further would be visible only at night. Becoming visible right when night falls would be the same distance
Total guess from non-astronomer
Closer to the sun/star would be visible only during the day, further would be visible only at night. Becoming visible right when night falls would be the same distance
Total guess from non-astronomer
It’s all about making that GTA
June 6th, 2020 at 8:52:33 AM
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300. When the sun is just setting, the zenith of the evening sky is pointing at the line tangent to the orbit of the planet. No other planet with an equal or smaller orbit could be on that tangent line, only a planet with a larger orbit.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
June 6th, 2020 at 10:10:00 AM
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I have no idea if this is the right answer, but sounds the most mathematical.
As I thought about it some more, I realized there is no mention of the speed of the other planets, so the answer has to be something unrelated to speed and where the other planets are in their orbit. Makes the answer by unJon sound correct.
As I thought about it some more, I realized there is no mention of the speed of the other planets, so the answer has to be something unrelated to speed and where the other planets are in their orbit. Makes the answer by unJon sound correct.
June 7th, 2020 at 3:07:43 PM
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300. As the sky just darkens they are seeing the face of a planet "lit" by the sun. That can only be the case if the planet is further from the sun.
Two planets cannot share a circular orbit at the same distance from the sun, unless they are 180 degrees apart -at exactly opposite poles of the orbit. And in that case they could not see each other -because the sun would block the line of view between the two planets.
Two planets cannot share a circular orbit at the same distance from the sun, unless they are 180 degrees apart -at exactly opposite poles of the orbit. And in that case they could not see each other -because the sun would block the line of view between the two planets.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
June 7th, 2020 at 5:44:32 PM
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Quote: gordonm888300. As the sky just darkens they are seeing the face of a planet "lit" by the sun. That can only be the case if the planet is further from the sun.
Two planets cannot share a circular orbit at the same distance from the sun, unless they are 180 degrees apart -at exactly opposite poles of the orbit. And in that case they could not see each other -because the sun would block the line of view between the two planets.
You can see Venus just as the sky darkens. Just not at the zenith of the sky ever!
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
June 8th, 2020 at 1:41:16 AM
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Mathematical approach
"Zenith" means looking directly upwards, so I don't know whether that means mathematically the other planet is considered as a point that is strictly on a line directly above the observer.
Suppose Xiddler has its pole always pointing along the path of the planet, and (say) London is at sunset, then all those on 0o would be seeing sunset (and 180o sunrise). Thus the points an observer having sunset form a half-disk that is along ther tangent of the planet's trajectory. So all points on that disk lie further from the Sun, so anything visible must be further from the Sun. This suggests the 300m answer as everything else is inside.
Suppose Xiddler has it pole at an angle to the path of itself. This means the line of sunsets is a curve, similar to ones we have on Earth. The question is whether this means someone could, at times, see another planet on the same orbit as ours. The "curve" on out Earth is only an illusion since it may form a curve on the map but actually is a great circle. This great circle will form a similar half-disk to the one discussed above.
As a Mathematical puzzle the answer is 300 (as you're not allowed to consider how the other planet manages to achieve its orbit).
Physics
My feeling was that for two things to have identical time round the Sun they would have to be at the same distance. There are some interesting articles that suggest this is possible, with some stable places and other unstable places. Also some of the unstable orbits are at a different distance but the same period.
According to Forbes there are even two moons (Saturn - Epimetheus and Janus) which swap orbits with one overtaking the other every two years.
https://www.forbes.com/sites/startswithabang/2019/08/13/yes-two-planets-can-both-share-the-same-orbit/#d380c322b8de
https://www.newscientist.com/article/dn20160-two-planets-found-sharing-one-orbit/
https://en.wikipedia.org/wiki/Lagrangian_point
Lagrangian Points
There are some points where it is possible for another planet to be in the same period. Some are stable and other unstable.
The stable ones are ahead (or behind) you forming an equilateral triangle. So the other planet would be at 60o to the Sun, so never directly above you at sunset (assuming moments after means a small time).
The unstable ones are on the extended line between you and the Sun, so either are towards the Sun (between or beyond) and opposite the Sun (so always in the shadow of the Sun, although you might be able to see it with a good telescope).
So there could be some planets but I don't think they're visible at sunset.
Other non mathematical solutions
I like the Saturn solution that the two go round together and occasionally swap places. This would be the first time they've passed each other. The answer would then be 150.
I also like the idea that the two planets have been in the same orbit L4 or L5 and somehow you can see it at sunset. The answer would then be 150.
Of course in the diagram we might at a Lagrange point and be able to see base planet (you can only have three using the logic earlier) but again it would still be on the extended line between us and the Sun or at 60o.
Finally there was mention that there might be other configurations where, due to the relative masses, the three go round each other (there is nothing that says their Sun is stationary), which would mean the 300 answer is technically possible.
Summary
Thus I think the practical answer is 300 and somehow they've found a way to be in perfect harmony together. (c.f. Casino Royale film)
"Zenith" means looking directly upwards, so I don't know whether that means mathematically the other planet is considered as a point that is strictly on a line directly above the observer.
Suppose Xiddler has its pole always pointing along the path of the planet, and (say) London is at sunset, then all those on 0o would be seeing sunset (and 180o sunrise). Thus the points an observer having sunset form a half-disk that is along ther tangent of the planet's trajectory. So all points on that disk lie further from the Sun, so anything visible must be further from the Sun. This suggests the 300m answer as everything else is inside.
Suppose Xiddler has it pole at an angle to the path of itself. This means the line of sunsets is a curve, similar to ones we have on Earth. The question is whether this means someone could, at times, see another planet on the same orbit as ours. The "curve" on out Earth is only an illusion since it may form a curve on the map but actually is a great circle. This great circle will form a similar half-disk to the one discussed above.
As a Mathematical puzzle the answer is 300 (as you're not allowed to consider how the other planet manages to achieve its orbit).
Physics
My feeling was that for two things to have identical time round the Sun they would have to be at the same distance. There are some interesting articles that suggest this is possible, with some stable places and other unstable places. Also some of the unstable orbits are at a different distance but the same period.
According to Forbes there are even two moons (Saturn - Epimetheus and Janus) which swap orbits with one overtaking the other every two years.
https://www.forbes.com/sites/startswithabang/2019/08/13/yes-two-planets-can-both-share-the-same-orbit/#d380c322b8de
https://www.newscientist.com/article/dn20160-two-planets-found-sharing-one-orbit/
https://en.wikipedia.org/wiki/Lagrangian_point
Lagrangian Points
There are some points where it is possible for another planet to be in the same period. Some are stable and other unstable.
The stable ones are ahead (or behind) you forming an equilateral triangle. So the other planet would be at 60o to the Sun, so never directly above you at sunset (assuming moments after means a small time).
The unstable ones are on the extended line between you and the Sun, so either are towards the Sun (between or beyond) and opposite the Sun (so always in the shadow of the Sun, although you might be able to see it with a good telescope).
So there could be some planets but I don't think they're visible at sunset.
Other non mathematical solutions
I like the Saturn solution that the two go round together and occasionally swap places. This would be the first time they've passed each other. The answer would then be 150.
I also like the idea that the two planets have been in the same orbit L4 or L5 and somehow you can see it at sunset. The answer would then be 150.
Of course in the diagram we might at a Lagrange point and be able to see base planet (you can only have three using the logic earlier) but again it would still be on the extended line between us and the Sun or at 60o.
Finally there was mention that there might be other configurations where, due to the relative masses, the three go round each other (there is nothing that says their Sun is stationary), which would mean the 300 answer is technically possible.
Summary
Thus I think the practical answer is 300 and somehow they've found a way to be in perfect harmony together. (c.f. Casino Royale film)
June 8th, 2020 at 3:53:19 AM
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Charlie, love your answers! But I’m missing the part of the puzzle that says the two planets have the same period to revolve around the sun? It compares Xiddler to Earth, but Earth isn’t one of the two planets.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
June 8th, 2020 at 4:21:08 AM
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I was probably confused by the other comments, in which case the answer is obviously 300 as you had stated earlier.Quote: unJonCharlie, love your answers! But I’m missing the part of the puzzle that says the two planets have the same period to revolve around the sun? It compares Xiddler to Earth, but Earth isn’t one of the two planets.
June 8th, 2020 at 6:19:59 AM
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Charliepatrick, thanks for mentioning the LaGrange points. I didn't feel like yet another argument with Gordon...lol. I did think there were commonly understood to be 6 of them, though each 60° around the circular orbit, and you have them at 120°. Has that been re-calculated for the maximum stable orbits, or are we just reading different things?
If the House lost every hand, they wouldn't deal the game.
June 8th, 2020 at 10:12:59 AM
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I'd never heard of them before. There's another great article at NASA where they show the five points and actually use the L1 and L2 ones to place telescopes etc. (it saves fuel as they only need to adjust occasionally to keep near the LaGrange point). They also explain that Trojans can be found at L4 and L5, e.g. the Earth has an asteroid at its leading point.Quote: beachbumbabsCharlie...are we just reading different things?
https://solarsystem.nasa.gov/resources/754/what-is-a-lagrange-point/
