Can a monkey outrun a bullet and still save her litter?

  • Thread starter powergirl
  • Start date
If you drive me to the airport.2. I am afraid of going up in elevators.3. Drinking cocoa keeps me warm on long winter nights.4. I hope the map leads us to buried treasure.5. "Eat another bonbon," said our charming hostess.6. Nepal may be the most interesting place I have ever visited.7. Remember to fold the map, please.8. I feel many lumps in this mattress.9. Word processing is not as useful as pens and paper for creative brainstorming.1. Willow2. Fir3. Oak4. Pine5. Chestnut6. Elm7. Maple8. Apple9. Ash
  • #176


unscientific said:
a bus conductor accidentally knocked down and killed 50 people while driving his bus in a drunken state.
The judge then sentenced him to death penalty by the electric chair.

On his execution date, he requested to have bananas before he died, and so the judge passed him a bundle of fresh bananas. Surprisingly the electric chair did not kill him.

So the next day, he was tried for execution again, and the same, he requested an apple, so the judge gave him a ripe apple. Amazingly, he survived.

Finally, the judge let him go.

Question: how did he survive?

My attempt:

I'm confused because it says "tried for execution" in the third paragraph, but says it was his "execution date" in the second paragraph. Is he being tried or executed?

The electric chair can't kill him if he's not sitting in it...
 
Physics news on Phys.org
  • #177


Need a hint.
My first suspicion is that the bananas and apple are red herrings.
Correct - Hint
What kind of conductor was he?


edit - damn now I'm hooked on this thread, as if the famous landmark one isn't bad enough.
 
  • #178


mgb_phys said:
Correct - Hint
What kind of conductor was he?

Wow, that's awful. It's not really a brain teaser, it's just a bad pun.

DaveE
 
  • #179


Pythagorean said:
My attempt:

Well, I thought you might be onto something for a second.

At first I thought 'it doesn't actually say he was convicted, or that he's going to the chair anytime soon, so it's not surprising he;s still alive (why wouldn't he be?)'

Except it does say 'on his execution date', which means he was, in fact, supposed to be executed on the first day and it didn't happen for some reason.
 
  • #180


I got a brilliant few puzzles here:

Hint: All the questions only have one solution.

1: I have a number that is less than one million. Putting a 1 after it makes it three times as large as putting a 1 before it. What is my number?

2: If Adam and Ben worked together to paint a house they would take 12 hours. Adam and Colin as a team would take 15 hours to paint the house and Ben and Colin as a pair would take 20 hours.
Assuming that the rate at which each painter works is not affected by whom he works with, how long would it take to paint the house if they all worked together?

3: Given any three-digit number, x, define f(x) to be
x minus the sum of the squares of the digits of x
What is the maximum possible value of f(x)?

4. A king has some (finite) number of subjects. One day he lines them all up in a row, facing towards the front of the row. He goes from the front of the row to the back, giving each subject a hat - either red or blue.

No subject can see his/her own hat, nor can any subject see behind himself/herself. But a subject can see ALL of the others that come before him/her in line.

So the king has made his way to the back of the line, and now he will ask the person at the back of the line "What color is your hat?" while brandishing a big sword. If answered correctly, the person is freed. If answered incorrectly, the person is beheaded immediately. No communication between subjects is allowed, although they can hear each others' guesses.

You have a chance to meet with the subjects ahead of time - the night before this incident - to give them a strategy that will save the most number of lives (guaranteed, not a probability or expected outcome).
 
Last edited:
  • #181


mgb_phys said:
Correct - Hint
What kind of conductor was he?
I still don't get it. Is it an electronics pun?
 
  • #182


DaveC426913 said:
I still don't get it. Is it an electronics pun?

The punch line (I looked this up elsewhere) goes something like:

A reporter asked him upon his release whether or not he had found some secret relating fruit to the electric chair.
"No," he replied, "I've just always been a bad conductor."

Very disappointing if you were looking for this to be a bona fide brain teaser.

DaveE
 
  • #183


Georgepowell said:
Hint: All the questions only have one solution.

1: I have a number that is less than one million. Putting a 1 after it makes it three times as large as putting a 1 before it. What is my number?

I already found 2 solutions.

Georgepowell said:
2: If Adam and Ben worked together to paint a house they would take 12 hours. Adam and Colin as a team would take 15 hours to paint the house and Ben and Colin as a pair would take 20 hours.
Assuming that the rate at which each painter works is not affected by whom he works with, how long would it take to paint the house if they all worked together?

Quick algebra yields that the answer is the 2nd digit of the answer to question 1 times the 4th digit of question 1.

Georgepowell said:
3: Given any three-digit number, x, define f(x) to be
x – the sum of the squares of the digits of x
What is the maximum possible value of f(x)?

I'm not sure I understand what the trick is here. Maximizing the sum is as easy as maximizing each term within the sum. Each term is a square of a digit of x, which means maximizing each digit's square. Maximizing a square of a digit simply means maximizing that digit. Hence, 999 = x, making f(x) = 9^2 + 9^2 + 9^2 = 243.

Georgepowell said:
4. A king has some (finite) number of subjects. One day he lines them all up in a row, facing towards the front of the row. He goes from the front of the row to the back, giving each subject a hat - either red or blue.

No subject can see his/her own hat, nor can any subject see behind himself/herself. But a subject can see ALL of the others that come before him/her in line.

So the king has made his way to the back of the line, and now he will ask the person at the back of the line "What color is your hat?" while brandishing a big sword. If answered correctly, the person is freed. If answered incorrectly, the person is beheaded immediately. No communication between subjects is allowed, although they can hear each others' guesses.

You have a chance to meet with the subjects ahead of time - the night before this incident - to give them a strategy that will save the most number of lives (guaranteed, not a probability or expected outcome).

I assume the question is "what is the strategy that guarantees the maximum number of lives saved"? This has been asked before here, but the answer is:
Each person in line counts the number of blue hats they see in front of them. God forbid any of them are colorblind. The last person in line (the first person to be chosen) says "blue" if the number he counts is odd, and "red" if it's even. He has a 50% chance to survive. Following this, each person can deduce the color of their hat. If the 2nd to last person's count was odd, and the last person's count was also odd, he knows he has a red hat. If the last person's count was even, then he knows he has a blue hat. And so forth down the line.

DaveE
 
  • #184


davee123 said:
Quick algebra yields that the answer is the 2nd digit of the answer to question 1 times the 4th digit of question 1.
[/SPOILER]DaveE
You sure about that? I don't have the answer, but even I know there's more to the answer than meets the eye.
The more painters, the less time it will take. i.e. <10.
 
  • #185


davee123 said:
I already found 2 solutions.
DaveE

What are they?... I don't think you have.
 
  • #186


DaveC426913 said:
You sure about that? I don't have the answer, but even I know there's more to the answer than meets the eye.

How so? That is, what's the more? I got an answer that works using simple algebra-- why ought I doubt it?

Georgepowell said:
What are they?... I don't think you have.

Well, I went outside the box. I was mildly annoyed that the question implied (but didn't stipulate) that it was "a positive integer less than one million". So I found the solution that I think was implied, then found another one assuming it was "a number less than one million".

[edit]Admittedly, in thinking about it, one could argue that the implications of placing a 1 before or after a number imply different mathematical functions-- in which case I could arguably have 0, 1, or 2 alternate solutions. And I'm not talking about using alternate bases or anything, although that could make for quite a large number of alternate solutions.[/edit]

DaveE
 
Last edited:
  • #187


davee123 said:
Admittedly, in thinking about it, one could argue that the implications of placing a 1 before or after a number imply different mathematical functions-- in which case I could arguably have 0, 1, or 2 alternate solutions. And I'm not talking about using alternate bases or anything, although that could make for quite a large number of alternate solutions.

DaveE
Here is how I did it
10x + 1 = 3(x+10^y) where y is an integer between 1 and 6, and x is an integer between 1 and a million... when y = 5, x = 42857

another solution that is too large (bigger than 1,000,000) is 4,285,714,286 ! I don't want to find the next highest one...
 
Last edited:
  • #188


davee123 said:
I'm not sure I understand what the trick is here. Maximizing the sum is as easy as maximizing each term within the sum. Each term is a square of a digit of x, which means maximizing each digit's square. Maximizing a square of a digit simply means maximizing that digit. Hence, 999 = x, making f(x) = 9^2 + 9^2 + 9^2 = 243.

DaveE

Given any three-digit number, x, define f(x) to be
x minus the sum of the squares of the digits of x
What is the maximum possible value of f(x)?

*[the '-' sign means 'minus' in this context, sorry for being unclear, I have changed it now.]

so if x = 999, then f(x) would equal 999-243 = 756
if x = 990 then f(x) would equal 999-162 = 828 (higher, but I don't think is the highest)

now you see the problem is a little more difficult...
 
  • #189


Georgepowell said:
*[the '-' sign means 'minus' in this context, sorry for being unclear, I have changed it now.]

so if x = 999, then f(x) would equal 999-243 = 756
if x = 990 then f(x) would equal 999-162 = 828 (higher, but I don't think is the highest)

now you see the problem is a little more difficult...

Ahhh, that makes more sense then.

f(950) = f(951) = 844

[edit]For the record on #1:
I found 42857 as the answer, but also -42857 and -3/2 arguably work. You could argue that if the number is "-42857" that adding a 1 *before* the number would mean "1-42857" instead of "-142857", but if the latter is accepted, then it's also a solution. -3/2 could (again arguably) work since 1-3/2 is -1/2, and -3/2 with a "1" after it may imply a multiplication by 1 (obviously if it implies an addition of 1, then it's unchanged from 1-3/2). I haven't checked, but in typing out this response, it's possible that adding a 1 after a fraction, when written in a text editor might imply a denominator that's ten times as much, plus 1. There could be an answer there, too. Hm... I didn't even try complex numbers, but that's another thought.
[/edit]

DaveE
 
Last edited:
  • #190


davee123 said:
How so? That is, what's the more? I got an answer that works using simple algebra-- why ought I doubt it?
Again: is your answer less than 10? It should be. If not, you've fallen for the deception.
 
  • #191


[edit]wrapped spoilers[/edit]
DaveC426913 said:
Again: is your answer less than 10? It should be. If not, you've fallen for the deception.

You think it should be less than 10 hours? Why? Clearly it should be less than 12, because that's how long it took Adam and Ben to do a whole house-- so unless Colin is actively UNPAINTING the house, the answer must be less than 12. But where are you getting 10 from?

Let:
A = the number of houses Adam can paint in an hour
B = the number of houses Ben can paint in an hour
C = the number of houses Colin can paint in an hour

A*12 + B*12 = 1
A*15 + C*15 = 1
B*20 + C*20 = 1

Solve for A, B, and C. Next, you want to know how many hours (H) it'll take for all 3 of them to paint 1 house:

A*H + B*H + C*H = 1

Solve for H.

Did I miss something?

DaveE
 
  • #192


davee123 said:
I found 42857 as the answer, but also -42857 and -3/2 arguably work. You could argue that if the number is "-42857" that adding a 1 *before* the number would mean "1-42857" instead of "-142857", but if the latter is accepted, then it's also a solution. -3/2 could (again arguably) work since 1-3/2 is -1/2, and -3/2 with a "1" after it may imply a multiplication by 1 (obviously if it implies an addition of 1, then it's unchanged from 1-3/2). I haven't checked, but in typing out this response, it's possible that adding a 1 after a fraction, when written in a text editor might imply a denominator that's ten times as much, plus 1. There could be an answer there, too. Hm... I didn't even try complex numbers, but that's another thought.

DaveE

I think your looking into it a bit too deeply :p

2: If Adam and Ben worked together to paint a house they would take 12 hours. Adam and Colin as a team would take 15 hours to paint the house and Ben and Colin as a pair would take 20 hours.
Assuming that the rate at which each painter works is not affected by whom he works with, how long would it take to paint the house if they all worked together?


I got 10 hours exactly, can anyone confirm this?
 
Last edited:
  • #193


Georgepowell said:
I think your looking into it a bit too deeply :p

I agree.

Often times, people write incomplete or malformed questions. Particularly in this forum, people post things like word problems with huge gaping holes, and then claim that solutions presented are wrong simply because answerers are able to dodge the exact wording of the problem. As a computer programmer, I'm frequently reminded of the unexpected boundaries of inputs, so I'm perhaps a bit more anal than most.

Sometimes (more frequently in this forum than others online) it's because English isn't the person's first language. So many times, I have to be forgiving because the translation is imperfect, and I can't tell with what accuracy the person intended to write the problem.

But honestly, the real temptation was the fact that you stipulated with seeming assurity that "All the questions only have one solution". That just beckoned me-- something like a dare. "I dare you to find an error with my problems!" If it weren't for that baiting me, I probably would've ignored the post all together, since I think that all the problems you posted (other than the 3rd one) have been posted here before.

So I looked at the first problem, and almost immediately concluded that "a number less than 1 million" would almost assuredly provide for *TWO* solutions, the answer you expected and the negative counterpart of the answer you expected. Immediately, I started thinking about other interpretations as well, like decimals (although I believe I logically eliminated that possibility), fractions, bases, and so forth.

Anyway, with one error in hand, and a lack of understanding of the 3rd problem, I decided to go ahead and post. Actually, I should've pointed out some errors in your wording on problem #4 as well, since you didn't say that the king will ask the question to each person in the line starting from the back and working up to the front. You only said that the last person gets asked the question. You also said "guaranteed, not a probability or expected outcome", which is wrong. The last person's chance of survival are 50/50, no matter what. Hence, if the number of subjects is 1, the answer is necessarily a probability. And if the number of subjects is 2, then there are multiple strategies that result in the same survival rate, each of which in turn has a 50% chance of happening (either 1 person is killed, or 0 are killed).

And to go off on a tangent, when this (#4) has been asked before, very frequently people suggest that the strategy should be some alternate form of communication-- something like "if the next person's hat is red, you state the color of your hat quickly in a squeaky voice, and if it's blue, you state the color of your hat slowly in a deep voice". Which then means that you (as the asker of the question) have to say "no, you can't say 'red' or 'blue' any differently". And that sometimes continues for a dozen posts or so, with people trying to find loopholes. Instead, if you plan on asking it again elsewhere, you can avoid this by suggesting that each person must point to the color hat that they believe they were given, which the king will dutifully announce, such that the next people in line can hear.

Anyway, remember to clearly spell out your assumptions when writing word problems!

DaveE
 
  • #194


davee123 said:
[edit]wrapped spoilers[/edit]


You think it should be less than 10 hours? Why? Clearly it should be less than 12

DaveE

OK, well that's what I was getting at. I just got a little overzealous. :redface:
 
  • #195


Georgepowell said:
[edit]spoiler tags corrected[/edit]
another solution that is too large (bigger than 1,000,000) is 4,285,714,286 ! I don't want to find the next highest one...

That's not correct-- clearly any answer must end in a 7, since that's the only digit that, when multiplied by 3, will give you a product whose last digit is 1. In fact, it's pretty easy to find the next largest number, it's 42857142857. And the next one after that will be 42857142857142857, followed by 42857142857142857142857. The tricky part is finding that first pattern. Once you've got it, the rest of the solutions unfold easily.

DaveE
 
  • #196


davee123 said:
That's not correct-- clearly any answer must end in a 7, since that's the only digit that, when multiplied by 3, will give you a product whose last digit is 1. In fact, it's pretty easy to find the next largest number, it's 42857142857. And the next one after that will be 42857142857142857, followed by 42857142857142857142857. The tricky part is finding that first pattern. Once you've got it, the rest of the solutions unfold easily.

DaveE

sorry that was just a miss-type, you are right. I did find the next answer like this:

I found it by dividing 2999... etc. by 7, I kept adding another 9 until I found another multiple of 7, 42857142857*7 = 299,999,999,999, and 42857142857142857142857*7 = 29999 bla bla bla!
 
  • #197


davee123 said:
I agree.

Often times, people write incomplete or malformed questions. Particularly in this forum, people post things like word problems with huge gaping holes, and then claim that solutions presented are wrong simply because answerers are able to dodge the exact wording of the problem. As a computer programmer, I'm frequently reminded of the unexpected boundaries of inputs, so I'm perhaps a bit more anal than most.

-...-

Anyway, remember to clearly spell out your assumptions when writing word problems!

DaveE

I (partly) agree, but I did not write the questions, so I am not to blame :D

I got them from:

http://www.fmnetwork.org.uk/files/GRfin0708.pdf
and
http://www.fmnetwork.org.uk/files/GRreg0708.pdf
and
http://answers.yahoo.com/question/i...ePCbVMPsy6IX;_ylv=3?qid=20070810141316AA4644N

The first two links are questions from a maths challenge that I was a part of (our team of four came 6th) although that was this year, and the questions where not the same.
 
  • #198


The easiest way to solve it (I think) is by taking it one step at a time in reverse. I can't use pre-formated text AND spoilers at the same time, so bearing with that:

1abcdef
x3
abcdef1

Since the ones place of f*3 is 1, that means f=7:

1abcde7
x3
abcde71

Now, since 3*7 has a tens place of 2, and the ones place of 3*e + 2 = 7, we know that 3*e's ones place is a 5. Hence, e=5:

1abcd57
x3
abcd571

Same process, 3*5 has a tens place of 1, and the ones place of 3*d + 1 = 5, which means that 3*d's ones place is a 4. Hence, d=8.

1abc857
x3
abc8571

Next, again, 3*8 has a tens place of 2, and ones place of 3*c+2 = 8, therefore 3*c ends in a 6, making c=2.

1ab2857
x3
ab28571

Now, one more time, 3*2 has a tens place of 0, so 3*b+0 = 2, which means b=4.

1a42857
x3
a428571

And FINALLY, 3*4 has a tens place of 1, which means 3*a+1 = 4. So a=1. So since the 1st digit must be a 1, we can choose to stop there (at b), or keep going. However, since we want the smallest value possible, we use 42857 rather than 42857142857, which would be the next number we'd hit if we continued.

[edit]
Georgepowell said:
I (partly) agree, but I did not write the questions, so I am not to blame :D

I disagree that you're not to blame, although as stated earlier, the blame is small. You posted the problem, therefore you have a responsibility to ensure that the problem is correctly stated, regardless of how the problem was relayed to you. Spreading a faulty problem doesn't absolve you of responsibility, it just means that someone else is just as at fault as you are. But again, this is a pretty minor transgression. It's not like you're publishing it in a math textbook or something-- it's a forum, after all. But I still say the onus is on you as a poster to ensure the accuracy of problems presented.
[/edit]

DaveE
 
Last edited:
  • #199


new puzzle, I just made it up and i don't know the correct answer, if there is one...

I have 2, 3 digit numbers, a and b. what are the two values for a and b so that (a/b)'s decimal expansion is the longest? (not including reccuring decimals like 333/999 = 0.33333 etc.)

ie. if a and b where one digit numbers, then it would be something like 1/7 = 0.142857 blablabla
 
  • #200


Georgepowell said:
I have 2, 3 digit numbers, a and b. what are the two values for a and b so that (a/b)'s decimal expansion is the longest? (not including reccuring decimals like 333/999 = 0.33333 etc.)

Hmm... Effectively, you want a denominator whose only factors are 2 and 5 (which can occur multiple times). Otherwise, you get something like 1/7, which is a repeating decimal:

0.1428571428571428571428571428571428571428571428...

Georgepowell said:
if a and b where one digit numbers, then it would be something like 1/7 = 0.142857 blablabla

If they're 1-digit integers (not decimals, since "3.93" is a 3-digit number and ".3" is similarly a 1-digit number, and I assume you mean integers, although you should probably specify that), then the viable answers would be:

1/8 = 0.125
3/8 = 0.375
5/8 = 0.625
7/8 = 0.875

If we include 1-digit decimal values like ".4" or ".9", then we can get an extra digit of extension:

.1/8 = 0.0125
.3/8 = 0.0375
.5/8 = 0.0625
.7/8 = 0.0875

So my preliminary guess will be:

n/512 where n is not divisible by 2.

For example:
371/512 = 0.724609375

My offhand guess is that you want to cram in as many factors as you can to the denominator, and since 2 is smaller than 5, you just find the highest power of 2 such that it's an integer less than 1000, or 512. The numerator shouldn't really matter-- it might serve to *reduce* the number of digits in the result (because it may be a reducible fraction), so using a numerator of 1 ought to be just as effective or more effective than any other numerator. I may need to ponder on this more to make sure, but I think that's correct.

[edit]
Verified. n/512 (n not divisible by 2) is the longest length decimal with a non-repeating decimal portion (assuming "0000000000..." counts as non-repeating).

And if you meant the longest repeating portion of a decimal, then it's n/983 where n is not divisible by 983. The repeating decimal portion of 1/983 is 982 digits long:

0010172939979654120040691759918616480162767039674465920651068158697863682604272634791454730417090539165818921668362156663275686673448626653102746693794506612410986775178026449643947100712105798575788402848423194303153611393692777212614445574771108850457782299084435401831129196337741607324516785350966429298067141403865717192268565615462868769074262461851475076297049847405900305188199389623601220752797558494404883011190233977619532044760935910478128179043743641912512716174974567650050864699898270600203458799593082400813835198372329603255340793489318413021363173957273652085452695829094608341810783316378433367243133265513733468972533062054933875890132248219735503560528992878942014242115971515768056968463886063072227873855544252288911495422177009155645981688708036622583926754832146490335707019328585961342828077314343845371312309257375381485249237029501525940996948118006103763987792472024415055951169888097660223804679552390640895218718209562563580874872838250254323499491353

[/edit]

DaveE
 
Last edited:
  • #201


Another puzzle here:

You have the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9. Create two integers from these digits so that when you take one from the other you get 33,333.

eg.
12,345-6,789 = 5,556

As an extra challenge:

After you have found the solution(s), prove that they are the only solution(s).
 
  • #202


Georgepowell said:
You have the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9. Create two integers from these digits so that when you take one from the other you get 33,333.

41,268 - 7,935 = 33,333
41,286 - 7,953 = 33,333

Georgepowell said:
After you have found the solution(s), prove that they are the only solution(s).

Hmmm...

DaveE
 
  • #203


Those were the solutions I got. And they ARE the only solutions.

I have proved it using funny diagrams and logic, but I can't draw these diagrams and show my proof formally on this site.

The basic idea of the proof (don't read if you want the complete challenge for yourself):

First you need to prove that all solutions involve a 5 digit number and a 4 digit number. This is easy.

Then you prove that the 5 digit number starts in either 3 or 4. (easy)

Then you assume that the first number is 3, then show that there is no possible solutions with the first digit being 3. This is slightly longer, but still easy.

Then I drew a table of all the possible pairs of number that their difference is 3 or 4, and grouped them into four groups:
(4,X). remainder four, the first number in the pair is the larger one. i.e. (9 - 5)
(4,Y). remainder four, the first number is smaller. i.e. (3 - 9)
(3,X). remainder three, first is larger. i.e. (9 - 6)
(3,Y). remainder three, first is smaller. i.e. (2 - 9)

You need to pick four pairs that do not repeat a digit and do not include zero or 4 (because this is the first digit of the 5 digit number).

http://i349.photobucket.com/albums/q390/Georgepowell77/DSC00137.jpg

That is an image of my basic idea.

There are now 8 situations that could lead to a solution. I went through each one and proved that 7 of them could not yeild a solution (I did this by using weird clock-like diagrams).

Then there is one combination of group choices that could lead to a solution. and it is very easy to show that there are only two combinations of pairs from these groups that leads to a solution. And they are the solutions that we got.
 
  • #204


powergirl said:
NOt right;
Can anyone answer this?
2) A murderer is condemned to death. He has to choose between three rooms: The first is full of raging fires, the second is full of assassins with loaded guns, and the third is full of lions that haven't eaten in 3 years. Which room is safest for him?

the third room is the safest to choose as the lions would be dead if they hadnt eaten for 3 years
 
  • #205


Joe has been showing Penny a few optical experiments. In one experiment he placed six mirrors vertically to form a regular hexagon with small gaps between the mirrors. Through one gap he shone a laser beam so it emerged straight from the gap diametrically opposite.

Penny then had to work out the smallest angle through which the beam must be rotated so that it emerged from the same gap as before, after being reflected just once by all six mirrors.

What was that small angle (rounded to the nearest degree)?
 
  • #206


Georgepowell said:
Joe has been showing Penny a few optical experiments. In one experiment he placed six mirrors vertically to form a regular hexagon with small gaps between the mirrors. Through one gap he shone a laser beam so it emerged straight from the gap diametrically opposite.

Penny then had to work out the smallest angle through which the beam must be rotated so that it emerged from the same gap as before, after being reflected just once by all six mirrors.

What was that small angle (rounded to the nearest degree)?

Upon what axis is the beam rotated? I presume the rotation axis can't be in line with the centre of the hexagon (i.e. spin the mirror contraption rather than the laser), so the axis of rotation must be the point where the laser beam first enters the hexagon?
 
  • #207


DaveC426913 said:
Upon what axis is the beam rotated? I presume the rotation axis can't be in line with the centre of the hexagon (i.e. spin the mirror contraption rather than the laser), so the axis of rotation must be the point where the laser beam first enters the hexagon?

It is rotated about the point where it first enters the hexagon.
 
  • #208


I don't do answers. But I like diagramming questions!
 

Attachments

  • PF20090214_laser.gif
    PF20090214_laser.gif
    5.1 KB · Views: 526
  • #209


So is my question too difficult for the genius's of PhysicsForums.com ?
 
  • #210


Georgepowell said:
Joe has been showing Penny a few optical experiments. In one experiment he placed six mirrors vertically to form a regular hexagon with small gaps between the mirrors. Through one gap he shone a laser beam so it emerged straight from the gap diametrically opposite.

Penny then had to work out the smallest angle through which the beam must be rotated so that it emerged from the same gap as before, after being reflected just once by all six mirrors.

What was that small angle (rounded to the nearest degree)?

9 degrees.
In fact, atan(sqrt(3)/11).

:smile:
 

Similar threads

Replies
6
Views
542
  • General Discussion
Replies
1
Views
795
  • Classical Physics
Replies
10
Views
1K
  • General Discussion
2
Replies
43
Views
4K
  • General Discussion
3
Replies
102
Views
8K
  • Astronomy and Astrophysics
2
Replies
39
Views
2K
Replies
5
Views
936
Replies
1
Views
717
Replies
19
Views
1K
  • STEM Academic Advising
Replies
7
Views
1K
Back
Top