## Assassin’s Bullet

For the 2nd challange, we had a bit of dropoff. Nice job to those who were able to figure it out: *Hex, Lograh, Michaelc, Shawn, Joe, Mashplum, Bizarette18, Aaronlau, Brianu, Ludiavolo, Roel, Fuzzy, Diegote, Bach, Billypilgrim, Apex.JP, and EddB.*

OK, Here’s what Bilbao has uncovered for the 3rd puzzle in our clock-time series:

The bullet of the assassin hit the centre of the clock, got through and stopped the clock.

The hour and minute hands welded in a straight line, pointing at opposite directions, and went round freely (as shown in the picture) after the impact.

What time did the bullet break the clock?

Have also put this game up to play today: Totem Destoyer

BillyPilgrim| Profile March 2nd, 2009 - 2:03 pmThe hour and minute hand being opposite each other happens first at 360/11 minutes past 12 and every 720/11 minutes after that. To identify the exact time the “assassination” happened you have to look at the second hand, which reads somewhere between 45 and 50 seconds. This means that the only possible time is 10:21:49.09.

hex| PUZZLE MASTER | Profile March 2nd, 2009 - 2:58 pmThe angular position of the hours hand = 2 x Pi / 12 x (h + m / 60)

The angular position of the minutes hand = 2 x Pi x m / 60

The condition is:

angular position of the hours hand = (2 x k + 1 ) x Pi + angular position of the minutes hand

The conditions on the hours and minutes dictate that k=-1 for h=0..4 and k=0 for h=6..11 (h=5 is impossible)

Solving the above gives 11 distinct answers. Bilbao requires one so we’re missing something!

The supposedly seconds hand at the bottom of the clock shows around 49 seconds. The only possible answer would then be:

10:21:49.09090909

bizarette18| PUZZLE MASTER | Profile March 2nd, 2009 - 3:03 pm10:21:49

michaelc| Profile March 2nd, 2009 - 3:21 pmWell, we want the hour hand exactly half way around the clock from the minute hand.

This happens 11 times in a 12 hour period.

My original formula…

(h + m/60)/12 – (m/60) = 1/2

Simplifies to

m = (60h – 360)/11

You have do some “borrowing” for hours 1-5 to do away with the negatives, but you wind up with…

h=2 and m = 480/11

or roughly 2:43::38.18

I’m guessing the hands didn’t go “round freely” far from the picture? If it got shot in the middle, I wouldn’t think the hands would stray very much from their original positions, however the second hand doesn’t add up quite exactly for this time.

It looks like the second hand is on about 49 seconds. The closest time for the second hand to be on 49 seconds and the hands to be in a straight line from each other is ~10:21::49.09. h=10, and m = 240/11.

I’m guessing my first answer is probably what we’re looking for.

So roughly 2:43::38.18.

Until the answer is posted, I suppose there’s time to wonder…

Shawn| PUZZLE GRANDMASTER | Profile March 2nd, 2009 - 4:50 pm10:21:49.0909

The subdial at the bottom of the clock face is measuring seconds, and is stopped at ~49.

After some trial and error calculating, the only time for which both conditions could occur (hands at 180 degrees AND 49 seconds after the minute) is 10:21:49 plus 1/11 of a second.

APEX.JP| Profile March 2nd, 2009 - 8:01 pmThe bullet broke the clock at 10:21:49.09

scottk| Profile March 2nd, 2009 - 8:19 pm6 oclock.

bach| Profile March 2nd, 2009 - 10:01 pmthis one was more conceptually challenging than the last two.

i went about it in a somewhat “brute force” way; there’s probably a more efficient way to solve this, but let me go ahead and explain:

first, i found all the times during normal clock operation at which the minute hand and hour hand were 180 degrees apart, i.e., “pointing in opposite directions”. needless to say, the times are the same whether it’s ante or post meridiem. the equation for representing 12:00 was x+0=(1/12)x+30, where the left side represents the minute hand and the right side represents the hour hand, giving you minutes elapsed after 12:00. for 1:00, it would be x+0=(1/12)x+5+30, giving you minutes elapsed after 1:00, and so on.

here are those times. note that 5:00 does not have a time at which minute hand and hour hand directly oppose each other.

seconds are in decimal form, e.g., .727272= ~44 seconds

12:32.7272727 (beginning of cycle)

1:38.18181818

2:43.63636364

3:49.09090909

4:54.54545455

—

6:00

7:05.45454545

8:10.90909090

9:16.36363636

10:21.8181818

11:27.2727272 (end of cycle)

12:32.7272727 (beginning of the cycle)

at this point, it was necessary to observe the picture and note the approximate time, which i guessed was 2:44:49. then, it was just a matter of matching one of the possible times with the time on picture, paying close attention to the second hand value. so, 10:21 and 9/11 seconds, or 10:21:49 was a prime candidate.

i then sought to “prove” this with equations modeled for how the clock operates during “post-gunshot” speed. after the gunshot, it would only take 4 minutes to represent the picture above- if the gunshot occurred at 10:21:49 AM, then 4 minutes later it would show 2:44:49 PM, and obviously if the gunshot occurred at 10:21:49 PM, then 4 minutes later it would show 2:44:49 AM.

the equation would be x+(number of ticks at current time)=x+(number of ticks at current time+30), with minutes on the left side and hours on the right side.

we can represent 10:21:49 as:

x+(50+9/11)=x+((21+9/11)+30)

factoring in the 4 minutes that must logically elapse to get the picture, we get

4+(50+9/11)=4+((21+9/11)+30)

and, since 55 + 9/11 = 55 + 9/11, this means that at both times, the minute hand and hour hand are 30 “ticks” apart, thus the answer is conclusively (i hope) 10:21:49.

ludiavolo| Profile March 2nd, 2009 - 10:26 pmThe present location of the hour and minute hands in the picture is not important at all–all we need to know is that the clock broke when the hands were 180 degrees apart. Another key piece to the puzzle is noticing that the second hand (on the smaller dial) is frozen at about 49 seconds. Putting these two conditions together, I searched for a time of the day when the hour and minute hands were 180 deg apart with the additional condition that this time must occur when about 49 seconds had elapsed after the last full minute. The equation that I solved is

n*30 + 360*t/12 + 180 == 360*t

where the left hand side of the equation represents the angular position of the hour hand + 180 deg, and the right hand side is the angular position of the minute hand. The variable “n” represents the hour number. The solution for t in terms of n is:

t = (6 + n)/11

I rifled through values of n and checked to see which one worked, and n = -2 seemed to do the trick. (n = -2 corresponds to 10 o’clock.) Plugging in n = -2 makes t = .363636… hours which corresponds to 21 minutes and 49 seconds (bingo!).

The time when the assassin killed the clock is therefore 10:21:49

brianu| Profile March 2nd, 2009 - 11:31 pmI believe your are missing some vital info in the question

bach| Profile March 3rd, 2009 - 3:02 amjust noticed something relatively minor-

23 minutes must logically elapse, not 4… I was thinking about something to do with the # of hour markers that would be passed. (44+9/11)-(21+9/11)= 23

so the equation would be

23+(51+9/11)=23+((21+9/11)+30)

74+9/11=74+9/11, the answer remains 10:21:49.

aaronlau| Profile March 3rd, 2009 - 4:59 amThe seconds’ hand is at 49 sec.

it will be opp dir 11 times in 12 hours => 60/11 additional minutes from 60mins to reach. => 49sec will be at the 4th multiple.

assuming starting at 6 o’clock sharp, the 4th multiple will be at the 10:20 position.

Therefore the exact time is 10:21 and 49 seconds.

Roel| Profile March 3rd, 2009 - 7:16 amThere’s a 180 degrees difference between the hour and minute handle.

HOUR = MINUTE – 180

But the minute handle will have turned more times than the hour handle, so we have to deduct some turns from it.

HOUR = MINUTE – 180 – 360.N

The minute handle goes 12 times the speed of the hour handle.

HOUR = 12*HOUR – 180 – 360.N (1)

And the HOUR handle goes with a speed of 360 degrees per 12.60.60 seconds.

HOUR = 360/(12.60.60) . TIME (2)

So the two equations become:

TIME = (1/2 + N) . 12.60.60/11

Solving this for N=0, N=1, …, N=11 gives

TIME=1963,63636363636

TIME=5890,90909090909

TIME=9818,18181818182

TIME=13745,4545454545

TIME=17672,7272727273

TIME=21600

TIME=25527,2727272727

TIME=29454,5454545455

TIME=33381,8181818182

TIME=37309,0909090909

TIME=41236,3636363636

Which translates into the following:

TIME=0h 32min 43,6363636363637s

TIME=1h 38min 10,909090909091s

TIME=2h 43min 38,181818181818s

TIME=3h 49min 5,45454545454595s

TIME=4h 54min 32,7272727272721s

TIME=6h 0min 0s

TIME=7h 5min 27,2727272727279s

TIME=8h 10min 54,5454545454559s

TIME=9h 16min 21,8181818181838s

TIME=10h 21min 49,0909090909117s

TIME=11h 27min 16,3636363636397s

So the bullet broke the clock at one of the above times.

joe| Profile March 3rd, 2009 - 8:29 amFor this last part once again we use the fact that the minute hand travels 6 degrees every minute and that the hour hand will travel 0.5 degrees every minute.

We need to find out at what time the hour hand is 180 degrees more than the minute hand when the time of before 6 oclock or 180 degrees less after 6 oclock. It makes sense because at say, around quarter to three the hour hand is at a smaller angle past the 12 than the minute hand and like wise at say, around the 25 to 1 mark..

But it could be one of many times, 12 in fact where one hand is 180 degrees more than the other!! However we note that the clock stopped when the second hand (incorporated in the lower part of the clockface) was at approximately 49 seconds, or for calculation purposes at around the 0.81 in decimal form).

I did it on a trial and error basis until my calculation for the correct number of minutes gave me a x.81… number.

For each hour hand BEFORE 6 oclock using the above formula:

(0.5 x h x 60) + (0.5 x m) + 180 = 6 x m

where m is the number of minutes past the minute hand, and h is the number of hours past on the hour hand

AFTER 6 oclock( which is exactly straight line hands aswell)

(0.5 x h x 60) + (0.5 x m) – 180 = 6 x m

At the 10 oclock and some unknown minutes we hit bingo!

(0.5 x 10 x 60) + (0.5 m) -180 = 6 m

solving for m = 21.818181

So the clock stopped at 10:21.818181

or 10:21 and 49 seconds

diegote| Profile March 3rd, 2009 - 11:41 amI say the clock stopped at 10:21:49.

I made this calculations:

The difference between the angles of both hands is 30min, so:

|-(X-vh*T)+vm*T|=30

angles are expressed in time minutes

where

vh: speed of hour hand

vm: speed of minute hand

X: position where hour hand started

30: angle between both hands

operating:

T=(30+X)*12/11

X advances in 5 min steps, so X=n*5.

Using excel I found that with n=9 I can match the 49″ shown in the picture at T=81′ 49″. It means the clock stopped 1 hour 21′ and 49″ after 9 o´clock.

Migrated| Profile March 3rd, 2009 - 1:40 pmJust want to clarify a statement in the puzzle: “The hour and minute hands welded in a straight line, pointing at opposite directions, and went round freely (as shown in the picture) after the impact.”

Is that saying that after impact, the hour and minute hands that are welded together are still able to travel around the clock?

lograh| Profile March 3rd, 2009 - 2:05 pmEr, unless I’m missing something, there are 24 possible times (12 for AM and 12 for PM) that the bullet could have hit the clock. But the question clearly asks for a specific time, singular. Wouldn’t there be a point for each hour when the minute hand would be colinear with the hour hand? As an example, somewhere between 1:35 and 1:40 the hands are colinear, and again somewhere between 2:40 and 2:45.

I’ll think on getting the possible solutions but if the question could be clarified before I come back that’d be great.

alderney girl| Profile March 3rd, 2009 - 3:12 pmI think the clock stopped at 10 to 4 or thereabouts according to the smaller clock which would have stopped at moment of impact, and not have spun freely as the other one.

RK| Founder | Profile March 3rd, 2009 - 4:10 pmLograh had a similar concern Brainu, but I believe all the information is there to answer the question

Obiwan| Profile March 4th, 2009 - 12:48 amThis may be parsing words, but the statements are “hit the centre” and” “got through and stopped the clock.” From that description, the clock glass is smashed and there cannot be an intact mount for the hands to pivot about. Therefore the hands could not go around freely.

The only reasonable conclusion is that the the bullet stopped the clock as shown–approximately 16 minutes to 3.

bilbao| Profile March 4th, 2009 - 8:51 amClarification:

when the bullet hit the centre, the hour and minute hands were forming a straight line. Because of the impact the clock stops completely (and this is the time you must find).

Just after and as a consequence of the impact, both hands (since they are welded) go round madly/freely/loose until they finally stop at the time shown in the picture. Thus, the time shown in the picture does not correspond to the one you must find.

I hope this makes the puzzle clearer.

Val| Profile March 4th, 2009 - 11:53 am3:49?

michaelc| Profile March 4th, 2009 - 2:19 pmAha, thanks for the clarification.

That would mean the time would have had to have been my other answer.

~10:21::49.09. h=10, and m = 240/11.

Pretty clever.

Hendy| Profile March 4th, 2009 - 10:37 pmClue: 49 seconds.

Assuming the second hand is not also broken and free spinning and the clock is no longer running even one tick:

For the clock to have stoped with :49 seconds and with the hour and minute hands diametrically opposed, the time would have to be :

10:21:49 (a.m. or p.m. unknown).

I used Excel with cells for all the possible values for each second on the clock for 43200 seconds (12 hours x 60 minutes x 60 seconds) I calcuated the angular distance traveled around the face for hours and for minutes. I displayed the absolute value of the difference and looked for a time ending in :49 seconds that was near 180 (diametrically opposed) the actual value for 10:21:49 is 180.00833333 degrees of opposition.

I am sure you math wizzes could easily calculate this without seeing every value, but I am a software test engineer by trade and have to see everything to believe it. I gave up on formulas a long time ago. I don’t trust them to be honest, people who have trusted them sent rockets into the ground! However, I’d be quite happy to test your equation, if you’d like! ;-)

Diego| Profile March 4th, 2009 - 11:05 pm10:21 pm and 49 seconds

Although there is no mention of the seconds hand being paralyzed and not moving freely, Im assuming it is.

Calculating at which hours the minute and hour hands are at 180 degrees it gives me the only time at which it is at 49ish seconds.

Diego| Profile March 4th, 2009 - 11:08 pmUps, PM or AM

Migrated| Profile March 5th, 2009 - 1:37 amThanks bilbao for clearing it up.

So the most obvious answer is 6 o’clock.

But since this is a puzzle, I really have no idea on this one.

Jimmy Anders| PUZZLE MASTER | Profile March 5th, 2009 - 1:39 amMy answer is 10:21 + 9/11 min.

Because I don’t know the proper way to do it, I will use “D” for degrees. In one minute, the minute hand travels 1/60 of the way around the clock. The hour hand, which travels at 1/12 the speed, goes 1/720 of the way around. Therefore the distance between them after one minute is:

1/60 – 1/720 = 11/720 of the way around the clock.

Since all the way around is equal to 360 D, in degrees this distance is:

11/720 * 360 D = 11/2 D = 5.5 D.

So the hands part at a rate of 5.5 D/min. Then, the length of time required for the angle between the hands to change by 360 D is:

360 D / [5.5 D/min] = 65 5/11 min.

We know that the hands are at 180 D from eachother at 6:00, so all that is left is to successively add 65 5/11 min to that time and then compute those times in HH:MM:SS format (or else notice that the second hand is closest to 9/11 min), and then pick the time where the seconds match up with the second hand on the clock in the picture. That time is 10:21 plus 9/11 minutes, which is approx. 10:21:49. Most likely that would be the PM so the assassin has the night on his side, but it’s impossible to say for sure.

nightowl| Profile March 5th, 2009 - 2:11 am10:21:49.09 (given that if time stopped as pictured, the second hand had to be at 49 seconds)

fuzzy| Profile March 5th, 2009 - 10:27 amI’m getting either 1:38 or 2:43. I really have no idea why I should pick one over the other.

lograh| Profile March 5th, 2009 - 12:18 pmRight, I’m seeing my problem now. I had initially thought there would be 12 possible configurations of the clock hands which would be co-linear and opposing and thus be potential times. I now notice that the seconds hand is not on the same axle and thus is still fixed to the workings of the clock. This is vital information that is not mentioned in the words of the puzzle but is easily inferred when viewing the picture. I should have picked up on that quicker.

Using the fact that the seconds hand is stuck at about roughly 49 seconds in to the minute, we can narrow down the possible times considerably. For example, one possible time is 12:32:43, but since the clock’s second hand should be at roughly :49, this could not have been the correct time.

So, we just need to figure the twelve possible minute/hour configurations, and then eliminate all but the (hopefully only) one that has a seconds number somewhat close to 49.

A quick set-up, using M for “minutes since 12:00:00″ and m for “degrees between minute hand and ’12′” and h for “degrees between hour hand and ’12′” gives the following:

m=6(M%60)

h=.5M

m=h+180 // minute and hour hands are co-linear and opposing = there are 180 degrees between them

6(M%60)=.5M+180

6(M%60)-.5M=180

And this will generate all possible numbers of minutes since 12:00:00 where the two hands are as pictured. Since I’m really bad with modular arithmetic, this is where I split it up to multiple equations and started getting these:

(for between 12:00:00 and 13:00:00)

M(1-1/12)=30

M = 32.727272

time = 12:32:43.64

(for between 13:00:00 and 14:00:00)

M(1-1/12)-5=30

M = 38.181818

time = 13:38:10.91

(for between 14:00:00 and 15:00:00)

M(1-1/12)-10=30

M = 43.636363

time = 14:43:38.18

And I noticed the really handy pattern of “-0″, “-5″, “-10″ and decided to skip the messy manipulations and just hope that pattern holds for the rest.

15:49:05.45

16:54:32:73

18:00:00

And here I noticed I was getting numbers for the minute hand at 60 and higher, so I paused and noticed that the hour hand had passed the bottom of the clock and was working its way back up, so the minute hand was up at the top again. Not a big deal, but was certainly a brief amusement.

19:05:27.27

20:10:54.55

21:16:21.82

22:21:49.09 (I think this will be the winner, but there are only two more to check anywho)

23:27:16.36

00:32:43.64 (I did this one do double-check that the pattern did hold, it should (and did!) match the first answer I got, mod 12 hours)

So I wind up with the only time with the seconds close to what I expected of 22:21:49 (or possibly 10:21:49). That’s the time the clock was broken at.

Fun puzzle! I am rather ashamed I did not pick up on the seconds hand more quickly!

Roel| Profile March 5th, 2009 - 4:55 pmOk, I didn’t look at the seconds hand which shows about 50 seconds. So the only one from my previous list of possibilities that is has around 50 seconds is this one:

TIME=10h 21min 49,0909090909117s

Very clever riddle! Love it.

brianu| Profile March 5th, 2009 - 8:33 pm4:54:32.4

x=minutes minute hand = 6 degrees per minute

y=hours hour hand = .5 degrees per minute

the amount of minutes (x) in degrees (6x) minus 180 degrees (6x-180) is equal to the number of hours (y) multiplied by 60 minutes (60y) plus the amount of minutes indicated by minute hand (60y+x) multiplied by .5 degrees .5(60y+x)

6x-180=.5(60y+x)

trial and error with whole numbers for y and if y=4 then x=54.54repeating which works. this is an ugly way to do solve this, maybe I’ll figure out a cleaner version soon.

suineg| PUZZLE MASTER | Profile March 6th, 2009 - 9:00 ammy answer: 10 hours 21 minutes and 49,09 seconds

first if you see the picture there is a little cicle inside the larger circle with the seconds and with eye guessing(jajaja was little, you tricky ones jajajaja) I assume the time in seconds was 49 when the clock stops so this is my explanation( I guess):

M= degrees of the minute hand

H= degrees of the hour hand

you know that every clock has 180 degrees at 6:00:00 so that is my statring moment for the analysis

Now you have to get the ratios per second of the hour hand and the minute hand of the clock that was the easy part jajajaja, this was diabolic hard man jajaja.

ratio of the minute hand:

5min—> 30 degrees(360/12,cool)

1min—> X —> 6 degrees per minute—> 0.1 degrees per second

ratio of the second hand:

60min—> 30 degrees

1min—-> X—-> 0.5 degrees per minute—> 0.00833333 degrees per second

Ok now first I tried to put on an equation that when I substract both equation the result were 180 degrees that did not work but I like to share my failures jajaja: there were to cases: minute hand to the left, minute hand to the right( from 6:00:00 to 12:30:??) and a second case, the inverse of the first (12:30 to 6:59 you know jaja)

M=0.1* seconds: H= 0.00833333*second

H-M=180( because the minute hand is in the left so the degree of the minute hand is greater but not in all tha cases that is why this does not work) has to do equations by intervals(6, 6:30;7,7:30 and so on) and this really upset me so I stopped but was pretty cool jaja.

I solved in a pretty easy but not much elegant way, inductively thats right:

at 6:00:00 the angle betwwen the hands is 180 degrees

not try 7:05:00; M= 300 second *0.1=30 degrees; H=210+300/120

M-H=180–> but M= 180 + 30 + 300/120; now you create an equality identity( what a corn name)

M= second*0.1; H=second/120+300/120 you need to know when both angles are the same( opossite0, angles wish I could draw here jaja)

11/120 * second= 2,5–> seconds are equal to 27,272727..

so there is pattern every 1:05:27,272727272 you get and angle of 180 degrees so at 8:10:54; 9:16:22 ;10:21:49, that is my answer, cooooooooooooollllll, hope I have this time the answer, great puzzle indeed could not find a general formula for it if I wanna skip working by segments, but not always you can get what you want.

bilbao| Profile March 9th, 2009 - 1:38 amCongratulations!!! Correct answer is 10:21:49.0909

Get ready for the 4th one. It is much harder than the first three (at least if you want to get the exact elegant mathematical solution) but it is worth a try.

RK| Founder | Profile March 9th, 2009 - 10:07 amyes, very good: 10 h 21 min 49 1/11 s

michaelc| Profile March 9th, 2009 - 10:41 amHa! Hendy, “sent rockets into the ground”. Must have been a “-” sign misplaced in the equation somewhere! lol [Seriously, hope no one was on the rocket at the time. ]

Formulas are great, and are like a form of art work to me. But like a great painting, sometimes they need some “touch up” paint.

It’s just silly to put everything in a formula. I had a little help from Excel with my solution as well.