## The right time to advertise

It is well known that in commercial advertising of clocks a certain position of the hands is usually shown (see picture above). The reason might be aesthetic, equilibrium, ok shape, smile shape, etc.

What would be the real exact time shown in the picture, so that the angle between the hour hand and 12 o’clock is the same as the angle between the minute hand and 12 o’clock?

*This is Challenge #2 in Bilbao’s Clock Puzzle Series. **Answers to be unmasked Monday am, at which time the 3rd will go up!*

**Puzzle #1 Solvers**: Migrated, Cheddarmelt, Benson, Joe, BillyPilgrim, SharonH, Shawn, Michaelc, oneiric, Alderney Girl, Doar823, Sue, Zenith, Obiwan, Aaronlau, Bach, Deadskin, Deejam, Lograh, Fuzzy, Hex, Hendy, Roel, Mashplum, Ludiaolo, Apex.JP

BillyPilgrim| Profile February 26th, 2009 - 12:33 amWell, the minute and hour hand being at the same angle with respect to 12 o’clock on opposite sides happens many times in a 12 hour span, the most obvious being 12:00.

The time that this happens closest to the time shown in the picture is 10:09:13.85

I did this using trig functions to get x- and y-coordinates of each hand w.r.t. time (assuming they’re the same length). Where y_minute = y_hour and x_minute = – x_hour is where the criterion is satisfied.

bach| Profile February 26th, 2009 - 1:14 amif i’m understanding this exercise correctly, we have to find the time after 10:00 at which the 12:00 marker bisects the angle between the minute hand and the hour hand; that is, whether the clock given by the answer perfectly resembles the picture is irrelevant.

if that assumption is fair, then i think it just becomes a simple absolute value problem that builds off of the equation i made for the last problem.

for simplicity’s sake, we start the time at 10:00, which would be represented in terms of “ticks” (ticks= 1/60th the circumference of the clock) as -10 for the hour hand and 0 for the minute hand.

adding in rates, we get the equation: |(1/12x)-10|=x+0

when we solve for x, we get -120/11 and 120/13. since time can’t be negative, x=120/13, where x is the time elapsed after 10:00.

thus, the answer is 10:09:13 and 846/1000 of a second.

doar823| Profile February 26th, 2009 - 1:57 amI think 12:55:30 is pretty darn close to the same angles

doar823| Profile February 26th, 2009 - 1:59 amBut after re-reading the question my answer is 10:09:45

hex| PUZZLE MASTER | Profile February 26th, 2009 - 6:47 amThe 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 x Pi – angular position of the minutes hand

The conditions on the hours and minutes dictate that k=1 for h=1..11 and k=0 or 1 for h=0

Since h=10 in the picture, we solve for m with k=1, and hence:

m=120 / 13 = ~ 9.23 minutes

Shawn| PUZZLE GRANDMASTER | Profile February 26th, 2009 - 8:58 am10:09:13.85

Okay, so this builds on the research from problem #1. Using the same approach:

X = amount of time elapsed

Hour hand moves 1/12 of a “hash-mark” on the face in one minute

Minute hand moves 1 “hash-mark” on the face in one minute

Hour hand distance traveled = (1/12)X

Minute hand distance traveled = (1)X

At 10:00, the hour hand is pointed directly at “10″ and the minute hand is pointed directly at “12″, so we need to calculate at what point the minute hand will have moved to within the same distance from the “2″ as the hour hand has moved away from the “10.” Since there are 10 hash-marks between the “12″ and the “2″, the formula would be:

(1/12)X = 10 – (1)X

X = 9.2308 minutes, or 9 min., 13.85 sec.

Exact angles need not be calculated.

deejam| Profile February 26th, 2009 - 9:02 amwhenever the hands are superimposed, their angles are same…

SharonH| Profile February 26th, 2009 - 9:18 am11:04:30 and/or 9:14:30 without elaborate calculation

michaelc| Profile February 26th, 2009 - 9:55 amFrom the picture the time is 10:10::31, which I don’t think is the question that is asked.

It looks like the question is what is the precise time when the hour hand and the minute hand are at equal angles from 12:00, completely ignoring the second hand.

If we solve for some short time after 10:00, we get an equation like this…

m/60 = 1 – ( 10/12 + (m/60)/12)

Solving for m you see m = 120/13.

That time is ~10:09::14.

This is not unique, you could have other angles that are equal from 12:00, and could solve them readily by substituting the desired hour (h) into the equation.

m/60 = 1 – ( h/12 + (m/60)/12)

I think we’re wanting the one right after 10:00 however based upon the picture. So..

~10:09::14

joe| Profile February 26th, 2009 - 10:01 amMy effort:

Dividing the clock face (360 deg) by 60 minutes, each minute equates to 6 degrees. So this is how many degrees the minute hand does per minute.

In the same time, the hour hand which travels at 1/12th the speed (and distance) will therefore be 0.5 degrees. ( 1/12th of 6)

For this problem the degrees from the hour hand to the 12 and from the 12 to the minute hand will be equal and around the 10 oclock mark as the advertisement)

Say then that the minute hand travel 6degrees x minutes, m

The hour hand will have travelled 0.5degrees x the 10 hours x 60minutes + 0.5 degrees per minutes, m.

Minute hand: 6m

Hour hand: 0.5x10x60 + 0.5 m

The hour hand will have then: 360 -(0.5x10x60 + 0.5m) degrees to go till the 12 oclock mark

The minute hand 6m degrees past the 12oclock mark.

Equating these two angles

360 – ( 0.5x10x60 + 0.5m) = 6m

Solving , m = 9.230769923

So these are the minutes past 12 which marks the minute hand.

So the time will be 10:92307

OR 10: 09 13.8 secs approximately.

SharonH| Profile February 26th, 2009 - 10:58 amOops…a few more…

10:04 also

12:49

1:50 and 2:45

again, without elaborate calculation

jdesi15564| Profile February 26th, 2009 - 1:01 pm10:09:50:48

Use least squares to get slope/intercept eq. Pick a starting point for minutes, say 8 min. At 8 min after the hour, min hand is @ 48°. We know min hand moves 6°/min = 0.1°/sec. For Hr hand, start at 10:00. Angel is 60°. The two equations are a) for hr hand y=-0.0083x+60. For minute hand y=0.1x+48. Set them = to eachother and solve. x=110.803 seconds. Since we started the minutes at 8 add 110.803 seconds. We also started the hr at 10. so you get 10:09:50:48.

lograh| Profile February 26th, 2009 - 1:04 pmRight, this one is worded like it wants a slightly more precise answer than my last go. Again, I’ll be assuming the movement is smooth. Also, I’ll be doing all the numbers in degrees to make the calculations easier, rather than radians.

At the “X” mark, the hour hand makes an even 60 degree angle with the “XII” mark. Also, for each minute that passes, that angle shrinks by .5 degrees, to eventually hit a 30 degree angle at the “XI” mark. So we can describe the angle between the hour hand and the “XII” mark with the simple formula of “60-(.5*m)” where ‘m’= minutes since the hour. This being in degrees and only valid for times between 10:00 and 11:00 (but we only care about the solution within that range anyhow, even though there are solutions for every hour, since the prompt said “time shown in the picture”.

Moving on, we similarly describe the angle the minute hand makes with the “XII” mark in terms of “minutes since the hour” and come up with “6*m”, 360 degrees to the face, 60 minutes to the hour divide the former by the latter and we get 6 degrees per minute.

Setting the two equal and solving for minutes we get :

6m = 60-(.5m) -> 6m+.5m = 60 -> m(6+.5) = 60 -> (6.5)m = 60 -> m = 60/(6.5) = 9.23076…

So we see that at roughly 10:09:13.84 the angle between the hour hand and the “XII” mark will be equal to the angle between the minute hand and the “XII” mark. Alternatively, it could also be 22:09:13.84, but since I go to bed around 21:00 I’ll stick with the earlier answer (otherwise I’m up way too late and really should stop whatever it is I’m doing).

Unless I botched something, that is. Could be it really is 22:09 and I’m getting tired and making mistakes.

As an aside, looking carefully at the picture one observes that the hour hand is between the 10 and 11 marks, but the minute hand is between the 2 and 3 marks, or at best directly on the 2 mark. Hence the angles they form aren’t precisely equal. But if they were, it’d be just shortly after 10:09.

fuzzy| Profile February 26th, 2009 - 2:29 pmApproximately 10:09:15.

diegote| Profile February 26th, 2009 - 3:34 pmInstead of angles I used clock mins, to easier the calculation.

This happens once per hour, so is right to say that the sum of angles between each hand and 12:00 is 15 mins.

And the speed of each hand is:

vm=1 (one lap per hour or 60 min marks after 60 minutes transcurred)

vh=1/12 (one lap after 12 hours or 60 min marks after 720 minutes transcurred)

So:

hour hand angle:

5mins + 5 * T

mins hand angle:

1 * T

Putting all together

5mins + 5 * T + T = 15

T = 9,23 minutes = 9 mins 14 secs

Roel| Profile February 26th, 2009 - 7:08 pmLet’s call the angle the hour handle travelled from the 12 o’clock position HOUR, and the angle the minute hand travelled MINUTE.

Or said a bit differently, the HOUR handle is 360 – HOUR degrees away from 12 o’clock.

About the minute handle. A little after 10 o’clock, the minute handle made 10 complete turns and a little more. So substracting those 10 turns from the minute handle, it’s clear that for the hands to be symmetrical

360 – HOUR = MINUTE – 360.10

But since the minute handle turns 12 times faster than the hour handle, this becomes:

360 – HOUR = 12.HOUR – 360.10

Or put more simply:

HOUR = 360.11/13

Now, the hour handle goes with a speed of 360 degrees per 12 hours. This means that after some given TIME, the hour handle is in position

HOUR = TIME . 360/(12.60.60)

From the first equation we know the position of the hour handle, so the formule becomes:

360.11/13 = TIME . 360/(12.60.60)

Or better

TIME = 11.12.60.60/13 = 36553,84… seconds

So converting these seconds in hours, minutes and seconds becomes:

TIME = 10h (36000s) + 9min (540s) + 13.84…s

Mashplum| PUZZLE MASTER | Profile February 26th, 2009 - 9:18 pm10:09:13 (and 11/13 seconds)

brianu| Profile February 26th, 2009 - 9:50 pm10:09:12

aaronlau| Profile February 26th, 2009 - 10:16 pmInteresting point. Googled and found this “TIMEX SAYS THE HANDS ON TIMEPIECES ARE PLACED AT TEN-TEN SO THE COMPANY LOGO ON THE FACE WILL BE FRAMED AND NOT BLOCKED BY THE HANDS. THE INDUSTRY STANDARD USED TO BE EIGHT-TWENTY BUT THAT LOOKED TOO MUCH LIKE A FROWN AND CREATED AN UNHAPPY LOOK. TIMEX SAYS IN ITS ADS, THE CLOCK HANDS ARE PLACED AT TEN-NINE AND THIRTY SIX SECONDS, EXACTLY.”

60 – 360t/720 = 360t/60

=> t = 60/6.5 = 9.23 min

Therefore, time is 10 o’clock, 9 minutes and 13.846 seconds

zenitth| Profile February 26th, 2009 - 11:13 pmMy completely uneducated and non-technical GUESS is:

10:08:45

Migrated| Profile February 27th, 2009 - 6:28 amAfter some ponderance and calculations late at night…I got the answer of 10:10 and 17 seconds (rounded up). Or 10:10; and 16 and two-thirds seconds.

Benson| Profile February 27th, 2009 - 8:04 am2:06

APEX.JP| Profile February 27th, 2009 - 2:17 pmThe real exact time shown in the picture would be 1:50:46.15

ludiavolo| Profile February 27th, 2009 - 7:35 pmStarting the problem from 10 o’clock, the (smaller) angle between the hour hand and 12 o’clock can be expressed as:

360-(300+360*t/12) = 60-360*t/12

The minute hand makes an angle relative to 12 o’clock that can be expressed as:

360*t/1

The equation that needs to be solved is therefore:

60-360*t/12 = 360*t/1

whose solution is t = 0.153846 hrs

doing some conversions to minutes and seconds the correct time is approximately

10:09:14

brianu| Profile February 27th, 2009 - 8:55 pmIgnore the previous post.

Assuming the only hour is concerning the ad placement (10:00):

Every minute the minute had moves 6 degrees

Every minute the hour hand moves 0.5 degrees

The amount of minutes the minute hand moves during the ten o’clock hour can be expressed 120-x, if x is the number of minutes into the hour.

The degrees of rotation are the same so:

(120-x)(0.5 degrees)=(x)(6 degrees)

Solve for x and we get 9.230769, the amount of minutes into the hour, or 9 minutes and 13.846153 seconds

The time is 10:09:13:85

To figure other times the equation would have to solve for degrees of rotation which is 55.3846 degrees from top or bottom center.

brianu| Profile February 27th, 2009 - 8:56 pmsorry again, 120-x is the amount of minutes the hour hand is from twelve

bizarette18| PUZZLE MASTER | Profile February 28th, 2009 - 7:28 amThe hour and minute hands would be mirror symmetrical at 9 and 3/13 minutes past 10. But personally, I find rotational symmetry (especially 3x) more aesthetic than mirror symmetry, so I prefer the time shown on the Rolex: 10 minutes and 31 seconds past 10. You know, 120 degree angles are relaxed without being too relaxed. And that slight deviation to clockwise gives just a little bit of leniency and flexibility.

RK| Founder | Profile March 2nd, 2009 - 1:11 pmThe Brain from Spain tells me the correct answer to this one is:

10 h 9 m 13 s 11/13

3rd will be up soon…