Kvant Math Problem 6

Consider a standard 12-hour analog clock with an hour hand and a minute hand.

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Problem

You are given a clock. How many positions of the hands exist for which it is impossible to determine the time without knowing which hand is the hour hand and which is the minute hand?

(It is assumed that the position of each hand can be determined exactly, but observing their movement is not allowed.)

Exploration

Consider a standard 12-hour analog clock with an hour hand and a minute hand. The minute hand moves uniformly around the circle, completing a full rotation in 60 minutes. The hour hand moves uniformly as well, completing a full rotation in 12 hours.

Positions where the hands coincide, such as both at 12, are clearly symmetric: swapping the hands does not change the visual configuration. For other positions, one might imagine a few scenarios: when the hands are directly opposite each other, when the angle between them is 30 degrees, 60 degrees, and so on. To explore systematically, assign coordinates: let the minute hand point to $m/60 \cdot 360^\circ$ for minute $m$, and the hour hand point to $h/12 \cdot 360^\circ + (m/60) \cdot 30^\circ$ for hour $h$.

The problem asks for configurations where the time cannot be determined without knowing which hand is which. That occurs exactly when swapping the hands results in a valid time: the position of one hand can be interpreted as the hour hand, and the other as the minute hand, producing a different—but still valid—hour and minute. We should enumerate the times that satisfy the equation $$ \text{angle of hour hand} = \frac{360}{60} \cdot \text{minutes assigned from minute hand}, \quad \text{and} \quad \text{angle of minute hand} = \frac{360}{12} \cdot \text{hours assigned from hour hand} + \frac{30}{60} \cdot \text{minutes from hour hand}. $$ Exploring small integer multiples suggests a regular pattern. For example, 12:00 and 6:30 immediately work: swapping hands produces a valid time. Some other candidate times include 1:05 5/11, but we must check carefully whether the swapped hands produce integer hours and minutes. The crucial step is solving a Diophantine-like equation connecting $h$ and $m$ to $h'$ and $m'$ under the swap.

Problem Understanding

The problem asks to count the number of distinct hand positions for which the observer cannot identify which hand is the hour hand and which is the minute hand. This is a Type A problem: we must classify all positions satisfying this ambiguity. The core difficulty is solving for all $(h, m)$ in integer hours and minutes where swapping hands produces another valid $(h', m')$ with $0 \le h' \le 11$ and $0 \le m' \le 59$. Intuitively, ambiguity arises only when the hands are either exactly opposite each other or coincide, producing reflection symmetry across the line $y=x$ if we map hours to minute positions. Candidate times include 12:00, 6:30, and their symmetric counterparts.

Proof Architecture

Lemma 1: If a clock position is ambiguous, the angle between the hands must be a multiple of $30^\circ$. This is because only multiples of 30 degrees correspond to integer hours.

Lemma 2: Let $h$ and $m$ be the hour and minute. Ambiguity occurs exactly when swapping hands $(h,m) \mapsto (h', m')$ satisfies $$ \frac{360}{12}h + \frac{30}{60}m = \frac{360}{60}m', \quad \frac{360}{60}m = \frac{360}{12}h' + \frac{30}{60}m'. $$ Solving these yields $m = 12h/11$ or $m = (12h + 6)/11$, the only integer solutions give the possible positions.

Lemma 3: Only integer solutions with $0 \le m \le 59$ and $0 \le h \le 11$ count as valid ambiguous positions.

The hardest step is Lemma 2, solving the fractional system to find integer solutions; a careless step could miss or introduce spurious times.

Solution

Denote by $H$ the hour hand and $M$ the minute hand. Let $h \in {0,1,\dots,11}$ and $m \in {0,1,\dots,59}$ be the hour and minute. Let $A_H$ be the angle of the hour hand from 12: $$ A_H = 30h + \frac{m}{2}, $$ and let $A_M$ be the angle of the minute hand: $$ A_M = 6m. $$ Suppose swapping the hands yields another valid time $(h', m')$. Then $$ A_H = 6m', \quad A_M = 30h' + \frac{m'}{2}. $$ Substituting $A_H = 30h + m/2$ and $A_M = 6m$, we have the system $$ 30h + \frac{m}{2} = 6 m', \quad 6m = 30 h' + \frac{m'}{2}. $$ Multiply the first equation by 2: $60 h + m = 12 m'$, thus $m' = 5h + m/12$. Multiply the second equation by 2: $12 m = 60 h' + m'$, thus $h' = (12m - m')/60$. For integer solutions, $m/12$ must be integer, so $m$ divisible by 12. Then $m = 0,12,24,36,48$. Testing each case:

For $m=0$, $m' = 5h$, $0 \le m' \le 59$ gives $h = 0,1,2,3,4,5,11$. Check corresponding $h'$ integer:

  • $h=0$, $m'=0$, $h' = (0-0)/60 = 0$, valid.
  • $h=1$, $m'=5$, $h' = (12-5)/60 = 7/60$, not integer.
  • $h=2$, $m'=10$, $h' = (24-10)/60 = 14/60$, not integer.
  • $h=3$, $m'=15$, $h' = (36-15)/60 = 21/60$, not integer.
  • $h=4$, $m'=20$, $h' = 4/3$, invalid.
  • $h=5$, $m'=25$, $h' = 11/12$, invalid.
  • $h=11$, $m'=55$, $h' = (132-55)/60 = 77/60$, invalid.

Thus only $h=0$, $m=0$ works: 12:00.

For $m=12$, $m' = 5h + 1$, $0 \le 5h+1 \le 59$, $h=0,1,2,3,4,5,11$:

  • $h=6$, $m'=31$, etc. Then $h' = (1212 - m')/60 = (144-31)/60 = 113/60$, not integer. Only $m=30$ yields $m/12 = 2.5$, impossible. Then $m=36$, $m/12=3$, $m'=5h+3$. $h=0$ gives $m'=3$, $h'=(1236 -3)/60= (432-3)/60=429/60$, not integer. Only $h=6$, $m'=33$, $h'=(1236-33)/60=(432-33)/60=399/60=6.65$, not integer. Only $m=30$ with $h=6$ works: then $m'=56 + 2.5 = 32.5$, not integer. Actually only 6:30 works if you check exact fractional calculations. Verify 6:30: $h=6$, $m=30$, $A_H=306 + 30/2=180+15=195$, $A_M=630=180$, swap: $A_H=195=6m'$ gives $m'=32.5$, non-integer. Wait, we need integer minutes; only