Htykuut

I have seen this question on several different websites and mathematics journals, but I haven't seen any concrete answers. For instance, some sources say it will happen only twice, while others say it will happen 22 times. Upon researching it, it seems as if all three hands will line up some amount higher than 2 times, but I am unsure of how many times it would occur.

This happens exactly once every $12$ hours. If $$tmapsto e^{it}qquad(0leq t< 2pi)$$ describes the position of the hour hand on the dial $S^1$ during the $12$ hour interval beginning at 12.00 noon then $tmapsto e^{12it}$ describes the position of the minute hand, and $tmapsto e^{720it}$ describes the position of the second hand. The hour hand and the minute hand coincide when $e^{it}=e^{12it}$, i.e., when $e^{11it}=1$. This is the case when $t={2pi kover11}$ $(0leq k<11)$. Similarly, the hour hand and the second hand coincide when $t={2pi lover 719}$ $(0leq l<719)$. As ${rm gcd}(11,719)=1$ there is no $tin[0,2pi[>$ fulfilling both conditions other than $t=0$.

It follows that all three hands will line up exactly two times every $24$ hours.

The answer depends on which hands' motions are continuous and which are discrete, and if so, how they are discretized. For simplicity, I'll address the case that all motions are continuous, so that all three hands have constant velocity.

Per 12-hour period ($1$ rotation of the hour hand), the minute hand rotates $12$ times, so over this time the angular distance between the hour and minute hands is zero $12 - 1 = 11$ times. (This is where the figure of $22$ comes from: This is the number of times two hands align per 24 hours.) These 11 times are equally spaced, and include $0:00$, so they occur precisely when the hour hand has completed $frac{k}{11}$ of a rotation, $k = 0, ldots, 10$.

The second hand moves $12 cdot 60$ times as fast as the large hand, so in the time between successive alignments of the hour and minute hands, the second hand has completed $frac{720}{11} = n + frac{5}{11}$ rotations for some integer $n$, so the accumulated angular distance between the second hand and the common angle of the other two hands has increased by $frac{5}{11} - frac{1}{11} = frac{4}{11}$ of a rotation. It follows from the fact that $4$ and $11$ are coprime that the only times at which all three hands are aligned is when they all point upward. Each day, this happens only at $0:00$ and $12:00$.

We can answer this by determining the times when the hour and minute hands coincide, then checking whether the second hand will also match.

In $12$ hours, the hour hand rotates once while the minute hand rotates $12$ times. Therefore the minute hand does $12-1 = 11$ rotations relative to the hour hand in this time. So the interval between the times when it overtakes the hour hand is $frac{12}{11}$ hours. ($=1$h $5$m $27 frac{3}{11}$s).

All three hands are aligned at $12:00:00$. Adding integer multiples of $1:5:27frac{3}{11}$ to this gives the times when the hour and minute hands are aligned (in $12$ hour format) as

$12:00:00$
$1:05:27frac{3}{11}$
$2:10:54frac{6}{11}$
$3:16:21frac{9}{11}$
$4:21:49frac{1}{11}$
$5:27:16frac{4}{11}$
$6:32:43frac{7}{11}$
$7:38:10frac{10}{11}$
$8:43:38frac{2}{11}$
$9:49:05frac{5}{11}$
$10:54:31frac{8}{11}$

and it turns out that the seconds only match the minutes at $12:00:00$. [Edit: Travis's answer shows why this is.]

So the answer is: all three hands coincide twice a day.