Distance of time

You started off with

You ended with

Which do you wish to maximize?

Calculate the distance between the tips of the hands at the moment they are moving the fastest toward each other.

So I need to find the distance between the tips of the hands when they are moving the fastest toward each other.

Ok so I have been looking a bit at this, and say that both hands are moving. Well they are both always moving towards each other, and so I'm just not sure, but I think it has something to do with their angle they form...

they are closing the distance the fastest when the tangent to the hour hand circle intercepts the minute hands tip, so the tangent that intercepts the hour and minute hand is what I want

Not true - they'd eventually meet.

At 12:00, the minute hand will start moving away from the hour hand. It reaches the greatest distance a bit past 12:30. Then, it starts moving toward the hour hand.

At 12:30 the hands are still moving apart, don't forget the hour hand is going to be half way between 12 and 1 by then, you are right shortly after 12:30. I haven't an idea on how to figure out this formula or answer though.

There is probably a clever solution involving vectors (which may be where you were headed). I'll give that approach some thought later.

In the meantime, here's an approach that will yield a solution (you can approximate it in Excel).

t = 0 to 2*pi*12/11 radians
Mx = 4cos(t)
My = 4sin(t)
Hx = 3cos(t/12)
Hy = 3sin(t/12)
D(t) = sqrt((Mx-Hx)^2 + (My-Hy)^2)
= sqrt((4cos(t)-3cos(t/12))^2 + (4sin(t)-3sin(t/12))^2)

We want to minimize D'(t) (derivative of D(t)). That represents the position when the distance is decreasing at the fastest rate.

Therefore, we want D''(t) = 0 (second derivative of D(t)). There will be two values of t that satisfy this equation. We want the one where D'(t) is negative.

Once we know t, we can compute the positions of the hands and the angle between them.

By the way, 6.0 < t < 6.2

They also tell me:

well ... you're thinking much too complicated ... just think of my hint of _simple_ maths.
with this hint, there are just a few possible solutions:
with which angles and only simple math (no sin, cos or tan needed ... in germany you learn it after 7 years school) can the triangle be calculated?
and then try _every_ solution!

Well, it's not clear to my why the angle opposite the minute hand is a right angle, but it is.

Therefore, the distance is sqrt(4^2 - 3^2) = sqrt(7)

Very nice, you are right. And for the explaination:

At any time, the distance between the two tips is

Suppose that the clock is rotating counter-clockwise at the speed of the minute hand. Then the minute hand is effectively stationary while the hour hand is still moving at a constant speed. The point at which the hour hand's tip is moving most quickly towards the minute hand's tip is the same as the point at which the hour hand's tip is moving directly towards the minute hand's tip. If you draw a circle to represent the hour hand's tip as it moves, then you just need to draw a tangent to the circle which passes through the point representing the minute hand's tip. Now you have a right-angled triangle (with the clock's centre)

sqr(X) = sqr(A) + sqr ( - 2 A B cos (alpha) with A and B the lengths of the tips and alpha the corner between the hands.
This means that X = sqrt ( 25 - 24 cos (alpha)) which is an equation in one parameter, hence easy to derive.

Then just find the point of maximum accelaration (deriving twice) and solve it to cos (alpha)

Yes indeed, no need to actually calculate cos (alpha), cause you get an equation of the second order in cos(alpha), which tells you that cos(alpha) = 3/4

Substitute this in the equation above and you get X = sqrt (25 - 18 ) = sqrt(7)

After the nice explanation that leads to a right triangle, I don't understand why the rest of it is needed. Pythagoras give us a very straightforward calculation.

It's what the guy said that wrote this riddle . Just thought it may be interesting. Thanks again though markr, you are on top of your game.