Tick-Tock Triangles

A clock has a 3-inch hour-hand and a 4-inch minute-hand. A is the point at the tip of the hour hand, B is the point at the tip of the minute hand, and O is the point at the centre of the clock.
Given that the triangle AOB has integer length sides, find the probability that the triangle is (i) right angled, (ii) isosceles.

If the length of the hour hand is h and the length of the minute hand is m, find the probability that triangle, AOB, with integral length sides, is isosceles. (Assume m > h).

Originally posted by THUDandBLUNDER
A clock has a 3-inch hour-hand and a 4-inch minute-hand. A is the point at the tip of the hour hand, B is the point at the tip of the minute hand, and O is the point at the centre of the clock.
Given that the triangle AOB has integer length sides, find the probability that the triangle is (i) right angled, (ii) isosceles.

If the length of the h ...[text shortened]... nd the probability that triangle, AOB, with integral length sides, is isosceles. (Assume m > h).

The third side of the triangle is equally likely to take the values 1,2,...,m+h-1 (assuming we insist AOB is a proper triangle). So (i) has probability 1/6, (ii) has probability 1/3. In general with hands of length m and h, m > h, the probability that AOB is isosceles is 2/(m+h-1).

Acolyte, I have never seen much use for degenerate triangles.
Now that Bush has been re-elected, we must be on-guard against such subversive UNAMERIKAAAHN tendencies.

😀

Given that you seem to feel the same way, your answers are wrong. 😛

Originally posted by Acolyte
In general with hands of length m and h, m > h, the probability that AOB is isosceles is 2/(m+h-1).
For example, m = 4, h = 3 gives what?

Oh, I thought you had 2/(m-h-1)

But its still wrong.

If I've read it properly I agree with Acolyte. By triangle inequality, there are m+h-1 integer values the other side can take, of which two (namely m and h) give an isoceles triangle and at most 2 (namely (m^2+h^2)^1/2, (m^2-h^2)^1/2) give a right triangle, (m can be a leg or a hypotenuse) but in this case only one of them works. Finally each is equally likely.

Originally posted by royalchicken
If I've read it properly I agree with Acolyte.

Two wrongs don't make a right. 😛

In working on this, I realized that I am confused - partially cuz I'm an idiot, and partially, b/c I don't understand how you can get a probability out of it.

First thing I was trying to do was set boundaries - like - 1 hr time increment - b/c that seems like the logical place to start.

Second, I thought - hey, there are only two times within the hour that the hour and minute hand will form a 90 degree angle, so that should be 2/60 or 1/30. And that's where I seem to get lost. if anyone has a clock, similar to mine, the hour hand travels a certain distance every 60 minutes, as well as the minute hand travels the same distance in 300 seconds. (I'm referring to the arc length - I'm sure it's obvious). So, you have a velocity of 30 deg/sec for the minute hand and 30/360 deg/sec for the hour hand. That's only if it's a continually moving entity, but my minute hand only moves after the second hand has gone for 60 seconds, it moves 6 degrees every 60 seconds, and it takes one second on average to move the 6 degrees. So the problem I am seeing, is the time calculation at which the hour and minute hand actually form the 90 degree angle and for how long.

Then I thought - hey I'm making this problem way too hard, b/c there will only be 2 times each hour when this will happen, but the hands do move, and there are a certain number of times within the hour. But how many times are there?

Someone please shed some light for me!

one more thing - if you say the sides are of integer length - do you mean the hypotenuse is included as well.


If so, then I realize where I made my mistake. Reading the problem... 🙁

Originally posted by ShadowGnu
one more thing - if you say the sides are of integer length - do you mean the hypotenuse is included as well.
🙁

Yes.

Assuming this works like a normal clock, then the second hand moves (instantly) in 6 degree increments, the minute hand moves in 1/10th's of a degree increments and the hour hand moves in 1/120th's of a degree increments. Therefore we cannot form a right angled triangle of sides 3, 4 and sqrt(7) because the angle between the hands would be
acos(3/4), which is not a multiple of 120th's of a degree. Therefore the right angled triangle must have sides 3,4 and 5 and there must be exactly 90 degrees between the hour and the minute hand.

There can only be an integral number of degrees between the hour and the minute hand every 120 seconds, and the number of degrees difference generally increases by exactly 11 degrees every 120 seconds but gets shifted every hour when the minute hand jumps from 360 degrees to 0. A bit of experimentation shows that the only times when the angle is exactly 90 degrees are 3:00 and 9:00, therefore the odds that the hands form a right angled triangle are 2/43200 = 1/21600, slightly less than acolyte's estimate.

Originally posted by iamatiger
Assuming this works like a normal clock, then the second hand moves (instantly) in 6 degree increments, the minute hand moves in 1/10th's of a degree increments and the hour hand moves in 1/120th's of a degree increments. Therefore we cannot form a right angled triangle of sides 3, 4 and sqrt(7) because the angle between the hands would be
acos(3/4), w ...[text shortened]... ands form a right angled triangle are 2/43200 = 1/21600, slightly less than acolyte's estimate.

What happens if the minute hand and hour hand move continuously, and not in discrete increments?

Originally posted by The Plumber
What happens if the minute hand and hour hand move continuously, and not in discrete increments?

It's a different puzzle then - I have never seen a clock that worked like that.

Originally posted by iamatiger
Assuming this works like a normal clock, then the second hand moves (instantly) in 6 degree increments, the minute hand moves in 1/10th's of a degree increments and the hour hand moves in 1/120th's of a degree increments. Therefore we cannot form a right angled triangle of sides 3, 4 and sqrt(7) because the angle between the hands would be
acos(3/4), w ...[text shortened]... ands form a right angled triangle are 2/43200 = 1/21600, slightly less than acolyte's estimate.

Excellent deduction iamatiger (assuming we're not all missing something)... it seems both royalchicken and I forgot how real clocks 'tick'. 😳

Originally posted by iamatiger
It's a different puzzle then - I have never seen a clock that worked like that.

I'm no clock expert, just a simple worker of pipes, 🙂 but when I stare at my watch the minute hand seems to move fairly continuously and not in discrete increments (I didn't stare long enough to see if that is the case with the hour hand as well). Of course, that's a watch and not a clock, and I am afterall not a clockmaker....

Originally posted by The Plumber
I'm no clock expert, just a simple worker of pipes, 🙂 but when I stare at my watch the minute hand seems to move fairly continuously and not in discrete increments (I didn't stare long enough to see if that is the case with the hour hand as well). Of course, that's a watch and not a clock, and I am afterall not a clockmaker....

If you had a watch with a really long minute hand, and a really really long hour hand, then you would see that the move in jerks. There is no practical way to make a watch or clock where the hands move smoothly, although one might reduce the jerks to less that 1 seconds worth - in which case the second hand would move faster than once per second (but in jumps of less than a second of course), and the minute and hour hands would then move in proportionally smaller steps.