wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> Workmates' Lunchtime
        
(Message started by: THUDandBLUNDER on Apr 8^th, 2005, 2:49am)

Title: Workmates' Lunchtime
Post by THUDandBLUNDER on Apr 8^th, 2005, 2:49am

Two workmates each take one lunch break per day. They arrive at the restaurant independently at (uniformly) random times between 12pm and 1pm, and each stays for exactly x minutes. The probability that either workmate arrives while the other is in the restaurant is 2/5.

If x = a - bc^1/2 find the value of a + b + c
where a,b,c are positive integers and c is indivisible by the square of any prime.

Title: Re: Workmates' Lunchtime
Post by Sjoerd Job Postmus on Apr 8^th, 2005, 3:31am

A start:

awowip:Arrives while other workmate is present
P_{A_awowip} = 1 / 5
P_{B_awowip} = 1 / 5

From this follows that they both stay exactly [hide]12[/hide] minutes. ( A enters while B present + B enters while A present )

Now, to fill in the formula... :(

Modification: Not divisable by the square of any prime. So c != dp², right? but c might be the square of a prime... (same goes for a and b, but I'm more concerned about c)

Title: Re: Workmates' Lunchtime
Post by Grimbal on Apr 8^th, 2005, 9:04am

1. [hideb]
If x and y are the arrival time in hours, then the "collisions" cases look like this:


...../####
.../#####/
./######/.
/#####/...
####/.....

You can join the 2 empty areas to get:


##########
##########
####......
####......
####......

So the collision surface is 1²-(1-x)²
Solving 1²-(1-x)² = 2/5 gives
x² - 2x + 2/5 = 0
or
x = (2 - [sqrt](4-8/5))/2
x = 1 - sqrt(3/5) = 0.225566 hours = 13 min 32 s

[/hideb]

Title: Re: Workmates' Lunchtime
Post by THUDandBLUNDER on Apr 8^th, 2005, 9:07am

on 04/08/05 at 03:31:58, Sjoerd Job Postmus wrote:

A start:
From this follows that they both stay exactly [hide]12[/hide] minutes.

Modification: Not divisable by the square of any prime. So c != dp², right? but c might be the square of a prime.

PSB (and Grimbal?), it is not stipulated that they must leave before 1pm.

If c is a square of a prime then it is divisible by itself, a square of a prime. ::)

Title: Re: Workmates' Lunchtime
Post by Grimbal on Apr 8^th, 2005, 9:17am

If they had to leave before 1pm, they couldn't stay x minutes and arrive uniformly between 12 and 1.
[hide]I only consider arrival times. If the difference between their arrival times is less than x, then they meet.[/hide].

Title: Re: Workmates' Lunchtime
Post by Grimbal on Apr 8^th, 2005, 9:22am

AAAh. Now I understand what was asked.
[hide]x = 1 - sqrt(3/5) hours. = 60 - 12*sqrt(15) minutes[/hide]
a+b+c = [hide]87[/hide]

Title: Re: Workmates' Lunchtime
Post by Eigenray on Apr 8^th, 2005, 12:25pm

(If you want n coworkers to all meet, see [link=http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1088600571]Clustered Points[/link].)