There are four apples on a plate, of weights $ 600,400,300,$ and $ 250$ grammes. Two pupils take an apple each (one by one) and start eating simultaneously. A pupil can take another apple only after he has eaten the previous apple. It is assumed that they equaly fast (measured in grammes per second). How should the first pupil do in order to eat as large amount as possible?
Problem
Source: Ukraine 1997 grade 9
Tags: combinatorics unsolved, combinatorics
Mij
18.07.2009 21:21
If the first pupil starts by eating the 600 g apple, then the second pupil can eat the 250 g apple and the 300 g apple and start eating the 400 g apple by the time the first pupil is done. The first pupil will only eat 600 g of apples. If the first pupil starts by eating the 400 g apple, then the second pupil can eat at most one of the lighter apples and start to eat the 600 g apple by the time the first pupil is done. The second pupil will start with the 300 g apple. Then, by the time the second pupil is done withe 600 g apple, the first will have to eat the 250 g apple. The first pupil will have eaten 650 g of apples. If the first pupil starts by eating the 300 g apple, then the second pupil can eat the 250 g apple and then start to eat the 600 g apple by the time the first pupil is done. The first pupil will then eat the 400 g apple, ending up with 700 g of apple eaten. If the first pupil starts by eating the 250 g apple, then he can finish eating it before the second pupil finishes his apple. If the second pupil eats the 600 g apple, the first pupil can eat the 300 g and 400 g apples, ending up with 950 g. If the second pupil eats the 400 g or the 300 g apple, then the first pupil can start eating at most one of the apples left before the second pupil finishes. The first pupil will eat the 600 g apple, leaving the other to the second pupil. The first pupil will have eaten 850 g of apple. The second pupil will either eat the 400 g or 300 g apple because that will minimize the amount of apple the first pupil eats. The first pupil will, under perfect play by the second pupil, eat 850 g of apple if he starts with the 250 g apple. If he starts with a different apple, the second pupil can make it so the first one eats less than 850 g. Thus the first pupil must eat the 250 g apple first. By perfect play, the second pupil will eat either the 300 g or the 400 g apple, so the first pupil must then eat the 600 g apple. The first pupil will end up with 850 g, the maximum possible amount under perfect play by the second pupil.