Spring 2018, problem 66

By the pigeon hole principle (PHP) we know there are at least 17 balls of the same color among the 65 (16 * 4 + 1 = 65). WLOG, assume there are 17 red balls. Since there are 2 boxes, again by PHP we know one box must have at least 9 red balls. Among these 9, assume no three were the same color (if they were we're done). Take 5 from the 9. Then 2 of them are of the same radius, r1. Take these 2 and set them aside and then grab 2 more balls from the remaining pile of 4 to replace them. Again we have 5 balls so 2 of them must have the same radius, r2. Take these 2 and set them aside, again replacing them with 2 more from the remaining pile of 2. Having 5 balls we know 2 have the same radius, r3. Take these and set them aside and replace them with the last 2 balls. Since we have 5 we know 2 have the same radius, r4. Our 9 red balls have radii r1, r1, r2, r2, r3, r3, r4, r4, r5 where r1, r2, r3, r4, r5 are all distinct values. However r1, r2, r3, r4, r5 is a group of 5 balls which contradicts the condition that 2 of them must have the same radius. Therefore our assumption is wrong and there must be three of the same size radius.

We have $65$ balls and $4$ colors. Then $65\div 4=16.25$ ,and so there is a set $C$ with at least $17$ balls of the same color. Let $c=|C|\geq 17$. On the other hand, in each set of balls with the same color there can only be, at most $ 4 $ different radius, because " any five balls of the same color, at least two of them will always be of the same size(radius)" . Let $r$ be the number of different radius in $C$.If all balls have the same radius consider $r=1$. As $r\leq 4$ then $c\div r \geq 17\div 4=4.25$, we have that at least 5 balls of the set $C$ have the same radius. So at least $3 $ of these balls are in the same box.