Difference between revisions of "2001 USAMO Problems/Problem 1"
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=== Solution 2 === | === Solution 2 === | ||
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+ | Similar to the above solution, no color can appear 4 times or more. | ||
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+ | We can use PIE. Let <math>S_1,S_2,\cdots ,S_8</math> be the colors in each of the <math>8</math> boxes. Then <math>n=|\cup_{i=1}^8S_i|</math>. By PIE we know that<cmath>|\cup_{i=1}^8S_i|=\sum_{i=1}^8 |S_i|-\sum_{1\leq i<j\leq 8}|S_i\cap S_j|+\sum_{1\leq i<j<k\leq 8}|S_i\cap S_j\cap S_k|-\sum_{1\leq i<j<k<l\leq 8}|S_i\cap S_j\cap S_k\cap S_l|\cdots.</cmath>Note however that no color can appear 4 times or more, so all the items after <math>\sum_{1\leq i<j<k\leq 8}|S_i\cap S_j\cap S_k|</math> must be equal to 0, so<cmath>|\cup_{i=1}^8S_i|=\sum_{i=1}^8 |S_i|-\sum_{1\leq i<j\leq 8}|S_i\cap S_j|+\sum_{1\leq i<j<k\leq 8}|S_i\cap S_j\cap S_k|.</cmath>Now note that for each <math>i</math>, <math>|S_i|=6,</math> so <math>\sum_{i=1}^8 |S_i|=48.</math> Also no pair of colors appears twice, so for any <math>i,j, |S_i\cap S_j\leq 1|,</math> and so <math>\sum_{1\leq i<j\leq 8}|S_i\cap S_j|\leq \binom82=28.</math> Thus, <math>n\geq 48-28+\sum_{1\leq i<j<k\leq 8}|S_i\cap S_j\cap S_k|.</math> Given that there are <math>n</math> colors, to minimize the number of colors in 3 boxes means each color is in either 2 or 3 boxes, and since there are 48 total spots, this means there are <math>48-2n</math> colors in 3 boxes and <math>3n-48</math> colors in 2. Thus <math>n\geq 20+48-2n,</math> so <math>3n\geq 68</math>, and <math>n\geq23</math> and we are done. | ||
=== Solution 3 === | === Solution 3 === |
Revision as of 13:55, 3 June 2021
Problem
Each of eight boxes contains six balls. Each ball has been colored with one of colors, such that no two balls in the same box are the same color, and no two colors occur together in more than one box. Determine, with justification, the smallest integer for which this is possible.
Solution 1
We claim that is the minimum. Consider the following construction (replacing colors with numbers) which fulfills this:
Suppose a configuration exists with .
Suppose a ball appears or more times. Then the remaining balls of the boxes must be distinct, so that there are at least balls, contradiction. If a ball appears or more times, the remaining balls of the boxes must be distinct, leading to balls. The fifth box can contain at most four balls from the previous boxes, and then the remaining two balls must be distinct, leading to , contradiction.
However, by the Pigeonhole Principle, at least one ball must appear times. Without loss of generality suppose that appears three times, and let the boxes that contain these have balls with colors . Each of the remaining five boxes can have at most balls from each of these boxes. Thus, each of the remaining five boxes must have additional balls . Thus, it is necessary that we use balls to fill a grid by the same rules.
Again, no balls may appear times, but by Pigeonhole, one ball must appear times. Without loss of generality, let this ball have color ; then the three boxes containing must have at least balls, contradiction.
Therefore, is the minimum.
Solution 2
Similar to the above solution, no color can appear 4 times or more.
We can use PIE. Let be the colors in each of the boxes. Then . By PIE we know thatNote however that no color can appear 4 times or more, so all the items after must be equal to 0, soNow note that for each , so Also no pair of colors appears twice, so for any and so Thus, Given that there are colors, to minimize the number of colors in 3 boxes means each color is in either 2 or 3 boxes, and since there are 48 total spots, this means there are colors in 3 boxes and colors in 2. Thus so , and and we are done.
Solution 3
Let be the number of balls which are the same color as the ball in box (including that ball). For a fixed box , , consider the sums For each fixed , since no pair of colors is repeated, each of the remaining seven boxes can contribute at most one ball to . Thus, . It follows by the convexity of that is minimized when one of the is equal to 3 and the other five equal to 2. Hence . Note that Hence there must be 23 colors. The construction for is above.
See also
2001 USAMO (Problems • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.