Difference between revisions of "2017 AIME II Problems/Problem 5"
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Note that if <math>a>b>c>d</math> are the elements of the set, then <math>a+b>a+c>b+c,a+d>b+d>c+d</math>. Thus we can assign <math>a+b=x,a+c=y,b+c=320,a+d=287,b+d=234,c+d=189</math>. Then <math>x+y=(a+b)+(a+c)=\left[(a+d)-(b+d)+(b+c)\right]+\left[(a+d)-(c+d)+(b+c)\right]=791</math>. | Note that if <math>a>b>c>d</math> are the elements of the set, then <math>a+b>a+c>b+c,a+d>b+d>c+d</math>. Thus we can assign <math>a+b=x,a+c=y,b+c=320,a+d=287,b+d=234,c+d=189</math>. Then <math>x+y=(a+b)+(a+c)=\left[(a+d)-(b+d)+(b+c)\right]+\left[(a+d)-(c+d)+(b+c)\right]=791</math>. | ||
− | ==Solution 4 ( Short Casework )= | + | ==Solution 4 ( Short Casework )== |
There are two cases we can consider. Let the elements of our set be denoted <math>a,b,c,d</math>, and say that the largest sums <math>x</math> and <math>y</math> will be consisted of <math>b+d</math> and <math>c+d</math>. Thus, we want to maximize <math>b+c+2d</math>, which means <math>d</math> has to be as large as possible, and <math>a</math> has to be as small as possible to maximize <math>b</math> and <math>c</math>. So, the two cases we look at are: | There are two cases we can consider. Let the elements of our set be denoted <math>a,b,c,d</math>, and say that the largest sums <math>x</math> and <math>y</math> will be consisted of <math>b+d</math> and <math>c+d</math>. Thus, we want to maximize <math>b+c+2d</math>, which means <math>d</math> has to be as large as possible, and <math>a</math> has to be as small as possible to maximize <math>b</math> and <math>c</math>. So, the two cases we look at are: | ||
Latest revision as of 13:01, 30 April 2021
Contents
Problem
A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are , , , , , and . Find the greatest possible value of .
Solution 1
Let these four numbers be , , , and , where . needs to be maximized, so let and because these are the two largest pairwise sums. Now needs to be maximized. Notice that . No matter how the numbers , , , and are assigned to the values , , , and , the sum will always be . Therefore we need to maximize . The maximum value of is achieved when we let and be and because these are the two largest pairwise sums besides and . Therefore, the maximum possible value of .
Solution 2
Let the four numbers be , , , and , in no particular order. Adding the pairwise sums, we have , so . Since we want to maximize , we must maximize .
Of the four sums whose values we know, there must be two sums that add to . To maximize this value, we choose the highest pairwise sums, and . Therefore, .
We can substitute this value into the earlier equation to find that .
Solution 3
Note that if are the elements of the set, then . Thus we can assign . Then .
Solution 4 ( Short Casework )
There are two cases we can consider. Let the elements of our set be denoted , and say that the largest sums and will be consisted of and . Thus, we want to maximize , which means has to be as large as possible, and has to be as small as possible to maximize and . So, the two cases we look at are:
[b]Case 1:[/b]
[b]Case 2:[/b]
Note we have determined these cases by maximizing the value of determined by our previous conditions. So, the answers for each ( after some simple substitution ) will be:
[b]Case 1:[/b]
[b]Case 1:[/b]
See the first case has our largest , so our answer will be
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.